Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers

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Big Ideas Math Book Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers

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• Quadratic Equations and Complex Numbers Maintaining Mathematical Proficiency – Page 91
• Quadratic Equations and Complex Numbers Mathematical Practices – Page 92
• Lesson 3.1 Solving Quadratic Equations – Page(94-102)
• Solving Quadratic Equations 3.1 Exercises – Page(99-102)
• Lesson 3.2 Complex Numbers – Page(104-110)
• Complex Numbers 3.2 Exercises – Page(108-110)
• Lesson 3.3 Completing the Square – Page(112-118)
• Completing the Square 3.3 Exercises – Page(116-118)
• Quadratic Equations and Complex Numbers Study Skills: Creating a Positive Study Environment – Page 119
• Quadratic Equations and Complex Numbers 3.1–3.3 Quiz – Page 120
• Lesson 3.4 Using the Quadratic Formula – Page(122-130)
• Using the Quadratic Formula 3.4 Exercises – Page(127-130)
• Lesson 3.5 Solving Nonlinear Systems – Page(132-138)
• Solving Nonlinear Systems 3.5 Exercises – Page(136-138)
• Lesson 3.6 Quadratic Inequalities – Page(140-146)
• Quadratic Inequalities 3.6 Exercises – Page(144-146)
• Quadratic Equations and Complex Numbers Performance Task – Page 147
• Quadratic Equations and Complex Numbers Chapter Review – Page(148-150)
• Quadratic Equations and Complex Numbers Chapter Test – Page 151
• Quadratic Equations and Complex Numbers Cumulative Assessment – Page(152-153)

Quadratic Equations and Complex Numbers Maintaining Mathematical Proficiency

Simplify the expression.
Question 1.
\(\sqrt{27}\)
Answer:
Given expression is \(\sqrt{27}\)
27 can be written as 3³

\(\sqrt{27}\) = 3.

Question 2.
–\(\sqrt{112}\)
Answer:
Given expression is –\(\sqrt{112}\)

–\(\sqrt{112}\) = -4-\(\sqrt{7}\)

Question 3.
\(\sqrt{\frac{11}{64}}\)
Answer:
Given expression is \(\sqrt{\frac{11}{64}}\)
\(\sqrt{\frac{11}{8²}}\)
\(\sqrt{\frac{11}{64}}\) = \(\sqrt{11}\)/8

Question 4.
\(\sqrt{\frac{147}{100}}\)
Answer:
Given expression is \(\sqrt{\frac{147}{100}}\)
\(\sqrt{\frac{147}{100}}\) = \(\sqrt{\frac{147}{10²}}\)
\(\sqrt{\frac{147}{10²}}\) = \(\sqrt{147}\)/10
So, \(\sqrt{\frac{147}{10²}}\) = \(\sqrt{147}\)/10

Question 5.
\(\sqrt{\frac{18}{49}}\)
Answer:
Given expression is \(\sqrt{\frac{18}{49}}\)

\(\sqrt{\frac{18}{49}}\) = \(\sqrt{18}\)/7 = 6\(\sqrt{2}\)/10 = 3/5 . \(\sqrt{2}\)

Question 6.
–\(\sqrt{\frac{65}{121}}\)
Answer:
Given expression is –\(\sqrt{\frac{65}{121}}\)
121 = 11²
–\(\sqrt{\frac{65}{121}}\) = –\(\sqrt{65}\)/11

Question 7.
\(\sqrt{80}\)
Answer:
Given expression is \(\sqrt{80}\)

So, \(\sqrt{80}\) = 4\(\sqrt{5}\)

Question 8.
\(\sqrt{32}\)
Answer:
Given expression is \(\sqrt{32}\)

So, \(\sqrt{32}\) = 4\(\sqrt{2}\)

Factor the polynomial.
Question 9.
x² − 36
Answer:
Given,
x² – 36
x² – 6²
(x +6) (x- 6)
So, x² − 36 = (x +6) (x- 6)

Question 10.
x² − 9
Answer:
Given,
x² – 9
x² – 3²
(x – 3) ( x+3)
So, x² − 9 = (x – 3) (x + 3)

Question 11.
4x² − 25
Answer:
Given,
4x² – 25
= (2x)²  –  5²
= (2x +5)( 2x – 5)
So, 4x² – 25 = (2x +5)(2x – 5)

Question 12.
x² − 22x + 121
Answer:
Given,
x²– 22x +121
x² – 11x -11x +121
x(x-11)-11(x-11)
(x – 11)(x – 11)
So, x² − 22x + 121 = (x – 11)(x – 11)

Question 13.
x² + 28x + 196
Answer:
Given,
X² + 28x +196x
x²+14x+14x+196
x(x+14)+14(x+14)
(x+14)(x+14)
So, x² + 28x + 196 = (x+14)(x+14)

Question 14.
49x² + 210x + 225
Answer:
Given,
49x² + 210x + 225
x = (-b ± √b² – 4ac)/2a
x = (-210 ± √210² – 4(49)(225))/2(49)
x = -210/98
b² – 4ac = 0
x = -210/98
x = -15/7 or -2.14
The solution is x = -15/7 or -2.14

Question 15.
ABSTRACT REASONING
Determine the possible integer values of a and c for which the trinomial ax²+ 8x+c is factorable using the Perfect Square Trinomial Pattern. Explain your reasoning.
Answer:
The term ‘a’ is referred to as the leading coefficient, while ‘c’ is the absolute term of f(x). Every quadratic equation has two values of the unknown variable.

Quadratic Equations and Complex Numbers Mathematical Practices

Mathematically proficient students recognize the limitations of technology

Monitoring Progress

Question 1.
Explain why the second viewing window in Example 1 shows gaps between the upper and lower semicircles, but the third viewing window does not show gaps.
Answer:

Use a graphing calculator to draw an accurate graph of the equation. Explain your choice of viewing window.

Question 2.
y = \(\sqrt{x^{2}-1.5}\)
Answer:
We can use the graphing calculator to draw y = \(\sqrt{x^{2}-1.5}\).
Write the equation y = \(\sqrt{x^{2}-1.5}\) in the input.

Question 3.
y = \(\sqrt{x-2.5}\)
Answer:
We can use the graphing calculator to draw y = \(\sqrt{x-2.5}\)
Write the equation y = \(\sqrt{x-2.5}\) in the input.

Question 4.
x² + y²= 12.25
Answer:
We can use the graphing calculator to draw x² + y²= 12.25
Write the equation x² + y²= 12.25 in the input.

Question 5.
x² + y² = 20.25
Answer:
We can use the graphing calculator to draw x² + y²= 12.25
Write the equation x² + y²= 12.25 in the input.

Question 6.
x² + 4y² = 12.25
Answer:
We can use the graphing calculator to draw x² + 4y² = 12.25
Write the equation x² + 4y² = 12.25 in the input.

Question 7.
4x² + y² = 20.25
Answer:
We can use the graphing calculator to draw 4x² + y² = 20.25
Write the equation 4x² + y² = 20.25 in the input.

Lesson 3.1 Solving Quadratic Equations

Essential Question How can you use the graph of a quadratic equation to determine the number of real solutions of the equation?

EXPLORATION 1

Matching a Quadratic Function with Its Graph
Work with a partner. Match each quadratic function with its graph. Explain your reasoning. Determine the number of x-intercepts of the graph.
a. f(x) = x² − 2x
b. f(x) = x² − 2x + 1
c. f(x) = x² − 2x + 2
d. f(x) = −x² + 2x
e. f(x) = −x² + 2x − 1
f. f(x) = −x² + 2x − 2

EXPLORATION 2

Solving Quadratic Equations
Work with a partner. Use the results of Exploration 1 to find the real solutions (if any) of each quadratic equation.
a. x² − 2x = 0
b. x² − 2x + 1 = 0
c. x² − 2x + 2 = 0
d. −x² + 2x = 0
e. −x² + 2x − 1 = 0
f. −x² + 2x − 2 = 0

Communicate Your Answer

Question 3.
How can you use the graph of a quadratic equation to determine the number of real solutions of the equation?

Answer:
If the graph of a quadratic function crosses the x-axis at two points, then the has two real rational solutions. The number of real solutions of a quadratic equation can be found in the number of x-intercepts on the graph. If the curve passes through the x-axis twice, then the equation has two real solutions. If it passes through once, (the vertex is on the x-axis), then there is one real solution, and if it does not touch the x-axis at all, then it has no real solutions.

Question 4.
How many real solutions does the quadratic equation x² + 3x + 2 = 0 have? How do you know? What are the solutions?
Answer:
Given equation is x² + 3x + 2 = 0
x² + 3x + 2 = 0
x² + 1x + 2x + 2 = 0
x(x + 1) +2 (x + 1) = 0
(x + 2) (x + 1) = 0
x = -2 or x = -1
There are 2 solutions.
Parabolas have a highest or the lowest point called the Vertex

Monitoring Progress

Solve the equation by graphing.
Question 1.
x² − 8x + 12 = 0
Answer:

Question 2.
4x² − 12x + 9 = 0
Answer:

Question 3.
\(\frac{1}{2}\)x² = 6x − 20
Answer:

Solve the equation using square roots.
Question 4.
\(\frac{2}{3}\)x² + 14 = 20
Answer:

Question 5.
−2x² + 1 = −6
Answer:

Question 6.
2(x − 4)² = −5
Answer:

Solve the equation by factoring.
Question 7.
x² + 12x + 35 = 0
Answer:

Question 8.
3x² − 5x = 2
Answer:

Find the zero(s) of the function.
Question 9.
f(x) = x² − 8x
Answer:

Question 10.
f(x) = 4x² + 28x + 49
Answer:

Question 11.
WHAT IF?
The magazine initially charges $21 per annual subscription. How much should the magazine charge to maximize annual revenue? What is the maximum annual revenue?
Answer:
The magazine initially charges $21 per annual subscription.
Let x be the increase in price and R(x) is annual revenue with initial subscribers of 48,000.
Annual revenue = Number of subscribers + subscription price
Number of subscribers = Initial subscribers – Number of subscribers lost × x
= 48,000 – 2000x
Subscription price = initial price + increase price
= 21 + x
R(x) = (48000 – 2000x)(21 + x)
R(x) = -2000(x – 24)(x + 21)
x – 24 = 0
x + 21 = 0
x = 24, -21
24-21/2 = 3/2
Thus to maximize profit it must cost 21 + 1.5 = 22.5
R(x) = (48,000 – 2000x)(21 + x)
R(x) = -2000 (x – 24)(x + 21)
Substitute x = 1.5 in the equation.
R(1.5) = -2000(1.5 – 2.4) (1.5 + 21)
= $1,012,500
The maximum annual revenue is $1,012,500

Question 12.
WHAT IF?
The egg container is dropped from a height of 80 feet. How does this change your answers in parts (a) and (b)?
Answer:
Given,
h = 80 ft
h = -16t² + h
h = -16t² + 80
0 = -16t² + 80
-16t² + 80 = 0
-16t² = -80
t² = -80/-16
t² = 5
t = ± \(\sqrt{5}\)
For h(1) and h(1.5) we will replace t by 1and 1.5
h(1) = -16(1)² + 80
h(1) = 64
h(1.5) = -16(1.5)² + 80
= -36 + 80  = 44

Solving Quadratic Equations 3.1 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Explain how to use graphing to find the roots of the equation ax² + bx + c = 0.
Answer:

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

Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, solve the equation by graphing.
Question 3.
x² + 3x + 2 = 0
Answer:

Question 4.
−x² + 2x + 3 = 0
Answer:

Question 5.
y = x² − 9
Answer:

Question 6.
−8 = −x² − 4
Answer:

Question 7.
8x = −4 − 4x²
Answer:

Question 8.
3x² = 6x − 3
Answer:

Question 9.
7 = −x² − 4x
Answer:

Question 10.
2x = x² + 2
Answer:
Given,
2x = x² + 2
x² – 2x + 2 = 0
Now the graph the related function.

