Big Ideas Math Algebra 1 Answers Chapter 9 Solving Quadratic Equations

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Big Ideas Math Book Algebra 1 Answer Key Chapter 9 Solving Quadratic Equations

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• Solving Quadratic Equations Maintaining Mathematical Proficiency – Page 477
• Solving Quadratic Equations Mathematical Practices – Page 478
• Lesson 9.1 Properties of Radicals – Page(480-488)
• Properties of Radicals 9.1 Exercises – Page(485-488)
• Lesson 9.2 Solving Quadratic Equations by Graphing – Page(490-496)
• Solving Quadratic Equations by Graphing 9.2 Exercises – Page(494-496)
• Lesson 9.3 Solving Quadratic Equations Using Square Roots – Page(498-502)
• Solving Quadratic Equations Using Square Roots 9.3 Exercises – Page(501-502)
• Solving Quadratic Equations Study Skills: Keeping a Positive Attitude – Page 503
• Solving Quadratic Equations 9.1 – 9.3 Quiz – Page 504
• Lesson 9.4 Solving Quadratic Equations by Completing the Square – Page(506-514)
• Solving Quadratic Equations by Completing the Square 9.4 Exercises – Page(511-514)
• Lesson 9.5 Solving Quadratic Equations Using the Quadratic Formula – Page(516-524)
• Solving Quadratic Equations Using the Quadratic Formula 9.5 Exercises – Page(521-524)
• Lesson 9.6 Solving Nonlinear Systems of Equations – Page(526-532)
• Solving Nonlinear Systems of Equations 9.6 Exercises – Page(530-532)
• Solving Quadratic Equations Performance Task: Form Matters – Page 533
• Solving Quadratic Equations Chapter Review – Page(534-536)
• Solving Quadratic Equations Chapter Test – Page 537
• Solving Quadratic Equations Cumulative Assessment – Page(538-539)

Solving Quadratic Equations Maintaining Mathematical Proficiency

Factor the trinomial.
Question 1.
x² + 10x + 25
Answer:
x² + 10x + 25
a² + 2ab + b² = (a + b )²
x² + 10x + 25 = x² + 2 (5) (x) + 5²
a = x and b = 5
x² + 2 (5) (x) + 5² = (x + 5)²

Question 2.
x² – 20x + 100
Answer:
x² – 20x + 100
a² – 2ab + b² = (a – b )²
x² – 20x + 100 = x² – 2 (10) (x) + 10²
a = x and b = 10
x² – 2 (10) (x) + 10² = (x – 10)²

Question 3.
x² + 12x + 36
Answer:
x² + 12x + 36
a² + 2ab + b² = (a + b )²
x² + 12x + 36 = x² + 2 (6) (x) + 6²
a = x and b = 6
x² + 2 (6) (x) + 6² = (x + 6)²

Question 4.
x² – 18x + 81
Answer:
x² – 18x + 81
a² – 2ab + b² = (a – b )²
x² – 18x + 81 = x² – 2 (9) (x) + 9²
a = x and b = 9
x² – 2 (9) (x) + 9² = (x – 9)²

Question 5.
x² + 16x + 64
Answer:
x² + 16x + 64
a² + 2ab + b² = (a + b )²
So,
x² + 16x + 64 = x² + 2 (8) (x) + 8²
a = x and b = 8
x² + 2 (8) (x) + 8² = (x + 8)²

Question 6.
x² – 30x + 225
Answer:
x² – 30x + 225
a² – 2ab + b² = (a – b )²
x² – 30x + 225
= x² – 2 (15) (x) + 15²
a = x and b = 15
x² – 2 (15) (x) + 15² = (x – 15)²

Solve the system of linear equations by graphing.
Question 7.
y = -5x + 3
y = 2x – 4
Answer:

From the graph, the intersection point of the given system of linear equations is (1, -2)

Question 8.
y = \(\frac{3}{2}\)x – 2
y = –\(\frac{1}{4}\)x + 5
Answer:

From the graph, the intersection point of the given system of linear equations is (4, 4)

Question 9.
y = \(\frac{1}{2}\)x + 4
y = -3x – 3
Answer:

From the graph, the intersection point of the given system of linear equations is (-2, 3)

Question 10.
ABSTRACT REASONING
What value of c makes x² + bx + c a perfect square trinomial?
Answer:
x² + bx + c
To make the given linear equation a perfect square trinomial,
The value of c must be a square
For perfect square trinomials,
We compare the equation with
a² + 2ab + b²
x² + 2 (\(\frac{b}{2}\)) (x) + c
c = \(\frac{b}{2}\)
The value of c to make the given linear equation a perfect square trinomial is \(\frac{b}{2}\)

Solving Quadratic Equations Mathematical Practices

Mathematically proficient students monitor their work and change course as needed.

Monitoring Progress

Question 1.
Use the graph in Example 1 to approximate the negative solution of the equation x² + x – 1 = 0 to the nearest thousandth.

Answer:
x² + x – 1 = 0
We can estimate the approximate solution of the given equation is 0.65
x = 0.65,
(0.65)² + 0.65 – 1 = 0.0725
For x = 0.64,
The approximate value of the given equation is
(0.64)² + 0.64 – 1 = 0.0496
For x = 0.63,
The approximate value of the given equation is
(0.63)² + 0.63 – 1 = 0.0269
For x = 0.62,
The approximate value of the given equation is
(0.62)² + 0.62 – 1 = 0.0044
For x = 0.61,
The approximate value of the given equation is
(0.61)² + 0.61 – 1 = -0.0179
For x = 0.60,
The approximate value of the given equation is
(0.60)² + 0.60 – 1 = -0.04
For x = 0.59,
The approximate value of the given equation is
(0.59)² + 0.59 – 1 = -0.0619
Thus the negative solution of the given equation is 0.61

Question 2.
The graph of y = x² + x – 3 is shown. Approximate both solutions of the equation x² + x – 3 = 0 to the nearest thousandth.

Answer:
y = x² + x – 3
The approximate positive solution of the given equation is 1.60
The approximate negative solution of the given equation is 2.2
For the positive solution of the given equation
For x = 1.60,
The approximate value of the given equation is
(1.60)² + 1.60 – 3 = 1.162
For x = 1.59,
The approximate value of the given equation is
(1.59)² + 1.59 – 3 = 1.118
For x = 1.30,
The approximate value of the given equation is
(1.30)² + 1.30 – 3 = -0.01
The positive solution of the given equation is 1.30
For the negative solution of the given equation
For x = 2.20,
The approximate value of the given equation is
(2.20)² + 2.20 – 3 = 4.040
For x = 2.10,
The approximate value of the given equation is
(2.10)² + 2.10 – 3 = 3.510
For x = 1.90,
The approximate value of the given equation is
(1.90)² + 1.90 – 3 = 2.510
For x = 1.20,
The approximate value of the given equation is
(1.20)² + 1.20 – 3 = -0.360
Thus the negative solution of the given equation is: 1.90

Lesson 9.1 Properties of Radicals

Essential Question
How can you multiply and divide square roots?
Answer:
You can multiply and divide square roots by using the “Product Rule” and “Quotient Rule”
\(\sqrt{a}\) ⋅ \(\sqrt{b}\) = \(\sqrt{{a}\cdot b}\)
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)

EXPLORATION 1

Operations with Square Roots
Work with a partner. For each operation with square roots, compare the results obtained using the two indicated orders of operations. What can you conclude?
a. Square Roots and Addition
Is \(\sqrt{36}\) + \(\sqrt{64}\) equal to \(\sqrt{36+64}\)?
In general, is \(\sqrt{a}\) + \(\sqrt{b}\) equal to \(\sqrt{a+b}\)? Explain your reasoning.
Answer:
\(\sqrt{36}\) + \(\sqrt{64}\) = \(\sqrt{36 + 64}\)
\(\sqrt{a}\) + \(\sqrt{b}\) and \(\sqrt{a + b}\)
a = 36 and b = 64
\(\sqrt{36}\) + \(\sqrt{64}\)
= 6 + 8
= 14
\(\sqrt{36 + 64}\)
= \(\sqrt{100}\)
= 10
\(\sqrt{a}\) + \(\sqrt{b}\) ≠ \(\sqrt{a + b}\)

b. Square Roots and Multiplications
\(\sqrt{4}\) • \(\sqrt{9}\) equal to \(\sqrt{{4} \cdot 9}\)?
In general, is \(\sqrt{a}\) • \(\sqrt{b}\) equal to \(\sqrt{{a} \cdot b}\)? Explain your reasoning.
Answer:
\(\sqrt{4}\) ⋅ \(\sqrt{9}\)
\(\sqrt{{4}\cdot9}\)
\(\sqrt{a}\) ⋅ \(\sqrt{b}\) and \(\sqrt{{a}\cdot b}\)
a = 4 and b = 9
\(\sqrt{4}\) ⋅ \(\sqrt{9}\)
= 2 ⋅ 3
= 6
\(\sqrt{{4}\cdot 9}\)
= \(\sqrt{36}\)
= 6
\(\sqrt{a}\) ⋅ \(\sqrt{b}\) = \(\sqrt{{a}\cdot b}\)

c. Is \(\sqrt{64}\) – \(\sqrt{36}\) equal to \(\sqrt{64 – 36}\)?
In general, is \(\sqrt{a}\) – \(\sqrt{b}\) equal to \(\sqrt{a-b}\)? Explain your reasoning.
Answer:
\(\sqrt{64}\) – \(\sqrt{36}\)
\(\sqrt{64 – 36}\)
\(\sqrt{a}\) – \(\sqrt{b}\) and \(\sqrt{a – b}\)
a = 64 and b = 36
\(\sqrt{64}\) – \(\sqrt{36}\)
= 8 – 6
= 2
\(\sqrt{64 – 36}\)
= \(\sqrt{28}\)
= 5.291
\(\sqrt{a}\) – \(\sqrt{b}\) ≠ \(\sqrt{a – b}\)

d. Square Roots and Division
Is \(\frac{\sqrt{100}}{\sqrt{4}}\) equal to \(\sqrt{\frac{100}{4}}\)?
In general, is \(\frac{\sqrt{a}}{\sqrt{b}}\) equal to \(\sqrt{\frac{a}{b}}\)? Explain your reasoning.
Answer:
\(\frac{\sqrt{100}}{\sqrt{4}}\)
\(\sqrt{\frac{100}{4}}\)
\(\frac{\sqrt{a}}{\sqrt{b}}\)  and \(\sqrt{\frac{a}{b}}\)
a = 100 and b = 4
\(\frac{\sqrt{100}}{\sqrt{4}}\)
= \(\frac{10}{2}\)
= 5
\(\sqrt{\frac{100}{4}}\)
= \(\sqrt{25}\)
= 5
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)

EXPLORATION 2

Writing Counterexamples
Work with a partner. A counterexample is an example that proves that a general statement is not true. For each general statement in Exploration 1 that is not true, write a counterexample different from the example given.

Communicate Your Answer

Question 3.
How can you multiply and divide square roots?
Answer:
You can multiply and divide square roots by using the “Product Rule” and “Quotient Rule”
\(\sqrt{a}\) ⋅ \(\sqrt{b}\) = \(\sqrt{{a}\cdot b}\)
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)

Question 4.
Give an example of multiplying square roots and an example of dividing square roots that are different from the examples in Exploration 1.
Answer:
We use the “Product Rule” for the multiplication of square roots
\(\sqrt{a}\) ⋅ \(\sqrt{b}\) = \(\sqrt{{a}\cdot b}\)
The values of a and b are:
a = 144 and b = 36
\(\sqrt{144}\) ⋅ \(\sqrt{36}\)
= 12 ⋅ 6
= 72
\(\sqrt{{144}\cdot 36}\)
= \(\sqrt{5,184}\)
= 72
The quotient rule is:
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)
a = 144 and b = 36
\(\frac{\sqrt{144}}{\sqrt{36}}\)
= \(\frac{12}{6}\)
= 2
\(\sqrt{\frac{144}{36}}\)
= \(\sqrt{\frac{12}{3}}\)
= \(\sqrt{4}\)
= 2

Question 5.
Write an algebraic rule for each operation.
a. the product of square roots
b. the quotient of square roots
Answer:
The algebraic rule for the product of square roots
\(\sqrt{a}\) ⋅ \(\sqrt{b}\) = \(\sqrt{{a}\cdot b}\)
The algebraic rule for the division of square roots is
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)

9.1 Exercise

Monitoring Progress

Simplify the expression.
Question 1.
\(\sqrt{24}\)
Answer:
\(\sqrt{24}\)
\(\sqrt{{a}\cdot b}\) = \(\sqrt{a}\) ⋅ \(\sqrt{b}\)
\(\sqrt{24}\)
= \(\sqrt{{4}\cdot 6}\)
= \(\sqrt{4}\) ⋅ \(\sqrt{6}\)
= 2 \(\sqrt{6}\)
\(\sqrt{24}\) = 2 \(\sqrt{6}\)

Question 2.
–\(\sqrt{80}\)
Answer:
–\(\sqrt{80}\)
\(\sqrt{{a}\cdot b}\) = \(\sqrt{a}\) ⋅ \(\sqrt{b}\)
–\(\sqrt{80}\)
= –\(\sqrt{{16}\cdot 5}\)
= –\(\sqrt{16}\) ⋅ \(\sqrt{5}\)
= -4 \(\sqrt{5}\)
–\(\sqrt{80}\) = -4 \(\sqrt{5}\)

Question 3.
\(\sqrt{49 x^{3}}\)
Answer:
\(\sqrt{49 x^{3}}\)
\(\sqrt{{a}\cdot b}\) = \(\sqrt{a}\) ⋅ \(\sqrt{b}\)
\(\sqrt{49 x^{3}}\)
= \(\sqrt{{49 x^{2}}\cdot x}\)
= \(\sqrt{49 x^{2}}\) ⋅ \(\sqrt{x}\)
= 7x \(\sqrt{x}\)

Question 4.
\(\sqrt{49 n^{5}}\)
Answer:
\(\sqrt{49 n^{5}}\)
\(\sqrt{{a}\cdot b}\) = \(\sqrt{a}\) ⋅ \(\sqrt{b}\)
\(\sqrt{49 n^{5}}\)
= \(\sqrt{{49 n^{4}}\cdot n}\)
= \(\sqrt{49 n^{4}}\) ⋅ \(\sqrt{n}\)
= 7n² \(\sqrt{n}\)

Simplify the expression.
Question 5.
\(\sqrt{\frac{23}{9}}\)
Answer:
\(\sqrt{\frac{23}{9}}\)
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)
\(\sqrt{\frac{23}{9}}\)
= \(\sqrt{\frac{23}{9}}\)
= \(\frac{\sqrt{23}}{\sqrt{9}}\)
= \(\frac{\sqrt{23}}{3}\)

Question 6.
–\(\sqrt{\frac{17}{100}}\)
Answer:
–\(\sqrt{\frac{17}{100}}\)
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)
–\(\sqrt{\frac{17}{100}}\)
= –\(\sqrt{\frac{17}{100}}\)
= –\(\frac{\sqrt{17}}{\sqrt{100}}\)
= –\(\frac{\sqrt{17}}{10}\)

Question 7.
\(\sqrt{\frac{36}{z^{2}}}\)
Answer:
\(\sqrt{\frac{36}{z^{2}}}\)
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)
\(\sqrt{\frac{36}{z^{2}}}\)
= \(\sqrt{\frac{36}{z^{2}}}\)
= \(\frac{\sqrt{36}}{\sqrt{z^{2}}}\)
= \(\frac{6}{z}\)

Question 8.
\(\sqrt{\frac{4 x^{2}}{64}}\)
Answer:
\(\sqrt{\frac{4 x^{2}}{64}}\)
\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\sqrt{\frac{a}{b}}\)
\(\sqrt{\frac{4 x^{2}}{64}}\)
= \(\sqrt{\frac{4 x^{2}}{64}}\)
= \(\frac{\sqrt{4 x^{2}}}{\sqrt{64}}\)
= \(\frac{2x}{8}\)
= \(\frac{x}{4}\)

Question 9.
\(\sqrt[3]{54}\)
Answer:
\(\sqrt[3]{54}\)
\(\sqrt[3]{a}{b}\) = \(\sqrt[3]{a}\) ⋅ \(\sqrt[3]{b}\)
\(\sqrt[3]{54}\)
= \(\sqrt[3]{54}{1}\)
= \(\sqrt[3]{54}\) ⋅ \(\sqrt[3]{1}\)
= \(\sqrt[3]{27 × 2}\)
= \(\sqrt[3]{27}\) ⋅ \(\sqrt[3]{2}\)
= 3\(\sqrt[3]{2}\)

Question 10.
\(\sqrt[3]{16 x^{4}}\)
Answer:
\(\sqrt[3]{16 x^{4}}\)
\(\sqrt[3]{a}{b}\) = \(\sqrt[3]{a}\) ⋅ \(\sqrt[3]{b}\)
\(\sqrt[3]{16 x^{4}}\)
= \(\sqrt[3]{x^{3}}{16x}\)
= \(\sqrt[3]{x^{3}}\) ⋅ \(\sqrt[3]{16x}\)
= x\(\sqrt[3]{16x}\)

Question 11.
\(\sqrt[3]{\frac{a}{-27}}\)
Answer:
\(\sqrt[3]{\frac{a}{-27}}\)
\(\sqrt[3]{\frac{a}{b}}\) = \(\sqrt[3]{a}\) / \(\sqrt[3]{b}\)
\(\sqrt[3]{\frac{a}{-27}}\)
= \(\sqrt[3]{a}\) / \(\sqrt[3]{-27}\)
=\(\sqrt[3]{a}\) / \(\sqrt[3]{-3^{3}}\)
= –\(\sqrt[3]{a}\) / 3

Question 12.
\(\sqrt[3]{\frac{25 c^{7} d^{3}}{64}}\)
Answer:
\(\sqrt[3]{\frac{25 c^{7} d^{3}}{64}}\)
\(\sqrt[3]{\frac{a}{b}}\) = \(\sqrt[3]{a}\) / \(\sqrt[3]{b}\)
\(\sqrt[3]{\frac{25 c^{7} d^{3}}{64}}\)
= \(\sqrt[3]{25 c^{7} d^{3}}\) / \(\sqrt[3]{64}\)
=\(\sqrt[3]{25 c^{3} c^{4} d^{3}}\) / \(\sqrt[3]{4^{3}}\)
= c²d\(\sqrt[3]{25c}\) / 4

Simplify the expression.
Question 13.
\(\frac{1}{\sqrt{5}}\)
Answer:
\(\frac{1}{\sqrt{5}}\)
= \(\frac{1}{\sqrt{5}}\) × (\(\sqrt{5}\) / \(\sqrt{5}\))
= \(\sqrt{5}\) / (\(\sqrt{5}\) × \(\sqrt{5}\))
= \(\sqrt{5}\) / 5

Question 14.
\(\frac{\sqrt{10}}{\sqrt{3}}\)
Answer:
\(\frac{\sqrt{10}}{\sqrt{3}}\)
\(\frac{\sqrt{10}}{\sqrt{3}}\)
= \(\frac{\sqrt{10}}{\sqrt{3}}\) × (\(\sqrt{3}\) / \(\sqrt{3}\))
= \(\sqrt{10 × 3}\) / (\(\sqrt{3}\) × \(\sqrt{3}\))
= \(\sqrt{30}\) / 3

Question 15.
\(\frac{7}{\sqrt{2 x}}\)
Answer:
\(\frac{7}{\sqrt{2 x}}\)
\(\frac{7}{\sqrt{2 x}}\)
= \(\frac{7}{\sqrt{2 x}}\) × (\(\sqrt{2 x}\) / \(\sqrt{2 x}\))
= 7\(\sqrt{2 x}\) / (\(\sqrt{2 x}\) × \(\sqrt{2 x}\))
= 7\(\sqrt{2 x}\) / 2x

Question 16.
\(\sqrt{\frac{2 y^{2}}{3}}\)
Answer:
\(\sqrt{\frac{2 y^{2}}{3}}\)
\(\sqrt{\frac{2 y^{2}}{3}}\)
= \(\frac{\sqrt{2 y^{2}}}{\sqrt{3}}\)
\(\frac{\sqrt{2 y^{2}}}{\sqrt{3}}\)
= \(\frac{\sqrt{2 y^{2}}}{\sqrt{3}}\) × (\(\sqrt{3}\) / \(\sqrt{3}\))
= \(\sqrt{2 y^{2} × 3}\) / (\(\sqrt{3}\) × \(\sqrt{3}\))
= y\(\sqrt{6}\) / 3

Question 17.
\(\frac{5}{\sqrt[3]{32}}\)
Answer:
\(\frac{5}{\sqrt[3]{32}}\)
= \(\frac{5}{\sqrt[3]{32}}\) × (\(\sqrt[3]{2}\) / \(\sqrt[3]{2}\))
= 5\(\sqrt[3]{2}\) / (\(\sqrt[3]{32}\) × \(\sqrt[3]{2}\))
= 5\(\sqrt[3]{2}\) / 4

Question 18.
\(\frac{8}{1+\sqrt{3}}\)
Answer:
\(\frac{8}{1 + \sqrt{3}}\)
\(\frac{8}{1 + \sqrt{3}}\)
= \(\frac{8}{1 + \sqrt{3}}\) × (\(1 – \sqrt{3}\) / \(1 – \sqrt{3}\))
= 8\((1 – \sqrt{3})\) / (\(1 + \sqrt{3}\) × \(1 – \sqrt{3}\))
= 8\((1 – \sqrt{3})\) / -2
= – 8\((1 – \sqrt{3})\) / 2

Question 19.
\(\frac{\sqrt{13}}{\sqrt{5}-2}\)
Answer:
The given expression is:
\(\frac{\sqrt{13}}{\sqrt{5}-2}\)
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{5} + 2\)
\(\frac{\sqrt{13}}{\sqrt{5}-2}\)
= \(\frac{\sqrt{13}}{\sqrt{5}-2}\) × (\( \sqrt{5} – 2\) / \( \sqrt{5} – 2\))
= 13(\(\sqrt{{5}} – 2\)) / (\( \sqrt{5} – 2\) × \( \sqrt{5} + 2\))
= 13(\(\sqrt{{5}} – 2\)) / -1
= – 13(\(\sqrt{{5}} – 2\))

Question 20.
\(\frac{12}{\sqrt{2}+\sqrt{7}}\)
Answer:
\(\frac{12}{\sqrt{2}+\sqrt{7}}\)
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{2} -\sqrt{7} \)
\(\frac{12}{\sqrt{2}+\sqrt{7}}\)
= \(\frac{12}{\sqrt{2}+\sqrt{7}}\) × (\(\sqrt{2} -\sqrt{7} \) / \(\sqrt{2} -\sqrt{7} \))
= 12(\(\sqrt{2} -\sqrt{7} \)) / (\(\sqrt{2} +\sqrt{7} \) × \(\sqrt{2} -\sqrt{7} \))
= 12(\(\sqrt{2} -\sqrt{7}\)) / -5
= – 12(\(\sqrt{2} -\sqrt{7} \))/ 5

Question 21.
WHAT IF?
In Example 6, how far can you see when your eye level is 35 feet above the water?
Answer:
In Example 6,
It is given that
d = \(\sqrt{\frac{3h}{2}}\)
Where,
‘h’ is the height that your eye can see above the water
‘d’ is the distance your eye can see
h = 35 feet
d = \(\sqrt{\frac{3 ×105}{2}}\)
d = \(\sqrt{\frac{315}{2}}\)
\(\sqrt{\frac{a}{b}}\) = \(\frac{\sqrt{a}}{\sqrt{b}}\)
\(\sqrt{\frac{315}{2}}\)
= \(\frac{\sqrt{315}}{\sqrt{2}}\)
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{2}\)
\(\sqrt{\frac{315}{2}}\)
= \(\frac{\sqrt{315}}{\sqrt{2}}\) × (\(\sqrt{2}\) / \(\sqrt{2}\))
= \(\sqrt{315 × 2}\) / (\(\sqrt{2}\) × \(\sqrt{2}\))
= \(\sqrt{630}\) / 2
= 12.54 feet

Question 22.
The dimensions of a dance floor form a golden rectangle. The ratio of the length to the width of the golden rectangle is \(1 + \sqrt{5}\) : 2. The longer side of the dance floor is 50 feet. What is the length of the shorter side of the dance floor?
Answer:
The dimensions of a dance floor form a golden rectangle.
The ratio of the length to the width of the golden rectangle is \(1 + \sqrt{5}\) : 2.
The longer side of the dance floor is 50 feet.
\(\frac{1+\sqrt{5}}{2}\) = \(\frac{50}{h}\)
h (\(1 + \sqrt{5}\)) = 50 × 2
h (\(1 + \sqrt{5}\)) = 100
h = \(\frac{100}{1 + \sqrt{5}}\)
To rationalize the denominator, multiply and divide the given fraction with \(1 – \sqrt{5}\)
\(\frac{100}{1 + \sqrt{5}}\)
= \(\frac{100}{1 + \sqrt{5}}\) × (\(1 – \sqrt{5}\) / \(1 – \sqrt{5}\))
= 100\((1 – \sqrt{5})\) / (\(1 + \sqrt{5}\) × \(1 – \sqrt{5}\))
= 100\((1 – \sqrt{5})\) / -4
= – 100\((1 – \sqrt{5})\) / 4
= 30.90 feet

Simplify the expression.
Question 23.
3\(\sqrt{2}\) – \(\sqrt{6}\) + 10\(\sqrt{2}\)
Answer:
3\(\sqrt{2}\) – \(\sqrt{6}\) + 10\(\sqrt{2}\)
= \(\sqrt{2}\) (3 + 10) – \(\sqrt{6}\)
= \(\sqrt{2}\) (13) – \(\sqrt{6}\)
= 13\(\sqrt{2}\) – \(\sqrt{6}\)

Question 24.
4\(\sqrt{7}\) – 6\(\sqrt{63}\)
Answer:
4\(\sqrt{7}\) – 6\(\sqrt{63}\)
= 4\(\sqrt{7}\) – 6\(\sqrt{7 × 9}\)
= 4\(\sqrt{7}\) – 6 (3)\(\sqrt{7}\)
= 4\(\sqrt{7}\) – 18\(\sqrt{7}\)
= \(\sqrt{7}\) ( 4 – 18)
= \(\sqrt{7}\) (-14)
= -14\(\sqrt{7}\)

Question 25.
4\(\sqrt [3]{ 5x }\) – 11\(\sqrt [3]{ 5x }\)
Answer:
4\(\sqrt [3]{ 5x }\) – 11\(\sqrt [3]{ 5x }\)
= \(\sqrt [3]{ 5x }\) (4 – 11)
= \(\sqrt [3]{ 5x }\) (-7)
= -7\(\sqrt [3]{ 5x }\)

Question 26.
\(\sqrt{3}\)(8\(\sqrt{2}\) + 7\(\sqrt{32}\))
Answer:
\(\sqrt{3}\)(8\(\sqrt{2}\) + 7\(\sqrt{32}\))
= \(\sqrt{3}\)(8\(\sqrt{2}\) + 7\(\sqrt{16 × 2}\))
= \(\sqrt{3}\)(8\(\sqrt{2}\) + 7 (4)\(\sqrt{2}\))
= \(\sqrt{3}\)(8\(\sqrt{2}\) + 28\(\sqrt{2}\))
= \(\sqrt{3}\)(\(\sqrt{2}\) (8 + 28))
= \(\sqrt{3}\)(\(\sqrt{2}\) (36))
= 36(\(\sqrt{3}\))(\(\sqrt{2}\) )
= 36\(\sqrt{6}\)

Question 27.
(2\(\sqrt{5}\) – 4)²
Answer:
(2\(\sqrt{5}\) – 4)²
(a – b)² = a² – 2ab + b²
a = 2\(\sqrt{5}\) and b = 4
(2\(\sqrt{5}\) – 4)²
= (2\(\sqrt{5}\))² – 2 (2\(\sqrt{5}\)) (4) + 4²
= 2² (\(\sqrt{5}\))² – 16\(\sqrt{5}\) + 16
= 4 (5) – 16\(\sqrt{5}\) + 16
= 20 – 16\(\sqrt{5}\) + 16
= 20 – 16 (1 – \(\sqrt{5}\))

Question 28.
\(\sqrt [3]{ -4 }\)(\(\sqrt [3]{ 2 }\) – \(\sqrt [3]{ 16 }\))
Answer:
\(\sqrt [3]{ -4 }\)(\(\sqrt [3]{ 2 }\) – \(\sqrt [3]{ 16 }\))
= \(\sqrt [3]{-4 × 2}\) – \(\sqrt [3]{-4 × 16}\)
= \(\sqrt [3]{-8}\) – \(\sqrt [3]{-64}\)
= -2 – (-4)
= -2 + 4
= 2

Properties of Radicals

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The process of eliminating a radical from the denominator of a radical expression is called _______________.
Answer:

Question 2.
VOCABULARY
What is the conjugate of the binomial \(\sqrt{6}\) + 4?
Answer:
\(\sqrt{6}\) + 4
The binomial conjugate of \(\sqrt{6}\) + 4 is \(\sqrt{6}\) – 4

Question 3.
WRITING
Are the expressions \(\frac{1}{3} \sqrt{2 x}\) and \(\sqrt{\frac{2 x}{9}}\) equivalent? Explain your reasoning.
Answer:

Question 4.
WHICH ONE DOESN’T BELONG?
Which expression does not belong with the other three? Explain your reasoning.

Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, determine whether the expression is in simplest form. If the expression is not in simplest form, explain why.
Question 5.
\(\sqrt{19}\)
Answer:

Question 6.
\(\sqrt{\frac{1}{7}}\)
Answer:
\(\sqrt{\frac{1}{7}}\) is not in its simplest form

Question 7.
\(\sqrt{48}\)
Answer:

Question 8.
\(\sqrt{34}\)
Answer:
\(\sqrt{34}\) is in its simplest form because the radicand has not any perfect square number

Question 9.
\(\frac{5}{\sqrt{2}}\)
Answer:

Question 10.
\(\frac{3 \sqrt{10}}{4}\)
Answer:
\(\frac{3 \sqrt{10}}{4}\) is in its simplest form

Question 11.
\(\frac{1}{2+\sqrt[3]{2}}\)
Answer:

Question 12.
\(6-\sqrt[3]{54}\)
Answer:
\(6-\sqrt[3]{54}\) is not in its simplest form because the radicand contains the perfect cube root

In Exercises 13–20, simplify the expression.
Question 13.
\(\sqrt{20}\)
Answer:

Question 14.
\(\sqrt{32}\)
Answer:
\(\sqrt{32}\)
= \(\sqrt{2 × 16}\)
= \(\sqrt{2}\) × \(\sqrt{16}\)
= \(\sqrt{2}\) × 4
= 4\(\sqrt{2}\)

Question 15.
\(\sqrt{128}\)
Answer:

Question 16.
–\(\sqrt{72}\)
Answer:
–\(\sqrt{72}\)
= –\(\sqrt{36 × 2}\)
= –\(\sqrt{36}\) × \(\sqrt{2}\)
= -6 × \(\sqrt{2}\)
= -6\(\sqrt{2}\)

Question 17.
\(\sqrt{125b}\)
Answer:

Question 18.
\(\sqrt{4 x^{2}}\)
Answer:
\(\sqrt{4 x^{2}}\)
= \(\sqrt{(2 x)^{2}}\)
= 2x

Question 19.
\(-\sqrt{81 m^{3}}\)
Answer:

Question 20.
\(\sqrt{48 n^{5}}\)
Answer:
\(\sqrt{48 n^{5}}\)
= \(\sqrt{16 n^{4} × 3 n}\)
= \(\sqrt{16 n^{4}}\) × \(\sqrt{3 n}\)
= 4 n² × \(\sqrt{3 n}\)
= 4n²\(\sqrt{3 n}\)

In Exercises 21–28, simplify the expression.
Question 21.
\(\sqrt{\frac{4}{49}}\)
Answer:

Question 22.
\(-\sqrt{\frac{7}{81}}\)
Answer:
\(-\sqrt{\frac{7}{81}}\)
= –\(\frac{\sqrt{7}}{\sqrt{81}}\)
= –\(\frac{\sqrt{7}}{9}\)

Question 23.
\(-\sqrt{\frac{23}{64}}\)
Answer:

Question 24.
\(\sqrt{\frac{65}{121}}\)
Answer:
\(\sqrt{\frac{65}{121}}\)
= \(\frac{\sqrt{65}}{\sqrt{121}}\)
= \(\frac{\sqrt{65}}{11}\)

Question 25.
\(\sqrt{\frac{a^{3}}{49}}\)
Answer:

Question 26.
\(\sqrt{\frac{144}{k^{2}}}\)
Answer:
\(\sqrt{\frac{144}{k^{2}}}\)
= \(\frac{\sqrt{144}}{\sqrt{k^{2}}}\)
= \(\frac{12}{k}\)

Question 27.
\(\sqrt{\frac{100}{4x^{2}}}\)
Answer:

Question 28.
\(\sqrt{\frac{25 v^{2}}{36}}\)
Answer:
\(\sqrt{\frac{25v^{2}}{36}}\)
= \(\frac{\sqrt{25v^{2}}}{\sqrt{36}}\)
= \(\frac{5v}{6}\)

In Exercises 29–36, simplify the expression.
Question 29.
\(\sqrt [3]{ 16 }\)
Answer:

Question 30.
\(\sqrt [3]{ -108 }\)
Answer:
\(\sqrt [3]{ -108 }\)
= \(\sqrt [3]{ -27 × 4 }\)
= \(\sqrt [3]{ -27 }\) × \(\sqrt [3]{ 4 }\)
= -3 × \(\sqrt [3]{ 4 }\)
= -3\(\sqrt [3]{ 4 }\)

Question 31.
\(\sqrt[3]{-64 x^{5}}\)
Answer:

Question 32.
–\(\sqrt[3]{343 n^{2}}\)
Answer:
–\(\sqrt[3]{343 n^{2}}\)
= –\(\sqrt[3]{343 }\) × \(\sqrt[3]{ n^{2}}\)
= -7 × \(\sqrt[3]{ n^{2}}\)
= -7\(\sqrt[3]{ n^{2}}\)

Question 33.
\(\sqrt[3]{\frac{6 c}{-125}}\)
Answer:

Question 34.
\(\sqrt[3]{\frac{8 h^{4}}{27}}\)
Answer:
\(\sqrt[3]{\frac{8 h^{4}}{27}}\)
= \(\frac{\sqrt[3]{8 h^{4}}}{\sqrt[3]{27}}\)
= \(\frac{\sqrt[3]{8 h^{3} × h}}{\sqrt[3]{27}}\)
= \(\frac{2\sqrt[3]{ h}}{3}\)

Question 35.
\(-\sqrt[3]{\frac{81 y^{2}}{1000 x^{3}}}\)
Answer:

Question 36.
\(\sqrt[3]{\frac{21}{-64 a^{3} b^{6}}}\)
Answer:
\(\sqrt[3]{\frac{21}{-64 a^{3} b^{6}}}\)
= \(\frac{\sqrt[3]{21}}{\sqrt[3]{-64 a^{3} b^{6}}}\)
= \(\frac{\sqrt[3]{21}}{\sqrt[3]{-64 a^{3} b^{3} b^{3}}}\)
= \(\frac{\sqrt[3]{ 21}}{-4ab²}\)

ERROR ANALYSIS In Exercises 37 and 38, describe and correct the error in simplifying the expression.
Question 37.

Answer:

Question 38.

Answer:
\(\sqrt[3]{\frac{128 y^{3}}{125}}\)
= \(\frac{\sqrt[3]{128 y^{3}}}{\sqrt[3]{125}}\)
= \(\frac{\sqrt[3]{64 y^{3} × 2}}{\sqrt[3]{125}}\)
= \(\frac{4y\sqrt[3]{ 2}}{5}\)

In Exercises 39–44, write a factor that you can use to rationalize the denominator of the expression.
Question 39.
\(\frac{4}{\sqrt{6}}\)
Answer:

Question 40.
\(\frac{1}{\sqrt{13 z}}\)
Answer:
\(\frac{1}{\sqrt{13 z}}\)
To rationalize the given expression,
We have to multiply the given expression by a factor \(\frac{\sqrt{13 z}}{\sqrt{13 z}}\)

Question 41.
\(\frac{2}{\sqrt[3]{x^{2}}}\)
Answer:

Question 42.
\(\frac{3 m}{\sqrt[3]{4}}\)
Answer:
\(\frac{3 m}{\sqrt[3]{4}}\)
We have to multiply the given expression by a factor \(\frac{\sqrt[3]{2}}{\sqrt[3]{2}}\)

Question 43.
\(\frac{\sqrt{2}}{\sqrt{5}-8}\)
Answer:

Question 44.
\(\frac{5}{\sqrt{3}+\sqrt{7}}\)
Answer:
\(\frac{5}{\sqrt{3}+\sqrt{7}}\)
We have to multiply the given expression by a factor \(\frac{\sqrt{3} – \sqrt{7}}{\sqrt{3} – \sqrt{7}}\)

In Exercises 45–54, simplify the expression.
Question 45.
\(\frac{2}{\sqrt{2}}\)
Answer:

Question 46.
\(\frac{4}{\sqrt{3}}\)
Answer:
\(\frac{4}{\sqrt{3}}\)
= \(\frac{4}{\sqrt{3}}\) × (\(\sqrt{3}\) / \(\sqrt{3}\))
= 4\(\sqrt{3}\) / (\(\sqrt{3}\) × \(\sqrt{3}\))
= 4\(\sqrt{3}\) / 3

Question 47.
\(\frac{\sqrt{5}}{\sqrt{48}}\)
Answer:

Question 48.
\(\sqrt{\frac{4}{52}}\)
Answer:
\(\sqrt{\frac{4}{52}}\)
= \(\frac{\sqrt{4}}{\sqrt{52}}\)
= \(\frac{\sqrt{4}}{\sqrt{4 × 13}}\)
= \(\frac{2}{2\sqrt{13}}\)

Question 49.
\(\frac{3}{\sqrt{a}}\)
Answer:

Question 50.
\(\frac{1}{\sqrt{2 x}}\)
Answer:
\(\frac{1}{\sqrt{2 x}}\)
= \(\frac{1}{\sqrt{2 x}}\) × (\(\sqrt{2 x}\) / \(\sqrt{2 x}\))
= \(\sqrt{2 x}\) / (\(\sqrt{2 x}\) × \(\sqrt{2 x}\))
= \(\sqrt{2 x}\) / 2x

Question 51.
\(\sqrt{\frac{3 d^{2}}{5}}\)
Answer:

Question 52.
\(\frac{\sqrt{8}}{\sqrt{3 n^{3}}}\)
Answer:
\(\frac{\sqrt{8}}{\sqrt{3 n^{3}}}\)
= \(\frac{\sqrt{4 × 2}}{\sqrt{3 n^{2} × n}}\)
= \(\frac{2\sqrt{2}}{n\sqrt{3 n}}\)

Question 53.
\(\frac{4}{\sqrt[3]{25}}\)
Answer:

Question 54.
\(\sqrt[3]{\frac{1}{108 y^{2}}}\)
Answer:
\(\sqrt[3]{\frac{1}{108 y^{2}}}\)
= \(\frac{\sqrt[3]{1}}{\sqrt[3]{27 × 4 y^{2}}}\)
= \(\frac{2\sqrt[3]{1}}{3\sqrt[3]{4 y^{2}}}\)
\(\sqrt[3]{\frac{1}{108 y^{2}}}\) = \(\frac{2\sqrt[3]{1}}{3\sqrt[3]{4 y^{2}}}\)

In Exercises 55–60, simplify the expression.
Question 55.
\(\frac{1}{\sqrt{7}+1}\)
Answer:

Question 56.
\(\frac{2}{5-\sqrt{3}}\)
Answer:
\(\frac{2}{5 – \sqrt{3}}\)
= \(\frac{2}{5 – \sqrt{3}}\) × (\(5 + \sqrt{3}\) / \(5 + \sqrt{3}\))
= 2\((5 + \sqrt{3})\) / (\(5 + \sqrt{3}\) × \(5 – \sqrt{3}\))
= 2\((5 + \sqrt{3})\) / 22
= \((5 + \sqrt{3})\) / 11

Question 57.
\(\frac{\sqrt{10}}{7-\sqrt{2}}\)
Answer:

Question 58.
\(\frac{\sqrt{5}}{6+\sqrt{5}}\)
Answer:
\(\frac{\sqrt{5}}{6 + \sqrt{5}}\)
= \(\frac{\sqrt{5}}{6 + \sqrt{5}}\) × (\( 6 – \sqrt{5} \) / \(6 – \sqrt{5}\))
= \(\frac{\sqrt{5}}{6 – \sqrt{5}}\) / (\(6 +  \sqrt{5} \) × \( 6 – \sqrt{5} \))
= \(\frac{\sqrt{5}}{6 + \sqrt{5}}\) / 31

Question 59.
\(\frac{3}{\sqrt{5}-\sqrt{2}}\)
Answer:

Question 60.
\(\frac{\sqrt{3}}{\sqrt{7}+\sqrt{3}}\)
Answer:
\(\frac{\sqrt{3}}{\sqrt{7}+\sqrt{3}}\)
= \(\frac{\sqrt{3}}{\sqrt{7}+\sqrt{3}}\) × (\(\sqrt{7} -\sqrt{3} \) / \(\sqrt{7} -\sqrt{3} \))
= \(\frac{\sqrt{3}}{\sqrt{7}-\sqrt{3}}\) / (\(\sqrt{7} +\sqrt{3} \) × \(\sqrt{7} -\sqrt{3} \))
= \(\frac{\sqrt{3}}{\sqrt{7}+\sqrt{3}}\) / 4

Question 61.
MODELING WITH MATHEMATICS
The time t (in seconds) it takes an object to hit the ground is given by t = \(\sqrt{\frac{h}{16}}\), where his the height (in feet) from which the object was dropped.

a. How long does it take an earring to hit the ground when it falls from the roof of the building?
b. How much sooner does the earring hit the ground when it is dropped from two stories (22 feet) below the roof?
Answer:

Question 62.
MODELING WITH MATHEMATICS
The orbital period of a planet is the time it takes the planet to travel around the Sun. You can find the orbital period P (in Earth years) using the formula P = \(\sqrt{d^{3}}\), where d is the average distance (in astronomical units, abbreviated AU) of the planet from the Sun.

a. Simplify the formula.
Answer:
P = \(\sqrt{d^{3}}\)
where d is the average distance
P = \(\sqrt{d^{3}}\)
P = \(\sqrt{d^{2} × d}\)
P = d\(\sqrt{d}\)

b. What is Jupiter’s orbital period?
Answer:
P = d\(\sqrt{d}\)
The distance of Jupiter from the sun is: 5.2 AU
The Orbital Period of Jupiter (P) is:
P = \(\sqrt{5.2^{3}}\)
P = 11.85 AU

Question 63.
MODELING WITH MATHEMATICS
The electric current I (in amperes) an appliance uses is given by the formula I = \(\sqrt{\frac{P}{R}}\), where P is the power (in watts) and R is the resistance (in ohms). Find the current an appliance uses when the power is 147 watts and the resistance is 5 ohms.

Answer:

Question 64.
MODELING WITH MATHEMATICS
You can find the average annual interest rate r (in decimal form) of a savings account using the formula r = \(\sqrt{\frac{V_{2}}{V_{0}}}\) – 1, where V₀ is the initial investment and V₂ is the balance of the account after 2 years. Use the formula to compare the savings accounts. In which account would you invest money? Explain.

Answer:
r = \(\sqrt{\frac{V_{2}}{V_{0}}}\) – 1
where,
V₀ is the initial investment
V₂ is the balance of the account after 2 years
For \(\frac{V_{2}}{V_{0}}\) = \(\frac{293}{275}\),
r = \(\sqrt{\frac{293}{275}}\) – 1
r = 3.19%
For \(\frac{V_{2}}{V_{0}}\) = \(\frac{382}{361}\),
r = \(\sqrt{\frac{382}{361}}\) – 1
r = 2.85%
For \(\frac{V_{2}}{V_{0}}\) = \(\frac{214}{199}\),
r = \(\sqrt{\frac{214}{199}}\) – 1
r = 3.68%
For \(\frac{V_{2}}{V_{0}}\) = \(\frac{272}{254}\),
r = \(\sqrt{\frac{272}{254}}\) – 1
r = 3.44%
For \(\frac{V_{2}}{V_{0}}\) = \(\frac{406}{386}\),
r = \(\sqrt{\frac{406}{386}}\) – 1
r = 2.51%

In Exercises 65–68, evaluate the function for the given value of x. Write your answer in simplest form and in decimal form rounded to the nearest hundredth.
Question 65.
h(x) = \(\sqrt{5x}\); x = 10
Answer:

Question 66.
g(x) = \(\sqrt{3x}\); x = 60
Answer:
g(x) = \(\sqrt{3x}\)
x = 60
g(60) = \(\sqrt{3 × 60}\)
g(60) = \(\sqrt{9 × 20}\)
g(60) = \(\sqrt{9 × 5 × 4}\)
g(60) = \(\sqrt{36 × 5}\)
g(60) = 6\(\sqrt{5}\)
g(60) = 13.41

Question 67.
r(x) = \(\sqrt{\frac{3 x}{3 x^{2}+6}}\); x = 4
Answer:

Question 68.
p(x) = \(\sqrt{\frac{x-1}{5 x}}\); x = 8
Answer:
p(x) = \(\sqrt{\frac{x-1}{5 x}}\) with x = 8
p(8) = \(\sqrt{\frac{8-1}{5 × 8}}\)
p(8) = \(\sqrt{\frac{7}{40}}\)
p(8) = 0.41

In Exercises 69–72, evaluate the expression when a = −2, b = 8, and c = \(\frac{1}{2}\). Write your answer in simplest form and in decimal form rounded to the nearest hundredth.
Question 69.
\(\sqrt{a^{2}+b c}\)
Answer:

Question 70.
\(-\sqrt{4 c-6 a b}\)
Answer:
\(-\sqrt{4 c-6 a b}\)
= \(-\sqrt{4 × (1/2) – 6 (-2) (8)}\)
= \(-\sqrt{2- 6 (-16)}\)
= \(-\sqrt{2 + 96}\)
= \(-\sqrt{98}\)
= -9.89

Question 71.
\(-\sqrt{2 a^{2}+b^{2}}\)
Answer:

Question 72.
\(\sqrt{b^{2}-4 a c}\)
Answer:
\(\sqrt{b^{2}-4 a c}\)
= \(\sqrt{8^{2}-4 (-2) × (1/2)}\)
= \(\sqrt{64 + 4}\)
= \(\sqrt{68}\)
= 8.24

Question 73.
MODELING WITH MATHEMATICS
The text in the book shown forms a golden rectangle. What is the width w of the text?

Answer:

Question 74.
MODELING WITH MATHEMATICS
The flag of Togo is approximately the shape of a golden rectangle. What is the width w of the flag?

Answer:
The flag of Togo is approximately the shape of a golden rectangle
The length of the golden rectangle is 42 in.
The width of the given rectangle is w inch
The ratio of the length and the width in a golden rectangle = \(\frac{1+\sqrt{5}}{2}\)
\(\frac{1+\sqrt{5}}{2}\) = \(\frac{42}{w}\)
w (\(1 + \sqrt{5}\)) = 42 × 2
h (\(1 + \sqrt{5}\)) = 84
h = \(\frac{84}{1 + \sqrt{5}}\)
\(\frac{84}{1 + \sqrt{5}}\)
= \(\frac{84}{1 + \sqrt{5}}\) × (\(1 – \sqrt{5}\) / \(1 – \sqrt{5}\))
= 84\((1 – \sqrt{5})\) / (\(1 + \sqrt{5}\) × \(1 – \sqrt{5}\))
= 84\((1 – \sqrt{5})\) / -4
= – 84\((1 – \sqrt{5})\) / 4
= 25.95 in
Hence, from the above,

In Exercises 75–82, simplify the expression.
Question 75.
\(\sqrt{2}\) – 2\(\sqrt{2}\) + 6\(\sqrt{2}\)
Answer:

Question 76.
\(\sqrt{5}\) – 5\(\sqrt{13}\) – 8\(\sqrt{5}\)
Answer:
\(\sqrt{5}\) – 5\(\sqrt{13}\) – 8\(\sqrt{5}\)
= \(\sqrt{5}\) (1 – 8) – 5\(\sqrt{13}\)
= \(\sqrt{5}\) (-7) – 5\(\sqrt{13}\)
= -7\(\sqrt{5}\)  – 5\(\sqrt{13}\)

Question 77.
2\(\sqrt{6}\) – 5\(\sqrt{54}\)
Answer:

Question 78.
9\(\sqrt{32}\) + \(\sqrt{2}\)
Answer:
9\(\sqrt{32}\) + \(\sqrt{2}\)
= 9\(\sqrt{16 × 2}\) + \(\sqrt{2}\)
= 9 (4)\(\sqrt{2}\) + \(\sqrt{2}\)
= 36\(\sqrt{2}\) + \(\sqrt{2}\)
= \(\sqrt{2}\) (6 + 1)
= \(\sqrt{2}\) (7)
= 7\(\sqrt{2}\)

Question 79.
\(\sqrt{12}\) + 6\(\sqrt{3}\) + 2\(\sqrt{6}\)
Answer:

Question 80.
3\(\sqrt{7}\) – 5\(\sqrt{14}\) + 2\(\sqrt{28}\)
Answer:
3\(\sqrt{7}\) – 5\(\sqrt{14}\) + 2\(\sqrt{28}\)
= 3\(\sqrt{7}\) – 5\(\sqrt{14}\) + 2\(\sqrt{7 × 4}\)
= 3\(\sqrt{7}\) – 5\(\sqrt{14}\) + 2 (2)\(\sqrt{7}\)
= 3\(\sqrt{7}\) – 5\(\sqrt{14}\) + 4\(\sqrt{7}\)
= \(\sqrt{7}\) (3 + 4) – 5\(\sqrt{14}\)
= \(\sqrt{7}\) (7) – 5\(\sqrt{14}\)
= 7\(\sqrt{7}\) – 5\(\sqrt{14}\)

Question 81.
\(\sqrt[3]{-81}\) + 4\(\sqrt[3]{3}\)
Answer:

Question 82.
6\(\sqrt[3]{128 t}\) – 2\(\sqrt[3]{2 t}\)
Answer:
6\(\sqrt[3]{128 t}\) – 2\(\sqrt[3]{2 t}\)
= 6\(\sqrt[3]{64 × 2 t}\) – 2\(\sqrt[3]{2 t}\)
= 6 (4)\(\sqrt[3]{2 t}\) – 2\(\sqrt[3]{2 t}\)
= 24\(\sqrt[3]{2 t}\) – 2\(\sqrt[3]{2 t}\)
= \(\sqrt[3]{2 t}\) (24 – 2)
= \(\sqrt[3]{2 t}\) (22)
= 22\(\sqrt[3]{2 t}\)

In Exercises 83–90, simplify the expression.
Question 83.
\(\sqrt{2}\)(\(\sqrt{45}\) + \(\sqrt{5}\))
Answer:

Question 84.
\(\sqrt{3}\)(\(\sqrt{72}\) – 3\(\sqrt{2}\))
Answer:
\(\sqrt{3}\)(\(\sqrt{72}\) – 3\(\sqrt{2}\))
= \(\sqrt{3}\)(\(\sqrt{36 × 2}\) – 3\(\sqrt{2}\))
= \(\sqrt{3}\)(6\(\sqrt{2}\) – 3\(\sqrt{2}\))
= \(\sqrt{3}\)(\(\sqrt{2}\) (6 – 3))
= \(\sqrt{3}\)(\(\sqrt{2}\) (3))
= \(\sqrt{3}\)(3\(\sqrt{2}\))
= 3 ×  \(\sqrt{3}\) × \(\sqrt{2}\)
= 3 \(\sqrt{6}\)

Question 85.
\(\sqrt{5}\)(2\(\sqrt{6x}\) – \(\sqrt{96x}\))
Answer:

Question 86.
\(\sqrt{7y}\)(\(\sqrt{27y}\) + 5\(\sqrt{12y}\))
Answer:
\(\sqrt{7y}\)(\(\sqrt{27y}\) + 5\(\sqrt{12y}\))
= \(\sqrt{7y}\)(\(\sqrt{9 × 3y}\) + 5\(\sqrt{4 × 3y}\))
= \(\sqrt{7y}\)(3\(\sqrt{3y}\) + 5 (2)\(\sqrt{3y}\))
= \(\sqrt{7y}\)(3\(\sqrt{3y}\) + 10\(\sqrt{3y}\))
= \(\sqrt{7y}\)(\(\sqrt{3y}\) (3 + 10))
= \(\sqrt{7y}\)(\(\sqrt{3y}\) (13))
= 13 × \(\sqrt{7y}\) × \(\sqrt{3y}\)
= 13(\(\sqrt{21y}\)

Question 87.
(4\(\sqrt{2}\) – \(\sqrt{98}\))²
Answer:

Question 88.
(\(\sqrt{3}\) + \(\sqrt{48}\)) (\(\sqrt{20}\) – \(\sqrt{5}\))
Answer:
(\(\sqrt{3}\) + \(\sqrt{48}\)) (\(\sqrt{20}\) – \(\sqrt{5}\))
= (\(\sqrt{3}\) + \(\sqrt{16 × 3}\)) (\(\sqrt{5 × 4}\) – \(\sqrt{5}\))
= (\(\sqrt{3}\) + 4\(\sqrt{3}\)) (2\(\sqrt{5}\) – \(\sqrt{5}\))
= (\(\sqrt{3}\) (1 + 4)) (\(\sqrt{5}\) (2 – 1))
= (\(\sqrt{3}\) (5)) (\(\sqrt{5}\) (1))
= 5\(\sqrt{3}\) × \(\sqrt{5}\)
= 5\(\sqrt{15}\)

Question 89.
\(\sqrt[3]{3}(\sqrt[3]{4}+\sqrt[3]{32})\)
Answer:

Question 90.
\(\sqrt[3]{2}(\sqrt[3]{135}-4 \sqrt[3]{5})\)
Answer:
\(\sqrt[3]{2}(\sqrt[3]{135}-4 \sqrt[3]{5})\)
= \(\sqrt[3]{2}(\sqrt[3]{5 × 27}-4 \sqrt[3]{5})\)
= \(\sqrt[3]{2}(3\sqrt[3]{5}-4 \sqrt[3]{5})\)
= \(\sqrt[3]{2}(\sqrt[3]{5} (3 – 4)\)
= \(\sqrt[3]{2}(\sqrt[3]{5} (-1)\)
= –\(\sqrt[3]{2}(\sqrt[3]{5}) \)
= –\(\sqrt[3]{10}\)

Question 91.
MODELING WITH MATHEMATICS
The circumference C of the art room in a mansion is approximated by the formula C ≈ \(\sqrt\frac{a^{2}+b^{2}}{2}\). Approximate the circumference of the room.

Answer:

Question 92.
CRITICAL THINKING
Determine whether each expression represents a rational or an irrational number. Justify your answer.

Answer:
a.
4 + \(\sqrt{6}\)
Any fraction that we can write in the form of \(\frac{p}{q}\) is called a “Rational Number”. Otherwise, that number is an “Irrational number”
We can say that the given expression can’t be expressed in the form of \(\frac{p}{q}\)
We can conclude that 4 + \(\sqrt{6}\) is an irrational number
b.
\(\frac{\sqrt{48}}{\sqrt{3}}\)
Any fraction that we can write in the form of \(\frac{p}{q}\) is called a “Rational Number”. Otherwise, that number is an “Irrational number”
We can say that the given expression can be expressed in the form of \(\frac{p}{q}\)
We can conclude that \(\frac{\sqrt{48}}{\sqrt{3}}\) is a rational number
c.
\(\frac{8}{\sqrt{12}}\)
Any fraction that we can write in the form of \(\frac{p}{q}\) is called a “Rational Number”. Otherwise, that number is an “Irrational number”
We can say that the given expression can’t be expressed in the form of \(\frac{p}{q}\)
We can conclude that \(\frac{8}{\sqrt{12}}\) is an irrational number
d.
\(\frac{\sqrt{3}}{\sqrt{7}}\)
Any fraction that we can write in the form of \(\frac{p}{q}\) is called a “Rational Number”. Otherwise, that number is an “Irrational number”
We can say that the given expression can’t be expressed in the form of \(\frac{p}{q}\)
We can conclude that\(\frac{\sqrt{3}}{\sqrt{7}}\) is an irrational number
e.
\(\frac{a}{\sqrt{10} – \sqrt{2}}\)
Any fraction that we can write in the form of \(\frac{p}{q}\) is called a “Rational Number”. Otherwise, that number is an “Irrational number”
We can say that the given expression can be expressed in the form of \(\frac{p}{q}\)
We can conclude that\(\frac{a}{\sqrt{10} – \sqrt{2}}\) is a rational number
f.
\(\frac{2 + \sqrt{5}}{2 b + \sqrt{5 b^{2}}}\)
Any fraction that we can write in the form of \(\frac{p}{q}\) is called a “Rational Number”. Otherwise, that number is an “Irrational number”
We can say that the given expression can be expressed in the form of \(\frac{p}{q}\)
We can conclude that\(\frac{2 + \sqrt{5}}{2 b + \sqrt{5 b^{2}}}\) is a rational number

In Exercises 93–98, simplify the expression.
Question 93.
\(\sqrt[5]{\frac{13}{5 x^{5}}}\)
Answer:

Question 94.
\(\sqrt[4]{\frac{10}{81}}\)
Answer:
\(\sqrt[4]{\frac{10}{81}}\)
= \(\frac{\sqrt[4]{10}}{\sqrt[4]{81}}\)
= \(\frac{\sqrt[4]{10}}{\sqrt[4]{3^{4}}}\)
= \(\frac{\sqrt[4]{10}}{3}\)

Question 95.
\(\sqrt[4]{256 y}\)
Answer:

Question 96.
\(\sqrt[5]{160 x^{6}}\)
Answer:
\(\sqrt[5]{160 x^{6}}\)
= \(\sqrt[5]{40 (4) x^{5} × x}\)
= \(\sqrt[5]{20 (8) x^{5} × x}\)
= \(\sqrt[5]{32 (5) x^{5} × x}\)
= \(\sqrt[5]{2^{5} × 5  x^{5} × x}\)
= 2x\(\sqrt[5]{5 x}\)

Question 97.
\(6 \sqrt[4]{9}-\sqrt[5]{9}+3 \sqrt[4]{9}\)
Answer:

Question 98.
\(\sqrt[5]{2}(\sqrt[4]{7}+\sqrt[5]{16})\)
Answer:
\(\sqrt[5]{2}(\sqrt[4]{7}+\sqrt[5]{16})\)
= (\(\sqrt[5]{2}\) × \(\sqrt[4]{7}\)) + (\(\sqrt[5]{2}\) × \(\sqrt[5]{16}\))
= (\(\sqrt[5]{2}\) × \(\sqrt[4]{7}\)) + (\(\sqrt[5]{32}\)
= (\(\sqrt[5]{2}\) × \(\sqrt[4]{7}\)) + 2

REASONING In Exercises 99 and 100, use the table shown.