There is no x-intercept. So, there is no solution.

Question 11.
\(\frac{1}{5}\)x² + 6 = 2x
Answer:

Question 12.
3x = \(\frac{1}{4}\)x² + 5
Answer:

In Exercises 13–20, solve the equation using square roots.
Question 13.
s² = 144
Answer:

Question 14.
a² = 81
Answer:

Question 15.
(z − 6)² = 25
Answer:
Given,

Question 16.
(p − 4)² = 49
Answer:

Question 17.
4(x − 1)² + 2 = 10
Answer:
Given,

Question 18.
2(x + 2)² − 5 = 8
Answer:
Given

Question 19.
\(\frac{1}{2}\)r² − 10 = \(\frac{3}{2}\)r²
Answer:

Question 20.
\(\frac{1}{5}\)x² + 2 = \(\frac{3}{5}\)x²
Answer:

Question 21.
ANALYZING RELATIONSHIPS
Which equations have roots that are equivalent to the x-intercepts of the graph shown?

A. −x² − 6x − 8 = 0
B. 0 = (x + 2)(x + 4)
C. 0 = −(x + 2)² + 4
D. 2x² − 4x − 6 = 0
E. 4(x + 3)² − 4 = 0
Answer:

Question 22.
ANALYZING RELATIONSHIPS
Which graph has x-intercepts that are equivalent to the roots of the equation (x − \(\frac{3}{2}\))² = \(\frac{25}{4}\)? Explain your reasoning.

Answer:

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in solving the equation.
Question 23.

Answer:

Question 24.

Answer:
-2x² – 8 = 0
x² + 8 = 0
x² = -8
x = -2√2

Question 25.
OPEN-ENDED
Write an equation of the form x² = d that has (a) two real solutions, (b) one real solution, and (c) no real solution.
Answer:

Question 26.
ANALYZING EQUATIONS
Which equation has one real solution? Explain.
A. 3x² + 4 = −2(x² + 8)
B. 5x² − 4 = x² − 4
C. 2(x + 3)² = 18
D. \(\frac{3}{2}\)x² − 5 = 19
Answer:

In Exercises 27–34, solve the equation by factoring.
Question 27.
0 = x² + 6x + 9
Answer:
Given,

Question 28.
0 = z² − 10z + 25
Answer:

Question 29.
x² − 8x = −12
Answer:

Question 30.
x² − 11x = −30
Answer:

Question 31.
n² − 6n = 0
Answer:

Question 32.
a² − 49 = 0
Answer:

Question 33.
2w² − 16w = 12w − 48
Answer:

Question 34.
−y + 28 + y² = 2y + 2y²
Answer:

MATHEMATICAL CONNECTIONS In Exercises 35–38, find the value of x.
Question 35.
Area of rectangle = 36

Answer:

Question 36.
Area of circle = 25π

Answer:

Question 37.
Area of triangle = 42

Answer:

Question 38.
Area of trapezoid = 32

Answer:

In Exercises 39–46, solve the equation using any method. Explain your reasoning.
Question 39.
u² = −9u
Answer:

Question 40.
\(\frac{t^{2}}{20}\) + 8 = 15
Answer:

Question 41.
−(x + 9)² = 64
Answer:

Question 42.
−2(x + 2)² = 5
Answer:

Question 43.
7(x − 4)² − 18 = 10
Answer:

Question 44.
t² + 8t + 16 = 0
Answer:

Question 45.
x² + 3x + \(\frac{5}{4}\) = 0
Answer:

Question 46.
x² − 1.75 = 0.5
Answer:

In Exercises 47–54, find the zero(s) of the function.
Question 47.
g(x) = x² + 6x + 8
Answer:

Question 48.
f(x) = x² − 8x + 16
Answer:

Question 49.
h(x) = x² + 7x − 30
Answer:

Question 50.
g(x) = x² + 11x
Answer:

Question 51.
f(x) = 2x² − 2x − 12
Answer:

Question 52.
f(x) = 4x² − 12x + 9
Answer:

Question 53.
g(x) = x² + 22x + 121
Answer:

Question 54.
h(x) = x² + 19x + 84
Answer:

Question 55.
REASONING
Write a quadratic function in the form f(x) = x² + bx + c that has zeros 8 and 11.
Answer:

Question 56.
NUMBER SENSE
Write a quadratic equation in standard form that has roots equidistant from 10 on the number line.
Answer:

Question 57.
PROBLEM SOLVING
A restaurant sells 330 sandwiches each day. For each $0.25 decrease in price, the restaurant sells about 15 more sandwiches. How much should the restaurant charge to maximize daily revenue? What is the maximum daily revenue?

Answer:

Question 58.
PROBLEM SOLVING
An athletic store sells about 200 pairs of basketball shoes per month when it charges $120 per pair. For each $2 increase in price, the store sells two fewer pairs of shoes. How much should the store charge to maximize monthly revenue? What is the maximum monthly revenue?
Answer:
Given,
An athletic store sells about 200 pairs of basketball shoes per month when it charges $120 per pair.
For each $2 increase in price, the store sells two fewer pairs of shoes.
Revenue = (120+2x)(200-2x)
Revenue formula
Set 120+2x = 0
and 200-2x = 0
Find the zeros of the equation 3
p = -60
and q = 100
f(x) = (x-p)(x-q) = 0
maximum at x = 20

Charge = $160
Charge = 120+2(20) = $160
Maximum Revenue = $25,600
Revenue = 160(200-2*20)= 160*160 = $25,600

Question 59.
MODELING WITH MATHEMATICS
Niagara Falls is made up of three waterfalls. The height of the Canadian Horseshoe Falls is about 188 feet above the lower Niagara River. A log falls from the top of Horseshoe Falls.
a. Write a function that gives the height h (in feet) of the log after t seconds. How long does the log take to reach the river?
b. Find and interpret h(2) − h(3).
Answer:

Question 60.
MODELING WITH MATHEMATICS
According to legend, in 1589, the Italian scientist Galileo Galilei dropped rocks of different weights from the top of the Leaning Tower of Pisa to prove his conjecture that the rocks would hit the ground at the same time. The height h (in feet) of a rock after t seconds can be modeled by h(t) = 196 − 16t².

a. Find and interpret the zeros of the function. Then use the zeros to sketch the graph.
b. What do the domain and range of the function represent in this situation?
Answer:

Question 61.
PROBLEM SOLVING
You make a rectangular quilt that is 5 feet by 4 feet. You use the remaining 10 square feet of fabric to add a border of uniform width to the quilt. What is the width of the border?

Answer:

Question 62.
MODELING WITH MATHEMATICS
You drop a seashell into the ocean from a height of 40 feet. Write an equation that models the height h (in feet) of the seashell above the water after t seconds. How long is the seashell in the air?
Answer:

Question 63.
WRITING
The equation h = 0.019s² models the height h (in feet) of the largest ocean waves when the wind speed is s knots. Compare the wind speeds required to generate 5-foot waves and 20-foot waves.

Answer:

Question 64.
CRITICAL THINKING
Write and solve an equation to find two consecutive odd integers whose product is 143.
Answer:

Question 65.
MATHEMATICAL CONNECTIONS
A quadrilateral is divided into two right triangles as shown in the figure. What is the length of each side of the quadrilateral?

Answer:

Question 66.
ABSTRACT REASONING
Suppose the equation ax² + bx + c = 0 has no real solution and a graph of the related function has a vertex that lies in the second quadrant.
a. Is the value of a positive or negative? Explain your reasoning.
b. Suppose the graph is translated so the vertex is in the fourth quadrant. Does the graph have any x-intercepts? Explain.
Answer:

Question 67.
REASONING
When an object is dropped on any planet, its height h (in feet) after t seconds can be modeled by the function h = −\(\frac{g}{2}\)t² + h₀, where h₀ is the object’s initial height and g is the planet’s acceleration due to gravity. Suppose a rock is dropped from the same initial height on the three planets shown. Make a conjecture about which rock will hit the ground first. Justify your answer.

Answer:
Given,
When an object is dropped on any planet, its height h (in feet) after t seconds can be modeled by the function h = −\(\frac{g}{2}\)t² + h₀, where h₀ is the object’s initial height and g is the planet’s acceleration due to gravity.
We need to make a conjecture about which rock will hit the ground first.
The rock will hit the ground on Jupiter first. Because the first term is negative, the height of the falling object will decrease faster as ga gets larger.

Question 68.
PROBLEM SOLVING
A café has an outdoor, rectangular patio. The owner wants to add 329 square feet to the area of the patio by expanding the existing patio as shown. Write and solve an equation to find the value of x. By what distance should the patio be extended?

Answer:

Question 69.
PROBLEM SOLVING
A flea can jump very long distances. The path of the jump of a flea can be modeled by the graph of the function y = −0.189x² + 2.462x, where x is the horizontal distance (in inches) and y is the vertical distance (in inches). Graph the function. Identify the vertex and zeros and interpret their meanings in this situation.
Answer:

Question 70.
HOW DO YOU SEE IT?
An artist is painting a mural and drops a paintbrush. The graph represents the height h (in feet) of the paintbrush after t seconds.

a. What is the initial height of the paintbrush?
b. How long does it take the paintbrush to reach the ground? Explain.
Answer:

Question 71.
MAKING AN ARGUMENT
Your friend claims the equation x² + 7x =−49 can be solved by factoring and has a solution of x = 7. You solve the equation by graphing the related function and claim there is no solution. Who is correct? Explain.
Answer:
Given,
x² + 7x =−49
x = 7

Question 72.
ABSTRACT REASONING
Factor the expressions x² − 4 and x² − 9. Recall that an expression in this form is called a difference of two squares. Use your answers to factor the expression x² − a². Graph the related function y = x² − a². Label the vertex, x-intercepts, and axis of symmetry.
Answer:

Question 73.
DRAWING CONCLUSIONS
Consider the expression x² + a², where a > 0.
a. You want to rewrite the expression as (x + m)(x + n). Write two equations that m and n must satisfy.
b. Use the equations you wrote in part (a) to solve for m and n. What can you conclude?
Answer:

Question 74.
THOUGHT PROVOKING
You are redesigning a rectangular raft. The raft is 6 feet long and 4 feet wide. You want to double the area of the raft by adding to the existing design. Draw a diagram of the new raft. Write and solve an equation you can use to find the dimensions of the new raft.

Answer:

Question 75.
MODELING WITH MATHEMATICS
A high school wants to double the size of its parking lot by expanding the existing lot as shown. By what distance x should the lot be expanded?

Answer:

Maintaining Mathematical Proficiency

Find the sum or difference.
Question 76.
(x² + 2) + (2x² − x)
Answer:

Question 77.
(x³ + x² − 4) + (3x² + 10)
Answer:

Question 78.
(−2x + 1) − (−3x² + x)
Answer:

Question 79.
(−3x³ + x² − 12x) − (−6x² + 3x − 9)
Answer:

Find the product.
Question 80.
(x + 2)(x − 2)
Answer:

Question 81.
2x(3 − x + 5x²)
Answer:

Question 82.
(7 − x)(x − 1)
Answer:
Given,
(7 − x)(x − 1)
(7−x)(x−1)​=7x+7(−1)+(−x)x+(−x)(−1)=7x−7−x²+x=−x²+(7x+x)−7=−x²+8x−7​​Use (1)SimplifyGroup like terms Reduce like terms​

Question 83.
11x(−4x² + 3x + 8)
Answer:

Lesson 3.2 Complex Numbers

Essential Question What are the subsets of the set of complex numbers?
In your study of mathematics, you have probably worked with only real numbers, which can be represented graphically on the real number line. In this lesson, the system of numbers is expanded to include imaginary numbers. The real numbers and imaginary numbers compose the set of complex numbers.

EXPLORATION 1

Classifying Numbers
Work with a partner. Determine which subsets of the set of complex numbers contain each number.
a. \(\sqrt{9}\)
b. \(\sqrt{0}\)
c. −\(\sqrt{4}\)
d. \(\sqrt{\frac{4}{9}}\)
e. \(\sqrt{2}\)
f. \(\sqrt{-1}\)

EXPLORATION 2

Complex Solutions of Quadratic Equations

Work with a partner. Use the definition of the imaginary unit i to match each quadratic equation with its complex solution. Justify your answers.
a. x² − 4 = 0
b. x² + 1 = 0
c. x² − 1 = 0
d. x² + 4 = 0
e. x² − 9 = 0
f. x² + 9 = 0
A. i
B. 3i
C. 3
D. 2i
E. 1
F. 2

Communicate Your Answer

Question 3.
What are the subsets of the set of complex numbers? Give an example of a number in each subset.
Answer:
Real numbers are the subset of complex numbers. A combination of a real and an imaginary number in the form a + bi, where a and b are real, and i is imaginary.
The values a and b can be zero, so the set of real numbers and the set of imaginary numbers are subsets of the set of complex numbers.
Example:
-3.5i is an example of the subset of the imaginary number.
9 is an example of real numbers subset.
2 + i, 3 – 6i combination of real and imaginary numbers

Question 4.
Is it possible for a number to be both whole and natural? natural and rational? rational and irrational? real and imaginary? Explain your reasoning.
Answer:

• A natural number is a subset of a whole number all natural numbers are whole numbers. But all the whole numbers are not natural numbers except zero.
• A number cannot be rational or irrational. All rational numbers can be written as a fraction with an integer as the numerator and a non-zero integer as the denominator.
• All natural numbers are not rational numbers.
• A square of an imaginary number is an imaginary number but not an actual number. The 0 is considered both real and imaginary.

Monitoring Progress

Find the square root of the number.
Question 1.
\(\sqrt{-4}\)
Answer:

Question 2.
\(\sqrt{-12}\)
Answer:

Question 3.
−\(\sqrt{-36}\)
Answer:

Question 4.
2\(\sqrt{-54}\)
Answer:

Find the values of x and y that satisfy the equation.
Question 5.
x + 3i = 9 − yi
Answer:
Given,
x + 3i = 9 − yi
Equate the equation to find the values of x and y.
x = 9 and 3i = -yi
3i = -yi
y = -3
So, x = 9 and y = -3

Question 6.
9 + 4yi = −2x + 3i
Answer:
9 + 4yi = −2x + 3i
Equate the equation to find the values of x and y.
-2x = 9 and 4yi = 3i
x = -9/2 and 4y = 3
y = 3/4
So, x = -9/2 and y = 3/4

Question 7.
WHAT IF?
In Example 4, what is the impedance of the circuit when the capacitor is replaced with one having a reactance of 7 ohms?
Answer:

Perform the operation. Write the answer in standard form.
Question 8.
(9 − i ) + (−6 + 7i )
Answer:
Given,
(9 − i ) + (−6 + 7i )
9 – i – 6 – 7i
Combine the like terms to write the given expression in the standard form.
3 – 8i
So, (9 − i ) + (−6 + 7i ) = 3 – 8i

Question 9.
(3 + 7i ) − (8 − 2i )
Answer:
Given,
(3 + 7i ) − (8 − 2i )
Combine the like terms to write the given expression in the standard form.
3 + 7i – 8 + 2i
-5 + 9i
So, (3 + 7i ) − (8 − 2i ) = -5 + 9i

Question 10.
−4 − (1 + i) − (5 + 9i)
Answer:
Given,
−4 − (1 + i) − (5 + 9i)
-4 – 1 – i – 5 – 9i
-5 – i – 5 – 9i
Combine the like terms to write the given expression in the standard form.
-10 – 10i
Take -10 as common.
-10(1 + i)
So, −4 − (1 + i) − (5 + 9i) = -10(1 + i)

Question 11.
(−3i)(10i)
Answer:
Given,
(−3i)(10i)
-30i²
We know that
i² = -1
-30(-1)
= 30
So, (−3i)(10i) = 30

Question 12.
i(8 − i)
Answer:
Given,
i(8 − i)
8i – i²
We know that
i² = -1
8i – (-1)
8i + 1
i(8 − i) = 8i + 1

Question 13.
(3 + i)(5 −i)
Answer:
Given,
(3 + i)(5 −i)
3(5 – i) + i(5 – i)
15 – 3i + 5i – i²
We know that
i² = -1
15 + 2i + 1
16 + 2i
Thus, (3 + i)(5 −i) = 16 + 2i

Solve the equation.
Question 14.
x² = −13
Answer:
Given,
x² = −13
Applying square root on both sides
x = \(\sqrt{-13}\)

Question 15.
x²= −38
Answer:
Given,
x²= −38
Applying square root on both sides
x = \(\sqrt{-38}\)

Question 16.
x² + 11 = 3
Answer:
Given,
x² + 11 = 3
x² = 3 – 11
x² = -8
Applying square root on both sides
x = 2√-2

Question 17.
x² − 8 = −36
Answer:
Given,
x² − 8 = −36
x² = -36 + 8
x² = -28
Applying square root on both sides
x = 2√-7
So, x = 2√-7

Question 18.
3x² − 7 = −31
Answer:
Given,
3x² − 7 = −31
3x² = −31 + 7
3x² = -24
x² = -24/3
x² = -8
Applying square root on both sides
x = 2√-2

Question 19.
5x² + 33 = 3
Answer:
Given,
5x² + 33 = 3
5x² = 3 – 33
5x² = -30
x² = -30/5
x² = -6
Applying square root on both sides
x = √-6

Find the zeros of the function.
Question 20.
f(x) = x² + 7
Answer:
Given,
f(x) = x² + 7
f(x) = 0
Equate the function to zero.
x² + 7 = 0
x(x + 7) = 0
x = 0 or x + 7 = 0
x = 0 or x = -7

Question 21.
f(x) = −x² − 4
Answer:
Given,
f(x) = −x² − 4
f(x) = 0
Equate the function to zero.
−x² − 4 = 0
x² + 4 = 0
(x + 2) (x + 2) = 0
x + 2 = 0 or x + 2 = 0
x = -2 or x = -2

Question 22.
f(x) = 9x² + 1
Answer:
f(x) = 9x² + 1
f(x) = 0
Equate the function to zero.
9x² + 1 = 0
(3x + 1) (3x + 1) = 0
3x + 1 = 0 or 3x + 1 = 0
3x = -1 or 3x = -1
x = -1/3 or x = -1/3

Complex Numbers 3.2 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
What is the imaginary unit i defined as and how can you use i?
Answer:

Question 2.
COMPLETE THE SENTENCE
For the complex number 5 + 2i, the imaginary part is ____ and the real part is ____.
Answer: For the complex number 5 + 2i, the imaginary part is 2i and the real part is 5.

Question 3.
WRITING
Describe how to add complex numbers.
Answer: To add two complex numbers, add the real parts and the imaginary parts separately.

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

Answer:
3 + 0i = 3 real number
2 + 5i is a complex number
\(\sqrt{3}\) + 6i is a complex number
0 – 7i is a complex number
3 + 0i does not belong to the other three because it is a real number.

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, find the square root of the number.
Question 5.
\(\sqrt{-36}\)
Answer:

Question 6.
\(\sqrt{-64}\)
Answer:
Given,
\(\sqrt{-64}\)
= \(\sqrt{64}\) . \(\sqrt{-1}\)
= 8i
So, \(\sqrt{-64}\) = 8i

Question 7.
\(\sqrt{-18}\)
Answer:

Question 8.
\(\sqrt{-24}\)
Answer:
Given,
\(\sqrt{-24}\)
= \(\sqrt{24}\) . \(\sqrt{-1}\)
= \(\sqrt{4}\) . \(\sqrt{6}\) . \(\sqrt{-1}\)
= 2\(\sqrt{6}\) . i
= 2i\(\sqrt{6}\)
So, \(\sqrt{-24}\) = 2i\(\sqrt{6}\)

Question 9.
2\(\sqrt{-16}\)
Answer:

Question 10.
−3\(\sqrt{-49}\)
Answer:
Given,
−3\(\sqrt{-49}\)
= -3\(\sqrt{49}\) . \(\sqrt{-1}\)
-3 × 7i
= -21i
So, −3\(\sqrt{-49}\) = -21i

Question 11.
−4\(\sqrt{-32}\)
Answer:

Question 12.
6\(\sqrt{-63}\)
Answer:
Given,
6\(\sqrt{-63}\)
= 6 \(\sqrt{9}\) . \(\sqrt{7}\) . \(\sqrt{-1}\)
= 6 × 3 \(\sqrt{7}\)i
= 18 × \(\sqrt{7}\)i
= 18i . \(\sqrt{7}\)
So, 6\(\sqrt{-63}\) = 18i . \(\sqrt{7}\)

In Exercises 13–20, find the values of x and y that satisfy the equation.
Question 13.
4x + 2i = 8 + yi
Answer:

Question 14.
3x + 6i = 27 + yi
Answer:
Given
3x + 6i = 27 + yi
3x = 27 and 6i = yi
x = 27/3 and 6 = y
x = 9 and y = 6

Question 15.
−10x + 12i = 20 + 3yi
Answer:

Question 16.
9x − 18i = −36 + 6yi
Answer:
Given,
9x − 18i = −36 + 6yi
9x = -36 and -18i = 6yi
x = -36/9 and y = -18/6
x = -4 and -3 = y
So, x = -4 and y = -3

Question 17.
2x − yi = 14 + 12i
Answer:

Question 18.
−12x + yi = 60 − 13i
Answer:
Given,
−12x + yi = 60 − 13i
-12x = 60 and yi = -13i
x = 60/-12 and y = -13
x = -5 and y = -13i
So, the real part x = -5 and imaginary part y = -13i

Question 19.
54 − \(\frac{1}{7}\)yi = 9x− 4i
Answer:

Question 20.
15 − 3yi = \(\frac{1}{2}\)x + 2i
Answer:
Given,
15 − 3yi = \(\frac{1}{2}\)x + 2i
15 = \(\frac{1}{2}\)x and -3yi = 2i
x = 15 × 2 and y = -2/3
30 = x and y = -2/3
So, x = 30 and y = -2/3
So, the real part x = 30 and imaginary part y = -2/3

In Exercises 21–30, add or subtract. Write the answer in standard form.
Question 21.
(6 − i) + (7 + 3i)
Answer:

Question 22.
(9 + 5i) + (11 + 2i )
Answer:
Given,
(9 + 5i) + (11 + 2i )
Combine the real parts and imaginary parts.
9 + 5i + 11 + 2i
20 + 7i
So, (9 + 5i) + (11 + 2i ) = 20 + 7i

Question 23.
(12 + 4i) − (3 − 7i)
Answer:

Question 24.
(2 − 15i) − (4 + 5i)
Answer:
Given,
(2 − 15i) − (4 + 5i)
Combine the real parts and imaginary parts.
2 – 15i – 4 – 5i
-2 – 20i
-2(1 + 10i)

Question 25.
(12 − 3i) + (7 + 3i)
Answer:

Question 26.
(16 − 9i) − (2 − 9i)
Answer:
Given,
(16 − 9i) − (2 − 9i)
Combine the real parts and imaginary parts.
16 – 9i – 2 + 9i
16 – 2 = 14
So, (16 − 9i) − (2 − 9i) = 14

Question 27.
7 − (3 + 4i) + 6i
Answer:

Question 28.
16 − (2 − 3i) − i
Answer:
Given,
16 − (2 − 3i) − i
Combine the real parts and imaginary parts.
16 – 2 + 3i  – i
= 14 + 2i
= 2(7 + i)
So, 16 − (2 − 3i) − i = 2(7 + i)

Question 29.
−10 + (6 − 5i) − 9i
Answer:

Question 30.
−3 + (8 + 2i) + 7i
Answer:
Given,
−3 + (8 + 2i) + 7i
Combine the real parts and imaginary parts.
-3 + 8 + 2i + 7i
5 + 9i
So, −3 + (8 + 2i) + 7i = 5 + 9i

Question 31.
USING STRUCTURE
Write each expression as a complex number in standard form.
a. \(\sqrt{-9}+\sqrt{-4}-\sqrt{16}\)
b. \(\sqrt{-16}+\sqrt{8}+\sqrt{-36}\)
Answer:

Question 32.
REASONING
The additive inverse of a complex number z is a complex number z_a such that z + z_a = 0. Find the additive inverse of each complex number.
a. z = 1 + i
b. z = 3 − i
c. z = −2 + 8i
Answer:
a. z = 1 + i
The additive inverse of a complex number is changing the sign of the number
-z = -1 – i
b. z = 3 − i
The additive inverse of a complex number is changing the sign of the number
z = 3 − i = -z = -3 + i
c. z = −2 + 8i
The additive inverse of a complex number is changing the sign of the number
z = −2 + 8i
-z = 2 – 8i

In Exercises 33–36, find the impedance of the series circuit.
Question 33.

Answer:

Question 35.

Answer:

In Exercises 37–44, multiply. Write the answer in standard form.
Question 37.
3i(−5 + i)
Answer:

Question 38.
2i(7 − i)
Answer:
Given,
2i(7 − i) = 14i – 2i²
We know that,
i² = -1
= 14i – 2(-1)
= 14i + 2
So, 2i(7 − i) = 14i + 2

Question 39.
(3 − 2i)(4 + i)
Answer:

Question 40.
(7 + 5i)(8 − 6i)
Answer:
Given,
(7 + 5i)(8 − 6i)
7(8 – 6i) + 5i(8 – 6i)
56 – 42i + 40i – 30i²
We know that,
i² = -1
56 – 2i – 30(-1)
56 – 2i + 30
26 – 2i
So, (7 + 5i)(8 − 6i) = 26 – 2i

Question 41.
(4 − 2i)(4 + 2i)
Answer:

Question 42.
(9 + 5i)(9 − 5i)
Answer:
Given,
(9 + 5i)(9 − 5i)
9² – (5i)²
81 – 25i²
We know that,
i² = -1
81 – 25(-1)
81 + 25 = 106
So, (9 + 5i)(9 − 5i) = 106

Question 43.
(3 − 6i)²
Answer:

Question 44.
(8 + 3i)²
Answer:
Given,
(8 + 3i)²
8² + 2 (8)(3i) + (3i)²
64 + 48i + 9 i²
We know that,
i² = -1
64 + 48i + 9(-1)
= 64 + 48i – 9
= 55 + 48i
So, (8 + 3i)² = 55 + 48i

JUSTIFYING STEPS In Exercises 45 and 46, justify each step in performing the operation.
Question 45.
11 − (4 + 3i) + 5i

Answer:

Question 46.
(3 + 2i)(7 − 4i)

Answer:

REASONING In Exercises 47 and 48, place the tiles in the expression to make a true statement.
Question 47.
(____ − ____i) – (____ − ____i ) = 2 − 4i

Answer:

Question 48.
____i(____ + ____i ) = −18 − 10i

Answer:
2i(-5 + 9i)
= -10i + 18i²
We know that,
i² = -1
= -10i + 18(-1)
= -10i – 18
So, 2i(-5 + 9i) = -10i – 18

In Exercises 49–54, solve the equation. Check your solution(s).
Question 49.
x² + 9 = 0
Answer:

Question 50.
x² + 49 = 0
Answer:
Given,
x² + 49 = 0
x² + 7² = 0
x² = -7²
Applying square root on both sides
x = -7
x = 7i
So, the value of x is ±7i

Question 51.
x² − 4 = −11
Answer:

Question 52.
x² − 9 = −15
Answer:
Given,
x² − 9 = −15
x² = −15 + 9
x² = −6
x = √-6
x = i√6
So, the value of x is i√6.

Question 53.
2x² + 6 = −34
Answer:

Question 54.
x² + 7 = −47
Answer:
Given,
x² + 7 = −47
x² = -47 – 7
x² = -54
x = √-54
x = 3i√6
So, the value of x is 3i√6

In Exercises 55–62, find the zeros of the function.
Question 55.
f(x) = 3x² + 6
Answer:

Question 56.
g(x) = 7x² + 21
Answer:
Given,
g(x) = 7x² + 21
7x² + 21 = 0
7x² = -21
x² = -21/7
x² = -3
x = i√3
So, the zeros of g is i√3

Question 57.
h(x) = 2x² + 72
Answer:

Question 58.
k(x) = −5x² − 125
Answer:
Given,
k(x) = −5x² − 125
−5x² − 125 = 0
-5(x² – 25) =0
x² – 25 = 0
x² = 25
x = 5
So, the zeros of k are 5 or -5.

Question 59.
m(x) = −x² − 27
Answer:

Question 60.
p(x) = x² + 98
Answer:
Given,
p(x) = x² + 98
x² + 98 = 0
x² = -98
x =7 i√14
So, the zeros of p is 7 i√14

Question 61.
r(x) = − \(\frac{1}{2}\)x² − 24
Answer:

Question 62.
f(x) = −\(\frac{1}{5}\)x² − 10
Answer:
Given,
f(x) = −\(\frac{1}{5}\)x² − 10
f(x) = 0
−\(\frac{1}{5}\)x² − 10 = 0
\(\frac{1}{5}\)x² + 10 = 0
\(\frac{1}{5}\)x² = -10
x² = -50
x = √-50
x = 5i√2
So, the zeros of f is 5i√2

ERROR ANALYSIS In Exercises 63 and 64, describe and correct the error in performing the operation and writing the answer in standard form.
Question 63.

Answer:

Question 64.

Answer:
Given,
(4 + 6i)² = (4)² + 2(4)(6i) + (6i)²
= 16 + 48i + 36(-1)
= 16 + 48i – 36
= -20 + 48i

Question 65.
NUMBER SENSE
Simplify each expression. Then classify your results in the table below.
a. (−4 + 7i) + (−4 − 7i)
b. (2 − 6i) − (−10 + 4i)
c. (25 + 15i) − (25 − 6i)
d. (5 + i)(8 − i)
e. (17 − 3i) + (−17 − 6i)
f. (−1 + 2i)(11 − i)
g. (7 + 5i) + (7 − 5i)
h. (−3 + 6i) − (−3 − 8i)

Answer:

Question 66.
MAKING AN ARGUMENT
The Product Property ofSquare Roots states \(\sqrt{a}\) • \(\sqrt{b}\) = \(\sqrt{ab}\) . Your friend concludes \(\sqrt{-4}\) • \(\sqrt{-9}\) = \(\sqrt{36}\) = 6. Is your friend correct? Explain.
Answer:
Given,
The Product Property of Square Roots states \(\sqrt{a}\) • \(\sqrt{b}\) = \(\sqrt{ab}\) .
\(\sqrt{-4}\) • \(\sqrt{-9}\)
2i × 3i = 6i²
We know that,
i² = -1
= 6(-1) = -6
So, your friend is not correct.

Question 67.
FINDING A PATTERN
Make a table that shows the powers of i from i¹ to i⁸ in the first row and the simplified forms of these powers in the second row. Describe the pattern you observe in the table. Verify the pattern continues by evaluating the next four powers of i.
Answer:

Question 68.
HOW DO YOU SEE IT?
The graphs of three functions are shown. Which function(s) has real zeros? imaginary zeros? Explain your reasoning.

Answer:
The graphs of three quadratic functions.
From the above graph, we can see only f and g make intercepts on the x-axis.
So, only f and g have real zeros.
The function h makes no intercepts.
From this, we can conclude that h has only imaginary roots.
Also, the function f only intersects the axis once so it has only one real zero.

In Exercises 69–74, write the expression as a complex number in standard form.
Question 69.
(3 + 4i) − (7 − 5i) + 2i(9 + 12i)
Answer:

Question 70.
3i(2 + 5i) + (6 − 7i) − (9 + i)
Answer:
Given,
3i(2 + 5i) + (6 − 7i) − (9 + i)
Combine the real parts and imaginary parts.
= 6i + 30i² + 6 – 7i – 9 – i
= 6i + 30(-1) + 6 – 7i – 9 – i
= 6i – 30 + 6 – 7i – 9 – i
= -2i -33
= -33 – 2i
So, 3i(2 + 5i) + (6 − 7i) − (9 + i) = -33 – 2i

Question 71.
(3 + 5i)(2 − 7i⁴)
Answer:

Question 72.
2i³(5 − 12i )
Answer:
Given
2i³(5 − 12i )
= 10i³ – 24i⁴
= 10i(-1) – 24(1)
= -10i – 24
= -2(5i + 12)
So, 2i³(5 − 12i ) = -2(5i + 12)

Question 73.
(2 + 4i⁵) + (1 − 9i⁶) − (3 +i⁷)
Answer:

Question 74.
(8 − 2i⁴) + (3 − 7i⁸) − (4 + i⁹)
Answer:
Given,
(8 − 2i⁴) + (3 − 7i⁸) − (4 + i⁹)
8 + 3 – 4 – 2i⁴ – 7i⁸ – i⁹
7 – i⁴(2 + 7i⁴ + i⁵)
i⁴ = 1
7 – 1(2 + 7 + 1i)
7 – 2 – 7 – i
-2 – i
So, (8 − 2i⁴) + (3 − 7i⁸) − (4 + i⁹) = -2 – i

Question 75.
OPEN-ENDED
Find two imaginary numbers whose sum and product are real numbers. How are the imaginary numbers related?
Answer:

Question 76.
COMPARING METHODS
Describe the two different methods shown for writing the complex expression in standard form. Which method do you prefer? Explain.