Question 99.
Copy and complete the table by (a) finding each sum ( 2 + 2, 2 + \(\frac{1}{4}\), etc. ) and (b) finding each product ( 2 • 2, 2 • \(\frac{1}{4}\), etc. )
Answer:

Question 100.
Use your answers in Exercise 99 to determine whether each statement is always, sometimes, or never true. Justify your answer.
a. The sum of a rational number and a rational number is rational.
b. The sum of a rational number and an irrational number is irrational.
c. The sum of an irrational number and an irrational number is irrational.
d. The product of a rational number and a rational number is rational.
e. The product of a nonzero rational number and an irrational number is irrational.
f. The product of an irrational number and an irrational number is irrational.
Answer:
From Exercise 99,
a. The completed table of the sum of rational numbers and irrational numbers is:

b. The completed table of the product of rational numbers and irrational numbers is:

The given statements are:
a. The sum of a rational number and a rational number is rational.
b. The sum of a rational number and an irrational number is irrational.
c. The sum of an irrational number and an irrational number is irrational.
d. The product of a rational number and a rational number is rational.
e. The product of a nonzero rational number and an irrational number is irrational.
f. The product of an irrational number and an irrational number is irrational.
So,
a.
The given statement is sometimes true
b.
The given statement is sometimes true
c.
The given statement is always true
d.
The given statement is always true
e.
The given statement is sometimes true
f.
The given statement is sometimes true

Question 101.
REASONING
Let m be a positive integer. For what values of m will the simplified form of the expression \(\sqrt{2^{m}}\) contain a radical? For what values will it not contain a radical? Explain.
Answer:

Question 102.
HOW DO YOU SEE IT?
The edge length s of a cube is an irrational number, a surface area is an irrational number, and a volume is a rational number. Give a possible value of s.

Answer:
The length (s) of a cube is: An irrational number
The surface area of a cube is: An irrational number
The volume of a cube is: A rational number
The surface area of a cube = 6 (Side)²
The volume of a cube = (Side)³
Let the length of the side of the cube = \(\sqrt{2}\)
The surface area of a cube = 6 (√2)²
= 6 (2)
= 12
The volume of a cube = (√2)³
The surface area of a cube will always be rational irrespective the number is rational or irrational
There is no possible value of s

Question 103.
REASONING
Let a and b be positive numbers. Explain why \(\sqrt{ab}\) lies between a and b on a number line. (Hint: Let a< b and multiply each side of a < b by a. Then let a < b and multiply each side by b.)
Answer:

Question 104.
MAKING AN ARGUMENT
Your friend says that you can rationalize the denominator of the expression \(\frac{2}{4+\sqrt[3]{5}}\) by multiplying the numerator and denominator by 4 – \(\sqrt[3]{5}\). Is your friend correct? Explain.
Answer:
You can rationalize the denominator of the expression \(\frac{2}{4+\sqrt[3]{5}}\) by multiplying the numerator and denominator by 4 – \(\sqrt[3]{5}\).
We will rationalize a radical to make that radical a rational number
A “Rational number” is a number that can be written in the form of \(\frac{p}{q}\)
We will see whether we can rationalize the denominator or not in the given fraction
(4 – \(\sqrt[3]{5}\)) × (4 + \(\sqrt[3]{5}\))
= 4² – (\(\sqrt[3]{5}\) × \(\sqrt[3]{5}\))
= 16 – (\(\sqrt[3]{5}\) × \(\sqrt[3]{5}\))
From the above value,
We can observe that the denominator can’t be rationalized
We can conclude that your friend is not correct

Question 105.
PROBLEM -SOLVING
The ratio of consecutive terms \(\frac{a_{n}}{a_{n}-1}\) in the Fibonacci sequence gets closer and closer to the golden ratio \(\frac{1+\sqrt{5}}{2}\) as n increases. Find the term that precedes 610 in the sequence.
Answer:

Question 106.
THOUGHT-PROVOKING
Use the golden ratio \(\frac{1+\sqrt{5}}{2}\) and the golden ratio conjugate \(\frac{1-\sqrt{5}}{2}\) for each of the following.
a. Show that the golden ratio and golden ratio conjugate are both solutions of x² – x – 1 = 0.
Answer:
x² – x – 1 = 0
Solution1, Solution 2 = \(\frac{-b + \sqrt{b² – 4 a c}}{2 a}\), \(\frac{-b – \sqrt{b² – 4 a c}}{2 a}\)
Compare the give equation with
ax² +  x + c = 0
a = 1, b = -1, and c = -1
Solution1, Solution 2 = \(\frac{1 + \sqrt{(-1)² – 4 (1) (-1)}}{2 (1)}\), \(\frac{1 – \sqrt{(-1)² – 4 (1) (-1)}}{2 (1)}\)
Solution1, Solution 2 = \(\frac{1 + \sqrt{1 + 4}}{2}\), \(\frac{1 – \sqrt{1 + 4}}{2}\)
Solution1, Solution 2 = \(\frac{1 + \sqrt{5}}{2}\), \(\frac{1 – \sqrt{5}}{2}\)
The golden ratio and golden ratio conjugate are both solutions of the given equation

b. Construct a geometric diagram that has the golden ratio as the length of a part of the diagram.
Answer:
The golden ratio is: \(\frac{1 + \sqrt{5}}{2}\)
\(\frac{1 + \sqrt{5}}{2}\) = 1.61
The representation of the golden ratio as the length of the part of the diagram is:

Question 107.
CRITICAL THINKING
Use the special product pattern (a + b)(a² – ab + b²) = a³ + b³ to simplify the expression \(\frac{2}{\sqrt[3]{x}+1}\). Explain your reasoning.
Answer:

Maintaining Mathematical Proficiency

Graph the linear equation. Identify the x-intercept.
Question 108.
y = x – 4
Answer:
y = x – 4
y = mx + c
m is the slope
c is the y-intercept
To find the x-intercept,put y = 0
x – 4 = 0
x = 4

The x-intercept of the linear equation is: 4

Question 109.
y = -2x + 6
Answer:

Question 110.
y = –\(\frac{1}{3}\)x – 1
Answer:
y = –\(\frac{1}{3}\)x – 1
y = mx + c
Where,
m is the slope
c is the y-intercept
To find the x-intercept,put y = 0
So,
–\(\frac{1}{3}\)x – 1 = 0
–\(\frac{1}{3}\)x = 1
x = -3

The x-intercept of the linear equation is: -3

Question 111.
y = \(\frac{3}{2}\)x + 6
Answer:

Solve the equation. Check your solution.
Question 112.
32 = 2^x
Answer:
32 = 2^x
2^5 = 2^x
x = 5

Question 113.
27^x = 3^{x – 6}
Answer:

Question 114.
(\(\frac{1}{6}\))^2x = 216^{1 – x}
Answer:
(\(\frac{1}{6}\))^2x = 216^{1 – x}
\(\frac{1}{x}\) = x^-1
[6^-1]^2 x = [6^3]^1 – x
6^[-2x] = 6^[3(1 – x)]
-2x = 3 (1 – x)
-2x = 3 – 3x
-2x + 3x = 3
x = 3

Question 115.
625^x = (\(\frac{1}{25}\))^{x + 2}
Answer:

Lesson 9.2 Solving Quadratic Equations by Graphing

Essential Question How can you use a graph to solve a quadratic equation in one variable?

Based on what you learned about the x-intercepts of a graph in Section 3.4, it follows that the x-intercept of the graph of the linear equation
y = ax + b —-> 2 variables
is the same value as the solution of
ax + b = 0 —–> 1 variable
You can use similar reasoning to solve quadratic equations.

EXPLORATION 1

Solving a Quadratic Equation by Graphing
Work with a partner.
a. Sketch the graph of y = x² – 2x.

Answer:

b. What is the definition of an x-intercept of a graph? How many x-intercepts does this graph have? What are they?
Answer:
The x-intercept is the point where a line crosses the x-axis,
There are 2 points where a line crosses the x-axis
The 2 points that a line crosses the x-axis are 0, 2

c. What is the definition of a solution of an equation in x? How many solutions does the equation x² – 2x = 0 have? What are they?
Answer:
x² – 2x = 0
x (x – 2) = 0
x = 0 or x – 2 = 0
x = 0 or x = 0 + 2
x = 0 or x = 2
The 2 solutions of the given equation are 0 and 2

d. Explain how you can verify the solutions you found in part (c).
Answer:
Put x = 0 in the given equation
0 – 2 (0) = 0
0 = 0
Put x = 2 in the given equation
2² – 2 (2) = 0
4 – 4 = 0
0 = 0

EXPLORATION 2

Solving Quadratic Equations by Graphing
Work with a partner. Solve each equation by graphing.
a. x² – 4 = 0
Answer:

The solutions that satisfy the above equation i.e., x-intercepts are -2 and 2
b. x² + 3x = 0
Answer:

The solutions that satisfy the above equation i.e., x-intercepts are: 0 and -3

c. -x² + 2x = 0
Answer:

The solutions that satisfy the above equation i.e., x-intercepts are 0 and -2

d. x² – 2x + 1 = 0
Answer:

The solution that satisfies the above equation i.e., x-intercepts is: 1

e. x² – 3x + 5 = 0
Answer:

The approximate solutions that satisfy the above equation i.e., x-intercepts are 0.9 and 2.1

f. -x² + 3x – 6 = 0
Answer:

The solutions that satisfy the above equation i.e., x-intercepts are 1 and 2

Communicate Your Answer

Question 3.
How can you use a graph to solve a quadratic equation in one variable?

Answer:
y = ax + b —-> 2 variables
is the same value as the solution of
ax + b = 0 —–> 1 variable
You can use similar reasoning to solve quadratic equations.

Question 4.
After you find a solution graphically, how can you check your result algebraically? Check your solutions for parts (a)-(d) in Exploration 2 algebraically.
Answer:
a. x² – 4 = 0
The solutions of the given equation are -2 and 2
(-2)² -4 = 0
4 – 4 = 0
2² – 4 = 0
4 – 4 = 0
0 = 0
b. x² + 3x = 0
The solutions of the given equation are 0 and -3
0² + 3 (0) = 0
0 + 0 = 0
0 = 0
(-3)² + 3 (-3) = 0
9 – 9 = 0
0 = 0
c. -x² + 2x = 0
The solutions of the given equation are 0 and 2
0² + 2 (0) = 0
0 + 0 = 0
0 = 0
-(2)² + 2 (2) = 0
4 – 4 = 0
0 = 0
d. x² – 2x + 1 = 0
The solution of the given equation are 1
1² – 2 (1) + 1 = 0
1 – 2 + 1 = 0
2 – 2 = 0
0 = 0

Question 5.
How can you determine graphically that a quadratic equation has no solution?
Answer:
For any quadratic equation, the graph will be a parabola. If the parabola does not cross or intersect the x-axis, then the equation has no real solution. And no real solution does not mean that there is no solution, but that the solutions are not real numbers.

Monitoring Progress

Solve the equation by graphing. Check your solutions.
Question 1.
x² – x – 2 = 0
Answer:

We can observe the solutions that satisfy the given equation i.e., x-intercepts from the graph are -1 and 2
(-1)² – (-1) – 2 = 0
1 +1 – 2 = 0
2 – 2 = 0
0 = 0
(2)² – 2 + 2 = 0
4 – 2 + 2 = 0
2 – 2 = 0
0 = 0

Question 2.
x² + 7x = -10
Answer:

We can observe the solutions that satisfy the given equation i.e., x-intercepts from the graph are: -2 and -5
(-2)² + 7(-2) + 10 = 0
4 – 14 + 10 = 0
14 – 14 = 0
0 = 0
(-5)² + 7(-5) + 10 = 0
25 – 35 + 10 = 0
35 – 35 = 0
0 = 0

Question 3.
x² + x = 12
Answer:

We can observe the solutions that satisfy the given equation i.e., x-intercepts from the graph are -4 and 3
(-4)² -4 – 12 = 0
16 – 4 – 12 = 0
12 – 12 = 0
0 = 0
(3)² + 3 – 12 = 0
9 – 12 + 3 = 0
12 – 12 = 0
0 = 0

Solve the equation by graphing.
Question 4.
x² + 36 = 12x
Answer:

The solution of the given equation is 6

Question 5.
x² + 4x = 0
Answer:

The solutions of the given equation are 0 and -4

Question 6.
x² + 10x = -25
Answer:

The solution of the given equation is -5

Question 7.
x² = 3x – 3
Answer:

There are no real solutions for the given equation

Question 8.
x² + 7x = -6
Answer:

The solutions of the given equation are -1 and -6

Question 9.
2x + 5 = -x²
Answer:

There are no real solutions for the given equation

Question 10.
Graph f(x) = x² + x – 6. Find the zeros of f.
Answer:
f(x) = x² + x – 6
To find the zeroes of f(x), make f(x) = 0
x² + x – 6 = 0
x² + 3x – 2x – 6 = 0
x (x + 3) -2 (x + 3) = 0
(x – 2) (x + 3) = 0
x – 2 = 0 or x + 3 = 0
x = 2 or x = -3

The zeroes of the given equation are -3 and 2

Question 11.
Graph f(x) = -x² + 2x + 2. Approximate the zeros of f to the nearest tenth.
Answer:

Question 12.
WHAT IF?
After how many seconds is the football 65 feet above the ground?
Answer:
The function ‘h’ that represents the height (h) in feet of the football after t seconds is:
h = -16t² + 75t + 2
-16t² + 75t + 2 = 65
-16t² + 75t + 2 – 65 = 0
-16t² + 75t – 63 = 0

The football is 65 feet above the ground after 1 second and after about 3.8 seconds

Solving Quadratic Equations by Graphing 9.2 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What is a quadratic equation?
Answer:

Question 2.
WHICH ONE DOESN’T BELONG?
Which equation does not belong with the other three? Explain your reasoning.

Answer:

x² – 7x + 12 = 0 does not belong with the other three

Question 3.
WRITING
How can you use a graph to find the number of solutions of a quadratic equation?
Answer:

Question 4.
WRITING
How are solutions, roots, x-intercepts, and zeros related?
Answer:
Zeroes, roots, and x-intercepts are all names for values that make a function equal to zero. Zeros, roots, or x-intercepts of a function f(x) are the values of x where the graph of the function f crosses the x-axis.

Monitoring Progress and Modeling with Mathematics

In Exercises 5–8, use the graph to solve the equation.
Question 5.
-x² + 2x + 3 = 0

Answer:

Question 6.
x² – 6x + 8 = 0

Answer:

We can observe that there are two x-intercepts i.e., two solutions for the  given equation
The two x-intercepts of the given equation are: 2 and 4

Question 7.
x² + 8x + 16 = 0

Answer:

Question 8.
-x² – 4x – 6 = 0

Answer:
There are no x-intercepts i.e., no real solutions for the  given equation

In Exercises 9–12, write the equation in standard form.
Question 9.
4x² = 12
Answer:

Question 10.
-x² = 15
Answer:
-x² = 15
-x² – 15 = 0
– (x² + 15) = 0
x² + 15 = 0

Question 11.
2x – x² = 1
Answer:

Question 12.
5 + x = 3x²
Answer:
5 + x = 3x²
5 + x – 3x² = 0
– (3x² – x – 5) = 0
5 + x – 3x² = 0

In Exercises 13–24, solve the equation by graphing.
Question 13.
x² – 5x = 0
Answer:

Question 14.
x² – 4x + 4 = 0
Answer:

The solution of the given equation is 2

Question 15.
x² – 2x + 5 = 0
Answer:

Question 16.
x² – 6x – 7 = 0
Answer:

The solutions of the given equation are -1 and 7

Question 17.
x² – 6x = 9
Answer:

Question 18.
-x² = 8x + 20
Answer:

There are no real solutions for the given equation

Question 19.
x² = -1 – 2x
Answer:

Question 20.
x² = -x – 3
Answer:

There are no real solutions for the given equation

Question 21.
4x – 12 = -x²
Answer:

Question 22.
5x – 6 = x²
Answer:

The solutions of the given equation are 2 and 3

Question 23.
x² – 2 = -x
Answer:

Question 24.
16 + x² = -8x
Answer:
16 + x² = -8x
x² + 8x + 16 = 0

The solution of the given equation is -4

Question 25.
ERROR ANALYSIS
Describe and correct the error in solving x² + 3x = 18 by graphing.

Answer:

Question 26.
ERROR ANALYSIS
Describe and correct the error in solving x² + 6x + 9 = 0 by graphing.

Answer:

The solution of the given equation is x = -3

Question 27.
MODELING WITH MATHEMATICS
The height y (in yards) of a flop shot in golf can be modeled by y = -x² + 5x, where x is the horizontal distance (in yards).

a. Interpret the x-intercepts of the graph of the equation.
b. How far away does the golf ball land?
Answer:

Question 28.
MODELING WITH MATHEMATICS
The height h (in feet) of an underhand volleyball serve can be modeled by h = -16t² + 30t + 4, where t is the time (in seconds).
a. Do both t-intercepts of the graph of the function have meaning in this situation? Explain.
Answer:
h = -16t² + 30t + 4
Where,
t is the time in seconds

The two t-intercepts of the given equation are 0 and 2
From the two intercepts of t,
We can observe that both intercepts have the meaning
When t = 0,
The underhand volleyball can’t be served
When t = 2,
The underhand vollyball can be served
The two intercepts of t have the meaning in this situation

b. No one receives the serve. After how many seconds does the volleyball hit the ground?
Answer:
We can observe that the t-intercepts are 0 and 2
The time that the volleyball hit the ground = 2 – 0 = 2 seconds
After 2 seconds, the volleyball hit the ground

In Exercises 29–36, solve the equation by using Method 2 from Example 3.
Question 29.
x² = 10 – 3x
Answer:

Question 30.
2x – 3 = x²
Answer:
2x – 3 = x²
y = x²
y = 2x – 3

The given equation does not have any real solution

Question 31.
5x – 7 = x²
Answer:

Question 32.
x² = 6x – 5
Answer:
x² = 6x – 5
y = x²
y = 6x – 5

The graphs of two equations intersect at (1, 1)
The solution of the given equation is 1

Question 33.
x² + 12x = -20
Answer:

Question 34.
x² + 8x = 9
Answer:
x²  + 8x = 9
x² = -8x + 9
y = x²
y = -8x + 9

The graphs of two equations intersect at (1, 1)
The solution of the given equation is 1

Question 35.
-x² – 5 = -2x
Answer:

Question 36.
-x² – 4 = -4x
Answer:
– x² – 4 = -4x
-x² = -4x + 4
x² = 4x – 4
y = x²
y = 4x – 4

The graphs of two equations intersect at (2, -4)
The solution of the given equation is 2

In Exercises 37–42, find the zero(s) of f.
Question 37.

Answer:

Question 38.

Answer:
f(x) = (x + 1) (x² + 6x + 8)
f(x) = 0
(x + 1) (x² + 6x + 8) = 0
x + 1 = 0 (or)  x² + 6x + 8 = 0
x + 1 = 0 (or)  x² + 4x + 2x + 8 = 0
x + 1 = 0 (or)  x (x + 4) + 2 (x + 4) = 0
x + 1 = 0 (or) (x + 2) (x + 4) = 0
x + 1 = 0 (or) x + 2 = 0 (or) x + 4 = 0
x = -1 (or) x = -2 (or) x = -4
The zeroes of the given function are -1, -2, and -4

Question 39.

Answer:

Question 40.

Answer:
f(x) = (x – 5) (-x² + 3x – 3)
f(x) = 0
(x – 5) (-x² + 3x – 3) = 0
x – 5 = 0 (or)  -x² + 3x – 3 = 0
-x² + 3x – 3 = 0,
x – 5 = 0
x = 5
The zero of the given function is -5

Question 41.

Answer:

Question 42

Answer:
f(x) = (x² + 1) (x² – x – 2)
f(x) = 0
(x² + 1) (x² – x – 2) = 0
x² + 1 = 0 (or)  x² – x – 2 = 0
x² + 1 = 0 (or)  x² + x – 2x – 2 = 0
x² + 1 = 0 (or)  x (x + 1) – 2 (x + 1) = 0
x² + 1 = 0 (or) (x + 1) (x – 2) = 0
x² + 1 = 0 (or) x + 1 = 0 (or) x – 2 = 0
x² = -1 (or) x = -1 (or) x = 2
x² + 1 = 0,
The zeroes of the given function are -1, and 2

In Exercises 43–46, approximate the zeros of f to the nearest tenth.
Question 43.

Answer:

Question 44.

Answer:

Question 45.

Answer:

Question 46.

Answer:
There are two x intercepts, one between 0 and 1, and the other between 5 and 6.

X	0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9
F(x)	-1.41	-0.24	-0.29	0.24	0.75	1.24	1.71	2.16	2.59

Tabulating x values between 5 and 6

X	5.1	5.2	5.3	5.4	5.5	5.6	5.7	5.8	5.9
F(x)	2.59	2.16	1.71	1.24	0.75	0.24	-0.29	-0.84	-1.41

Thus the function has zeros at x = 0.4, 0.5 approximate to the nearest tenth.

In Exercises 47–52, graph the function. Approximate the zeros of the function to the nearest tenth, if necessary.
Question 47.
f(x) = x² + 6x + 1
Answer:

Question 48.
f(x) = x² – 3x + 2
Answer:

Question 49.
y = -x² + 4x – 2
Answer:

Question 50.
y = -x² + 9x – 6
Answer:

Question 51.
f(x) = \(\frac{1}{2}\)x² + 2x – 5
Answer:

Question 52.
f(x) = -3x² + 4x + 3
Answer:

Question 53.
MODELING WITH MATHEMATICS
At a Civil War reenactment, a cannonball is fired into the air with an initial vertical velocity of 128 feet per second. The release point is 6 feet above the ground. The function h = -16t² + 128t + 6 represents the height h (in feet) of the cannonball after t seconds.
a. Find the height of the cannonball each second after it is fired.
b. Use the results of part (a) to estimate when the height of the cannonball is 150 feet.
c. Using a graph, after how many seconds is the cannonball 150 feet above the ground?

Answer:

Question 54.
MODELING WITH MATHEMATICS
You throw a softball straight up into the air with an initial vertical velocity of 40 feet per second. The release point is 5 feet above the ground. The function h = -16t² + 40t + 5 represents the height h (in feet) of the softball after t seconds.

a. Find the height of the softball each second after it is released.
Answer:
b. Use the results of part (a) to estimate when the height of the softball is 15 feet.
c. Using a graph, after how many seconds is the softball 15 feet above the ground?
Answer:

MATHEMATICAL CONNECTIONS In Exercises 55 and 56, use the given surface area S of the cylinder to find the radius r to the nearest tenth.
Question 55.
S = 225 ft²

Answer:

Question 56.
S = 750 m²

Answer:
S = 750 m²
The surface area of a cylinder (S) = 2πr² + 2πrh
h = 13 m
750 = 2πr² + 2πrh
750 = 2πr² + 2πr (13)
750 = 2πr (r + 13)
375 = πr² + 13πr
πr² + 13πr – 375 = 0

The radius of the given cylinder is 6.21

Question 57.
WRITING
Explain how to approximate zeros of a function when the zeros are not integers.
Answer:

Question 58.
HOW DO YOU SEE IT?
Consider the graph shown.

a. How many solutions does the quadratic equation x² = -3x + 4 have? Explain.
Answer:
x² = -3x + 4
x² + 3x – 4 = 0

The solutions of the given equation are -4 and 1

b. Without graphing, describe what you know about the graph of y = x² + 3x – 4.
Answer:
y = x² + 3x – 4
y = 0
x² + 3x – 4 = 0
x² + 4x – x – 4 = 0
x (x + 4) – 1 (x + 4) = 0
(x – 1) (x + 4) = 0
x – 1 = 0 or x + 4 = 0
x = 1 or x = -4

Question 59.
COMPARING METHODS
Example 3 shows two methods for solving a quadratic equation. Which method do you prefer? Explain your reasoning.
Answer:

Question 60.
THOUGHT PROVOKING
How many different parabolas have -2 and 2 as x-intercepts? Sketch examples of parabolas that have these two x-intercepts.
Answer:
There is only 1 parabola with -2 and 2 as x-intercepts
If the equation has 2 solutions, then that equation will be a quadratic equation
The equation with -2 and 2 as x-intercepts are:
(x+ 2) (x – 2) = 0

Question 61.
MODELING WITH MATHEMATICS
To keep water off a road, the surface of the road is shaped like a parabola. A cross-section of the road is shown in the diagram. The surface of the road can be modeled by y = -0.0017x² + 0.041x, where x and y are measured in feet. Find the width of the road to the nearest tenth of a foot.

Answer:

Question 62.
MAKING AN ARGUMENT
A stream of water from a fire hose can be modeled by y = -0.003x² + 0.58x + 3,where x and y are measured in feet. A firefighter is standing 57 feet from a building and is holding the hose 3 feet above the ground. The bottom of a window of the building is 26 feet above the ground. Your friend claims the stream of water will pass through the window. Is your friend correct? Explain.
Answer:

y = -0.003x² + 0.58x + 3
y = -0.003(57)² + 0.58(57) + 3
y = -9.747 + 33.06 + 3
y = 26.313
So, 26.313 > 26 and the stream of water passes through the window.

REASONING In Exercises 63–65, determine whether the statement is always, sometimes, or never true. Justify your answer.
Question 63.
The graph of y = ax² + c has two x-intercepts when a is negative.
Answer:

Question 64.
The graph of y = ax² + c has no x-intercepts when a and c have the same sign.
Answer:
y = ax² + c

The given statement is always false

Question 65.
The graph of y = ax² + bx + c has more than two x-intercepts when a ≠ 0.
Answer:

Maintaining Mathematical Proficiency

Determine whether the table represents an exponential growth function, an exponential decay function, or neither. Explain.
Question 66.

Answer:

The given table is exponential

Question 67.

Answer:

Lesson 9.3 Solving Quadratic Equations Using Square Roots

Essential Question
How can you determine the number of solutions of a quadratic equation of the form ax² + c = 0?
Answer:
Quadratic functions can have 0,1, or 2 solutions. This depends on the discriminant.
x = (-b ±\(\sqrt{ b² – 4ac}\)/2a

EXPLORATION 1

The Number of Solutions of ax² + c = 0
Work with a partner. Solve each equation by graphing. Explain how the number of solutions of ax² + c = 0 relates to the graph of y = ax² + c.
a. x² – 4 = 0
Answer:
x² – 4 = 0
x² – 4 + 4 = 0 + 4
x² = 4
√x² = √4
x = 2 or x = -2

There are 2 real solutions for the given equation i.e., -2 and 2

b. 2x² + 5 = 0
Answer:
2x² + 5 = 0
2x² + 5 – 5 = 0 – 5
2x² = -5
x² = –\(\frac{5}{2}\)
√x² = -√\(\frac{5}{2}\)

The given equation does not have any real solutions.

c. x² = 0
Answer:
x² = 0
x = 0

The given equation has only one solution i.e., 0

d. x² – 5 = 0
Answer:
x² – 5 = 0
x² – 5 + 5 = 0 + 5
x² = 5
√x² = √5
x = √5 or x = -√5

There are 2 real solutions i.e., √5 and -√5

EXPLORATION 2

Estimating Solutions
Work with a partner. Complete each table. Use the completed tables to estimate the solutions of x² – 5 = 0. Explain your reasoning.

Answer:
The completed tables are:

x² – 5 = 0
The solutions of the given equation are: 2.24 and -2.24

EXPLORATION 3

Using Technology to Estimate Solutions
Work with a partner. Two equations are equivalent when they have the same solutions.

a. Are the equations x² – 5 = 0 and x² = 5 equivalent? Explain your reasoning
Answer:
x² – 5 = 0
x² – 5 + 5= 0 + 5
x²= 5
x² – 5 and x² = 5 are equivalent

b. Use the square root key on a calculator to estimate the solutions of x² – 5 = 0. Describe the accuracy of your estimates in Exploration 2.
Answer:

c. Write the exact solutions of x² – 5 = 0.
Answer:
x² – 5 = 0
The exact solutions of the given equation are: 2.24 and -2.24

Communicate Your Answer

Question 4.
How can you determine the number of solutions of a quadratic equation of the form ax² + c = 0?
Answer:
Quadratic functions can have 0,1, or 2 solutions. This depends on the discriminant.
x = (-b ±\(\sqrt{ b² – 4ac}\) / 2a

Question 5.
Write the exact solutions of each equation. Then use a calculator to estimate the solutions.
a. x² – 2 = 0
Answer:

b. 3x² – 18 = 0
Answer:

c. x² – 8 = 0
Answer:
x² – 8 = 0
x² – 8+ 8 = 0 + 8
x² = 8
√x² = √8
x = √8 or x = -√8
The value of √8 is: 2.83

Monitoring Progress

Solve the equation using square roots.
Question 1.
-3x² = -75
Answer:
-3x² = -75
3x² = 75
x² = \(\frac{75}{3}\)
x² = 25
√x² = √25
x = 5 or x = -5

Question 2.
x² + 12 = 10
Answer:
x² + 12 = 10
x² = 10 – 12
x² = -2
√x² = √-2
The given equation does not have any real solutions

Question 3.
4x² – 15 = -15
Answer:
4x² – 15 = -15
4x² = -15 + 15
4x² = 0
x² = 0
√x² = √0
x = 0

Question 4.
(x + 7)² = 0
Answer:
(x + 7)² = 0
√(x + 7)² = √0
x + 7 = 0
x = 0 – 7
x = -7

Question 5.
4(x – 3)² = 9
Answer:
4 (x – 3)² = 9
(x – 3)² = \(\frac{9}{4}\)
√(x – 3)² = √\(\frac{9}{4}\)
x – 3 = \(\frac{3}{2}\) or x – 3 = –\(\frac{3}{2}\)
x = \(\frac{3}{2}\) + 3 or x = –\(\frac{3}{2}\) + 3
x = \(\frac{9}{2}\) or x = \(\frac{3}{2}\)]

Question 6.
(2x + 1)² = 36
Answer:
(2x + 1)² = 36
√(2x + 1)² = √36
2x + 1 = 6 or 2x + 1 = -6
2x = 6 – 1 or 2x = -6 – 1
2x = 5 or 2x = -7
x = \(\frac{5}{2}\) or x = –\(\frac{7}{2}\)

Solve the equation using square roots. Round your solutions to the nearest hundredth.
Question 7.
x² + 8 = 19
Answer:
x² + 8 = 19
x² + 8 – 8 = 19 – 8
x² = 11
√x² = √11
x = √11 or x = -√11
The value of √11 is: 3.31

Question 8.
5x² – 2 = 0
Answer:
5x² – 2 = 0
5x² – 2 + 2 = 0 + 2
5x² = 2
x² = \(\frac{2}{5}\)
√x² = √\(\frac{2}{5}\)
x = √\(\frac{2}{5}\) or x = -√\(\frac{2}{5}\)
The value of \(\frac{2}{5}\) is: 0.40

Question 9.
3x² – 30 = 4
Answer:
3x² – 30 = 4
3x² = 4 + 30
3x²= 34
x² = \(\frac{34}{3}\)
√x² = √\(\frac{34}{3}\)
x = √\(\frac{34}{3}\) or x = -√\(\frac{34}{3}\)
The value of \(\frac{34}{3}\) is: 11.33

Question 10.
WHAT IF?
In Example 4, the volume of the tank is 315 cubic feet. Find the length and width of the tank.
Answer:
A touch tank has a height of 3 feet and its length is 3 times its width
The volume of the tank (V) = l × w × h
315 = 3w (w) (3)
315 = 9w²
w² = \(\frac{315}{9}\)
w² = 35
√w² = √35
w = √35 or w = -√35
The volume won’t be negative
The value of √35 is: 5.91
The width of the tank is: 5.91
The length of the tank is: 3 (5.91) = 17.74
The length of the tank is: 17.74
The width of the tank is 5.91

Question 11.
The surface area S of a sphere with radius r is given by the formula S = 4πr². Solve the formula for r. Then find the radius of a globe with a surface area of 804 square inches.