Answer: I prefer method 1 because it is somewhat easy for writing the expression in the standard form.

Question 77.
CRITICAL THINKING
Determine whether each statement is true or false. If it is true, give an example. If it is false, give a counterexample.
a. The sum of two imaginary numbers is an imaginary number.
b. The product of two pure imaginary numbers is a real number.
c. A pure imaginary number is an imaginary number.
d. A complex number is a real number.
Answer:

Question 78.
THOUGHT PROVOKING
Create a circuit that has an impedance of 14 − 3i.
Answer:
A resistor has a resistance of 14 ohms.
impedance = 14 ohms
An inductor having a reactance of 5i in ohms. So, its impedance is 5i ohms
A capacitor of having a reactance of 8i in ohms.
So its impedance is -8i ohms.
= 14 + 5i + (-8i)
= 14 + 5i – 8i
= 14 – 3i

Maintaining Mathematical Proficiency

Determine whether the given value of x is a solution to the equation.
Question 79.
3(x − 2) + 4x − 1 = x − 1; x = 1
Answer:

Question 80.
x³ − 6 = 2x² + 9 − 3x; x = −5
Answer:
Given,
x³ − 6 = 2x² + 9 − 3x
substitute x = -5 in the expression
(-5)³ − 6 = 2(-5)² + 9 − 3(-5)
-125 – 6 = 50 + 9 + 15
-131 = 74
-131 ≠ 74
So, x = -5 is not a solution to the equation

Question 81.
−x² + 4x = 19 — 3x²; x = −\(\frac{3}{4}\)
Answer:

Write a quadratic function in vertex form whose graph is shown.
Question 82.

Answer:
Given points (0, 3) and (1, 2)
(h, k) = (1, 2)
y = a(x – h)² + k
3 = a(0 – 1)² + 2
3 = a + 2
a = 3 – 2
a = 1
So, the equation of the parabola is y = 1(0 – 1)² + 2

Question 83.

Answer:

Question 84.

Answer:
The points are (3, -2) and (2, -1)
(h, k) = (3, -2)
x = 2 and y = -1
y = a(x – h)² + k
-1 = a(2 – 3)² + (-2)
-1 = -1a – 2
-a = 1
a = -1
So, the equation of the parabola is y = -1(2 – 3)² + (-2)

Lesson 3.3 Completing the Square

Essential Question How can you complete the square for a quadratic expression?

EXPLORATION 1

Using Algebra Tiles to Complete the Square
Work with a partner. Use algebra tiles to complete the square for the expression x² + 6x.

a. You can model x² + 6x using one x²-tile and six x-tiles. Arrange the tiles in a square. Your arrangement will be incomplete in one of the corners.
b. How many 1-tiles do you need to complete the square?
c. Find the value of c so that the expression x² + 6x + c is a perfect square trinomial.
d. Write the expression in part (c) as the square of a binomial.

EXPLORATION 2

Drawing Conclusions
Work with a partner.
a. Use the method outlined in Exploration 1 to complete the table.

b. Look for patterns in the last column of the table. Consider the general statement x² + bx + c = (x + d)². How are d and b related in each case? How are c and d related in each case?
c. How can you obtain the values in the second column directly from the coefficients of x in the first column?

Communicate Your Answer

Question 3.
How can you complete the square for a quadratic expression?
Answer:
i. Divide all the terms of the coefficient of x².
ii. Move the number c/a to the right side of the equation.
iii. 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.
Describe how you can solve the quadratic equation x² + 6x = 1 by completing the square.
Answer:
Given,
x² + 6x = 1
x² + 6x + 9 = 1 + 9
(x + 3)² = 10

Monitoring Progress

Solve the equation using square roots. Check your solution(s).
Question 1.
x² + 4x + 4 = 36
Answer:
Given,
x² + 4x + 4 = 36
The expression is in the form of a² + 2ab + b² = (a + b)²
(x + 2)² = 36
Applying square root on both sides
x + 2 = 36
x = 36 – 2
x = 34
The solution is x = 34.

Question 2.
x² − 6x + 9 = 1
Answer:
Given,
x² − 6x + 9 = 1
The expression is in the form of a² – 2ab + b² = (a – b)²
x² − 6x + 9 = 1
(x – 3)² = 1
Applying square root on both sides
x – 3 = 1
x = 1 + 3
x = 4
The solution is x = 4.

Question 3.
x² − 22x + 121 = 81
Answer:
Given,
x² − 22x + 121 = 81
This expression is in the form of a² – 2ab + b² = (a – b)²
(x – 11)² = 81
(x – 11)² = 9²
Applying square root on both sides
x – 11 = 9
x = 9 + 11
x = 20
The solution is x = 20.

Find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial.
Question 4.
x² + 8x + c
Answer:
Given,
x² + 8x + c
c = b/2a
b = 8 and a = 1
8/2 = 4
c = 4²
c = 16
x² + 8x + 16 = (x + 4)²

Question 5.
x² − 2x + c
Answer:

Question 6.
x² − 9x + c
Answer:

Solve the equation by completing the square.
Question 7.
x² − 4x + 8 = 0
Answer:

Question 8.
x² + 8x − 5 = 0
Answer:

Question 9.
−3x² − 18x − 6 = 0
Answer:

Question 10.
4x² + 32x = −68
Answer:

Question 11.
6x(x + 2) = −42
Answer:

Question 12.
2x(x − 2) = 200
Answer:

Write the quadratic function in vertex form. Then identify the vertex.
Question 13.
y = x² − 8x + 18
Answer:

Question 14.
y = x² + 6x + 4
Answer:

Question 15.
y = x² − 2x − 6
Answer:

Question 16.
WHAT IF?
The height of the baseball can be modeled by y = −16t² + 80t + 2. Find the maximum height of the baseball. How long does the ball take to hit the ground?
Answer:

Completing the Square 3.3 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
What must you add to the expression x² + bx to complete the square?
Answer:

Question 2.
COMPLETE THE SENTENCE
The trinomial x² − 6x + 9 is a ____ because it equals ____.
Answer:
The trinomial x² − 6x + 9 is a perfect square trinomial because it equals (x – 3)².

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, solve the equation using square roots. Check your solution(s).
Question 3.
x² − 8x + 16 = 25
Answer:

Question 4.
r² − 10r + 25 = 1
Answer:

Question 5.
x² − 18x + 81 = 5
Answer:

Question 6.
m² + 8m + 16 = 45
Answer:

Question 7.
y² − 24y + 144 = −100
Answer:

Question 8.
x² − 26x + 169 = −13
Answer:

Question 9.
4w² + 4w + 1 = 75
Answer:

Question 10.
4x² − 8x + 4 = 1
Answer:

In Exercises 11–20, find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial.
Question 11.
x² + 10x + c
Answer:

Question 12.
x² + 20x + c
Answer:

Question 13.
y² − 12y + c
Answer:

Question 14.
t² − 22t + c
Answer:

Question 15.
x² − 6x + c
Answer:

Question 16.
x² + 24x + c
Answer:

Question 17.
z² − 5z + c
Answer:

Question 18.
x² + 9x + c
Answer:

Question 19.
w² + 13w + c
Answer:

Question 20.
s² − 26s + c
Answer:

In Exercises 21–24, find the value of c. Then write an expression represented by the diagram.
Question 21.

Answer:

Question 22.

Answer:

Question 23.

Answer:

Question 24.

Answer:

In Exercises 25–36, solve the equation by completing the square.
Question 25.
x² + 6x + 3 = 0
Answer:

Question 26.
s² + 2s − 6 = 0
Answer:

Question 27.
x² + 4x − 2 = 0
Answer:

Question 28.
t² − 8t − 5 = 0
Answer:

Question 29.
z(z + 9) = 1
Answer:

Question 30.
x(x + 8) = −20
Answer:

Question 31.
7t² + 28t + 56 = 0
Answer:

Question 32.
6r² + 6r + 12 = 0
Answer:

Question 33.
5x(x + 6) = −50
Answer:

Question 34.
4w(w − 3) = 24
Answer:

Question 35.
4x² − 30x = 12 + 10x
Answer:

Question 36.
3s² + 8s = 2s − 9
Answer:

Question 37.
ERROR ANALYSIS
Describe and correct the error in solving the equation.

Answer:

Question 38.
ERROR ANALYSIS
Describe and correct the error in finding the value of c that makes the expression a perfect square trinomial.

Answer:

Question 39.
WRITING
Can you solve an equation by completing the square when the equation has two imaginary solutions? Explain.
Answer:

Question 40.
ABSTRACT REASONING
Which of the following are solutions of the equation x² − 2ax + a² = b²? Justify your answers.
A. ab
B. −a − b
C. b
D. a
E. a − b
F. a + b
Answer:

USING STRUCTURE In Exercises 41–50, determine whether you would use factoring, square roots, or completing the square to solve the equation. Explain your reasoning. Then solve the equation.
Question 41.
x² − 4x − 21 = 0
Answer:

Question 42.
x² + 13x + 22 = 0
Answer:
Given,
x² + 13x + 22 = 0

Question 43.
(x + 4)² = 16
Answer:

Question 44.
(x − 7)² = 9
Answer:

Question 45.
x² + 12x + 36 = 0
Answer:

Question 46.
x² − 16x + 64 = 0
Answer:

Question 47.
2x² + 4x − 3 = 0
Answer:

Question 48.
3x² + 12x + 1 = 0
Answer:

Question 49.
x² − 100 = 0
Answer:

Question 50.
4x² − 20 = 0
Answer:
Given,
4x² − 20 = 0
4x² = 20
x² = 20/4
x² = 5
x = \(\sqrt{5}\)
The solution is x = \(\sqrt{5}\)

MATHEMATICAL CONNECTIONS In Exercises 51–54, find the value of x.
Question 51.
Area of rectangle = 50

Answer:

Question 52.
Area of parallelogram = 48

Answer:

Question 53.
Area of triangle = 40

Answer:

Question 54.
Area of trapezoid = 20

Answer:

In Exercises 55–62, write the quadratic function in vertex form. Then identify the vertex.
Question 55.
f(x) = x² − 8x + 19
Answer:

Question 56.
g(x) = x² − 4x − 1
Answer:

Question 57.
g(x) = x² + 12x + 37
Answer:

Question 58.
h(x) = x² + 20x + 90
Answer:

Question 59.
h(x) = x² + 2x − 48
Answer:

Question 60.
f(x) = x² + 6x − 16
Answer:

Question 61.
f(x) = x² − 3x + 4
Answer:

Question 62.
g(x) = x² + 7x + 2
Answer:

Question 63.
MODELING WITH MATHEMATICS
While marching, a drum major tosses a baton into the air and catches it. The height h (in feet) of the baton t seconds after it is thrown can be modeled by the function h = −16t² + 32t + 6.
a. Find the maximum height of the baton.
b. The drum major catches the baton when it is 4 feet above the ground. How long is the baton in the air?
Answer:

Question 64.
MODELING WITH MATHEMATICS
A firework explodes when it reaches its maximum height. The height h (in feet) of the firework t seconds after it is launched can be modeled by h = \(-\frac{500}{9} t^{2}+\frac{1000}{3} t\) + 10. What is the maximum height of the firework? How long is the firework in the air before it explodes?

Answer:

Question 65.
COMPARING METHODS
A skateboard shop sells about 50 skateboards per week when the advertised price is charged. For each $1 decrease in price, one additional skateboard per week is sold. The shop’s revenue can be modeled by y = (70 − x)(50 + x).

a. Use the intercept form of the function to find the maximum weekly revenue.
b. Write the function in vertex form to find the maximum weekly revenue.
c. Which way do you prefer? Explain your reasoning.
Answer:

Question 66.
HOW DO YOU SEE IT?
The graph of the function f(x) = (x − h)² is shown. What is the x-intercept? Explain your reasoning.

Answer:

Question 67.
WRITING
At Buckingham Fountain in Chicago, the height h (in feet) of the water above the main nozzle can be modeled by h = −16² + 89.6t, where t is the time (in seconds) since the water has left the nozzle. Describe three different ways you could find the maximum height the water reaches. Then choose a method and find the maximum height of the water.
Answer:

Question 68.
PROBLEM SOLVING
A farmer is building a rectangular pen along the side of a barn for animals. The barn will serve as one side of the pen. The farmer has 120 feet of fence to enclose an area of 1512 square feet and wants each side of the pen to be at least 20 feet long.
a. Write an equation that represents the area of the pen.
b. Solve the equation in part (a) to find the dimensions of the pen.

Answer:

Question 69.
MAKING AN ARGUMENT
Your friend says the equation x² + 10x = −20 can be solved by either completing the square or factoring. Is your friend correct? Explain.
Answer:
x² + 10x = −20
x² + 10x + 20 = 0
Your friend is incorrect because x² + 10x = −20 does not factor into rational numbers.

Question 70.
THOUGHT PROVOKING
Write a function g in standard form whose graph has the same x-intercepts as the graph of f(x) = 2x² + 8x + 2. Find the zeros of each function by completing the square. Graph each function.
Answer:

Question 71.
CRITICAL THINKING
Solve x² + bx + c = 0 by completing the square. Your answer will be an expression for x in terms of b and c.
Answer:

Question 72.
DRAWING CONCLUSIONS
In this exercise, you will investigate the graphical effect of completing the square.
a. Graph each pair of functions in the same coordinate plane.
y = x² + 2x y = x² − 6x
y = (x + 1)² y = (x − 3)²
b. Compare the graphs of y = x² + bx and y = (x + \(\frac{b}{2}\))². Describe what happens to the graph of y = x² + bx when you complete the square.
Answer:

Question 73.
MODELING WITH MATHEMATICS
In your pottery class, you are given a lump of clay with a volume of 200 cubic centimeters and are asked to make a cylindrical pencil holder. The pencil holder should be 9 centimeters high and have an inner radius of 3 centimeters. What thickness x should your pencil holder have if you want to use all of the clay?

Answer:

Maintaining Mathematical Proficiency

Solve the inequality. Graph the solution.
Question 74.
2x − 3 < 5
Answer:

Question 75.
4 − 8y ≥ 12
Answer:

Question 76.
\(\frac{n}{3}\) + 6 > 1
Answer:

Question 77.
−\(\frac{2s}{5}\) ≤ 8
Answer:

Graph the function. Label the vertex, axis of symmetry, and x-intercepts.
Question 78.
g(x) = 6(x − 4)²
Answer:

Question 79.
h(x) = 2x(x − 3)
Answer:

Question 80.
f(x) = x² + 2x + 5
Answer:

Question 81.
f(x) = 2(x + 10)(x − 12)
Answer:

Quadratic Equations and Complex Numbers Study Skills: Creating a Positive Study Environment

3.1–3.3 What Did You Learn?

Core Vocabulary

Core Concepts

Mathematical Practices
Question 1.
Analyze the givens, constraints, relationships, and goals in Exercise 61 on page 101.
Answer:

Question 2.
Determine whether it would be easier to find the zeros of the function in Exercise 63 on page 117 or Exercise 67 on page 118.
Answer:

Study Skills: Creating a Positive Study Environment

• Set aside an appropriate amount of time for reviewing your notes and the textbook, reworking your notes, and completing homework.
• Set up a place for studying at home that is comfortable, but not too comfortable. The place needs to be away from all potential distractions.
• Form a study group. Choose students who study well together, help out when someone misses school, and encourage positive attitudes.
  

Quadratic Equations and Complex Numbers 3.1–3.3 Quiz

Solve the equation by using the graph. Check your solution(s).
Question 1.
x² − 10x + 25 = 0

Answer:
Given,
f(x) = x² − 10x + 25
x² − 10x + 25 = 0
(x – 5)² = 0
Applying square root on both sides
x – 5 = 0
x = 5
The solution is x = 5.

Question 2.
2x² + 16 = 12x

Answer:
Given,
g(x) = 2x² -12x + 16
2x² + 16 = 12x
2x² -12x + 16 = 0
2(x² – 6x + 8) = 0
x² – 6x + 8 = 0
x = -b ± √b² – 4ac/2a
x = -(-6) ± √(-6)² – 4(1)(8)/2
x = 6 ± √36-32/2
x = 6 ± √4/2
x = (6+2)/2 = 8/2 = 4
x = (6-2)/2 = 4/2 = 2
The solution is x = 2

Question 3.
x² = −2x + 8

Answer:
Given,
h(x) = -x² – 2x + 8
x² = −2x + 8
x² + 2x – 8 = 0
x = -b ± √b² – 4ac/2a
x = -(2) ± √(2)² – 4(1)(-8)/2
x = -2 ± √4+32/2
x = 4/2 or x = -8/2
x = 2 or -4
The solutions are x = 2 and x = -4.

Solve the equation using square roots or by factoring. Explain the reason for your choice.
Question 4.
2x² − 15 = 0
Answer:
Given,
2x² − 15 = 0
2x² = 15
x² = 15/2
x² = 7.5
Applying square root on both sides.
x = √7.5
The solution x = √7.5

Question 5.
3x² − x − 2 = 0
Answer:
Given,
3x² − x − 2 = 0
x = (-b ± √b² – 4ac)/2a
x = -(-1) ± √(-1)² – 4(3)(-2)/2(3)
x = 1 ± √1+24/6
x = 1 ± √1+24/6
x = (1 ± 5)/6
x = 6/6 or -4/6
x = 1 or -2/3
The solution is x = 1 or -2/3

Question 6.
(x + 3)² = 8
Answer:
Given,
x² + 6x + 9 = 8
x² + 6x + 1 = 0
x = (-b ± √b² – 4ac)/2a
b = 6, a = 1 and c = 1
x = -(6) ± √(6)² – 4(1)(1)/2(1)
x = -6 ± √36-4/2
x = -6 ± √32/2
x = -0.17
x = -2.82
The solution is x = -2.82 and -0.17

Question 7.
Find the values of x and y that satisfy the equation 7x − 6i = 14 + yi.
Answer:
Given,
7x − 6i = 14 + yi.
7x = 14 and -6i = yi
x = 14/7 and y = -6
x = 2 and y = -6

Perform the operation. Write your answer in standard form
Question 8.
(2 + 5i) + (−4 + 3i)
Answer:
Given,
(2 + 5i) + (−4 + 3i)
= 2 + 5i – 4 + 3i
= -2 + 8i
So, (2 + 5i) + (−4 + 3i) = -2 + 8i

Question 9.
(3 + 9i) − (1 − 7i)
Answer:
Given,
(3 + 9i) − (1 − 7i)
= 3 + 9i – 1 + 7i
= 2 + 16i
So, (3 + 9i) − (1 − 7i) = 2 + 16i

Question 10.
(2 + 4i)(−3 − 5i)
Answer:
Given,
(2 + 4i)(−3 − 5i)
= 2(-3 – 5i) + 4i(-3 – 5i)
= -6 – 15i – 12i – 20i²
= -6 – 27i + 20
= 14 – 27i
So, (2 + 4i)(−3 − 5i) = 14 – 27i

Question 11.
Find the zeros of the function f(x) = 9x² + 2. Does the graph of the function intersect the x-axis? Explain your reasoning.
Answer:
Given,
f(x) = 9x² + 2
f(x) = 0
9x² + 2 = 0
9x² = -2
x = √(-2/9)
x = i/3 √2

Solve the equation by completing the square.
Question 12.
x² − 6x + 10 = 0
Answer:
Given,
x² – 6x + 10 = 0
x = (-b ± √b² – 4ac)/2a
b = -6, a = 1 and c = 10
x = -(6) ± √(6)² – 4(1)(10)/2(1)
x = -6 ± √-4/2
x = -6 ± √32/2
x = -3 ± i
So, x = -3 + i or -3 – i

Question 13.
x² + 12x + 4 = 0
Answer:
Given,
x² + 12x + 4 = 0
x = -b ± √b² – 4ac/2a
a = 1, b = 12 and c = 4
x = -(12) ± √(12)² – 4(1)(4)/2(1)
x = (-12 ± √144 – 16)/2
x = (-12±8√2)/2
x = -6 ± 4√2
So, x = -6 + 4√2 and x = -6 – 4√2

Question 14.
4x(x + 6) = −40
Answer:
Given,
4x(x + 6) = −40
4x² + 24x + 40 = 0
4(x² + 6x + 10) = 0
x² + 6x + 10 = 0
x = -b ± √b² – 4ac/2a
a = 1, b = 6 and c = 10
x = -(6) ± √(6)² – 4(1)(10)/2(1)
x = -6 ± √-4/2
x = -6 ± √32/2
x = -3 ± i
So, x = -3 + i or -3 – i

Question 15.
Write y = x² − 10x + 4 in vertex form. Then identify the vertex.
Answer:
Given
y = x² − 10x + 4
(b/2)² = (-10/2)²
y + 25 = (x² – 10x + 25) + 4
y + 25 = (x – 5)² + 4
y = (x – 5)² – 21
The vertex is (5, -21)

Question 16.
A museum has a café with a rectangular patio. The museum wants to add 464 square feet to the area of the patio by expanding the existing patio as shown.

a. Find the area of the existing patio.
b. Write an equation to model the area of the new patio.
c. By what distance x should the length of the patio be expanded?
Answer:
a. Area of the rectangle = length × breadth
length = 30 ft
breadth = 20 ft
= 30 × 20
= 600 sq. ft
So, the area of the existing patio = 600 sq. ft
b. 600 + 464 = (30 + x) (20 + x)
1064 = 600 + 30x + 20x + x²
x² + 50x + 600 – 1064 = 0
x² + 50x – 464 = 0
c. x² + 50x – 464 = 0
(x + m)(x + n)
mn = -464 and m + n = 50
x² + 58x – 8x – 464 = 0
x(x + 58) – 8(x + 58) = 0
(x + 58) (x – 8) = 0
x + 58 = 0
x = -58
The dimensions cannot be negative.
x – 8 = 0
x = 8
So, x = 8

Question 17.
Find the impedance of the series circuit.