Answer:
The surface area S of a sphere is given by the formula
S = 4πr²
r² = \(\frac{S}{4π}\)
√r² = √\(\frac{S}{4π}\)
r = √\(\frac{S}{4π}\) or r = -√\(\frac{S}{4π}\)
The surface area S of the sphere won’t take negative values
r = √\(\frac{S}{4π}\)
r = √\(\frac{804}{4π}\)
r = 64.01 units
The radius of a globe with a surface area of 804 sq. units is: 64.01 units

Solving Quadratic Equations Using Square Roots 9.3 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The equation x² = d has ____ real solutions when d > 0.
Answer:

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers

Answer:
a, b, and d are the same because in these 3 equations,
x² = 144
But, in d,
x² = -144
The equations a, b, and d are the same whereas the equation c is different

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, determine the number of real solutions of the equation. Then solve the equation using square roots.
Question 3.
x² = 25
Answer:

Question 4.
x² = -36
Answer:
x² = -36
√x² = √-36
x = √-36
There are no real solutions for the given equation

Question 5.
x² = -21
Answer:

Question 6.
x² = 400
Answer:
x² = 400
√x² = √400
x = 20 or x = -20
There are 2 real solutions for the given equation i.e., 20 and -20

Question 7.
x² = 0
Answer:

Question 8.
x² = 169
Answer:
x² = 169
√x² = √169
x = 13 or x = -13
There are 2 real solutions for the given equation i.e., 13 and -13

In Exercises 9–18, solve the equation using square roots.
Question 9.
x² – 16 = 0
Answer:

Question 10.
x² + 6 = 0
Answer:
x² + 6 = 0
x² + 6 – 6 = 0 – 6
x² = -6
√x² = √-6
x = √-6
There are no real solutions for the given equation

Question 11.
3x² + 12 = 0
Answer:

Question 12.
x² – 55 = 26
Answer:
x² – 55 = 26
x² – 55 + 55 = 26 + 55
x² = 81
√x² = √81
x = 9 or x = -9

Question 13.
2x² – 98 = 0
Answer:

Question 14.
-x² + 9 = 9
Answer:
-x² + 9 = 9
-x² + 9 – 9 = 9 – 9
-x² = 0
x² = 0
√x² = √0
x = 0

Question 15.
-3x² – 5 = -5
Answer:

Question 16.
4x² – 371 = 29
Answer:
4x² – 371 = 29
4x² – 371 + 371 = 29 + 371
4x² = 400
x² = \(\frac{400}{4}\)
x² = 100
√x² = √100
x = 10 or x = -10

Question 17.
4x² + 10 = 11
Answer:

Question 18.
9x² – 35 = 14
Answer:
9x² – 35 = 14
9x² – 35 + 5 = 14 + 35
9x² = 49
x² = \(\frac{49}{9}\)
√x² = √\(\frac{49}{9}\)
x = \(\frac{7}{3}\) or x = –\(\frac{7}{3}\)

In Exercises 19–24, solve the equation using square roots.
Question 19.
(x + 3)² = 0
Answer:

Question 20.
(x – 1)² = 4
Answer:
(x – 1)² = 4
√(x – 1)² = √4
x – 1 = 2 or x – 1 = -2
x = 2 + 1 or x = -2 + 1
x = 3 or x = -1

Question 21.
(2x – 1)² = 81
Answer:

Question 22.
(4x + 5)² = 9
Answer:
(4x + 5)² = 9
√(4x + 5)² = √9
4x + 5 = 3 or 4x + 5 = -3
4x = 3 – 5 or 4x = -3 – 5
4x = -2 or 4x = -8
x = \(\frac{-2}{4}\) or x = \(\frac{-8}{4}\)
x = –\(\frac{1}{2}\) or x = -2

Question 23.
9(x + 1)² = 16
Answer:

Question 24.
4(x – 2)² = 25
Answer:
4 (x – 2)² = 25
(x – 2)² = \(\frac{25}{4}\)
√(x – 2)² = √\(\frac{25}{4}\)
x – 2 = \(\frac{5}{2}\) or x – 2 = –\(\frac{5}{2}\)
x= \(\frac{5}{2}\) + 2 or x = –\(\frac{5}{2}\) + 2
x = \(\frac{9}{2}\) or x = –\(\frac{1}{2}\)

In Exercises 25–30, solve the equation using square roots. Round your solutions to the nearest hundredth.
Question 25.
x² + 6 = 13
Answer:

Question 26.
x² + 11 = 24
Answer:
x² + 11 = 24
x² + 11 – 11 = 24 – 11
x² = 13
√x² = √13
x = √13 or x = -√13
The solutions of the given equation are: 3.60 and -3.60

Question 27.
2x² – 9 = 11
Answer:

Question 28.
5x² + 2 = 6
Answer:
5x² + 2 = 6
5x² + 2 – 2 = 6 – 2
5x² = 4
x² = \(\frac{4}{5}\)
√x² = √\(\frac{4}{5}\)
x = \(\frac{2}{√5}\) or x = –\(\frac{2}{√5}\)
\(\frac{2}{√5}\) = 0.89
The solutions of the given equation are 0.89 and -0.89

Question 29.
-21 = 15 – 2x²
Answer:

Question 30.
2 = 4x² – 5
Answer:
2 = 4x² – 5
2 + 5 = 4x² – 5 + 5
7 = 4x²
x² = \(\frac{7}{4}\)
√x² = √\(\frac{7}{4}\)
x = √\(\frac{7}{4}\) or x = -√\(\frac{7}{4}\)
The solutions of the given equation are 1.32 and -1.32

Question 31.
ERROR ANALYSIS
Describe and correct the error in solving the equation 2x² – 33 = 39 using square roots.

Answer:

Question 32.
MODELING WITH MATHEMATICS
An in-ground pond has the shape of a rectangular prism. The pond has a depth of 24 inches and a volume of 72,000 cubic inches. The length of the pond is two times its width. Find the length and width of the pond.

Answer:
The volume of a rectangular prism (V) = l × w × h
It is also given that the length of the pond is two times its width
V = (2w) × w × h
72,000 = 2w² × 24
2w² = \(\frac{72,000}{24}\)
2w² = 3,000
w² = \(\frac{3,000}{2}\)
w² = 1,500
√w² = √1,500
w = √1,500 or w = -√1,500
w = √1,500
The width of the rectangular prism = 38.72 inches
The length of the rectangular prism = 2 (38.72) = 77.45 inches

Question 33.
MODELING WITH MATHEMATICS
A person sitting in the top row of the bleachers at a sporting event drops a pair of sunglasses from a height of 24 feet. The function h = -16x² + 24 represents the height h (in feet) of the sunglasses after x seconds. How long does it take the sunglasses to hit the ground?
Answer:

Question 34.
MAKING AN ARGUMENT
Your friend says that the solution of the equation x² + 4 = 0 is x = 0. Your cousin says that the equation has no real solutions. Who is correct? Explain your reasoning.
Answer:
x² + 4 = 0
x² + 4 – 4 = 0- 4
x² = -4
The square of a real number can’t take any negative numbers
There are no real solutions for the given equation
Your cousin is correct

Question 35.
MODELING WITH MATHEMATICS
The design of a square rug for your living room is shown. You want the area of the inner square to be 25% of the total area of the rug. Find the side length x of the inner square.

Answer:

Question 36.
MATHEMATICAL CONNECTIONS
The area A of a circle with radius r is given by the formula A = πr².
a. Solve the formula for r.
Answer:
A = πr²
r² = \(\frac{A}{π}\)
√r² = √\(\frac{A}{π}\)
r = √\(\frac{A}{π}\) or r = -√\(\frac{A}{π}\)
The radius of the circle won’t take any negative values
r = √\(\frac{A}{π}\)

b. Use the formula from part (a) to find the radius of each circle.

Answer:
r = √\(\frac{A}{π}\)
For A = 113ft²,
r = √\(\frac{113}{π}\)
r = 5.99ft
For A = 810 in.²,
r = √\(\frac{1810}{π}\)
r = 24 in.
For A = 531m²,
r = √\(\frac{531}{π}\)
r = 13m

c. Explain why it is beneficial to solve the formula for r before finding the radius.
Answer:
It is beneficial to solve the formula for r before finding the radius because the formula for r is complicated due to the square root and it will become easy if we simplified the formula for r before finding the value of r

Question 37.
WRITING
How can you approximate the roots of a quadratic equation when the roots are not integers?
Answer:

Question 38.
WRITING
Given the equation ax² + c = 0, describe the values of a and c so the equation has the following number of solutions.
a. two real solutions
b. one real solution
c. no real solutions
Answer:
ax² + c = 0
ax²  + c – c = 0 – c
ax² = -c
a. Two real solutions
For the equations to have 2 real solutions,
a —-> Positive
c —–> Negative
b. One real solution
For the equation to have 1 real solution,
a —–> Positive
c ——> Zero
c. No real solutions
For the equation to have no real solutions,
a —–> Positive
c —–> Positive

Question 39.
REASONING
Without graphing, where do the graphs of y = x² and y = 9 intersect? Explain.
Answer:

Question 40.
HOW DO YOU SEE IT?
The graph represents the function f(x) = (x – 1)². How many solutions does the equation (x – 1)² = 0 have? Explain.

Answer:
f (x) = (x – 1)²
f (x) = 0
(x – 1)² = 0
√(x – 1)² = √0
x – 1 = 0
x = 0 + 1
x = 1

Question 41.
REASONING
Solve x² = 1.44 without using a calculator. Explain your reasoning.
Answer:

Question 42.
THOUGHT-PROVOKING
The quadratic equation ax² + bx + c = 0 can be rewritten in the following form. \(\left(x+\frac{b}{2 a}\right)^{2}=\frac{b^{2}-4 a c}{4 a^{2}}\) Use this form to write the solutions of the equation.
Answer:

Question 43.
REASONING
An equation of the graph shown is y = \(\frac{1}{2}\)(x – 2)² + 1. Two points on the parabola have y-coordinates of 9. Find the x-coordinates of these points.

Answer:

Question 44.
CRITICAL THINKING
Solve each equation without graphing.
a. x² – 12x + 36 = 64
Answer:

b. x² + 14x + 49 = 16
Answer:

Maintaining Mathematical Proficiency

Factor the polynomial.
Question 45.
x² + 8x + 16
Answer:

Question 46.
x² – 4x + 4
Answer:
x² – 4x + 4
= x² – 2x – 2x + 4
= x (x – 2) – 2 (x – 2)
= (x – 2) ( x – 2)
= (x – 2)²

Question 47.
x² – 14x + 49
Answer:

Question 48.
x² + 18x + 81
Answer:
x² + 18x + 81
= x² + 9x + 9x + 81
= x (x + 9) + 9 (x + 9)
= (x + 9) (x + 9)
= (x + 9)²

Question 49.
x² + 12x + 36
Answer:

Question 50.
x² – 22x + 121
Answer:
x² – 22x + 121
= x² – 11x – 11x + 121
= x (x – 11) – 11 (x – 11)
= (x – 11) (x – 11)
= (x – 11)²

Solving Quadratic Equations Study Skills: Keeping a Positive Attitude

9.1–9.3 What Did You Learn?

Core Vocabulary

Core Concepts

Mathematical Practices

Question 1.
For each part of Exercise 100 on page 488 that is sometimes true, list all examples and counterexamples from the table that represent the sum or product being described.
Answer:

Question 2.
Which Examples can you use to help you solve Exercise 54 on page 496?
Answer:
You can use Example 6 on page 493 to solve Exercise 54 on page 496

Question 3.
Describe how solving a simpler equation can help you solve the equation in Exercise 41 on page 502.
Answer:
The equation in Exercise 41 on page 502 is in the form of:
x² – a = 0
If d> 0 i.e., the square is a positive number, then the only square root exists
We can solve Exercise 41 on page 502

Study Skills: Keeping a Positive Attitude

Do you ever feel frustrated or overwhelmed by math? You’re not alone. Just take a deep breath and assess the situation. Try to find a productive study environment, review your notes and the examples in the textbook, and ask your teacher or friends for help.

Solving Quadratic Equations 9.1 – 9.3 Quiz

Simplify the expression.
Question 1.
\(\sqrt{112 x^{3}}\)
Answer:
\(\sqrt{112 x^{3}}\)
= \(\sqrt{{112 x^{2}}\cdot x}\)
= \(\sqrt{16 × 7 x^{2}}\) ⋅ \(\sqrt{x}\)
= 4x \(\sqrt{7x}\)

Question 2.
\(\sqrt{\frac{18}{81}}\)
Answer:
\(\sqrt{\frac{18}{81}}\)
= \(\sqrt{\frac{18}{81}}\)
= \(\frac{\sqrt{18}}{\sqrt{81}}\)
= \(\frac{3\sqrt{2}}{9}\)
= \(\frac{\sqrt{2}}{3}\)

Question 3.
\(\sqrt[3]{-625}\)
Answer:
\(\sqrt[3]{-625}\)
= \(\sqrt[3]{-625}{1}\)
= \(\sqrt[3]{-625}\) ⋅ \(\sqrt[3]{1}\)
= \(\sqrt[3]{-125 × 5}\)
= \(\sqrt[3]{-125}\) ⋅ \(\sqrt[3]{5}\)
= -5\(\sqrt[3]{5}\)

Question 4.
\(\frac{12}{\sqrt{32}}\)
Answer:
\(\frac{12}{\sqrt{32}}\)
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{32}\)
= \(\frac{12}{\sqrt{32}}\) × (\(\sqrt{32}\) / \(\sqrt{32}\))
= \(12\sqrt{32}\) / (\(\sqrt{32}\) × \(\sqrt{32}\))
= \(12\sqrt{32}\) / 32
= \(3\sqrt{32}\) / 8

Question 5.
\(\frac{4}{\sqrt{11}}\)
Answer:
\(\frac{4}{\sqrt{11}}\)
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{11}\)
= \(\frac{4}{\sqrt{11}}\) × (\(\sqrt{11}\) / \(\sqrt{11}\))
= \(4\sqrt{11}\) / (\(\sqrt{11}\) × \(\sqrt{11}\))
= \(4\sqrt{11}\) / 11

Question 6.
\(\sqrt{\frac{144}{13}}\)
Answer:
\(\sqrt{\frac{144}{13}}\)
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{13}\)
\(\frac{\sqrt{144}}{\sqrt{13}}\)
= \(\frac{\sqrt{144}}{\sqrt{13}}\) × (\(\sqrt{13}\) / \(\sqrt{13}\))
= \(\sqrt{144 × 13}\) / (\(\sqrt{13}\) × \(\sqrt{13}\))
= 12\(\sqrt{13}\) / 13

Question 7.
\(\sqrt[3]{\frac{54 x^{4}}{343 y^{6}}}\)
Answer:
\(\sqrt[3]{\frac{54 x^{4}}{343 y ^{6}}}\)
= \(\frac{\sqrt[3]{54 x^{4}}}{\sqrt[3]{343 y^{6}}}\)
= \(\frac{\sqrt[3]{27 x^{3} × 2x}}{\sqrt[3]{343 y^{6}}}\)
= \(\frac{3x\sqrt[3]{ 2x}}{7 y^{2}}\)

Question 8.
\(\sqrt{\frac{4 x^{2}}{28 y^{4} z^{5}}}\)
Answer:
\(\sqrt{\frac{4 x^{2}}{28 y^{4} z^{5}}}\)
= \(\frac{\sqrt{4 x^{2}}}{\sqrt{28 y^{4} z^{5}}}\)
= \(\frac{x}{y^{2} z^{2}\sqrt{7z}}\)

Question 9.
\(\frac{6}{5+\sqrt{3}}\)
Answer:
\(\frac{6}{5 + \sqrt{3}}\)
To rationalize the denominator, multiply and divide the given fraction with \(5 – \sqrt{3}\)
= \(\frac{6}{5 + \sqrt{3}}\) × (\(5 – \sqrt{3}\) / \(5 – \sqrt{3}\))
= 6\((5 – \sqrt{3})\) / (\(5 + \sqrt{3}\) × \(5 – \sqrt{3}\))
= 6\((5 – \sqrt{3})\) / 22
= 6\((5 – \sqrt{3})\) / 22

Question 10.
2\(\sqrt{5}\) + 7\(\sqrt{10}\) – 3\(\sqrt{20}\)
Answer:
2\(\sqrt{5}\) + 7\(\sqrt{10}\) – 3\(\sqrt{20}\)
= 2\(\sqrt{5}\) + 7\(\sqrt{10}\) – 3\(\sqrt{5 × 4}\)
= 2\(\sqrt{5}\) + 7\(\sqrt{10}\) – 3 (2)\(\sqrt{5}\)
= 2\(\sqrt{5}\) + 7\(\sqrt{10}\) – 6\(\sqrt{5}\)
= \(\sqrt{5}\) ( 2 – 6) + 7\(\sqrt{10}\)
= \(\sqrt{5}\) (-4) + 7\(\sqrt{10}\)
= -4\(\sqrt{5}\) + 7\(\sqrt{10}\)

Question 11.
\(\frac{10}{\sqrt{8}-\sqrt{10}}\)
Answer:
\(\frac{10}{\sqrt{8}-\sqrt{10}}\)
To rationalize the denominator, multiply and divide the given fraction with \(\sqrt{8} +\sqrt{10} \)
= \(\frac{10}{\sqrt{8}-\sqrt{10}}\) × (\(\sqrt{8} +\sqrt{10} \) / \(\sqrt{8} +\sqrt{10} \))
= 10(\(\sqrt{8} +\sqrt{10} \)) / (\(\sqrt{8} -\sqrt{10} \) × \(\sqrt{8} +\sqrt{10} \))
= 10(\(\sqrt{8} +\sqrt{10}\)) / -2
= – 10(\(\sqrt{8} +\sqrt{10} \))/ 2

Question 12.
\(\sqrt{6}\)(7\(\sqrt{12}\) – 4\(\sqrt{3}\))
Answer:
\(\sqrt{6}\)(7\(\sqrt{12}\) – 4\(\sqrt{3}\))
= \(\sqrt{6}\)(-4\(\sqrt{3}\) + 7\(\sqrt{4 × 3}\))
= \(\sqrt{6}\)(-4\(\sqrt{3}\) + 7 (2)\(\sqrt{3}\))
= \(\sqrt{6}\)(-4\(\sqrt{3}\) + 14\(\sqrt{3}\))
= \(\sqrt{6}\)(\(\sqrt{3}\) (14 – 4))
= \(\sqrt{6}\)(\(\sqrt{3}\) (12))
= 12(\(\sqrt{3}\))(\(\sqrt{6}\) )
= 12\(\sqrt{18}\)
= 12\(\sqrt{9 × 2}\)
= 36\(\sqrt{2}\)

Use the graph to solve the equation.
Question 13.
x² – 2x – 3 = 0

Answer:
The solutions of the equations are the x-intercepts of those equations when we draw the graph for them
The solutions of the given equation are -1 and 3

Question 14.
x² – 2x + 3 = 0

Answer:
The solutions of the equations are the x-intercepts of those equations when we draw the graph for them
There are no real solutions for the given equation

Question 15.
x² + 10x + 25 = 0

Answer:
The solutions of the equations are the x-intercepts of those equations when we draw the graph for them
The solution of the given equation is -5

Solve the equation by graphing.
Question 16.
x² + 9x + 14 = 0
Answer:

The solutions of the given equation are -2 and -7

Question 17.
x² – 7x = 8
Answer:

The solutions of the given equation are -1 and 8

Question 18.
x + 4 = -x²
Answer:

There are no real solutions for the given equation

Solve the equation using square roots.
Question 19.
4x² = 64
Answer:

Question 20.
-3x² + 6 = 10
Answer:

There are no real solutions for the given equation

Question 21.
(x – 8)² = 1
Answer:

Question 22.
Explain how to determine the number of real solutions of x² = 100 without solving.
Answer:
x² = 100
x² – 100 = 0
x² -a = 0
The number of real solutions for the given equation is: 2

Question 23.
The length of a rectangular prism is four times its width. The volume of the prism is 380 cubic meters. Find the length and width of the prism.

Answer:
The length of a rectangular prism is 4 times its width
The height of the rectangular prism is: 5m
l = 4w
The volume of the rectangular prism = l × w × h
380 = w × 4w × 5
4w² = \(\frac{380}{5}\)
4w² = 76
w² = \(\frac{76}{4}\)
w² = 19
√w² = √19
w = √19 or w = -√19
The width of the rectangular prism can’t be negative
w = √19
l = 4√19

Question 24.
You cast a fishing lure into the water from a height of 4 feet above the water. The height h (in feet) of the fishing lure after t seconds can be modeled by the equation h = -16t² + 24t + 4.

a. After how many seconds does the fishing lure reach a height of 12 feet?
Answer:
h = -16t² + 24t + 4
12 = -16t² + 24t + 4
-16t² + 24t = 12 – 4
-16t² + 24t = 8
-2t² + 3t = 1
2t² – 3t + 1 = 0
2t² – 2t – t + 1 = 0
2t (t – 1) – 1 (t – 1) = 0
(2t – 1) (t – 1) = 0
2t – 1 = 0 or t – 1 = 0
2t = 1 or t = 1
t = \(\frac{1}{2}\) or t = 1

b. After how many seconds does the fishing lure hit the water?
Answer:
From part (a),
The values of t are: [laex]\frac{1}{2}[/latex] and 1
So,
After 1 second, the fishing lure will hit the water

Lesson 9.4 Solving Quadratic Equations by Completing the Square

Essential Question How can you use “completing the square” to solve a quadratic equation?
Answer:
The steps to “Completing the Square” to solve a quadratic equation are:
1. Divide all terms by a (The coefficient of x²)
2. Move the number term (c / a) to the right side of the equation
3. Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation

EXPLORATION 1

Solving by Completing the Square
Work with a partner.

a. Write the equation modeled by the algebra tiles. This is the equation to be solved.
Answer:

x² + 4x = -2

b. Four algebra tiles are added to the left side to “complete the square.” Why are four algebra tiles also added to the right side?
Answer:

The four algebra tiles are added to the right side for the purpose of “Completing the square”
The same four algebra tiles also have to be added to the right side to balance the equation
We added four algebra tiles to the right side also

c. Use algebra tiles to label the dimensions of the square on the left side and simplify on the right side.
Answer:

Blue tile (Represents the coefficient of x²)
Green tile (Represents the coefficient of x)
Yellow and Red tiles (represent the constant terms)

d. Write the equation modeled by the algebra tiles so that the left side is the square of a binomial. Solve the equation using square roots.
Answer:
x² + 4x = -2
x² + 4x + 4 = -2 + 4
(x + 2)² = 2
√(x + 2)² = √2
x + 2 = √2 or x + 2 = -√2
x = √2 – 2 or x = -(√2 + 2)

EXPLORATION 2

Solving by Completing the Square
Work with a partner.
a. Write the equation modeled by the algebra tiles.

Answer:
From the given algebra tiles representation,
The equation that can be modeled is x² + 6x = -5

b. Use algebra tiles to “complete the square.”
Answer:
The equation modeled by algebra tiles is x² + 6x = -5
To solve the quadratic equation by using algebra tiles,
x² + 6x + 9 = -5 + 9
(x + 3)² = 4

c. Write the solutions of the equation.
Answer:
(x + 3)² = 4
√(x + 3)² = √4
x + 3 = 2 or x + 3 = -2
x = 2 – 3 or x = -2 – 3
x = -1 or x = -5

d. Check each solution in the original equation.
Answer:
(x + 3)² = 4
Put x = -1 and -5 in the above equation
For x = -1,
(-1 + 3)² = 4
(2)² = 4
4 = 4
For x = -5,
(-5 + 3)² = 4
(-2)² = 4
4 = 4

Communicate Your Answer

Question 3.
How can you use “completing the square” to solve a quadratic equation?
Answer:
The steps to “Completing the Square” to solve a quadratic equation are:
1. Divide all terms by a (The coefficient of x²)
2. Move the number term (c / a) to the right side of the equation
3. Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation

Question 4.
Solve each quadratic equation by completing the square.
a. x² – 2x = 1
Answer:
x² – 2x = 1
x² – 2x – 1 = 0
x² – 2x = 1
x² – 2x + 1 = 1 + 1
(x – 1)² = 2

b. x² – 4x = -1
Answer:
x² – 4x = -1
x² – 4x + 1 = 0
x² – 4x = -1
x² – 4x + 4 = -1 + 4
(x – 2)² = 3

c. x² + 4x = -3
Answer:
x² + 4x = -3
x² + 4x + 3 = 0
x² + 4x = -3
x² + 4x + 4 = -3 + 4
(x + 2)² = 1

Monitoring Progress

Complete the square for the expression. Then factor the trinomial.
Question 1.
x² + 10x
Answer:
x² + 10x
ax² + bx
a = 1 and b = 10
\(\frac{b}{2}\) = \(\frac{10}{2}\)
\(\frac{b}{2}\) = 5
\(\frac{b²}{4}\) = 5² = 25
x² + 10x + 25 = (x + 5)²

Question 2.
x² – 4x
Answer:
x² – 4x
ax² + bx
a = 1 and b = -4
\(\frac{b}{2}\) = \(\frac{-4}{2}\)
\(\frac{b}{2}\) = -2
\(\frac{b²}{4}\) = (-2)² = 4
x² – 4x + 4 = (x – 2)²

Question 3.
x² + 7x
Answer:
x² + 7x
ax² + bx
a = 1 and b = 7
\(\frac{b}{2}\) = \(\frac{7}{2}\)
\(\frac{b²}{4}\)
= \(\frac{49}{4}\)
x² + 7x + \(\frac{49}{4}\) = (x + \(\frac{7}{2}\))²

Solve the equation by completing the square. Round your solutions to the nearest hundredth, if necessary.
Question 4.
x² – 2x = 3
Answer:
x² – 2x = 3
x² + bx = d
b = -2 and d = 3
\(\frac{b²}{4}\) = \(\frac{4}{4}\) = 1
x² – 2x + 1 = 3 + 1
(x – 1)² = 4
√(x – 1)² = √4
x – 1 = 2 or x – 1 = -2
x = 2 + 1 or x = -2 + 1
x = 3 or x = -1

Question 5.
m² + 12m = -8
Answer:
m² + 12m = -8
x² + bx = d
b = 12 and d = -8
\(\frac{b²}{4}\) = \(\frac{144}{4}\) = 36
m² + 12m + 36 = -8 + 36
(m + 6)² = 28
√(m + 6)² = √28
m + 6 = √28 or m + 6 = -√28
m = √28 – 6 or m = -(√28 + 6)

Question 6.
3g² – 24g + 27 = 0
Answer:
3g² – 24g + 27 = 0
3(g² – 8g + 9) = 0
g² – 8g + 9 = 0
g² – 8g = -9
x² + bx = d
b = -8 and d = -9
\(\frac{b²}{4}\) = \(\frac{64}{4}\) = 16
g² – 8g + 16 = -9 + 16
(g – 4)² = 7
√(g – 4)² = √7
g – 4 = √7 or g – 4 = -√7
g = √7 + 4 or g = -√7 + 4

Determine whether the quadratic function has a maximum or minimum value. Then find the value.
Question 7.
y = -x² – 4x + 4
Answer:
y = -x² – 4x + 4
y – 4 = -x²  – 4x + 4 – 4
y – 4 = -x²  – 4x
y – 4 = – (x² + 4x)
y – 4 – 4 = -(x² + 4x + 4)
y  – 4= -(x + 2)²
y = -(x + 2)² + 4

The given equation will have a minimum value
The minimum value of the given equation is -4

Question 8.
y = x² + 12x + 40
Answer:
y = x² + 12x + 40
y – 40 = x²  + 12x + 40 – 40
y – 40 = x²  + 12x
y – 40 + 36 = x² + 12x + 36
y  – 4= (x + 6)²
y = (x + 6)² + 4

We can observe that the graph is in an open shape
The given equation will have a maximum value
The maximum value of the given equation is 4

Question 9.
y = x² – 2x – 2
Answer:
y = x² – 2x – 2
y + 2 = x²  – 2x – 2 + 2
y + 2 = x²  – 2x
y + 2 + 1 = x² – 2x + 1
y  + 3= (x – 1)²
y = (x – 1)² – 3

We can observe that the graph is in an open shape
The given equation will have a maximum value
The maximum value of the given equation is 3