Answer:
R = 7Ω
Li = 5i
C = 2
impedance = R + Li + (-Ci)
= 7 + 5i + (-2i)
= 7 + 3i

Question 18.
The height h (in feet) of a badminton birdie t seconds after it is hit can be modeled by the function h = −16t² + 32t + 4.
a. Find the maximum height of the birdie.
Answer:
h = −16t² + 32t + 4
-16 + h = -16(t² -2t + 1) + 4
h = -16(t – 1)² + 4 + 16
h = -16(t – 1)² + 20
The maximum height of the birdie is 20 feet

b. How long is the birdie in the air?
Answer:
0 = -16(t – 1)² + 20
-20 = -16(t – 1)²
1.25 = (t – 1)²
1.1 = t – 1
t = 2.1 sec

Lesson 3.4 Using the Quadratic Formula

Essential Question How can you derive a general formula for solving a quadratic equation?

EXPLORATION 1

Deriving the Quadratic Formula
Work with a partner. Analyze and describe what is done in each step in the development of the Quadratic Formula.

EXPLORATION 2

Using the Quadratic Formula
Work with a partner. Use the Quadratic Formula to solve each equation.
a. x² − 4x + 3 = 0
b. x² − 2x + 2 = 0
c. x² + 2x − 3 = 0
d. x² + 4x + 4 = 0
e. x² − 6x + 10 = 0
f. x² + 4x + 6 = 0

Communicate Your Answer

Question 3.
How can you derive a general formula for solving a quadratic equation?
Answer:
Start with an equation of the form ax² + bx + c = 0.
Rewrite the equation so that ax² + bx is isolated on one side.
Complete the square by adding b²/4a² to both sides.
Rewrite the perfect square trinomial as a square of a binomial.

Question 4.
Summarize the following methods you have learned for solving quadratic equations: graphing, using square roots, factoring, completing the square, and using the Quadratic Formula.
Answer:
The methods for solving quadratic equations are summarized as:
Method 1 is graphing
Method 2 is factoring
Method 3 is Completing the square
Graphing: The x-intercepts serve as the solution for the equation.
Using square roots: The properties of square roots are used to solve the problem.
Factoring: The factors are framed for the equation and the zero product property is used to solve the problem.
Completing the square: The square is completed so that the properties of the square root may be applied as illustrated in the solution by using square roots.
Quadratic formula: Start with an equation of the form ax² + bx + c = 0
x = (-b ± √b² – 4ac)/2a

Monitoring Progress

Solve the equation using the Quadratic Formula.
Question 1.
x² − 6x + 4 = 0
Answer:

Question 2.
2x² + 4 = −7x
Answer:

Question 3.
5x² = x + 8
Answer:

Solve the equation using the Quadratic Formula.
Question 4.
x² + 41 = −8x
Answer:
Given equation,
x² + 41 = −8x
x² + 41 + 8x = 0
x = -b ± √b² – 4ac/2a
a = 1, b = 8 and c = 41
x = -8 ± √8² – 4(1)(41)/2(1)
x = (-8 ± √64 – 164)/2
x = (-8 ± √-100)/2
x = -8/2 ± 10i/2
x = -4 ± 5i
The solution is x = -4 + 5i and x = -4 – 5i

Question 5.
−9x² = 30x + 25
Answer:
Given equation,
−9x² = 30x + 25
-9x² – 30x – 25 = 0
9x² + 30x + 25 = 0
a = 9, b = 30 and c = 25
Substitute the values of a, b and c in the formula.
x = -b ± √b² – 4ac/2a
x = -30 ± √30² – 4(9)(25)/2(9)
x = (-30 ± √900- 900)/2
x = (-30 ± √0)/18
x = -30/18
x = -5/3
The solution x = -5/3.

Question 6.
5x − 7x² = 3x + 4
Answer:
Given,
5x − 7x² = 3x + 4
-7x² + 5x – 3x – 4 = 0
-7x² + 2x – 4 = 0
7x² – 2x + 4 = 0
a = 7, b = -2 and c = 4
x = -b ± √b² – 4ac/2a
x = 2 ± √(-2)² – 4(7)(4)/2(7)
x = (2 ± √4- 112)/14
x = 1/7 ± 3i√3/7
The solution is x = 1/7 + 3i√3/7 and x = 1/7 – 3i√3/7

Find the discriminant of the quadratic equation and describe the number and type of solutions of the equation.
Question 7.
4x² + 8x + 4 = 0
Answer:
Given,
4x² + 8x + 4 = 0
The discriminant is b² – 4ac
d = b² – 4ac
b² – 4ac = 0
8² – 4(4)(4)
a = 4, b = 8 and c = 4
64 – 64 = 0
So, the discriminant is 0.

Question 8.
\(\frac{1}{2}\)x² + x − 1 = 0
Answer:
\(\frac{1}{2}\)x² + x − 1 = 0
The discriminant is b² – 4ac
d = b² – 4ac
b² – 4ac = 0
a = \(\frac{1}{2}\), b = 1, c = -1
1² – 4(\(\frac{1}{2}\))(-1)
1 + 2 = 3
The discriminant is 3.

Question 9.
5x² = 8x − 13
Answer:
Given,
5x² = 8x − 13
5x² – 8x + 13 = 0
The discriminant is b² – 4ac
b² – 4ac = 0
(-8)² – 4(5)(-13)
a = 5, b = -8 and c = -13
64 + 260 = 324
The discriminant is 324

Question 10.
7x² − 3x = 6
Answer:
Given,
7x² − 3x = 6
The discriminant is b² – 4ac
b² – 4ac = 0
7x² − 3x – 6 = 0
a = 7, b = -3 and c = -6
(3)² – 4(7)(-6)
9 + 168 = 177
The discriminant is 177.

Question 11.
4x² + 6x = −9
Answer:
Given,
4x² + 6x = −9
4x² + 6x + 9 = 0
The discriminant is b² – 4ac
a = 4, b = 6 and c = 9
6² – 4(4)(-9)
36 + 144 = 180
The discriminant is 180.

Question 12.
−5x² + 1x = 6 − 10x
Answer:
Given,
−5x² + 1 = 6 − 10x
-5x² + x – 6 + 10x
-5x² + 11x – 6 = 0
The discriminant is b² – 4ac
a = -5, b = 11 and c = -6
(11)² – 4(-5)(-6)
121 – 120 = 1
The discriminant is 1.

Question 13.
Find a possible pair of integer values for a and c so that the equation ax² + 3x + c = 0 has two real solutions. Then write the equation.
Answer:
Given,
ax² + 3x + c = 0
If b² – 4ac > 0 that means there are 2 real solutions
b² – 4ac = 0 there is 1 real solution
b² – 4ac < 0 that means there are imaginary solutions
ax² + 3x + c = 0
b = 3, a = a and c = c.
3² – 4ac = 0
9 – 4ac = 0
-4ac = -9
ac = 9/4
a = 9/4 c
So, there are 2 real solutions

Question 14.
WHAT IF?
The ball leaves the juggler’s hand with an initial vertical velocity of 40 feet per second. How long is the ball in the air?
Answer:

Using the Quadratic Formula 3.4 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
When a, b, and c are real numbers such that a ≠ 0, the solutions of the quadratic equation ax² + bx + c = 0 are x= ____________.
Answer:

Question 2.
COMPLETE THE SENTENCE
You can use the ____________ of a quadratic equation to determine the number and type of solutions of the equation.
Answer: You can use the discriminant of a quadratic equation to determine the number and type of solutions of the equation.

Question 3.
WRITING
Describe the number and type of solutions when the value of the discriminant is negative.
Answer: If the discriminant of a quadratic equation is negative, then the equation has two imaginary solutions.

Question 4.
WRITING
Which two methods can you use to solve any quadratic equation? Explain when you might prefer to use one method over the other.
Answer: There are three basic methods for solving quadratic equations:

1. Factoring,
2. Using the quadratic formula, and
3. Completing the square.
   The quadratic formula can be used to solve any quadratic equation and it is easy to just plug in the numbers. Graphing would be a little bit more complicated but if you have a graphing calculator, solving this equation would be easy.

Monitoring Progress and Modeling with Mathematics

In Exercises 5–18, solve the equation using the Quadratic Formula. Use a graphing calculator to check your solution(s).
Question 5.
x² − 4x + 3 = 0
Answer:

Question 6.
3x² + 6x + 3 = 0
Answer:
3x² + 6x + 3 = 0
3(x² + 2x + 1) = 0
x² + 2x + 1 = 0
x² + x + x + 1 = 0
x(x + 1) + 1 (x + 1) = 0
(x + 1) (x + 1) = 0
x = -1
The solution is x = -1
Check the solution with the graph.

From the above graph, we observe that the solution is -1.

Question 7.
x² + 6x + 15 = 0
Answer:

Question 8.
6x² − 2x + 1 = 0
Answer:
Given equation
6x² − 2x + 1 = 0
x = -b ± √b² – 4ac/2a
a = 6, b = -2 and c = 1
x = -(-2) ± √(-2)² – 4(6)(1)/2(6)
x = (2 ± √4 – 24)/12
x = (2 ± √-20)/12
x = 2/12 ± 2√5i/12
x = 1/6 ± √5/6i
x = 0.16 + 0.37i
x = 0.16 – 0.37i
The solution is x = 0.16 + 0.37i and x = 0.16 – 0.37i
Check the solution with the graph.

Question 9.
x² − 14x = −49
Answer:

Question 10.
2x² + 4x = 30
Answer:
Given,
2x² + 4x = 30
2(x² + 2x – 15) = 0
x² + 2x – 15 = 0
x = -b ± √b² – 4ac/2a
a = 2, b = 4 and c = -30
x = -(2) ± √(2)² – 4(1)(15)/2(1)
x = (-2 ± √4 – 60)/2
x = (-2 ± √-56)/2
x = -2/2 ± 2√14i/2
x = -1± √14i
x = -1 + 3.74i
x = -1 – 3.74i
The solution is x = -1 + 3.74i and x = -1 – 3.74i
Check the solution with the graph.

Question 11.
3x² + 5 = −2x
Answer:

Question 12.
−3x = 2x² − 4
Answer:
Given,
−3x = 2x² − 4
2x² − 4 + 3x = 0
x = -b ± √b² – 4ac/2a
a = 2, b = 3 and c = -4
x = -(3) ± √(3)² – 4(2)(-4)/2(2)
x = (-3 ± √9+32)/4
x = (-3 ± √41)/4
x = -3/4 ± √41/4
x = 0.85
x = -2.35
The solution is x = 0.85 and x = -2.35
Check the solution with the graph.

Question 13.
−10x = −25 − x²
Answer:

Question 14.
−5x² − 6 = −4x
Answer:
Given,
−5x² − 6 = −4x
5x² – 4x + 6 = 0
x = -b ± √b² – 4ac/2a
x = -(-4) ± √(-4)² – 4(5)(6)/2(5)
x = (4 ± √16-120)/10
x = (4 ± √-104)/10
x = (4 ± 2√26i)/10
x = 2/5 ± √26i/5
x = 0.4 + 1.0198i
x = 0.4 – 1.0198i
The solution is x = 0.4 + 1.0198i and x = 0.4 – 1.0198i
Check the solution with the graph.

Question 15.
−4x² + 3x = −5
Answer:

Question 16.
x² + 121 = −22x
Answer:
Given,
x² + 121 = −22x
x² + 22x + 121 = 0
b = a × c
x² + 11x + 11x + 121 = 0
x(x + 11) + 11(x + 11) = 0
(x + 11)(x + 11) = 0
x = -11
The solution is x = -11.
Check the solution with the graph.