Determine whether the function could be represented by the graph in Example 6. Explain.
Question 10.
h(x) = (x – 8)² + 10
Answer:
h (x) = (x – 8)² + 10
The vertex point that is the solution for the given equation is (8, 10)
Since the values of x and y are positive, the graph must be in the 1st quadrant

Question 11.
n(x) = -2(x – 5)(x – 20)
Answer:
n (x) = -2 (x – 5) (x – 20)
The vertex point that is the solution for the given equation is (-5, -20)
Since the values of x and y are positive, the graph must be in the 3rd quadrant

Question 12.
WHAT IF?
Repeat Example 7 when the function is y = -16x² + 128x.
Answer:
y = -16x² + 128x
y = -16(x² – 8x)
y – 16 = -16(x² – 8x + 16)
y – 16 = -16(x – 4)²
y = -16 (x – 4)² + 16
a) The maximum height of the rocket:
We can observe that the maximum value is 16
We can conclude that the model rocket reaches a maximum height of 16 feet
b) The axes of symmetry:
From the given equation that we obtained in the vertex form,
The vertex is (4, 16)
The axis of symmetry is x = 4

Question 13.
WHAT IF?
You want the chalkboard to cover 4 square feet. Find the width of the border to the nearest inch.
Answer:
From Example 8,

The length of the chalkboard = 7 – 2x
The width of the chalkboard = 3 – 2x
Area of the rectangle = Length × Width
4 = (7 – 2x) (3 – 2x)
21 – 14x – 6x + 4x² = 4
4x² – 20x + 21 = 4
4x² – 20x = 4 – 21
4x² – 20x = -17
4 (x² – 5x) = -17
x² – 5x = –\(\frac{17}{4}\)
x² – 5x + \(\frac{25}{4}\) = –\(\frac{17}{4}\) + \(\frac{25}{4}\)
x² – 5x + \(\frac{25}{4}\) = 2
(x – \(\frac{5}{2}\))² = 2
√(x – \(\frac{5}{2}\))² = √2
x – \(\frac{5}{2}\) = √2 or x – \(\frac{5}{2}\) = -√2
x = 1.414 + 2.5 or x = -1.414 + 2.5
x = 3.91 or x = 1.08
The width of the border = 1.08 foot
1.08 × 12 inches = 12.96 inches
The width of the border in inches is 12.96 inches

Solving Quadratic Equations by Completing the Square 9.4 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The process of adding a constant c to the expression x² + bx so that x² + bx + c is a perfect square trinomial is called ________________.
Answer:

Question 2.
VOCABULARY
Explain how to complete the square for an expression of the form x² + bx.
Answer:
The steps that are required to complete the square for an expression x² + bx are:
1. Find one half of b i.e., \(\frac{b}{2}\)
2. Find \(\frac{b²}{4}\) and add it to the expression x² + bx
x² + bx + \(\frac{b²}{4}\)

Question 3.
WRITING
Is it more convenient to complete the square for x² + bx when b is odd or when b is even? Explain.
Answer:

Question 4.
WRITING
Describe how you can use the process of completing the square to find the maximum or minimum value of a quadratic function.
Answer:
There are two ways to find the absolute maximum/minimum value for f(x) = ax² + bx + c
f(x) = a(x − h)² + k, and the absolute maximum/minimum value is k and it occurs at x = h.
If a > 0, then the parabola opens up, and it is a minimum functional value of f (x)

Monitoring Progress and Modeling with Mathematics

In Exercises 5–10, find the value of c that completes the square.
Question 5.
x² – 8x + c
Answer:

Question 6.
x² – 2x + c
Answer:
x² – 2x + c
x² + bx + c
b = -2
x² + bx + \(\frac{b²}{4}\)
c = \(\frac{b²}{4}\)
c = \(\frac{4}{4}\)
c = 1

Question 7.
x² + 4x + c
Answer:

Question 8.
x² + 12x + c
Answer:
x² + 12x + c
x² + bx + c
b = 12
x² + bx + \(\frac{b²}{4}\)
c = \(\frac{b²}{4}\)
c = \(\frac{144}{4}\)
c = 36

Question 9.
x² – 15x + c
Answer:

Question 10.
x² + 9x + c
Answer:
x² + 9x + c
x² + bx + c
b = 9
x² + bx + \(\frac{b²}{4}\)
c = \(\frac{b²}{4}\)
c = \(\frac{81}{4}\)

In Exercises 11–16, complete the square for the expression. Then factor the trinomial.
Question 11.
x² – 10x
Answer:

Question 12.
x² – 40x
Answer:
x² – 40x
ax² + bx
a = 1 and b = -40
\(\frac{b}{2}\) = \(\frac{-40}{2}\)
\(\frac{b}{2}\) = -20
\(\frac{b²}{4}\) = (-20)² = 400
x² – 40x + 200 = (x – 20)²

Question 13.
x² + 16x
Answer:

Question 14.
x² + 22x
Answer:
x² + 22x
ax² + bx
a = 1 and b = 22
\(\frac{b}{2}\) = \(\frac{22}{2}\)
\(\frac{b}{2}\) = 11
\(\frac{b²}{4}\) = 11² = 121
x² + 22x + 121 = (x + 11)²

Question 15.
x² + 5x
Answer:

Question 16.
x² – 3x
Answer:
x² – 3x
ax² + bx
a = 1 and b = -3
\(\frac{b}{2}\) = \(\frac{-3}{2}\)
\(\frac{b}{2}\) = –\(\frac{3}{2}\)
\(\frac{b²}{4}\) = \(\frac{9}{4}\)
x² – 3x + \(\frac{9}{4}\) = (x – \(\frac{3}{2}\))²

In Exercises 17–22, solve the equation by completing the square. Round your solutions to the nearest hundredth, if necessary.
Question 17.
x² + 14x = 15
Answer:

Question 18.
x² – 6x = 16
Answer:
x² – 6x = 16
x² – 6x + 9 = 16 + 9
(x – 3)² = 25
√(x – 3)² = √25
x – 3 = 5 or x – 3 = -5
x = 5 + 3 or x = -5 + 3
x = 8 or x = -2

Question 19.
x² – 4x = -2
Answer:

Question 20.
x² + 2x = 5
Answer:
x² + 2x = 5
x² + 2x + 1 = 5 + 1
(x + 1)² = 6
√(x + 1)² = √6
x + 1 = √6 or x + 1 = -√6
We know that,
√6 = 2.44
x = 2.44 – 1 or x = -2.44 – 1
x = 1.44 or x = -3.44

Question 21.
x² – 5x = 8
Answer:

Question 22.
x² + 11x = -10
Answer:
x² + 11x = -10
x² + 11x + \(\frac{121}{4}\) = -10 + \(\frac{121}{4}\)
(x + 11)² = \(\frac{81}{4}\)
√(x + 11)² = √\(\frac{81}{4}\)
x + 11 = √\(\frac{81}{4}\) or x + 11 = -√\(\frac{81}{4}\)
x = 20.25 – 11 or x = -20.25 – 11
x = 9.25 or x = -31.25

Question 23.
MODELING WITH MATHEMATICS
The area of the patio is 216 square feet.

a. Write an equation that represents the area of the patio.
b. Find the dimensions of the patio by completing the square.
Answer:

Question 24.
MODELING WITH MATHEMATICS
Some sand art contains sand and water sealed in a glass case, similar to the one shown. When the art is turned upside down, the sand, and waterfall to create a new picture. The glass case has a depth of 1 centimeter and a volume of 768 cubic centimeters.

a. Write an equation that represents the volume of the glass case.
Answer:
The volume of the glass case = L × W × H
768 = x (x – 8) × 1
x² – 8x = 768

b. Find the dimensions of the glass case by completing the square.
Answer:
x² – 8x = 768
x² – 8x + 16 = 768 + 16
(x – 4)² = 784
√(x – 4)² = √784
x – 4 = 28 or x – 4 = -28
x = 28 + 4 or x = -28 + 4
x = 32 or x = -24
x = 32
Length of the glass case = x – 8 = 32 – 8 =24 cm
Width of the glass case = x = 32 cm

In Exercises 25–32, solve the equation by completing the square. Round your solutions to the nearest hundredth, if necessary.
Question 25.
x² – 8x + 15 = 0
Answer:

Question 26.
x² + 4x – 21 = 0
Answer:
x² + 4x – 21 = 0
x² + 4x – 21 + 21 = 0 + 21
x² + 4x = 21
x² + 4x + 4 = 21 + 4
(x + 2)² = 25
√(x + 2)² = √25
x + 2 = 5 or x + 2 = -5
x = 5 – 2 or x = -5 – 2
x = 3 or x = -7

Question 27.
2x² + 20x + 44 = 0
Answer:

Question 28.
3x² – 18x + 12 = 0
Answer:
3x² – 18x + 12 = 0
3 (x² – 6x + 4) = 0
x² – 6x + 4 = 0
x² – 6x – 4 + 4 = 0 + 4
x² – 6x = 4
x² – 6x + 9 = 9 + 4
(x – 3)² = 13
√(x – 3)² = √13
x – 3 = 3.60 or x – 3 = -3.60
x = 3.60 + 3 or x = -3.60 + 3
x = 6.60 or x = -0.60

Question 29.
-3x² – 24x + 17 = -40
Answer:

Question 30.
-5x² – 20x + 35 = 30
Answer:

Question 31.
2x² – 14x + 10 = 26
Answer:

Question 32.
4x² + 12x – 15 = 5
Answer:

Question 33.
ERROR ANALYSIS
Describe and correct the error in solving x² + 8x = 10 by completing the square.

Answer:

Question 34.
ERROR ANALYSIS
Describe and correct the error in the first two steps of solving 2x² – 2x – 4 = 0 by completing the square.

Answer:
2x² – 2x – 4 = 0
ax² + bx + c = 0
x² – x – 2 = 0
x² – x = 2
x² – x + \(\frac{1}{4}\) = 2+ \(\frac{1}{4}\)
(x – \(\frac{1}{2}\))² = \(\frac{9}{4}\)

Question 35.
NUMBER SENSE
Find all values of b for which x² + bx + 25 is a perfect square trinomial. Explain how you found your answer.
Answer:

Question 36.
REASONING
You are completing the square to solve 3x² + 6x = 12. What is the first step?
Answer:
The given equation is:
3x² + 6x = 12
3(x² + 2x) = 3(4)
x² + 2x = 4

In Exercises 37–40, write the function in vertex form bycompleting the square. Then match the function with its graph.
Question 37.
y = x² + 6x + 3
Answer:

Question 38.
y = -x² + 8x – 12
Answer:
y = -x² + 8x – 12
y + 12 = -x² + 8x – 12+ 12
y + 12 = -x² + 8x
y + 12 = – (x² – 8x)
y + 12 – 16 = – (x² – 8x + 16)
y – 4 = – (x – 4)²
y = -(x – 4)² + 4
The vertex point is (-4, 4)
The vertex point matches with graph A

Question 39.
y = -x² – 4x – 2
Answer:

Question 40.
y = x² – 2x + 4
Answer:
y = x² – 2x + 4
y – 4 = x² – 2x + 4 – 4
y – 4 = x² – 2x
y – 4 + 1 = x² – 2x + 1
y – 3 = (x – 1)²
y = (x – 1)² + 3
The vertex point is (1, 3)
The vertex point matches with graph C

In Exercises 41–46, determine whether the quadratic function has a maximum or minimum value. Then find the value.
Question 41.
y = x² – 4x – 2
Answer:

Question 42.
y = x² + 6x + 10
Answer:
y = x² + 6x + 10
y – 10 = x² + 6x + 10 – 10
y – 10 = x² + 6x
y – 10 + 9 = x² + 6x + 9
y – 1 = (x + 3)²
y = (x + 3)² + 1
The vertex point is: (-3, 1)
From the above vertex point, the x-coordinate is negative
The parabola closes down and the y-coordinate will be the maximum value of the given equation.
The given equation has a maximum value and the maximum value of the given equation is: 1

Question 43.
y = -x² – 10x – 30
Answer:

Question 44.
y = -x² + 14x – 34
Answer:
y = -x² + 14x – 34
y + 34 = -x² + 14x + 34 – 34
y + 34 = -x² + 14x
y + 34 = – (x² – 14x)
y + 34 – 49 = – (x² – 14x + 49)
y – 15 = -(x – 7)²
y = -(x – 7)² + 15
The vertex point is (-7, 15)
From the above vertex point, the x-coordinate is negative
The parabola closes down and the y-coordinate will be the maximum value of the given equation
The given equation has a maximum value and the maximum value of the given equation is 15

Question 45.
f(x) = -3x² – 6x – 9
Answer:

Question 46.
f(x) = 4x² – 28x + 32
Answer:
f (x) = 4x² – 28x + 32
f (x) – 32 = 4x² – 28x + 32 – 32
f (x) – 32 = 4x² – 28x
f (x) – 32 = 4 (x² – 7x)
f (x) – 32 + \(\frac{49}{4}\) =4 ( x² – 7x + \(\frac{49}{4}\))
y – \(\frac{51}{4}\) = 4 (x – \(\frac{7}{2}\))²
y = 4(x – \(\frac{7}{2}\))² + \(\frac{51}{4}\)
The vertex point is (14, \(\frac{51}{4}\))
We can observe that the x-coordinate is positive
The parabola opens up and the y-coordinate will be the minimum value of the given equation
The given equation has a minimum value and the minimum value of the given equation is \(\frac{51}{4}\)

In Exercises 47–50, determine whether the graph could represent the function. Explain.
Question 47.
y = -(x + 8)(x + 3)

Answer:

Question 48.
y = (x – 5)²

Answer:
y = (x – 5)²
y = (x – 5) (x – 5)
The vertex point for the given equation is (5, 5)
The parabola closes up since the x-coordinate is positive and we will have a minimum value
The minimum value will be the y-coordinate of the equation in the vertex form
The given graph matches the given equation

Question 49.
y = \(\frac{1}{4}\)(x + 2)² – 4

Answer:

Question 50.
y = -2(x – 1)(x + 2)

Answer:
y = -2 (x – 1) (x + 2)
The vertex point for the given equation is (-2, 2)
We can observe that the parabola opens down since the x-coordinate is negative and we will have a maximum value.
The maximum value will be the y-coordinate of the equation in the vertex form

In Exercises 51 and 52, determine which of the functions could be represented by the graph. Explain.
Question 51.

Answer:

Question 52.

Answer:

Question 53.
MODELING WITH MATHEMATICS
The function h = -16t² + 48t represents the height h (in feet) of a kickball t seconds after it is kicked from the ground.
a. Find the maximum height of the kickball.
b. Find and interpret the axis of symmetry.
Answer:

Question 54.
MODELING WITH MATHEMATICS
You throw a stone from a height of 16 feet with an initial vertical velocity of 32 feet per second. The function h = -16t² + 32t + 16 represents the height h (in feet) of the stone after t seconds.

a. Find the maximum height of the stone.
Answer:
Hmax = h + \(\frac{v²}{2g}\)
g = 10 m/s
Hmax = 16 + \(\frac{1024}{20}\)
Hmax = 67.2 feet

b. Find and interpret the axis of symmetry.
Answer:
h = -16t² + 32t + 16
h – 16 = -16t² + 32t + 16 – 16
h – 16 = -16t² + 32t
h – 16 = -16 (t² – 2t)
h – 16 – 16 = -16 (t² – 2t + 1)
h – 32 = -16 (t – 1)²
h = -16 (t – 1)² + 32
t = 1

Question 55.
MODELING WITH MATHEMATICS
You are building a rectangular brick patio surrounded by a crushed stone border with a uniform width, as shown. You purchase patio bricks to cover 140 square feet. Find the width of the border.

Answer:

Question 56.
MODELING WITH MATHEMATICS
You are making a poster that will have a uniform border, as shown. The total area of the poster is 722 square inches. Find the width of the border to the nearest inch.

Answer:
The total area of the poster is 722 square inches
The width of the poster is (28 – 2x) inches
The length of the poster is (22 – 2x) inches
Area = Length × Width
722 = (28 – 2x) (22 – 2x)
722 = 616 – 56x – 44x + 4x²
722 – 616 = 4x² – 56x – 44x
106 = 4x² – 100x
\(\frac{53}{2}\) = x² – 25x
\(\frac{53}{2}\) + \(\frac{625}{4}\) = x² – 25x + \(\frac{625}{4}\)
\(\frac{731}{4}\) = (x – \(\frac{25}{2}\))²
√(x – \(\frac{25}{2}\))² = √\(\frac{731}{4}\)
x – \(\frac{25}{2}\) = √\(\frac{731}{4}\) or x – \(\frac{25}{2}\) = -√\(\frac{731}{4}\)
x = 13.62 + 12.5 or x = -13.62 + 12.5
x = 26.12 or x = -1.12
We can observe that the width of the poster is 28 inches and the width of the poster won’t be negative
The width of the border is -1.12 inches
The width of the border = 28 – 2x
= 28 – 2 (-1.12)
= 30.24 inches

MATHEMATICAL CONNECTIONS In Exercises 57 and 58, find the value of x. Round your answer to the nearest hundredth, if necessary.
Question 57.
A = 108 m²

Answer:

Question 58.
A = 288 in.²

Answer:
The area of the rectangle = Length × Width
288 = (2x + 10) (3x)
288 = 6x² + 30x
48 = x² + 5x
48 + \(\frac{25}{4}[latex] = x² + 5x + [latex]\frac{25}{4}\)
54.25 = (x + \(\frac{5}{2}\))²
√(x + \(\frac{5}{2}\))² = √54.25
√54.25 = 7.36
x + \(\frac{5}{2}\) = 7.36 or x + \(\frac{5}{2}\) = -7.36
x = 7.36 – 2.5 or x = -7.36 – 2.5
x = 4.86 or x = -9.86
The length and width won’t be negative
x = 4.86

In Exercises 59–62, solve the equation by completing the square. Round your solutions to the nearest hundredth, if necessary.
Question 59.
0.5x² + x – 2 = 0
Answer:

Question 60.
0.75x² + 1.5x = 4
Answer:

Question 61.
\(\frac{8}{3}\) x – \(\frac{2}{3}\)x² = –\(\frac{5}{6}\)
Answer:

Question 62.
\(\frac{1}{4}\)x² + \(\frac{1}{2}\)x – \(\frac{5}{4}\) = 0
Answer:

Question 63.
PROBLEM-SOLVING
The distance d (in feet) that it takes a car to come to a complete stop can be modeled by d = 0.05s² + 2.2s, where s is the speed of the car (in miles per hour). A car has 168 feet to come to a complete stop. Find the maximum speed at which the car can travel.
Answer:

Question 64.
PROBLEM SOLVING
During a “big air” competition, snowboarders launch themselves from a half-pipe, perform tricks in the air, and land back in the half-pipe. The height h (in feet) of a snowboarder above the bottom of the half-pipe can be modeled by h = -16t² + 24t + 16.4, where t is the time (in seconds) after the snowboarder launches into the air. The snowboarder lands 3.2 feet lower than the height of the launch. How long is the snowboarder in the air? Round your answer to the nearest tenth of a second.

Answer:

Question 65.
PROBLEM SOLVING
You have 80 feet of fencing to make a rectangular horse pasture that covers 750 square feet. A barn will be used as one side of the pasture, as shown.

a. Write equations for the amount of fencing to be used and the area enclosed by the fencing.
b. Use substitution to solve the system of equations from part (a). What are the possible dimensions of the pasture?
Answer:

Question 66.
HOW DO YOU SEE IT?
The graph represents the quadratic function y = x² – 4x + 6.

a. Use the graph to estimate the x-values for which y = 3.
Answer:
y = x² – 4x + 6
y = 3
x² – 4x + 6 = 3
x² – 4x = 3 – 6
x² – 4x = -3
x² – 4x + 4 = -3 + 4
(x – 2)² = 1
√(x – 2)² = √1
x – 2 = 1 or x – 2 = -1
x = 1 + 2 or x = -1 + 2
x = 3 or x = 1

b. Explain how you can use the method of completing the square to check your estimates in part (a).
Answer:
The steps to “Completing the Square” to check the estimates in part (a) are:
1. Divide all terms by a (The coefficient of x²)
2. Move the number term (c / a) to the right side of the equation
3. Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation

Question 67.
COMPARING METHODS
Consider the quadratic equation x² + 12x + 2 = 12.
a. Solve the equation by graphing.
b. Solve the equation by completing the square.
c. Compare the two methods. Which do you prefer? Explain.
Answer:

Question 68.
THOUGHT-PROVOKING
Sketch the graph of the equation x² – 2xy + y² – x – y = 0. Identify the graph.
Answer:
x² – 2xy + y² – x – y = 0
x² + y² + 2gx + 2fy + 2hxy + c = 0
The graph of the given equation looks like a circle

Question 69.
REASONING
The product of two consecutive even integers that are positive is 48. Write and solve an equation to find the integers.
Answer:

Question 70.
REASONING
The product of two consecutive odd integers that are negative is 195. Write and solve an equation to find the integers.
Answer:
(1 – n) (3 – n) = 195

Question 71.
MAKING AN ARGUMENT
You purchase stock for $16 per share. You sell the stock 30 days later for $23.50 per share. The price y (in dollars) of a share during the 30-day period can be modeled by y = -0.025x² + x + 16, where x is the number of days after the stock is purchased. Your friend says you could have sold the stock earlier for $23.50 per share. Is your friend correct? Explain.

Answer:

Question 72.
REASONING
You are solving the equation x² + 9x =18. What are the advantages of solving the equation by completing the square instead of using other methods you have learned?
Answer:
x² + 9x = 18
x² + 2( \(\frac{9}{2}\) )x = 18
To make the expression present on the left side a factor trinomial to get the exact values of x, we have to use the “Completing the squares” method
When we use the graphing method,
We won’t get the exact values of x for the equation
We can conclude that we have to solve the above equation by using the “Completing the squares” method to get the exact values of x when we compared it with the other methods

Question 73.
PROBLEM-SOLVING
You are knitting a rectangular scarf. The pattern results in a scarf that is 60 inches long and 4 inches wide. However, you have enough yarn to knit 396 square inches. You decide to increase the dimensions of the scarf so that you will use all your yarn. The increase in the length is three times the increase in the width. What are the dimensions of your scarf?

Answer:

Question 74.
WRITING
How many solutions does x² + bx = c have when c < -(\(\frac{b}{2}\))² ? Explain.
Answer:
x² + bx = c
x² + bx + \(\frac{b²}{4}\) = c + \(\frac{b²}{4}\)
(x + \(\frac{b}{2}\))² = c +c + \(\frac{b²}{4}\)
c < – \(\frac{b²}{4}\)
When c < – \(\frac{b²}{4}\), the square value will also be negative
But, we know that the square won’t take any negative values
The number of solutions of x² + bx = c when c <- \(\frac{b²}{4}\) are: 0

Maintaining Mathematical Proficiency

Write a recursive rule for the sequence.
Question 75.

Answer:

Question 76.

Answer:
The first term is 3 (When x = 1)
Each term is obtained by multiplying the previous term by 2
The recursive rule for the sequence is a1 = an-1 × 2

Question 77.

Answer:

Simplify the expression \(\sqrt{b^{2}-4 a c}\) for the given values.
Question 78.
a = 3, b = -6, c = 2
Answer:
\(\sqrt{b² – 4ac}\)
a = 3, b = -6, and c = 2
\(\sqrt{b² – 4ac}\)
= \(\sqrt{(-6)² – 4 (3) (2)}\)
= \(\sqrt{36 – 24}\)
= \(\sqrt{12}\)
= \(\sqrt{4 × 3}\)
= 2\(\sqrt{3}\)

Question 79.
a = -2, b = 4, c = 7
Answer:

Question 80.
a = 1, b = 6, c = 4
Answer:
\(\sqrt{b² – 4ac}\)
a = 1, b = 6, and c = 4
\(\sqrt{b² – 4ac}\)
= \(\sqrt{(6)² – 4 (1) (4)}\)
= \(\sqrt{36 – 16}\)
= \(\sqrt{20}\)
= \(\sqrt{4 × 5}\)
= 2\(\sqrt{5}\)

Lesson 9.5 Solving Quadratic Equations Using the Quadratic Formula

Essential Question How can you derive a formula that can be used to write the solutions of any quadratic equation in standard form?

EXPLORATION 1

Deriving the Quadratic Formula
Work with a partner. The following steps show a method of solving ax² + bx + c = 0. Explain what was done in each step.

Answer:
1. Write the equation
2. Multiply with “4a” on both sides
3. Add b² on both sides
4. Subtract with “4ac” on both sides
5. Write the expression that is present on the left side i.e., in the form of a trinomial i.e., (x ± a)²
6. Take “Square root” on both sides
7. Subtract with “b” on both sides
8. Divide by “2a” into both sides

EXPLORATION 2

Deriving the Quadratic Formula by Completing the Square
Work with a partner.
a. Solve ax² + bx + c = 0 by completing the square. (Hint: Subtract c from each side, divide each side by a, and then proceed by completing the square.)
Answer:
ax² + bx + c = 0
ax² + bx + c – c = 0 – c
ax² + bx = -c
x² + \(\frac{b}{a}\)x = –\(\frac{c}{a}\)
x² + \(\frac{b}{a}\) + \(\frac{b²}{4a²}\) = –\(\frac{c}{a}\) + \(\frac{b²}{4a²}\)
(x + \(\frac{b}{2a}\))² = \(\frac{-4c + b²}{4a²}\)
√(x + \(\frac{b}{2a}\))² = √\(\frac{-4c + b²}{4a²}\)
x + \(\frac{b}{2a}\) = \(\frac{\sqrt{-4c + b²}}{2a}\) or x + \(\frac{b}{2a}\) = –\(\frac{\sqrt{-4c + b²}}{2a}\)
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) or x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)

b. Compare this method with the method in Exploration 1. Explain why you think 4a and b² were chosen in Steps 2 and 3 of Exploration 1.
Answer:
In Exploration 1,
We add 4a to make the expression on the left side of a factor trinomial
We add b² to make the factor trinomial complete i.e., to make the expression on the left side in the form of a² + 2ab + b²

Communicate Your Answer

Question 3.
How can you derive a formula that can be used to write the solutions of any quadratic equation in standard form?
Answer:
The formula can be actually derived using the steps involved in completing the square.
It stems from the fact that any quadratic function or equation of the form
y = ax² + b x + c can be solved for its roots.

Question 4.
Use the Quadratic Formula to solve each quadratic equation.
a. x² + 2x – 3 = 0
Answer:
x² + 2x – 3 = 0
ax² + bx + c = 0
a = 1, b = 2, and c = -3
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{2}{2 (1)}\) + \(\frac{\sqrt{2² – 4 (1) (-3)}}{2 (1)}\)
x = -1 + \(\frac{\sqrt{4 + 12}}{2}\)
x= -1 + 2
x = 1
x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{2}{2 (1)}\) – \(\frac{\sqrt{2² – 4 (1) (-3)}}{2 (1)}\)
x = -1 – \(\frac{\sqrt{4 + 12}}{2}\)
x= -1 – 2
x = -3

b. x² – 4x + 4 = 0
Answer:
x² – 4x + 4 = 0
ax² + bx + c = 0
a = 1, b = -4, and c = 4
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{-4}{2 (1)}\) + \(\frac{\sqrt{(-4)² – 4 (1) (4)}}{2 (1)}\)
x = 2 + \(\frac{\sqrt{16 – 16}}{2}\)
x= 2
x =  –\(\frac{-4}{2 (1)}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = 2 – \(\frac{\sqrt{(-4)² – 4 (1) (4)}}{2 (1)1}\)
x = 2 – \(\frac{\sqrt{16 – 16}}{2}\)
x = 2

c. x² + 4x + 5 = 0
Answer:
x² + 4x + 5 = 0
ax² + bx + c = 0
a = 1, b = 4, and c = 5
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{4}{2 (1)}\) + \(\frac{\sqrt{(4)² – 4 (1) (5)}}{2 (1)}\)
x = -2 + \(\frac{\sqrt{16 – 20}}{2}\)
x= -2 + i
x =  –\(\frac{4}{2 (1)}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = -2 – \(\frac{\sqrt{(4)² – 4 (1) (5)}}{2 (1)}\)
x = -2 – \(\frac{\sqrt{16 – 20}}{2}\)
x= -2 – i
The solutions of the given equation are: -2 ± i

Question 5.
Use the Internet to research imaginary numbers. How are they related to quadratic equations?

Answer:
An “Imaginary number” is a number that, when squared, has a negative result. Essentially, an imaginary number is the square root of a negative number and does not have a tangible value.  Imaginary numbers are usually denoted by the symbol i
√-1 = i

Monitoring Progress

Solve the equation using the Quadratic Formula. Round your solutions to the nearest tenth, if necessary.
Question 1.
x² – 6x + 5 = 0
Answer:
x² – 6x + 5 = 0
ax² + bx + c = 0
a = 1, b = -6, and c = 5
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = \(\frac{6}{2 (1)}\) + \(\frac{\sqrt{(-6)² – 4 (1) (5)}}{2 (1)}\)
x = 3 + \(\frac{\sqrt{36 – 20}}{2}\)
x= 3 + 2
x = 5
x =  \(\frac{6}{2 (1)}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = 3 – \(\frac{\sqrt{(-6)² – 4 (1) (5)}}{2 (1)}\)
x = 3 – \(\frac{\sqrt{36 – 20}}{2}\)
x= 3 – 2
x = 1

Question 2.
\(\frac{1}{2}\)x² + x – 10 = 0
Answer:
\(\frac{1}{2}\)x² + x – 10 = 0
x² + 2x – 20 = 0
ax² + bx + c = 0
a = 1, b = 2, and c = -20
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{2}{2 (1)}\) + \(\frac{\sqrt{(2)² + 4 (1) (20)}}{2 (1)}\)
x = -1 + \(\frac{\sqrt{84}}{2}\)
x= 3.5
x =  –\(\frac{2}{2 (1)}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = -1 – \(\frac{\sqrt{(2)² + 4 (1) (20)}}{2 (1)}\)
x = -1 – \(\frac{\sqrt{84}}{2}\)
x= -4.5

Question 3.
-3x² + 2x + 7 = 0
Answer:
-3x² + 2x + 7 = 0
ax² + bx + c = 0
a = -3, b = 2, and c = 7
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = \(\frac{2}{2 (3)}\) + \(\frac{\sqrt{(2)² + 4 (3) (7)}}{2 (-3)}\)
x = \(\frac{1}{3}\) + \(\frac{\sqrt{88}}{-6}\)
x= -1.2
x =  \(\frac{2}{2 (3)}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = \(\frac{1}{3}\) – \(\frac{\sqrt{(2)² + 4 (3) (7)}}{2 (-3)}\)
x = \(\frac{1}{3}\) + \(\frac{\sqrt{88}}{6}\)
x= 1.8

Question 4.
4x² – 4x = -1
Answer:
4x² – 4x = -1
4x² – 4x + 1 = 0
ax² + bx + c = 0
a = 4, b = -4, and c = 1
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = \(\frac{4}{2 (4)}\) + \(\frac{\sqrt{(-4)² – 4 (1) (4)}}{2 (4)}\)
x = \(\frac{1}{2}\) + \(\frac{\sqrt{16 – 16}}{8}\)
x= \(\frac{1}{2}\)
x =  \(\frac{4}{2 (4)}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = \(\frac{1}{2}\) – \(\frac{\sqrt{(-4)² – 4 (1) (4)}}{2 (4)}\)
x = \(\frac{1}{2}\) – \(\frac{\sqrt{16 – 16}}{8}\)
x= \(\frac{1}{2}\)

Question 5.
WHAT IF?
When were there about 60 wolf breeding pairs?
Answer:
The given quadratic equation that used to model the wolf pairs for x years is:
y = 0.20x² + 1.8x – 3
0.20x² + 1.8x – 3 = 60
0.20x² + 1.8x – 63 =  0
x² + 9x – 315 = 0
ax² + bx + c = 0
a = 1, b = 9, and c = -315
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\) and x =  –\(\frac{b}{2a}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{b}{2a}\) + \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{9}{2 (1)}\) + \(\frac{\sqrt{(9)² + 4 (1) (315)}}{2 (1)}\)
x = –\(\frac{9}{2}\) + \(\frac{\sqrt{1,341}}{2}\)
x= 13.80
x = 14
x =  –\(\frac{9}{2 (1)}\) – \(\frac{\sqrt{b² – 4ac}}{2a}\)
x = –\(\frac{9}{2}\) – \(\frac{\sqrt{(9)² + 4 (1) (315)}}{2 (1)}\)
x = –\(\frac{9}{2}\) – \(\frac{\sqrt{1,341}}{2}\)
x= -22.8
= -23
Since x represents the number of years, the number of years won’t be negative
x = 14 years
After 14 years, i.e, in 2004, there were 60 wolf-breeding pairs

Question 6.
The number y of bald eagle nesting pairs in state x years since 2000 can be modeled by the function y = 0.34x² + 13.1x + 51.
a. When were there about 160 bald eagle nesting pairs?
Answer:
y = 0.34x² + 13.1x + 51
y =160
160 = 0.34x² + 13.1x + 51
0.34x² + 13.1x – 109 = 0
ax² + bx + c = 0

Since x is the number of years, the value of x won’t be negative
x = 7.03 years
x = 7 years
We can conclude that after 7 years from 2000, i.e., in 2007, there were 160 bald eagle nesting pairs

b. How many bald eagle nesting pairs were there in 2000?
Answer:
There are 160 bald eagle nesting pairs in 2007
To find the number of bald eagle nesting pairs in 2000,
Subtract the number of bald eagle nesting pairs in 2007 from the difference between 2000 and 2007
The number of bald eagles nesting pairs in 2000 = 160 – (2007 – 2000)
= 160 – 7
= 153

Determine the number of real solutions of the equation.
Question 7.
-x² + 4x – 4 = 0
Answer:
-x² + 4x – 4 = 0
ax² + bx + c = 0
a = -1, b = 4, and c = -4
If b² – 4ac > 0, then there will be 2 real solutions
If b² – 4ac = 0, then there will be one real solution
If b² – 4ac < 0, then there are no real solutions
b² – 4ac = 4² – 4 (-1) (-4)
= 16 – 16
= 0
There will be only 1 solution for the given equation

Question 8.
6x² + 2x = -1
Answer:
6x² + 2x = -1
6x² + 2x + 1 = 0
ax² + bx + c = 0
a = 6, b = 2, and c = 1
If b² – 4ac > 0, then there will be 2 real solutions
If b² – 4ac = 0, then there will be one real solution
If b² – 4ac < 0, then there are no real solutions
b² – 4ac = 2² – 4 (6) (1)
= 4 – 24
= -20
There are no real solutions for the given equation

Question 9.
\(\frac{1}{2}\)x² = 7x – 1
Answer:
\(\frac{1}{2}\)x² = 7x – 1
\(\frac{1}{2}\)x² – 7x + 1= 0
x² – 14x + 2 = 0
ax² + bx + c = 0
a = 1, b = -14, and c = 2
If b² – 4ac > 0, then there will be 2 real solutions
If b² – 4ac = 0, then there will be one real solution
If b² – 4ac < 0, then there are no real solutions
b² – 4ac = (-14)² – 4 (1) (2)
= 196 – 8
= 188
There are 2 solutions for the given equation

Find the number of x-intercepts of the graph of the function.
Question 10.
y = -x² + x – 6
Answer:
y = -x² + x – 6
-x² + x – 6 = 0
ax² + bx + c = 0
a = -1, b = 1, and c = -6
If b² – 4ac > 0, then there will be 2 x- intercepts
If b² – 4ac = 0, then there will be one x-intercept
If b² – 4ac < 0, then there are no x-intercepts
b² – 4ac = 1² – 4 (-1) (-6)
= 1 – 24
= -23
There are no x-intercepts for the given equation

Question 11.
y = x² – x
Answer:
y = x² – x
x² – x = 0
ax² + bx + c = 0
a = 1, b = -1, and c = 0
If b² – 4ac > 0, then there will be 2 x- intercepts
If b² – 4ac = 0, then there will be one x-intercept
If b² – 4ac < 0, then there are no x-intercepts
b² – 4ac = (-1)² – 4 (1) (0)
= 1 – 0
= 1
There are 2 x-intercepts for the given equation

Question 12.
f(x) = x² + 12x + 36
Answer:
f (x) = x² + 12x + 36
x² + 12x + 36 = 0
ax² + bx + c = 0
a = 1, b = 12, and c = 36
If b² – 4ac > 0, then there will be 2 x- intercepts
If b² – 4ac = 0, then there will be one x-intercept
If b² – 4ac < 0, then there are no x-intercepts
b² – 4ac = 12² – 4 (1) (36)
= 144 – 144
= 0
There is only 1 x-intercept for the given equation

Solve the equation using any method. Explain your choice of method.
Question 13.
x² + 11x – 12 = 0
Answer:
x² + 11x – 12 = 0
x² + 12x – x – 12 = 0
x (x + 12) – 1 (x + 12) = 0
(x – 1) (x + 2) = 0
x – 1 = 0 or x + 12 = 0
x = 1 or x = -12

Question 14.
9x² – 5 = 4
Answer:
9x² – 5 = 4
9x² = 4 + 5
9x² = 9
x² = \(\frac{9}{9}\)
x² = 1
x² = d
√x² = √1
x = 1 or x = -1

Question 15.
5x² – x – 1 = 0
Answer:
5x² – x – 1 = 0
ax² + bx + c = 0
a = 5, b = -1, and c = -1
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{1 ± \sqrt{(-1)² + 4 (1) (5)}}{2 (5)}\)
x = \(\frac{1 ± \sqrt{21}}{10}\)

Question 16.
x² = 2x – 5
Answer:
x² = 2x – 5
x² – 2x = 5
x² – 2x + 1 = 5 + 1
(x – 1)² = 6
√(x – 1)² = √6
x – 1 = √6 or x – 1 = -√6
x = √6 + 1 or x = -√6 + 1

Solving Quadratic Equations Using the Quadratic Formula 9.5 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What formula can you use to solve any quadratic equation? Write the formula.
Answer:

Question 2.
VOCABULARY
In the Quadratic Formula, what is the discriminant? What does the value of the discriminant determine?
Answer:
The quadratic formula is
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
The discriminant is: b² – 4ac
If b² – 4ac > 0, then there will be 2 real solutions
If b² – 4ac = 0, then there will be one real solution
If b² – 4ac < 0, then there are no real solutions

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, write the equation in standard form. Then identify the values of a, b, and c that you would use to solve the equation using the Quadratic Formula.
Question 3.
x² = 7x
Answer:

Question 4.
x² – 4x = -12
Answer:
x² – 4x = -12
x² – 4x + 12 = 12 – 12
x² – 4x + 12 = 0
ax² + bx + c = 0
a = 1, b = -4, and c = 12
x² – 4x + 4x = -12 + 4x
x² = -12 + 4x
x² – x² = -12 + 4x – x²
-12 + 4x – x² = 0
ax² + bx + c = 0
a = -1, b = 4, and c = -12

Question 5.
-2x² + 1 = 5x
Answer:

Question 6.
3x + 2 = 4x²
Answer:
Method 1:
3x + 2 = 4x²
3x + 2 – 2 = 4x² – 2
3x – 3x = 4x² – 2 – 3x
4x² – 3x – 2 = 0
ax² + bx + c = 0
a = 4, b = -3, and c = -2
Method 2:
3x + 2 – 4x² = 4x² – 4x²
-4x² + 3x + 2 = 0
ax² + bx + c = 0
a = -4, b = 3, and c = 2

Question 7.
4 – 3x = -x² + 3x
Answer:

Question 8.
-8x – 1 = 3x² + 2
Answer:
-8x – 1 = 3x² + 2
-8x -1 – 2 = 3x²
-8x – 3 = 3x²
Method 1:
-8x – 3 + 3 = 3x² + 3
-8x = 3x² + 3
8x – 8x = 3x² + 3 + 8x
3x² + 8x + 3 = 0
ax² + bx + c = 0
a = 3, b = 8, and c = 3
Method 2:
-8x – 3 – 3x² = 3x² – 3x²
-3x² – 8x – 3 = 0
3x² + 8x + 3 = 0
ax² + bx + c = 0
a = 3, b = 8, and c = 3

In Exercises 9–22, solve the equation using the Quadratic Formula. Round your solutions to the nearest tenth, if necessary.
Question 9.
x² – 12x + 36 = 0
Answer:

Question 10.
x² + 7x + 16 = 0
Answer:
x² + 7x + 16 = 0
ax² + bx + c = 0
a = 1, b = 7, and c = 16
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-7 ± \sqrt{(7)² – 4 (1) (16)}}{2 (1)}\)
x = \(\frac{-7 ± \sqrt{-15}}{2}\)
x = \(\frac{-7 ± 15i}{2}\)

Question 11.
x² – 10x – 11 = 0
Answer:

Question 12.
2x² – x – 1 = 0
Answer:
2x² – x – 1 = 0
ax² + bx + c = 0
a = 2, b = -1, and c = -1
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{1 ± \sqrt{(-1)² + 4 (1) (2)}}{2 (2)}\)
x = \(\frac{1 ± \sqrt{9}}{4}\)
x = \(\frac{1 ± 3}{4}\)
x = \(\frac{1 + 3}{4}\) or x = \(\frac{1 – 3}{4}\)
x = 1 or x = –\(\frac{1}{2}\)

Question 13.
2x² – 6x + 5 = 0
Answer:

Question 14.
9x² – 6x + 1 = 0
Answer:
9x² – 6x + 1 = 0
ax² + bx + c = 0
a = 9, b = -6, and c = 1
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{6 ± \sqrt{(-6)² – 4 (1) (9)}}{2 (9)}\)
x = \(\frac{6 ± \sqrt{0}}{18}\)
x = \(\frac{6 + 0}{18}\) or x = \(\frac{6 – 0}{18}\)
x = \(\frac{6}{18}\) or x = \(\frac{6}{18}\)
x = \(\frac{1}{3}\) or x = \(\frac{1}{3}\)

Question 15.
6x² – 13x = -6
Answer:

Question 16.
-3x² + 6x = 4
Answer:
-3x² + 6x = 4
-3x² + 6x – 4 = 0
3x² – 6x + 4 = 0
ax² + bx + c = 0
a = 3, b = -6, and c = 4
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{6 ± \sqrt{(-6)² – 4 (3) (4)}}{2 (3)}\)
x = \(\frac{6 ± \sqrt{-12}}{6}\)
x = \(\frac{6 ± 12i}{6}\)
x = 1 ± 2i

Question 17.
1 – 8x = -16x²
Answer:

Question 18.
x² – 5x + 3 = 0
Answer:
x² – 5x + 3 = 0
ax² + bx + c = 0
a = 1, b = -5, and c = 3
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{5 ± \sqrt{(-5)² – 4 (1) (3)}}{2 (1)}\)
x = \(\frac{5 ± \sqrt{13}}{2}\)
x = \(\frac{5 ± 3.6}{2}\)
x = \(\frac{5 + 3.6}{2}\) or x = \(\frac{5 – 3.6}{2}\)
x = 4.3 or x = 0.7

Question 19.
x² + 2x = 9
Answer:

Question 20.
5x² – 2 = 4x
Answer:
5x² – 2 = 4x
5x² – 4x – 2 = 0
ax² + bx + c = 0
a = 5, b = -4, and c = -2
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{4 ± \sqrt{(-4)² + 4 (5) (2)}}{2 (5}\)
x = \(\frac{4 ± \sqrt{56}}{10}\)
x = \(\frac{4 ± 7.4}{10}\)
x = \(\frac{4 + 7.4}{10}\) or x = \(\frac{4 – 7.4}{10}\)
x = 1.14 or x = -0.34

Question 21.
2x² + 9x + 7 = 3
Answer:

Question 22.
8x² + 8 = 6 – 9x
Answer:
8x² + 8 = 6 – 9x
8x² + 9x + 8 – 6 = 0
8x² + 9x + 2 = 0
ax² + bx + c = 0
a = 8, b = 9, and c = 2
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-9 ± \sqrt{(9)² – 4 (8) (2)}}{2 (8)}\)
x = \(\frac{-9 ± \sqrt{17}}{16}\)
x = \(\frac{-9 ± 4.1}{16}\)
x = \(\frac{-9 + 4.1}{16}\) or x = \(\frac{-9 – 4.1}{16}\)
x = -0.3 or x = -0.8

Question 23.
MODELING WITH MATHEMATICS
A dolphin jumps out of the water, as shown in the diagram. The function h = -16t² + 26t models the height h (in feet) of the dolphin after t seconds. After how many seconds is the dolphin at a height of 5 feet?

Answer:

Question 24.
MODELING WITH MATHEMATICS
The amount of trout y (in tons) caught in a lake from 1995 to 2014 can be modeled by the equation y = -0.08x² + 1.6x + 10, where x is the number of years since 1995.
a. When were about 15 tons of trout caught in the lake?
Answer:
y = -0.08x² + 1.6x + 10
y = 15
15 = -0.08x² + 1.6x + 10
-0.08x² + 1.6x – 5 = 0
ax² + bx + c = 0

After 4 years, i.e., in 1999, 15 ton of trout were caught in the lake

b. Do you think this model can be used to determine the amounts of trout caught in future years? Explain your reasoning.
Answer:
Yes, we can use this model to determine the amounts of trout caught in future years but the maximum of trout can be caught only in 16 years

In Exercises 25–30, determine the number of real solutions of the equation.
Question 25.
x² – 6x + 10 = 0
Answer:

Question 26.
x² – 5x – 3 = 0
Answer:
x² – 5x – 3 = 0
ax² + bx + c = 0
a = 1, b = -5, and c = -3
If b² – 4ac > 0, then there will be 2 real solutions
If b² – 4ac = 0, then there will be one real solution
If b² – 4ac < 0, then there are no real solutions
b² – 4ac = (-5)² – 4 (1) (-3)
= 25 + 12
= 37
There will be 2 solutions for the given equation

Question 27.
2x² – 12x = -18
Answer:

Question 28.
4x² = 4x – 1
Answer:
4x² = 4x – 1
4x² – 4x + 1 = 0
ax² + bx + c = 0
a = 4, b = -4, and c = 1
If b² – 4ac > 0, then there will be 2 real solutions
If b² – 4ac = 0, then there will be one real solution
If b² – 4ac < 0, then there are no real solutions
b² – 4ac = (-4)² – 4 (1) (4)
= 16 – 16
= 0
There will be only 1 solution for the given equation

Question 29.
–\(\frac{1}{4}\)x² + 4x = -2
Answer:

Question 30.
-5x² + 8x = 9
Answer:
-5x² + 8x = 9
-5x² + 8x – 9 = 0
ax² + bx + c = 0
a = -5, b = 8, and c = -9
If b² – 4ac > 0, then there will be 2 real solutions
If b² – 4ac = 0, then there will be one real solution
If b² – 4ac < 0, then there are no real solutions
b² – 4ac = 8² – 4 (-5) (-9)
= 64 – 180
= -116
There will be no  real solutions for the given equation

In Exercises 31–36, find the number of x-intercepts of the graph of the function.
Question 31.
y = x² + 5x – 1
Answer:

Question 32.
y = 4x² + 4x + 1
Answer:
y = 4x² + 4x + 1
4x² + 4x + 1 = 0
ax² + bx + c = 0
a = 4, b = 4, and c = 1
If b² – 4ac > 0, then there will be 2 x- intercepts
If b² – 4ac = 0, then there will be one x-intercept
If b² – 4ac < 0, then there are no x-intercepts
b² – 4ac = 4² – 4 (1) (4)
= 16 – 16
= 0
There will be 1 x-intercept for the given equation

Question 33.
y = -6x² + 3x – 4
Answer:

Question 34.
y = -x² + 5x + 13
Answer:
y = -x² + 5x + 13
-x² + 5x + 13 = 0
ax² + bx + c = 0
a = -1, b = 5, and c = 13
If b² – 4ac > 0, then there will be 2 x- intercepts
If b² – 4ac = 0, then there will be one x-intercept
If b² – 4ac < 0, then there are no x-intercepts
b² – 4ac = 5² – 4 (-1) (13)
= 25 + 52
= 77
There are 2 x-intercepts for the given equation

Question 35.
f(x) = 4x² + 3x – 6
Answer:

Question 36.
f(x) = 2x² + 8x + 8
Answer:
y = 2x² + 8x + 8
2x² +8x + 8 = 0
ax² + bx + c = 0
a = 2, b = 8, and c = 8
If b² – 4ac > 0, then there will be 2 x- intercepts
If b² – 4ac = 0, then there will be one x-intercept
If b² – 4ac < 0, then there are no x-intercepts
b² – 4ac = 8² – 4 (2) (8)
= 64 – 64
= 0
There is only 1 x-intercept for the given equation

In Exercises 37–44, solve the equation using any method. Explain your choice of method.
Question 37.
-10x² + 13x = 4
Answer:

Question 38.
x² – 3x – 40 = 0
Answer:
x² – 3x – 40 = 0
x² – 8x + 5x – 40 = 0
x (x + 5) – 8 (x + 5) = 0
(x – 8) (x + 5) = 0
x – 8 = 0 or x + 5 = 0
x = 8 or x = -5

Question 39.
x² + 6x = 5
Answer:

Question 40.
-5x² = -25
Answer:
-5x² = -25
5x² = 25
x² = \(\frac{25}{5}\)
x² = 5
x² = d
√x² = √5
x = ±√5
The solution for the above equation is ±√5

Question 41.
x² + x – 12 = 0
Answer:

Question 42.
x² – 4x + 1 = 0
Answer:
x² – 4x + 1 = 0
x² – 4x + 4 = -1 + 4
(x – 2)² = 3
√(x – 2)² = √3
x – 2 = −√3 or x – 2 = √3
x = -√3 + 2 or x = √3 + 2

Question 43.
4x² – x = 17
Answer:

Question 44.
x² + 6x + 9 = 16
Answer:
x² + 6x + 9 = 16
x² + 6x = 16 – 9
x² + 6x = 7
x² + 6x + 9 = 7 + 9
(x + 3)² = 16
√(x + 3)² = √16
x + 3 = −4 or x + 3 = 4
x = -4 – 3 or x = 4 – 3
x = -7 or x = 1

Question 45.
ERROR ANALYSIS
Describe and correct the error in solving the equation 3x² – 7x – 6 = 0 using the Quadratic Formula

Answer:

Question 46.
ERROR ANALYSIS
Describe and correct the error in solving the equation -2x² + 9x = 4 using the Quadratic Formula.

Answer:
-2x² + 9x = 4
-2x² + 9x – 4 = 0
2x² – 9x + 4 = 0
ax² + bx + c = 0
a = 2, b = -9, and c = 4
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{9 ± \sqrt{(-9)² – 4 (2) (4)}}{2 (2}\)
x = \(\frac{9 ± \sqrt{49}}{4}\)
x = \(\frac{9 ± 7}{4}\)
x = \(\frac{9 + 7}{4}\) or x = \(\frac{9 – 7}{4}\)
x = 4 or x = \(\frac{1}{2}\)
The solutions of the given equation are 4 and \(\frac{1}{2}\)

Question 47.
MODELING WITH MATHEMATICS
A fountain shoots a water arc that can be modeled by the graph of the equation y = -0.006x² + 1.2x + 10, where x is the horizontal distance (in feet) from the river’s north shore and y is the height (in feet) above the river. Does the water arc reach a height of 50 feet? If so, how far from the north shore is the water arc 50 feet above the water?

Answer:

Question 48.
MODELING WITH MATHEMATICS
Between the months of April and September, the number y of hours of daylight per day in Seattle, Washington, can be modeled by y = -0.00046x² + 0.076x + 13, where x is the number of days since April 1.
a. Do any of the days between April and September in Seattle have 17 hours of daylight? If so, how many?
Answer:

b. Do any of the days between April and September in Seattle have 14 hours of daylight? If so, how many?
Answer:

Question 49.
MAKING AN ARGUMENT
Your friend uses the discriminant of the equation 2x² – 5x – 2 = -11 and determines that the equation has two real solutions. Is your friend correct? Explain your reasoning.
Answer:

Question 50.
MODELING WITH MATHEMATICS
The frame of the tent shown is defined by a rectangular base and two parabolic arches that connect the opposite corners of the base. The graph of y = -0.18x² + 1.6x models the height y (in feet) of one of the arches x feet along the diagonal of the base. Can a child who is 4 feet tall walk under one of the arches without having to bend over? Explain.

Answer:


MATHEMATICAL CONNECTIONS In Exercises 51 and 52, use the given area A of the rectangle to find the value of x. Then give the dimensions of the rectangle.
Question 51.
A = 91 m²

Answer:

Question 52.
A = 209 ft²

Answer:
l = 4x – 5
w = 4x + 3
Area = Length × Width
209 = (4x – 5) (4x + 3)
16x² + 12x – 20x – 15 = 209
16x² – 8x – 15 – 209 = 0
16x² – 8x – 224 = 0
2x² – x – 28 = 0
2x² – 8x + 7x – 28 = 0
2x ( x – 4) + 7 (x – 4) = 0
(2x + 7) (x – 4) = 0
2x + 7 = 0 or x – 4 = 0
2x = -7 or x = 4
x = –\(\frac{7}{2}\) or x = 4
The dimensions of a rectangle won’t be negative
The value of x is 4
The length of rectangle = 4x + 3
= 4 (4) + 3
= 16 + 3 = 19 ft
The width of the rectangle = 4x – 5
= 4 (4) – 5
= 16 – 5
= 11 ft
The length of the rectangle is 19 ft
The width of the rectangle is 11 ft

COMPARING METHODS In Exercises 53 and 54, solve the equation by (a) graphing, (b) factoring, and (c) using the Quadratic Formula. Which method do you prefer? Explain your reasoning.
Question 53.
x² + 4x + 4 = 0
Answer:

Question 54.
3x² + 11x + 6 = 0
Answer:
3x² + 11x + 6 = 0
a) By using graphing method:

b) By using factoring method:
3x² + 11x + 6 = 0
3x² + 9x + 2x + 6 = 0
3x (x + 3) + 2 (x + 3) = 0
(3x + 2) (x + 3) = 0
3x + 2 = 0 or x + 3 = 0
3x = -2 or x = -3
x = –\(\frac{2}{3}\) or x = -3
The solutions for the given equation are -3 and –\(\frac{2}{3}\)
c) By using the quadratic formula:
3x² + 11x + 6 = 0
ax² + bx + c = 0
a = 3, b = 11, and c = 6
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-11 ± \sqrt{(11)² – 4 (3) (6)}}{2 (3}\)
x = \(\frac{-11 ± \sqrt{49}}{6}\)
x = \(\frac{-11 ± 7}{6}\)
x = \(\frac{-11 + 7}{6}\) or x = \(\frac{-11 – 7}{6}\)
x = –\(\frac{2}{3}\) or x = -3
The solutions of the given equation are -3 and –\(\frac{2}{3}\)
We can find the solutions for the given equation easily by using the factoring method

Question 55.
REASONING
How many solutions does the equation ax² + bx + c = 0 have when a and c have different signs? Explain your reasoning.
Answer:

Question 56.
REASONING
When the discriminant is a perfect square, are the solutions of ax² + bx + c = 0 rational or irrational? (Assume a, b, and c are integers.) Explain your reasoning.
Answer:
The discriminant is a perfect square
Discriminant = b² – 4ac
A perfect square means a number that is greater than zero
b² – 4ac > 0
The solutions of ax² + bx + c = 0 are rational

REASONING In Exercises 57–59, give a value of c for which the equation has (a) two solutions, (b) one solution, and (c) no solutions.
Question 57.
x² – 2x + c = 0
Answer:

Question 58.
x² – 8x + c = 0
Answer:
x² – 8x + c = 0
ax² + bx + c = 0
a = 1 and b = -8
Discriminant (d) = b² – 4ac
If d > 0, then the equation has 2 real solutions
If d = 0, then the equation has one solution
If d < 0, then the equation has no real solutions
a) For two solutions:
b² – 4ac > 0
(-8)² – 4 (1) c > 0
64 – 4c > 0
64 > 4c
\(\frac{64}{4}\) > c
16 > c
c < 16
b) For one solution:
b² – 4ac = 0
(-8)² – 4 (1) c = 0
64 – 4c = 0
64 = 4c
\(\frac{64}{4}\) = c
c = 16
c) For no solutions:
b² – 4ac < 0
(-8)² – 4 (1) c < 0
64 – 4c < 0
64 < 4c
\(\frac{64}{4}\) < c
16 < c
c > 16

Question 59.
4x² + 12x + c = 0
Answer:

Question 60.
REPEATED REASONING
You use the Quadratic Formula to solve an equation.
a. You obtain solutions that are integers. Could you have used factoring to solve the equation? Explain your reasoning.
Answer:
Yes, the solutions correspond to zeros. If the solutions are p and q, we can write the equation in the form
(x – p) (x – q) = 0
The equation is factorized with the solutions p and q
Example:
If  the solutions are
p = 2 and q = -3, then
(x – 2) (x + 3) = 0

b. You obtain solutions that are fractions. Could you have used factoring to solve the equation? Explain your reasoning.
Answer:
Yes, the reason is the same as in part (a)
Example:
If the solutions are p = \(\frac{1}{2}[/altex] and q = –[latex]\frac{4}{3}\)
(2x – 1) (3x + 4) = 0

c. Make a generalization about quadratic equations with rational solutions.
Answer:
The quadratic equations with the rational solutions can be easily solved using the “factoring” method of the quadratic equation

Question 61.
MODELING WITH MATHEMATICS
The fuel economy y(in miles per gallon) of a car can be modeled by the equation y = -0.013x² + 1.25x + 5.6, where 5 ≤ x ≤ 75 and x is the speed (in miles per hour) of the car. Find the speed(s) at which you can travel and have a fuel economy of 32 miles per gallon.
Answer:

Question 62.
MODELING WITH MATHEMATICS
The depth d (in feet) of a river can be modeled by the equation d = -0.25t² + 1.7t + 3.5, where 0 ≤ t ≤ 7 and t is the time (in hours) after a heavy rain begins. When is the river 6 feet deep?
Answer:

ANALYZING EQUATIONS In Exercises 63–68, tell whether the vertex of the graph of the function lies above, below, or on the x-axis. Explain your reasoning without using a graph.
Question 63.
y = x² – 3x + 2
Answer:

Question 64.
y = 3x² – 6x + 3
Answer:
y = 3x² – 6x + 3
3x² – 6x + 3 = 0
ax² + bx + c = 0
a = 3, b = -6, and c = 3
b² – 4ac = (-6)² – 4 (3) (3)
= 36 – 36
= 0
The discriminant is equal to 0
The vertex of the given function lies on x-axis

Question 65.
y = 6x² – 2x + 4
Answer:

Question 66.
y = -15x² + 10x – 25
Answer:
y = -15x² + 10x – 25
-15x² + 10x – 25 = 0
15x² – 10x + 25 = 0
ax² + bx + c = 0
a = 15, b = -10, and c = 25
b² – 4ac = (-10)² – 4 (15) (25)
= 100 – 1,500
= -1,400
The discriminant is less than 0
The graph of y has no x-intercepts and also because a > 0, the graph opens up and has a minimum value.
The vertex of the graph lies above the x-axis

Question 67.
f(x) = -3x² – 4x + 8
Answer:

Question 68.
f(x) = 9x² – 24x + 16
Answer:
f (x) = 9x² – 24x + 16
9x² – 24x + 16 = 0
ax² + bx + c = 0
a = 9, b = -24, and c = 16
b² – 4ac = (-24)² – 4 (9) (16)
= 576 – 576
= 0
The discriminant is equal to 0
The vertex of the given function lies on the x-axis

Question 69.
REASONING
NASA creates a weightless environment by flying a plane in a series of parabolic paths. The height h (in feet) of a plane after t seconds in a parabolic flight path can be modeled by h = -11t² + 700t + 21,000. The passengers experience a weightless environment when the height of the plane is greater than or equal to 30,800 feet. For approximately how many seconds do passengers experience weightlessness on such a flight? Explain.

Answer:

Question 70.
WRITING EQUATIONS
Use the numbers to create a quadratic equation with the solutions x = -1 and x = –\(\frac{1}{4}\)

Answer:
x = -1 and x = –\(\frac{1}{4}\)
The representation of the equation with the solutions are:
(x – p) (x – q) = 0(x + 1) (x + \(\frac{1}{4}\)) = 0
(x + 1) (4x + 1) = 0
4x² + x + 4x + 1 = 0
4x² + 5x + 1 = 0
The missing numbers of the given equation are: 4, 5, and 1

Question 71.
PROBLEM SOLVING
A rancher constructs two rectangular horse pastures that share a side, as shown. The pastures are enclosed by 1050 feet of fencing. Each pasture has an area of 15,000 square feet.

a. Show that y = 350 – \(\frac{4}{3}\)x.
b. Find the possible lengths and widths of each pasture.
Answer:

Question 72.
PROBLEM SOLVING
A kicker punts a football from a height of 2.5 feet above the ground with an initial vertical velocity of 45 feet per second.

a. Write an equation that models this situation using the function h = -16t² + v⁰t + s⁰, where h is the height (in feet) of the football, t is the time (in seconds) after the football is punted, v0 is the initial vertical velocity (in feet per second), and s0 is the initial height (in feet).
b. The football is caught 5.5 feet above the ground, as shown in the diagram. Find the amount of time that the football is in the air.
Answer:

Question 73.
CRITICAL THINKING
The solutions of the quadratic equation ax² + bx + c = 0 are x = \(\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\) and x = \(\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}\). Find the mean of the solutions. How is the mean of the solutions related to the graph of y = ax² + bx + c? Explain.
Answer:

Question 74.
HOW DO YOU SEE IT?
Match each graph with its discriminant. Explain your reasoning.

a. b² – 4ac > 0
Answer:
If b² – 4ac > 0, then the quadratic equation will have 2 real roots
When we observe the graphs, graph C matches the b² – 4ac > 0
Since a < 0, the quadratic equation will have the parabola closed down and have the maximum value
The graph of b² – 4ac > 0 will have the maximum value

b. b² – 4ac = 0
Answer:
If b² – 4ac = 0, then the quadratic equation will have 1 real root
When we observe the graphs, graph A matches the b² – 4ac = 0
Since a < 0, the quadratic equation will have the parabola closed down and have the maximum value
The graph of b² – 4ac = 0 will have the maximum value

c. b² – 4ac < 0
Answer:
If b² – 4ac< 0, then the quadratic equation will have no real roots
When we observe the graphs, graph B matches the b² – 4ac < 0
Since a > 0, the quadratic equation will have the parabola opens up and have the minimum imaginary value
The graph of b² – 4ac < 0 will have the minimum imaginary value

Question 75.
CRITICAL THINKING
You are trying to hang a tire swing. To get the rope over a tree branch that is 15 feet high, you tie the rope to a weight and throw it over the branch. You release the weight at a height s₀ of 5.5 feet. What is the minimum initial vertical velocity v₀ needed to reach the branch? (Hint: Use the equation h = -16t² + v₀ t + s₀)
Answer:

Question 76.
THOUGHT-PROVOKING
Consider the graph of the standard form of a quadratic function y = ax² + bx + c. Then consider the Quadratic Formula as given by
x = \(-\frac{b}{2 a} \pm \frac{\sqrt{b^{2}-4 a c}}{2 a}\)
Write a graphical interpretation of the two parts of this formula.
Answer:

Question 77.
ANALYZING RELATIONSHIPS
Find the sum and product of \(\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\) and \(\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}\). Then write a quadratic equation whose solutions have a sum of 2 and a product of \(\frac{1}{2}\).
Answer:

Question 78.
WRITING A FORMULA
Derive a formula that can be used to find solutions of equations that have the form ax² + x + c = 0. Use your formula to solve -2x² + x + 8 = 0.
Answer:
ax² + x + c = 0
ax² + bx + c = 0
a = a, b = 1, and c = c
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-1 ± \sqrt{1² – 4 (a) (c)}}{2a}\)
x = \(\frac{-1 ± \sqrt{1 – 4ac}}{2a}\)
The solutions of the equation ax² + x + c = 0 are: \(\frac{-1 ± \sqrt{1 – 4ac}}{2a}\)
-2x² + x + 8 = 0
ax² + x + c = 0
a = -2, b = 1, and c = 8
The solutions of the above equation are:
x = \(\frac{-1 ± \sqrt{1 – 4 (-2) (8)}}{2 (-2)}\)
x = \(\frac{-1 ± \sqrt{1 + 64}}{-4}\)
x = \(\frac{-1 ± \sqrt{65}}{-4}\)
x = \(\frac{-1 + \sqrt{65}}{-4}\) or x = \(\frac{-1 – \sqrt{65}}{-4}\)
x = -1.76 or x = 2.26

Question 79.
MULTIPLE REPRESENTATIONS
If p is a solution of a quadratic equation ax² + bx + c = 0, then (x – p) is a factor of ax² + bx + c.
a. Copy and complete the table for each pair of solutions.

b. Graph the related function for each equation. Identify the zeros of the function.
Answer:

CRITICAL THINKING In Exercises 80–82, find all values of k for which the equation has (a) two solutions, (b) one solution, and (c) no solutions.
Question 80.
2x² + x + 3k = 0
Answer:
2x² + x + 3k = 0
ax² + bx + c = 0
a = 2, b = 1, and c = 3k
If b² – 4ac > 0, then the quadratic equation has 2 soltions
If b² – 4ac = 0, then the quadratic equation has 1 solution
If b² – 4ac < 0, then the quadratic equation has no real solutions
a) For 2 solutions:
b² – 4ac > 0
1² – 4 (2) (3k) > 0
1 – 24k > 0
1 > 24k
24k < 1
k < \(\frac{1}{24}\)
b) For 1 solution:
b² – 4ac = 0
1² – 4 (2) (3k) = 0
1 – 24k = 0
1 = 24k
24k = 1
k = \(\frac{1}{24}\)
c) For no solutions:
b² – 4ac < 0
1² – 4 (2) (3k) < 0
1 – 24k < 0
1 < 24k
24k > 1
k > \(\frac{1}{24}\)

Question 81.
x² = 4kx + 36 = 0
Answer:

Question 82.
kx² + 5x – 16 = 0
Answer:
kx² + 5x – 16 = 0
ax² + bx + c = 0
a = k, b = 5, and c = -16
If b² – 4ac > 0, then the quadratic equation has 2 solutions
If b² – 4ac = 0, then the quadratic equation has 1 solution
If b² – 4ac < 0, then the quadratic equation has no real solutions
a) For 2 solutions:
b² – 4ac > 0
5² – 4 (k) (-16) > 0
25 + 64k > 0
25 > -64k
-64k < 25
k < –\(\frac{25}{64}\)
b) For 1 solution:
b² – 4ac = 0
5² – 4 (k) (-16) = 0
25 + 64k = 0
25 = -64k
-64k = 25
k = –\(\frac{25}{64}\)
c) For no solutions:
b² – 4ac < 0
5² – 4 (k) (-16) < 0
25 + 64k = 0
25 < -64k
-64k > 25
k > –\(\frac{25}{64}\)

Maintaining Mathematical Proficiency

Solve the system of linear equations using any method. Explain why you chose the method.
Question 83.
y = -x + 4
y = 2x – 8
Answer:

Question 84.
x = 16 – 4y
3x + 4y = 8
Answer:

The solution of the given system of linear equations is (-4, 5)

Question 85.
2x – y = 7
2x + 7y = 31
Answer:

Question 86.
3x – 2y = -20
x + 1.2y = 6.4
Answer:
3x – 2y = -20 —-(1)
x + 1.2y = 6.4 —(2)

The solution for the given system of linear equations is: (-2, 7)

Lesson 9.6 Solving Nonlinear Systems of Equations

Essential Question: How can you solve a system of two equations when one is linear and the other is quadratic?
Answer:
Step 1: Isolate one of the two variables in one of the equations
Step 2: Substitute the expression that is equal to the isolated variable from step 1 into the other equation
Step 3: Solve the resulting quadratic equation to find the x value(s) of the solution(s)

EXPLORATION 1
Solving a System of Equations
Work with a partner. Solve the system of equations by graphing each equation and finding the points of intersection.

System of Equations
y = x + 2
y = x² + 2x
Answer:

The intersection point of the given system of equations is (1, 3)

EXPLORATION 2

Analyzing Systems of Equations
Work with a partner. Match each system of equations with its graph. Then solve the system of equations.

Answer:
a.
y = x² – 4
y = -x – 2
x² – 4 = -x – 2
(x – 2) (x + 2) = -(x + 2)
x – 2 = -1
x = -1 + 2
x = 1
For x = 1,
y = -1 – 2
y = -3
The solution or intersection point of the given system of equations is (1, -3)
Graph A matches the solution of the given system of equations

b.
y = x² – 2x + 2
y = 2x – 2
x² – 2x + 2 = 2x – 2
x² – 2x – 2x + 2 + 2 = 0
x² – 4x + 4 = 0
(x – 2)² = 0
√x – 2 = √0
x – 2 = 0
x = 2
For x = 2,
y = 2 (2) – 2
= 4 – 2
= 2
The solution or intersection point of the given system of equations is (2, 2)
Graph C matches the solution of the given system of equations

c.
y = x² + 1
y = x – 1
x² + 1 = x – 1
x² – x + 1 + 1 = 0
x²- x + 2 = 0
There will be no solution or intersection point for the given system of equations
Graph B matches the solution of the given system of equations

d.
y = x² – x – 6
y = 2x – 2
x² – x – 6 = 2x – 2
x² – x – 2x – 6 + 2= 0
x² – 3x – 4 = 0
x² – 4x + x – 4 = 0
x (x – 4) + 1 ( x – 4) = 0
(x + 1) (x – 4) = 0
x + 1 = 0 or x – 4 = 0
x = -1 or x = 4
For x = -1,
y = 2 (-1) – 2
= -2 – 2
= -4
For x = 4,
y = 2 (4) – 2
= 8 – 2
= 6
The solutions or intersection points of the given system of equations are (-1, -4), and (4, 6)
Graph D matches with the solution of the given system of equations

Communicate Your Answer

Question 3.
How can you solve a system of two equations when one is linear and the other is quadratic?
Answer:
Step 1: Isolate one of the two variables in one of the equations
Step 2: Substitute the expression that is equal to the isolated variable from step 1 into the other equation
Step 3: Solve the resulting quadratic equation to find the x value(s) of the solution(s)

Question 4.
Write a system of equations (one linear and one quadratic) that has (a) no solutions, (b) one solution, and (c) two solutions. Your systems should be different from those in Explorations 1 and 2.

Answer:
The system of equations (one linear and one quadratic) has
a) No solutions:
y = x² – 5x + 7
y = x – 3

b) One solution:
y = x² – 12x + 36
y = x + 7

c) Two solutions:
y = 3x² + 11x + 6
y = x + 10

Monitoring Progress

Solve the system by graphing.
Question 1.
y = x² + 4x – 4
y = 2x – 5
Answer:

The intersection point of the given system of equations is (-6,-6)

Question 2.
y = -x + 6
y = -2x² – x + 3
Answer:

There is no intersection point of the 2 equations

Question 3.
y = 3x – 15
y = \(\frac{1}{2}\)x² – 2x – 7
Answer:

The intersection points of the given system of equations are (8, 9) and (2, -9)

Solve the system by substitution.
Question 4.
y = x² + 9 5.
y = 9
Answer:
y = x² + 9.5
y = 9
x² + 9.5 = 9
x² = 9 – 9.5
x² = -0.5
The given system of equations has no real solutions

Question 5.
y = -5x
y = x² – 3x – 3
Answer:
y = -5x
y = x² – 3x – 3
-5x = x² – 3x – 3
x² – 3x + 5x – 3 = 0
x² + 2x – 3 = 0
x² + 3x – x – 3 = 0
x (x + 3) – 1 ( x+ 3) = 0
(x – 1) (x + 3) = 0
x – 1 = 0 or x + 3 = 0
x = 1 or x = -3
For x = 1,
y = -5 (1)
= -5
For x = -3,
y = -5(-3)
= 15
The intersection points of the given system of equations are (1, -5) and (-3, 15)

Question 6.
y = -3x² + 2x + 1
y = 5 – 3x
Answer:
y = -3x² + 2x + 1
y = 5 – 3x
-3x² + 2x + 1 = 5 – 3x
-3x² + 2x + 3x + 1 – 5 = 0
-3x² + 5x – 4 = 0
3x² – 5x + 4 = 0
ax² + bx + c = 0
a = 3, b = -5, and c = 4
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{5 ± \sqrt{(-5)² – 4 (3) (4)}}{2(3)}\)
x = \(\frac{5 ± \sqrt{-23}}{6}\)
The given system of equations has no real solutions

Solve the system by elimination.
Question 7.
y = x² + x
y = x + 5
Answer:
y = x² + x —–(1)
y = x + 5 ——(2)
eq (1) – eq (2)
y – y = x² + x – x – 5
0 = x² – 5
x² = 5
√x² = √5
x = ±√5
For x = ±√5,
y = ±√5 + 5
The solutions for the given system of equations are: (±√5, ±√5 + 5)

Question 8.
y = 9x² + 8x – 6
y = 5x – 4
Answer:
y = 9x² + 8x – 6 —–(1)
y = 5x – 6 —–(2)
eq (1) – eq (2)
y – y = 9x² + 8x – 6 – 5x + 4
0 = 9x²+ 3x – 2
9x² + 3x – 2 = 0
9x² + 6x – 3x – 2 = 0
3x(3x – 1) + 2 (3x – 1) = 0
(3x + 2) (3x – 1) = 0
3x + 2 = 0 or 3x – 1 = 0
3x = -2 or 3x = 1
x = –\(\frac{2}{3}\) or x = \(\frac{1}{3}\)
For x = –\(\frac{2}{3}\),
y = 5 (-\(\frac{2}{3}\)) – 4
y = –\(\frac{22}{3}\)
For x = \(\frac{1}{3}\),
x = 5 (\(\frac{1}{3}\)) – 4
x = –\(\frac{7}{3}\)

Question 9.
y = 2x + 5
y = -3x² + x – 4
Answer:
y = 2x + 5 —-(1)
y = -3x² + x – 4 —–(2)
eq (1) – eq (2)
y – y = 2x + 5 + 3x² – x + 4
0 = 3x² + x + 9
3x² + x + 9 = 0
ax² + bx + c = 0
a = 3, b = 1, and c = 9
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-1 ± \sqrt{(1)² – 4 (3) (9)}}{2(3)}\)
x = \(\frac{-1 ± \sqrt{-107}}{6}\)
The given system of equations has no real solutions.

Use the method in Example 4 to approximate the solution(s) of the system to the nearest thousandth.
Question 10.
y = 4^x
y = x² + x + 3
Answer:

The solution of the given system of equations lies between -2 and -1

Question 11.
y = 4x² – 1
y = -2(3)^x + 4
Answer:
y = 4x² – 1
y = -2 (3^x) + 4

The solution for the given system of equations lies between 0 and 1

Question 12.
y = x² + 3x
y = -x² + x + 10
Answer:

The solution for the given system of equations lies between -3 and -2.5

Solve the equation. Round your solution(s) to the nearest hundredth.
Question 13.
3^x – 1 = x² – 2x + 5
Answer:

The solution for the given system of equations lies between 1 and 1.5

Question 14.
4x² + x = -2 (\(\frac{1}{2}\))^x + 5
Answer:

The solution for the given system of equations is (1, 4)

Solving Nonlinear Systems of Equations 9.6 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Describe how to use substitution to solve a system of nonlinear equations.
Answer:

Question 2.
WRITING
How is solving a system of nonlinear equations similar to solving a system of linear equations? How is it different?
Answer:
It is similar because you can use the same methods for linear and non-linear equations
However, the difference is that sometimes you also need to use different methods and you can obtain multiple solutions

In Exercises 3–6, match the system of equations with its graph. Then solve the system.
Question 3.
y = x² – 2x + 1
y = x + 1
Answer:

Question 4.
y = x² + 3x + 2
y = -x – 3
Answer:
y = x² + 3x + 2
y = -x – 3
x² + 3x + 2 = -x – 3
x² + 3x + 2 + x + 3 = 0
x²+ 4x + 5 = 0
There will be no solution or intersection point for the given system of equations
Graph D matches the solution of the given system of equations

Question 5.
y = x – 1
y = -x² + x – 1
Answer:

Question 6.
y = -x + 3
y = -x² – 2x + 5
Answer:
y = -x + 3
y = -x² – 2x + 5
-x + 3 = -x² – 2x + 5
x² + 2x – 5 – x + 3 = 0
x² + x – 2 = 0
x² + 2x – x – 2 = 0
x (x + 2) – 1 (x + 2) = 0
(x – 1) (x + 2) = 0
x – 1 = 0 or x + 2 = 0
x = 1 or x = -2
For x = 1,
y = -1 + 3
= 2
For x = -2,
y = 2 + 3
= 5
The solutions or the intersection points of the given system of equations are (1, 2), and (-2, 5)
Graph C matches the given system of equations

In Exercises 7–12, solve the system by graphing.
Question 7.
y = 3x² – 2x + 1
y = x + 7
Answer:

Question 8.
y = x² + 2x + 5
y = -2x – 5
Answer:

There are no solutions for the given system of equations

Question 9.
y = -2x² – 4x
y = 2
Answer:

Question 10.
y = \(\frac{1}{2}\)x² – 3x + 4
y = x – 2
Answer:

The solution of the given system of equations is (6, 4)

Question 11.
y = \(\frac{1}{3}\) x² + 2x – 3
y = 2x
Answer:

Question 12.
y = 4x² + 5x – 7
y = -3x + 5
Answer:

The solution of the given system of equations is (1, 2)

In Exercises 13–18, solve the system by substitution.
Question 13.
y = x – 5
y = x² + 4x – 5
Answer:

Question 14.
y = -3x²
y = 6x + 3
Answer:
y = -3x²
y = 6x + 3
-3x² = 6x + 3
3x² + 6x + 3 = 0
3x² + 3x + 3x + 3 = 0
3x (x + 1) + 3 (x + 1) = 0
(3x + 3) (x + 1) = 0
3x + 3 = 0 or x + 1 = 0
3x = -3 or x = -1
x = -1 or x = -1
For x = -1,
y = 6 (-1) + 3
= -6 + 3
= -3
The solution of the given system of equations is (-1, -3)

Question 15.
y = -x + 7
y = -x² – 2x – 1
Answer:

Question 16.
y = -x² + 7
y = 2x + 4
Answer:
y = -x² + 7
y = 2x + 4
-x² + 7 = 2x + 4
x² – 7 + 2x + 4 = 0
x² + 2x – 3 = 0
x² + 3x – x – 3 = 0
x (x + 3) – 1 (x + 3) = 0
(x – 1) (x + 3) = 0
x – 1 = 0 or x + 3 = 0
x = 1 or x = -3
For x = 1,
y = 2 (1) + 4
= 6
For x = -3,
y = 2 (-3) + 4
= -2
The solutions for the given system of equations are (1, 6), and (-3, -2)

Question 17.
y – 5 = -x²
y = 5
Answer:

Question 18.
y = 2x² + 3x – 4
y – 4x = 2
Answer:
y = 2x² + 3x – 4
y – 4x = 2
y = 4x + 2
2x² + 3x – 4 = 4x + 2
2x² + 3x – 4x – 4 – 2 = 0
2x² – x – 6 = 0
2x² – 4x + 3x – 6 = 0
2x (x – 2) + 3 (x – 2) = 0
(2x + 3) (x – 2) = 0
2x + 3 = 0 or x – 2 = 0
2x = -3 or x = 2
x = –\(\frac{3}{2}\) or x = 2
For x = –\(\frac{3}{2}\),
y = 4 (-\(\frac{3}{2}\)) + 2
= -4
For x = 2,
y = 4 (2) + 2
= 10
The solutions of the given system of equations are (-\(\frac{3}{2}\), -4), and (2, 10)

In Exercises 19–26, solve the system by elimination.
Question 19.
y = x² – 5x – 7
y = -5x + 9
Answer:

Question 20.
y = -3x² + x + 2
y = x + 4
Answer:
y = -3x² + x + 2 —- (1)
y = x + 4 —- (2)
eq (1) – eq (2)
-3x² + x + 2 – x – 4 = 0
-3x² + x – x = 4 – 2
-3x² = 2
3x² = -2
The given system of equations does not have any real solutions

Question 21.
y = -x² – 2x + 2
y = 4x + 2
Answer:

Question 22.
y = -2x² + x – 3
y = 2x – 2
Answer:
y = -2x² + x – 3 —-(1)
y = 2x – 2 —-(2)
eq (1) – eq (2)
So,
-2x² + x – 3 – 2x + 2 = 0
-2x² – x – 1 = 0
2x² + x + 1 = 0
ax² + bx + c = 0
a = 2, b = 1, and c = 1
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-1 ± \sqrt{1² – 4 (2) (1)}}{2 (2)}\)
x = \(\frac{-1 ± \sqrt{-7}}{4}\)
The given system of equations does not have any real solutions

Question 23.
y = 2x – 1
y = x²
Answer:

Question 24.
y = x² + x + 1
y = -x – 2
Answer:
y = x² + x + 1 —-(1)
y = -x – 2 —-(2)
eq (1) – eq (2)
x² + x + 1 + x + 2 = 0
x² + 2x + 3 = 0
ax² + bx + c = 0
a = 1, b = 2, and c = 3
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-2 ± \sqrt{2² – 4 (3) (1)}}{2 (1)}\)
x = \(\frac{-2 ± \sqrt{-8}}{2}\)
The given system of equations does not have any real solutions

Question 25.
y + 2x = 0
y = x² + 4x – 6
Answer:

Question 26.
y = 2x – 7
y + 5x = x² – 2
Answer:
y = 2x – 7 —– (1)
y = x² – 5x —–(2)
eq (1) – eq (2)
2x – 7 – x² + 5x = 0
-x² + 7x – 7 = 0
x² – 7x + 7 = 0
ax² + bx + c = 0
a = 1, b = -7, and c = 7
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{7 ± \sqrt{(-7)² – 4 (7) (1)}}{2 (1)}\)
x = \(\frac{7 ± \sqrt{21}}{2}\)
x = \(\frac{7 + \sqrt{21}}{2}\) or x = \(\frac{7 – \sqrt{21}}{2}\)
x = 5.79 or x = 1.20
The solutions for the given system of equations are 5.79 and 1.20

Question 27.
ERROR ANALYSIS
Describe and correct the error in solving the system of equations by graphing.

Answer:

Question 28.
ERROR ANALYSIS
Describe and correct the error in solving for one of the variables in the system.

Answer:
y = 3x² – 6x + 4
y = 4
3x² – 6x + 4 = 4
3x²- 6x = 4 – 4
3x²= 6x
3x = 6
x = \(\frac{6}{3}\)
x = 2
The solution of the given system of equations is (2, 4)

In Exercises 29–32, use the table to describe the locations of the zeros of the quadratic function f.
Question 29.

Answer:

Question 30.

Answer:
There is a sign change between x = 1 and x = 2 and f (1) and f (2) are in the same distance,
The first solution lies between 1 and 2
There is a sign change between x = 3 and x = 4 and f (3) and f (4) are in the same distance,
The second solution lies between 3 and 4
The solutions for the given function will lie between (1, 2) and (3, 4)

Question 31.

Answer:

Question 32.

Answer:
There is a sign change between x = 2 and x = 3 and f (2) and f (3) are in the same distance,
But, f (3) is very close to 0
The first solution is closer to 3 than 2
There is a sign change between x = 5 and x = 6 and f (5) and f (6) are in the same distance,
But, f (5) is very close to 0
The second solution is closer to 5 than 6
The solution for the given function will lie between (3, 5)

In Exercises 33–38, use the method in Example 4 to approximate the solution(s) of the system to the nearest thousandth.
Question 33.
y = x² + 2x + 3
y = 3^x
Answer:

Question 34.
y = 2^x + 5
y = x² – 3x + 1
Answer:

The solution for the given system of equations is approximately (1, 6)

Question 35.
y = 2(4)^x – 1
y = 3x² + 8x
Answer:

Question 36.
y = -x² – 4x – 4
y = -5^x – 2
Answer:

The solution for the given system of equations is approximately (-1, -3)

Question 37.
y = -x² – x + 5
y = 2x² + 6x – 3
Answer:

Question 38.
y = 2x² + x – 8
y = x² – 5
Answer:

The approximate solution for the given system of equations is (-1, 2)

In Exercises 39–46, solve the equation. Round your solution(s) to the nearest hundredth.
Question 39.
3x + 1 = x² + 7x – 1
Answer:

Question 40.
-x² + 2x = -2x + 5
Answer:

There are no solutions for the given system of equations

Question 41.
x² – 6x + 4 = -x² – 2x
Answer:

Question 42.
2x² + 8x + 10 = -x² – 2x + 5
Answer:

The approximate solution for the given equation is (-1, 6)

Question 43.
-4 (\(\frac{1}{2}\))^x = -x² – 5
Answer:

Question 44.
1.5(2)^x – 3 = -x² + 4x
Answer:

The approximate solutions for the given equation are (2, 4) and ((0, -2)

Question 45.
8^x-2 + 3 = 2 (\(\frac{3}{2}\))^x
Answer:

Question 46.
-0.5(4)^x = 5^x – 6
Answer:

The solution for the given equation is (1, -2)

Question 47.
COMPARING METHODS
Solve the system in Exercise 37 using substitution. Compare the exact solutions to the approximated solutions.
Answer:

Question 48.
COMPARING METHODS
Solve the system in Exercise 38 using elimination. Compare the exact solutions to the approximated solutions.
Answer:
y = 2x² + x – 8 —-(1)
y = x² – 5 —–(2)
eq (1) – eq(2)
2x² + x – 8 = x² – 5
2x² – x² + x – 8 + 5 = 0
x² + x – 3 = 0
ax² + bx + c = 0
a = 1, b= 1, and c = -3
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-1 ± \sqrt{1² + 4 (1) (3)}}{2 (1)}\)
x = \(\frac{-1 ± \sqrt{13}}{2}\)
x = \(\frac{-1 + \sqrt{13}}{2}\) or x = \(\frac{-1 – \sqrt{13}}{2}\)
We know that,
\(\sqrt{13}\) = 3.60
x = 1.30 or x = -2.30
The exact solutions and approximate solutions are the same

Question 49.
MODELING WITH MATHEMATICS
The attendances y for two movies can be modeled by the following equations, where x is the number of days since the movies opened.
y = -x² + 35x + 100 Movie A
y = -5x + 275 Movie B
When is the attendance for each movie the same?
Answer:

Question 50.
MODELING WITH MATHEMATICS
You and a friend are driving boats on the same lake. Your path can be modeled by the equation y = -x² – 4x – 1, and your friend’s path can be modeled by the equation y = -2x + 8. Do your paths cross each other? If so, what are the coordinates of the point(s) where the paths meet?

Answer:

The paths of yours and your friend did not cross each other

Question 51.
MODELING WITH MATHEMATICS
The arch of a bridge can be modeled by y = -0.002x² + 1.06x, where x is the distance (in meters) from the left pylons and y is the height (in meters) of the arch above the water. The road can be modeled by the equation y = 52. To the nearest meter, how far from the left pylons are the two points where the road intersects the arch of the bridge?

Answer:

Question 52.
MAKING AN ARGUMENT
Your friend says that a system of equations consisting of a linear equation and a quadratic equation can have zero, one, two, or infinitely many solutions. Is your friend correct? Explain.
Answer:

COMPARING METHODS In Exercises 53 and 54, solve the system of equations by (a) graphing, (b) substitution, and (c) elimination. Which method do you prefer? Explain your reasoning.
Question 53.
y = 4x + 3
y = x² + 4x – 1
Answer:

Question 54.
y = x² – 5
y = -x + 7
Answer:

The solution for the given system of equations is (3, 4)
b) By substitution:
x² – 5 = -x + 7
x² + x – 5 – 7 = 0
x² + x – 12 = 0
x² + 4x – 3x – 12 = 0
x (x – 3) + 4 (x – 3) = 0
(x + 4) (x – 3) = 0
x + 4 = 0 or x – 3 = 0
x = -4 or x = 3
For x = 3,
y = -3 + 7
= 4
For x = -4,
y = 4 + 7
= 11
The solutions for the given system of equations are (3, 4) and (-4, 11)
c) By elimination:
Since the variable on the left side of the given system of equations is equal,
eq (1) – eq (2) = 0
So,
x² + x – 5 – 7 = 0
x² + x – 12 = 0
x² + 4x – 3x – 12 = 0
x (x – 3) + 4 (x – 3) = 0
(x + 4) (x – 3) = 0
x + 4 = 0 or x – 3 = 0
x = -4 or x = 3
For x = 3,
y = -3 + 7
= 4
For x = -4,
y = 4 + 7
= 11
The solutions for the given system of equations are (3, 4) and (-4, 11)
We can prefer the substitution method because of the factorization method.

Question 55.
MODELING WITH MATHEMATICS
The function y = -x² + 65x + 256 models the number y of subscribers to a website, where x is the number of days since the website launched. The number of subscribers to a competitor’s website can be modeled by a linear function. The websites have the same number of subscribers on Days 1 and 34.
a. Write a linear function that models the number of subscribers to the competitor’s website.
b. Solve the system to verify the function from part (a).
Answer:

Question 56.
HOW DO YOU SEE IT?
The diagram shows the graphs of two equations in a system that has one solution.

a. How many solutions will the system have when you change the linear equation to y = c + 2?
Answer:
We can observe that y = c is a constant line that is parallel to the x-axis
y = c + 2 is also a constant line
We can conclude that the system will have only one solution when you change the linear equation to y = c + 2

b. How many solutions will the system have when you change the linear equation to y = c – 2?
Answer:
We can observe that y = c is a constant line that is parallel to the x-axis
y = c – 2 is also a constant line
We can conclude that the system will have only one solution when you change the linear equation to y = c – 2

Question 57.
WRITING
A system of equations consists of a quadratic equation whose graph opens up and a quadratic equation whose graph opens down. Describe the possible number of solutions of the system. Sketch examples to justify your answer.
Answer:

Question 58.
PROBLEM-SOLVING
The population of a country is 2 million people and increases by 3% each year. The country’s food supply is sufficient to feed 3 million people and increases at a constant rate that feeds 0.25 million additional people each year.
a. When will the country first experience a food shortage?
Answer:

b. The country doubles the rate at which its food supply increases. Will food shortages still occur? If so, in what year?
Answer:

Question 59.
ANALYZING GRAPHS
Use the graphs of the linear and quadratic functions.

a. Find the coordinates of point A.
b. Find the coordinates of point B.
Answer:

Question 60.
THOUGHT-PROVOKING
Is it possible for a system of two quadratic equations to have exactly three solutions? exactly four solutions? Explain your reasoning. (Hint: Rotations of the graphs of quadratic equations still represent quadratic equations.)
Answer:
The quadratic equation is in the form of
ax² + bx + c = 0
We can say that the quadratic equation will have only an even number of solutions even we consider the rotations of the graphs of the quadratic equations
The quadratic equation will have exactly four solutions

Question 61.
PROBLEM-SOLVING
Solve the system of three equations shown.
y = 2x – 8
y = x² – 4x – 3
y = -3(2)^x
Answer:

Question 62.
PROBLEM-SOLVING
Find the point(s) of intersection, if any, of the line y = -x – 1 and the circle x² + y² = 41.
Answer:
y = -x – 1 —-  (1)
x² + y² = 41 —- (2)
Substitute eq (1) into eq (2)
x² + (-x – 1)² = 41
x² + x² + 1 + 2x = 41
2x² + 2x – 40 = 0
x²+ x – 20 = 0
x² + 5x – 4x – 20 = 0
x ( x + 5 ) – 4 (x + 5) = 0
(x – 4) (x + 5) = 0
x – 4 = 0 or x + 5 = 0
x = 4 or x = -5
For x = 4,
y = -4 – 1
= -5
For x = -5,
y = 5 – 1,
= 4
The points of intersection for the given equations are (4, -5) and (-5, 4)

Maintaining Mathematical Proficiency

Graph the system of linear inequalities.
Question 63.
y > 2x
y > -x + 4
Answer:

Question 64.
y ≥ 4x + 1
y ≤ 7
Answer:

Question 65.
y – 3 ≤ -2x
y + 5 < 3x
Answer:

Question 66.
x + y > -6
2y ≤ 3x + 4
Answer:

Graph the function. Describe the domain and range.
Question 67.
y = 3x² + 2
Answer:

Question 68.
y = -x² – 6x
Answer:

The domain is the range of values of x that make the equation positive
The range is the range of values of y that make the equation positive
The domain of the given function is: 0 ≤ x ≤ -6
The range of the given function is: -10 ≤ y ≤ 9

Question 69.
y = -2x² + 12x – 7
Answer:

Question 70.
y = 5x² + 10x – 3
Answer:

The domain is the set of all the values of x that make the equation positive
The range is the set of all the values of y that make the equation positive
The approximate domain of the given function is: -2 ≤ x ≤ 0
The range of the given equation is: -8 ≤ y ≤ 10

Solving Quadratic Equations Performance Task: Form Matters

9.4–9.6 What Did You Learn?

Core Vocabulary

Core Concepts
Section 9.4
Completing the Square, p. 506

Section 9.5
Quadratic Formula, p. 516
Interpreting the Discriminant, p. 518

Section 9.6
Solving Systems of Nonlinear Equations, p. 526

Mathematical Practices
Question 1.
How does your answer to Exercise 74 on page 514 help create a shortcut when solving some quadratic equations by completing the square?
Answer:
For the equation that is of the form
ax² + bx + c
When we use the “Completing the squares” method,
c = (\(\frac{b}{2}\))² and in place of b,
b = \(\frac{b}{2}\)

Question 2.
What logical progression led you to your answer in Exercise 55 on page 522?
Answer:
In Exercise 55 on page 522,
We use the sign changes for a and c and find the discriminant to find the logical procession

Question 3.
Compare the methods used to solve Exercise 53 on page 532. Discuss the similarities and differences among the methods.
Answer:

Performance Task Form Matters

Each form of a quadratic function has its pros and cons. Using one form, you can easily find the vertex, but the zeros are more difficult to find. Using another form, you can easily find the y-intercept, but the vertex is more difficult to find. Which form would you use in different situations? How can you convert one form into another? To explore the answers to these questions and more, go to.

Solving Quadratic Equations Chapter Review

9.1 Properties of Radicals (pp. 479–488)

Simplify the expression.
Question 1.
\(\sqrt{72 p^{7}}\)
Answer:
\(\sqrt{72 p^{7}}\)
= \(\sqrt{72 p^{6} p^{1}}\)
= \(\sqrt{36 p^{6} × 2p^{1}}\)
= 6p³ \(\sqrt{2p}\)

Question 2.
\(\sqrt{\frac{45}{7 y}}\)
Answer:
\(\sqrt{\frac{45}{7 y}}\)
= \(\frac{\sqrt{45}}{\sqrt{7 y}}\)
= \(\frac{\sqrt{5 (9)}}{\sqrt{7 y}}\)
= 3\(\sqrt{\frac{5}{7 y}}\)

Question 3.
\(\sqrt[3]{\frac{125 x^{11}}{4}}\)
Answer:
\(\sqrt[3]{\frac{125 x^{11}}{4}}\)
= \(\sqrt[3]{\frac{5 ^{3} x^{9} x^{2}}{4}}\)
= 5x³\(\sqrt[3]{\frac{ x^{2}}{4}}\)

Question 4.
\(\frac{8}{\sqrt{6}+2}\)
Answer:
\(\frac{8}{\sqrt{6}+2}\)
Multiply and divide the given expression with \(\sqrt{6} – 2\)
= \(\frac{8}{\sqrt{6}+2}\) × \(\frac{\sqrt{6} – 2}{\sqrt{6} – 2}\)
= \(\frac{8 (\sqrt{6} – 2} {6 – 4}\)
= \(\frac{8\sqrt{6} – 2}{2}\)
= 4 \(\sqrt{6} – 2\)

Question 5.
4\(\sqrt{3}\) + 5\(\sqrt{12}\)
Answer:
4\(\sqrt{3}\) + 5\(\sqrt{12}\)
= 4\(\sqrt{3}\) + 5\(\sqrt{4 (3)}\)
= 4\(\sqrt{3}\) + 5 (2)\(\sqrt{3}\)
= 4\(\sqrt{3}\) + 10\(\sqrt{3}\)
= \(\sqrt{3}\) (5 + 4)
= 9\(\sqrt{3}\)

Question 6.
15\(\sqrt [ 3 ]{ 2 }\) – 2\(\sqrt [ 3 ]{ 54 }\)
Answer:
15\(\sqrt [ 3 ]{ 2 }\) – 2\(\sqrt [ 3 ]{ 54 }\)
= 15\(\sqrt [ 3 ]{ 2 }\) – 2\(\sqrt [ 3 ]{ 27 (2) }\)
= 15\(\sqrt [ 3 ]{ 2 }\) – 2 (3)\(\sqrt [ 3 ]{ 2 }\)
= 15\(\sqrt [ 3 ]{ 2 }\) – 6\(\sqrt [ 3 ]{ 2 }\)
= [/latex]\sqrt[3]{2}[/latex] (15 – 6)
= 9 \(\sqrt[3]{2}\)

Question 7.
(3\(\sqrt{7}\) + 5)²
Answer:
(3\(\sqrt{7}\) + 5)²
= (3\(\sqrt{7}\))² + 5² + 2 (3\(\sqrt{7}\)) (5)
= 9 (7) + 25 + 30\(\sqrt{7}\)
= 88 + 30\(\sqrt{7}\)

Question 8.
\(\sqrt{6}\)(\(\sqrt{18}\) + \(\sqrt{8}\))
Answer:
\(\sqrt{6}\)(\(\sqrt{18}\) + \(\sqrt{8}\))
= \(\sqrt{6}\)(\(\sqrt{9 (2)}\) + \(\sqrt{4 (2)}\))
= \(\sqrt{6}\)(\(3\sqrt{2}\) + \(2\sqrt{2}\))
= \(\sqrt{6}\)((3 + 2) \(\sqrt{2}\))
= \(\sqrt{6}\) (5\(\sqrt{2}\))
= 5 \(\sqrt{12}\)
= 5 \(\sqrt{4 (3)}\)
= 10 \(\sqrt{3}\)

9.2 Solving Quadratic Equations by Graphing (pp. 489–496)

Solve the equation by graphing

Question 9.
x² – 9x + 18 = 0
Answer:

The solutions of the given equation are 3 and 6

Question 10.
x² – 2x = -4
Answer:

There are no real solutions for the given equation as there are no x-intercepts

Question 11.
-8x – 16 = x²
Answer:

The solution for the given equation is -4

Question 12.
The graph of f(x) = (x + 1)(x² + 2x – 3) is shown. Find the zeros of f.

Answer:
f (x) = (x + 1) (x² + 2x – 3)
f (x) = 0
(x + 1) (x²  + 2x – 3) = 0
x + 1 = 0 or x² + 2x – 3 = 0
x + 1 = 0 or x² + 3x – x – 3 = 0
x = -1 or x (x + 3) – 1(x + 3) = 0
x = -1 or (x  1) (x + 3) = 0
x = -1 or x – 1 = 0 or x + 3 = 0
x = -1 or x = 1 or x = -3
The zeroes of the given function is -1, 1, and -3

Question 13.
Graph f(x) = x² + 2x – 5. Approximate the zeros of f to the nearest tenth.
Answer:

The approximate solution of the given function is (-4, 2)

9.3 Solving Quadratic Equations Using Square Roots (pp. 497–502)

Solve the equation using square roots. Round your solutions to the nearest hundredth, if necessary.
Question 14.
x² + 5 = 17
Answer:
x² + 5 = 17
x²+ 5 – 5 = 17 – 5
x² = 12
√x² = √12
x = ±√12
√12 = 3.46
x = ±3.46

Question 15.
x² – 14 = -14
Answer:
x² – 14 = -14
x² – 14 + 14 = -14 + 1 4
x² = 0
√x² = √0
x = 0

Question 16.
(x + 2)² = 64
Answer:
(x + 2)² = 64
√(x + 2)² = √64
x + 2 = ±8
x + 2 = 8 or x + 2 = -8
x = 8 – 2 or x = -8 – 2
x = 6 or x = -10

Question 17.
4x² + 25 = -75
Answer:
4x² + 25 = -75
4x² + 25 – 25 = -75 – 25
4x² = -100
x² = \(\frac{-100}{25}\)
The square of a real number will not take any negative number
There are no real solutions for the given equation

Question 18.
(x – 1)² = 0
Answer:
(x – 1)² = 0
√(x – 1)² = √0
x – 1 = 0
x = 0 + 1
x = 1

Question 19.
19 = 30 – 5x²
Answer:
19 = 30 – 5x²
19 – 19 = 30 – 19 – 5x²
0 = 11 – 5x²
5x² = 11
x² = \(\frac{11}{5}\)
√x² = √\(\frac{11}{5}\)
x = ±\(\frac{11}{5}\)
\(\frac{11}{5}\) = 2.20
x = ±2.20

9.4 Solving Quadratic Equations by Completing the Square (pp. 505–514)

Solve the equation by completing the square. Round your solutions to the nearest hundredth, if necessary.
Question 20.
x² + 6x – 40 = 0
Answer:
x² + 6x – 40 = 0
4² + 6x – 40 + 40 = 40 + 0
x² + 6x = 40
ax² + b = d
c = (\(\frac{b}{2}\))²
ax² + bx + c = d
x² + 6x + (\(\frac{6}{2}\))² = 40 + (\(\frac{6}{2}\))²
x² + 6x + 9 = 40 + 9
(x + 3)² = 49
√(x + 3)² = √49
x + 3 = 7 or x + 3 = -7
x = 7 – 3 or x = -7 – 3
x = 4 or x = -10

Question 21.
x² + 2x + 5 = 4
Answer:
x² + 2x + 5 = 4
x²+ 2x + 5 – 5 = 4 – 5
x²+ 2x = -1
ax² + bx = d
c = (\(\frac{b}{2}\))²
ax² + bx + c = d
x² + 2x + (\(\frac{2}{2}\))² = -1 + (\(\frac{2}{2}\))²
x² + 2x + 1 = -1 + 1
(x + 1)² = 0
√(x + 1)² = √0
x + 1 = 0
x = -1

Question 22.
2x² – 4x = 10
Answer:
2x² – 4x = 10
x² – 2x = 5
ax² + bx = d
c = (\(\frac{b}{2}\))²
ax² + bx + c = d
x² – 2x + 1 = 5 + 1
(x – 1)² = 6
√(x – 1)² = √6
x – 1 = ±√6
√6 = 2.44
x = 2.44 + 1 or x = -2.44 + 1
x = 3.44 or x = -1.44

Determine whether the quadratic function has a maximum or minimum value. Then find the value.
Question 23.
y = -x² + 6x – 1
Answer:
y = -x²+ 6x – 1
y + 1 = -x² + 6x – 1 + 1
y + 1 = -(x² – 6x)
c = (\(\frac{b}{2}\))²
y + 1 – 9 = -(x² – 6x + 9)
y – 8 = -(x – 3)²
y = -(x – 3)² + 8
The value of a is negative
Since a is negative, the parabola will be closed down and the value of y becomes the maximum value
The given function has a maximum value and the maximum value is 8

Question 24.
f(x) = x² + 4x + 11
Answer:
f (x) = x² + 4x + 11
f (x) – 11 = x² + 4x + 11 – 11
f (x) – 11 = x² + 4x
ax² + bx = d
c = (\(\frac{b}{2}\))²
f (x) – 11 + 4 = x² + 4x + 4
f (x) – 7 = (x + 2)²
f (x) = (x + 2)² + 7
The value of a is positive
Since a is positive, the parabola will be open up and the value of y becomes the minimum value
The given function has a minimum value and the minimum value is 7

Question 25.
y = 3x² – 24x + 15
Answer:
y = 3x² – 24x + 15
y – 15 = 3x² – 24x + 15 – 15
y – 15 = 3x² – 24x
y – 15 = 3 (x² – 8x)
ax² + bx = d
c = (\(\frac{b}{2}\))²
y – 15 + 144 = 3(x² – 8x + 48)
y + 129 = 3 (x – 12)²
y = 3 (x – 12)² – 129
The value of a is negative
Since a is negative, the parabola will be closed down and the value of y becomes the maximum value
The given function has a maximum value and the maximum value is -129

Question 26.
The width w of a credit card is 3 centimeters shorter than the length ℓ. The area is 46.75 square centimeters. Find the perimeter.
Answer:
Let the length of a credit card be x cm
The width of a credit card is: (x – 3) cm
46.75 = x (x – 3)
x² – 3x = 46.75
x² – 3x – 46.75 = 0

The solutions of x are -6 and 8
The length will not be negative,
The length of the credit card is 8 cm
The width of the credit card is 8 – 3 = 5 cm
The perimeter of the credit card = 2 (l + b)
= 2 (8 + 5) = 2 (13) = 26 cm
The perimeter of the credit card is 26 cm

9.5 Solving Quadratic Equations Using the Quadratic Formula (pp. 515–524)

Solve the equation using the Quadratic Formula. Round your solutions to the nearest tenth, if necessary.
Question 27.
x²+ 2x – 15 = 0
Answer;
x² + 2x – 15 = 0
ax²+ bx + c = 0
a = 1, b = 2, and c = -15
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-2 ± \sqrt{2² + 4 (1) (15)}}{2 (1)}\)
x = \(\frac{-2 ± \sqrt{64}}{2}\)
x = \(\frac{-2 ± 8}{2}\)
x = 3 or x = -5

Question 28.
2x² – x + 8 = 16
Answer:
2x² – x + 8 = 0
ax²+ bx + c = 0
a = 2, b = -1, and c = 8
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{1 ± \sqrt{(-1)² – 4 (2) (8)}}{2 (2)}\)
x = \(\frac{1 ± \sqrt{-63}}{4}\)
The square of a real number will not be a negative number
There are no real solutions for the given equation

Question 29.
-5x² + 10x = 5
Answer;
-5x² + 10x = 5
5x² – 10x + 5 = 0
x² – 2x + 1 = 0
ax²+ bx + c = 0
a = 1, b = -2, and c = 1
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{2 ± \sqrt{(-2)² – 4 (1) (1)}}{2 (1)}\)
x = \(\frac{2 ± \sqrt{0}}{2}\)
x = \(\frac{2}{2}\)
x = 1

Find the number of x-intercepts of the graph of the function.
Question 30.
y = -x² + 6x – 9
Answer:
y = -x² + 6x – 9
To find the x-interxept,
Put y = 0
-x² + 6x – 9 = 0
x² – 6x + 9 = 0
x² – 3x – 3x + 9 = 0
x (x – 3) – 3 (x – 3) = 0
(x – 3) (x – 3) = 0
x – 3 = 0 or x – 3 = 0
x = 3 or x = 3

Question 31.
y = 2x² + 4x + 8
Answer:
y = 2x² + 4x + 8
To find the x-intercept,
Put y = 0
2x² + 4x + 8 = 0
x² + 2x + 4 = 0
ax² + bx + c = 0
a = 1, b = 2, and c = 4
x = \(\frac{-b ± \sqrt{b² – 4ac}}{2a}\)
x = \(\frac{-2 ± \sqrt{2² – 4 (1) (4)}}{2 (1)}\)
x = \(\frac{-2 ± \sqrt{-12}}{2a}\)
The square of a real number will not take any negative values
There are no solutions for the given equation

Question 32.
y = – \(\frac{1}{2}\)x² + 2x
Answer:
y = –\(\frac{1}{2}\)x² + 2x
To find the x-intercept,
Put y = 0
– \(\frac{1}{2}\)x² + 2x = 0
– \(\frac{1}{2}\)x² = – 2x
\(\frac{1}{2}\)x² = 2x
x = 2 (2)
x = 4
The solution for the given equation is 4

9.6 Solving Nonlinear Systems of Equations (pp. 525–532)

Solve the system using any method.
Question 33.
y = x² – 2x – 4
y = -5
Answer:
y = x² – 2x – 4
y = 5
x² – 2x – 4 = 5
x² – 2x = 5 + 4
x² – 2x = 9

The solutions for the given system of equations are -2 and 4

Question 34.
y = x² – 9
y = 2x + 5
Answer:
y = x² – 9
y = 2x + 5
x² – 9 = 2x + 5
x² – 2x – 9 – 5 = 0
x² – 2x – 14 = 0

Question 35.
y = 2(\(\frac{1}{2}\))^x – 5
y = -x² – x + 4
Answer:

Solving Quadratic Equations Chapter Test

Solve the equation using any method. Explain your choice of method.
Question 1.
x² – 121 = 0
Answer:
x² = 121 = 0
x² – 121 + 121 = 0 + 121
x² = 121
√x² = √121
x = ±11

Question 2.
x² – 6x = 10
Answer:
x² – 6x = 10
x² – 6x + 9 = 10 + 9
(x – 3)² = 19
√(x – 3)² = √19
x – 3 = ±√19
We know that,
√19 = 4.35
So,
x = 4.35 + 3 or x = -4.35 + 3
x = 7.35 or x = -1.35

Question 3.
-2x² + 3x + 7 = 0
Answer:

Question 4.
x² – 7x + 12 = 0
Answer:
x² – 7x + 12 = 0
x² – 4x – 3x + 12 = 0
x (x – 4) – 3 (x – 4) = 0
(x – 3) (x – 4) = 0
x – 3 = 0 or x – 4 = 0
x = 3 or x = 4

Question 5.
5x² + x – 4 = 0
Answer:
5x² + x – 4 = 0
5x² + 5x – 4x – 4 = 0
5x (x + 1) – 4 (x + 1) = 0
(5x – 4) (x + 1) = 0
5x – 4 = 0 or x + 1 = 0
5x = 4 or x = -1
x = \(\frac{4}{5}\) or x = -1

Question 6.
(4x + 3)² = 16
Answer:
(4x + 3)² = 16
√(4x + 3)² = √16
4x + 3 = 4 or 4x + 3 = -4
4x = 4 – 3 or 4x = -4 – 3
4x = 1 or 4x = -7
x = \(\frac{1}{4}\) or x = –\(\frac{7}{4}\)

Question 7.
Describe how you can use the method of completing the square to determine whether the function f(x) = 2x² + 4x – 6 can be represented by the graph shown.

Answer:
f (x) = 2x² + 4x – 6
We have to make f (x) = 0
2x²+ 4x – 6 = 0
2x²+ 4x – 6 + 6 = 0 + 6
2x² + 4x = 6
x²+ 2x = 3
\(\frac{b}{2}\) = \(\frac{2}{2}\)
= 1
x²+ 2x + 1 = 3 + 1
(x + 1)² = 4
√(x + 1)² = √4
x + 1 = 2 or x + 1 = -2
x = 2 – 1 or x = -2 – 1
x = 1 or x = -3

Question 8.
Write an expression involving radicals in which a conjugate can be used to simplify the expression.
Answer:
The expression that involves radicals in which a conjugate can be used to simplify the expression is
\(\frac{8}{3 + \sqrt{6}}\)
The conjugate of the above expression is
\(\frac{3 – \sqrt{6}}\)

Solve the system using any method.
Question 9.
y = x² – 4x – 2
y = -4x + 2
Answer:
y = x² – 4x – 2
y = -4x + 2
x² – 4x – 2 = -4x + 2
x² – 4x – 2 + 4x – 2 = 0
x² – 4 = 0
x² = 4
√x² = √4
x = 2 or x = -2

Question 10.
y = -5x² + x – 1
y = -7
Answer:
y = -5x²+ x – 1
y = -7
-5x² + x – 1 = -7
-5x² + x – 1 + 7 = 0
-5x² +x + 6 = 0
-5x² + 6x – 5x + 6 = 0
-5x(x + 1) + 6 (x + 1) = 0
(-5x + 6) (x + 1) = 0
-5x + 6 = 0 or x + 1 = 0
-5x = -6 or x = -1
x = \(\frac{6}{5}\) or x = -1

Question 11.
y = \(\frac{1}{2}\)(4)^x + 1
y = x² – 2x + 4
Answer:

The solution for the given set of equations is (3, 1)

Question 12.
A skier leaves an 8-foot-tall ramp with an initial vertical velocity of 28 feet per second. The function h = -16t² + 28t + 8 represents the height h (in feet) of the skier after t seconds. The skier has a perfect landing. How many points does the skier earn?

Answer:

Question 13.
An amusement park ride lifts seated riders 265 feet above the ground. The riders are then dropped and experience free fall until the brakes are activated 105 feet above the ground. The function h = -16t² + 265 represents the height h (in feet) of the riders t seconds after they are dropped. How long do the riders experience free fall? Round your solution to the nearest hundredth.
Answer:
The function h = -16t² + 265 represents the height h (in feet) of the riders t seconds after they are dropped.
105 = -16t² + 265
0 = 16t² + 160
t² – 10 = 0
t = √10

Question 14.
Write an expression in the simplest form that represents the area of the painting shown.

Answer:
The area of the painting = \(\sqrt{30 x^{7}}\) ×\(\frac{36}{\sqrt{3}}\)
= \(\sqrt{10 (3) x^{6} x}\) ×\(\frac{36}{\sqrt{3}}\)
= x³ \(\sqrt{10}\) × \(\sqrt{3}\) × \(\frac{36}{\sqrt{3}}\)
= 36x³ \(\sqrt{10}\)

Question 15.
Explain how you can determine the number of times the graph of y = 5x² – 10x + 5 intersects the x-axis without graphing or solving an equation.
Answer:
y = 5x² – 10x + 5
We can determine the number of times the graph of the given equation intersects the x-axis by using the discriminant
Discriminant (d) = b² – 4ac
b² – 4ac > 0 ——> 2 intersection points
b² – 4ac = 0 ——-> 1 intersection point
b² – 4ac < 0 ——-> No intersection points
ax² + bx + c = 0
a = 5, b = -10, and c = 5
b² – 4ac = (-10)² – 4 (5) (5)
= 100 – 100
= 0

Question 16.
Consider the quadratic equation ax² + bx + c = 0 Find values of a, b, and c so that the graph of its related function has (a) two  x-intercepts, (b) one x-intercepts.and (c) no x-intercepts.
Answer:
ax² + bx + c = 0
b² – 4ac > 0 ——> 2 x-intercepts
b² – 4ac = 0 ——-> 1 x-intercept
b² – 4ac < 0 ——-> No x-intercepts
a) For 2 x-intercepts:
b² – 4ac > 0
b² > 4ac
4ac < b²
c < \(\frac{b²}{4a}\)
b) For 1 x-intercept:
b² – 4ac = 0
b² = 4ac
4ac = b²
c = \(\frac{b²}{4a}\)
c) For no x-intercepts:
b² – 4ac < 0
b² < 4ac
4ac > b²
c > \(\frac{b²}{4a}\)

Question 17.
The numbers y of two types of bacteria after x hours are represented by the models below.
y = 3x² + 8x + 20 Type A
y = 27x + 60 Type B
a. When are there 400 Type A bacteria?
b. When are the number of Type A and Type B bacteria the same?
c. When are there more Type A bacteria than Type B? When are there more Type B bacteria than Type A? Use a graph to support your answer.
Answer:

Solving Quadratic Equations Cumulative Assessment

Question 1.
The graphs of four quadratic functions are shown. Determine whether the discriminants of the equations formed by setting each function equal to zero are positive, negative, or zero.f(x) = 0, g(x) = 0, h(x) = 0, and j(x) = 0

Answer:
Discriminant (d) = b² – 4ac
d > 0, then the equation will have 2 solutions
If d = 0, then the equation will have 1 solution
If d < 0, then the equation will have no solutions
For f (x) = 0,
f (x) will not have any solutions since it did not touch the x-axis
For g (x) = 0,
g (x) will have 2 solutions since it touched the x-axis at 2 points
For h (x) = 0,
h (x) will have only 1 solution since it touched the x-axis at only 1 point
For j (x) = 0,
j (x) will have only 1 solution since it touched the x-axis at only 1 point

Question 2.
The function f(x) = a(1.08)^x represents the total amount of money (in dollars) in Account A after x years. The function g(x) = 600(b)^x represents the total amount of money (in dollars) in Account B after x years. Fill in values for a and b so that each statement is true.
a. When a = ____ and b = ____, Account B has a greater initial amount and increases at a faster rate than Account A.
b. When a = ____ and b = ____, Account B has a lesser initial amount than Account A but increases at a faster rate than Account A.
c. When a = ____ and b = ____, Account B and Account A have the same initial amount, and Account B increases at a slower rate than Account A.
Answer:

Question 3.
Your friend claims to be able to find the radius r of each figure, given the surface area S. Do you support your friend’s claim? Justify your answer.

Answer:
The claim of your friend is that we can find the radius r of each figure given the Surface area S
But,
For some Surface areas, the variable “Height (h)” will be present
We can find the surface area given radius r only in some cases
The claim of your friend is not correct

Question 4.
The tables represent the numbers of items sold at a concession stand on days with different average temperatures. Determine whether the data represented by each table show a positive, negative, or no correlation.

Answer:

Question 5.
Which graph shows exponential growth?

Answer:
The graph of an “Exponential function” will gradually increase or decrease
Graph A shows the gradual increase
Graph B shows the gradual decrease
Graph A and graph B show the exponential growth

Question 6.
Which statement best describes the solution(s) of the system of equations?
y = x² + 2x – 8
y = 5x + 2
A. The graphs intersect at one point, (-2, -8). So, there is one solution.
B. The graphs intersect at two points, (-2, -8) and (5, 27). So, there are two solutions.
C. The graphs do not intersect. So, there is no solution.
D. The graph of y = x² + 2x – 8 has two x-intercepts. So, there are two solutions.
Answer:
y = x² + 2x – 8
y = 5x + 2
x² + 2x – 8 = 5x + 2
x² + 2x – 5x – 8 – 2 = 0
x² – 3x – 10 = 0
x² – 5x + 2x – 10 = 0
x (x – 5) + 2 (x – 5) = 0
(x + 2) (x – 5) = 0
x + 2 = 0 or x – 5 = 0
x = -2 or x = 5
Option D best describes the solutions of the given system of equations

Question 7.
Which expressions are in the simplest form?

Answer:
Let the given expressions be named as a, b, c, d, e, f, g, h, i, and j respectively
The expressions that are in the simplest form are b, d, e, g, i, and j

Question 8.
The domain of the function shown is all integers in the interval -3 < x ≤ 3. Find all the ordered pairs that are solutions of the equation y = f(x).
f(x) = 4x – 5
Answer:
f (x) = 4x – 5
It is given that the domain of the given function is
-3 < x ≤ 3
The domain of the given function = (-2, -1, 0, 1, 2, 3)
For x = -2, f (-2) = -13
For x = -1, f (-1) = -9
For x = 0, f (0) = -5
For x = 1, f (1) = -1
For x = 2, f (2) = 3
For x = 3, f (3) = 7
The ordered pairs for the given function are (-2, 13), (-1, -9), (0, -5), (1, -1), (2, 3), and (3, 7)