Question 17.
−z² = −12z + 6
Answer:

Question 18.
−7w + 6 = −4w²
Answer:
−7w + 6 = −4w²
-4w² + 7w – 6 = 0
-(4w² – 7w + 6 = 0)
4w² – 7w + 6 = 0
x = -b ± √b² – 4ac/2a
x = -(-7) ± √(-7)² – 4(4)(6)/2(4)
x = (7 ± √49-96)/8
x = (7 ± √-47)/8
x = (7 ± √47i)/8
x = 7/8 ± √47i/8
x = 0.875 + 0.85i
x = 0.875 – 0.85i
The solution is x = 0.875 + 0.85i and x = 0.875 – 0.85i
Check the solution with the graph.

In Exercises 19–26, find the discriminant of the quadratic equation and describe the number and type of solutions of the equation.
Question 19.
x² + 12x + 36 = 0
Answer:

Question 20.
x² − x + 6 = 0
Answer:
Given equation
x² − x + 6 = 0
x = -b ± √b² – 4ac/2a
a = 1, b = -1 and c = 6
x = -(-1) ± √(-1)² – 4(1)(6)/2(1)
x = (1 ± √1-24)/2
x = (1 ± √-23)/2
x = (1 ± √23i)/2
x = 1/2 ± √23i/2
x = 0.5 + 2.39i
x = 0.5 – 2.39i
The solution is x = 0.5 + 2.39i and x = 0.5 – 2.39i
Check the solution with the graph.

Question 21.
4n² − 4n − 24 = 0
Answer:

Question 22.
−x² + 2x + 12 = 0
Answer:
Given,
x² – 2x – 12 = 0
x = -b ± √b² – 4ac/2a
a = 1, b = -2 and c = -12
x = -(-2) ± √(-2)² – 4(1)(12)/2(1)
x = (2 ± √4-48)/2
x = (2 ± √-44)/2
x = (2 ± √44i)/2
x = 1 ± √11i
x = 1 + 3.3i
x = 1 – 3.3i
The solution is x = 1 + 3.3i and x = 1 – 3.3i
Check the solution with the graph.

Question 23.
4x² = 5x − 10
Answer:

Question 24.
−18p = p² + 81
Answer:
Given,
p² + 18p + 81 = 0
p² + 9p + 9p + 81 = 0
p(p + 9) + 9(p + p)
(p + 9) (p + 9) = 0
p = -9
The solution is p = -9.
Check the solution with the graph.

Question 25.
24x = −48 − 3x²
Answer:

Question 26.
−2x² − 6 = x²
Answer:
Given,
−2x² − 6 = x²
x² + 2x² + 6 = 0
3x² + 6 = 0
3(x² + 2) = 0
x² + 2 = 0
x² = -2
x = √2i
The solution is x = √2i
Check the solution with the graph.

Question 27.
USING EQUATIONS
What are the complex solutions of the equation 2x²− 16x+ 50 = 0?
A. 4 + 3i, 4 − 3i
B. 4 + 12i, 4 − 12i
C. 16 + 3i, 16 − 3i
D. 16 + 12i, 16 − 12i
Answer:

Question 28.
USING EQUATIONS
Determine the number and type of solutions to the equation x² + 7x = −11.
A. two real solutions
B. one real solution
C. two imaginary solutions
D. one imaginary solution
Answer:
x² + 7x = −11
x² + 7x + 11 = 0
It has two imaginary solutions

ANALYZING EQUATIONS In Exercises 29–32, use the discriminant to match each quadratic equation with the correct graph of the related function. Explain your reasoning.
Question 29.
x² − 6x + 25 = 0
Answer:

Question 30.
2x² − 20x + 50 = 0
Answer:
Given,
2x² − 20x + 50 = 0
x² – 10x + 25 = 0
x² – 5x – 5x + 25 = 0
x(x – 5) -5(x – 5) = 0
(x – 5) (x – 5) = 0
x = 5
The discriminant is b² – 4ac
d = b² – 4ac
a = 1, b = 10 and c = 25
(10)² – 4(1)(25)
100 – 100 = 0

Question 31.
3x² + 6x − 9 = 0
Answer:

Question 32.
5x² − 10x − 35 = 0
Answer:
Given
5x² − 10x − 35 = 0
Take 5 as the common factor.
5(x² – 2x – 7) = 0
It becomes
x² – 2x – 7 = 0

ERROR ANALYSIS In Exercises 33 and 34, describe and correct the error in solving the equation.
Question 33.

Answer:

Question 34.

Answer:
Given
x² + 6x + 8 = 2
x² + 6x + 8 – 2 = 0
x² + 6x + 6 = 0
x = -b ± √b² – 4ac/2a
a = 1, b = 6 and c = 6
x = -(6) ± √(6)² – 4(1)(6)/2(1)
x = (-6 ± √36-24)/2
x = (-6 ± √12)/2
x = (-6 ± 2√3)/2
x = -3 ± √3
x = -1.26
x = -4.73
The solution is x = -1.26 and x = -4.73

OPEN-ENDED In Exercises 35–40, find a possible pair of integer values for a and c so that the quadratic equation has the given solution(s). Then write the equation.
Question 35.
ax² + 4x + c = 0; two imaginary solutions
Answer:

Question 36.
ax² + 6x + c = 0; two real solutions
Answer:
Given,
For a quadratic equation to have two imaginary solutions, the discriminant must be negative. So, the set the discriminant equal to any negative number.
ax² + 6x + c = 0
1. substitute a and c
2. check for discriminant
3. b² – 4ac ≥ 0
b = 6
4. 36 – 4ac ≥ 0
The ≥ sign will give us the result for two real unequal solutions and two real equal solutions. If we only need Real unequal solutions we only > sign instead of ≥.
36 – 4ac = 0
36 = 4ac
ac = 36/4
ac = 9

Question 37.
ax² − 8x + c = 0; two real solutions
Answer:

Question 38.
ax² − 6x + c = 0; one real solution
Answer:
Given,
ax² – 6x + c = 0
For a quadratic equation to have two imaginary solutions, the discriminant must be negative. So, they set the discriminant equal to any negative number.
1. substitute a and c
2. check for discriminant
3. b² – 4ac ≥ 0
4. 36 – 4ac ≥ 0
The ≥ sign will give us the result for two real unequal solutions and two real equal solutions. If we only need Real unequal solutions we only > sign instead of ≥.

Question 39.
ax² + 10x = c; one real solution
Answer:

Question 40.
−4x + c = −ax²; two imaginary solutions
Answer:
-4x + c + ax² = 0
ax² – 4x + c = 0
For a quadratic equation to have two imaginary solutions, the discriminant must be negative. So, they set the discriminant equal to any negative number.
d = b² – 4ac
b² – 4ac = 0
(-4)² – 4ac = 0
16 – 4ac = 0
4 – ac = 0
ac = 4

USING STRUCTURE In Exercises 41–46, use the Quadratic Formula to write a quadratic equation that has the given solutions.
Question 41.
x = \(\frac{-8 \pm \sqrt{-176}}{-10}\)
Answer:

Question 42.
x = \(\frac{15 \pm \sqrt{-215}}{22}\)
Answer:
Given,
The solution x = \(\frac{15 \pm \sqrt{-215}}{22}\)
22x = 15 ± √-215
22x – 15 = √-215
Apply Squaring on both sides
(22x – 15)² = (√-215)²
484x² + 225x – 660 = -215
484x² + 225x – 660 + 215 = 0
484x² + 225x – 445 = 0
Thus, the equation is 484x² + 225x – 445 = 0

Question 43.
x = \(\frac{-4 \pm \sqrt{-124}}{-14}\)
Answer:

Question 44.
x = \(\frac{-9 \pm \sqrt{137}}{4}\)
Answer:
Given
The solution is x = \(\frac{-9 \pm \sqrt{137}}{4}\)
4x = -9 ± √137
4x + 9 = √137
Apply Squaring on both sides to find the equation
(4x + 9)² = (√137)²
36x² + 81 + 72x = 137
36x² – 72x + 81 – 137 = 0
36x² – 72x – 56 = 0
Thus, the equation is 36x² – 72x – 56 = 0

Question 45.
x = \(\frac{-4 \pm 2}{6}\)
Answer:

Question 46.
x = \(\frac{2 \pm 4}{-2}\)
Answer:
Given the solution is
x = \(\frac{2 \pm 4}{-2}\)
-2x = 2 ± 4
-2x – 2 = 4
-x – 1 = 4
x + 1 = -4
Apply to square on both sides
(x + 1)² = 16
x² + 2x + 1 = 16
x² + 2x – 15 = 0
Thus, the equation is x² + 2x – 15 = 0

COMPARING METHODS In Exercises 47–58, solve the quadratic equation using the Quadratic Formula. Then solve the equation using another method. Which method do you prefer? Explain.
Question 47.
3x² − 21 = 3
Answer:

Question 48.
5x² + 38 = 3
Answer:
Given,
Use the factoring method to find the solution.
5x² + 38 = 3
5x² + 38 -3 = 0
5x² + 35 = 0
5(x² + 7) = 0
x² + 7 = 0
x² = -7
x = i√7
The solution is x = i√7

Question 49.
2x² − 54 = 12x
Answer:

Question 50.
x² = 3x + 15
Answer:
Given,
x² = 3x + 15
x² – 3x – 15 = 0
Use the quadratic equation formula to find the solution.
x = -b ± √b² – 4ac/2a
x = -(-3) ± √(-3)² – 4(1)(-5)/2(1)
x = (3 ± √9+20)/2
x = (3 ± √29)/2
x = 3/2 ± √29/2
x = 4.19
x = -1.19
The solution is x = 4.19 and x = -1.19

Question 51.
x² − 7x + 12 = 0
Answer:

Question 52.
x² + 8x − 13 = 0
Answer:
Given,
x² + 8x − 13 = 0
Use the quadratic equation formula to find the solution.
x = -b ± √b² – 4ac/2a
x = -(8) ± √(8)² – 4(1)(-13)/2(1)
a = 1, b = 8 and c = -13
x = (-8 ± √64+54)/2
x = (-8 ± √116)/2
x = -8/2 ± √116/2
x = -4 ± √29
x = 1.38
x = -9.38
The solution is x = 1.38 and -9.38

Question 53.
5x² − 50x = −135
Answer:

Question 54.
8x² + 4x + 5 = 0
Answer:
Given,
8x² + 4x + 5 = 0
Use the quadratic equation formula to find the solution.
x = -b ± √b² – 4ac/2a
a = 8, b = 4 and c = 5
x = -(4) ± √(4)² – 4(8)(-5)/2(8)
x = (-4 ± √16+160)/16
x = (-4 ± √176)/16
x = -4/16 ± √176/16
x = -1/4 ± √11/4
x = 0.57
x = -1.07
The solution is x = 0.57 and x = -1.07

Question 55.
−3 = 4x² + 9x
Answer:

Question 56.
−31x + 56 = −x²
Answer:
Given,
−31x + 56 = −x²
x² – 31x + 56 = 0
Use the quadratic equation formula to find the solution.
x = -b ± √b² – 4ac/2a
a = 1, b = -31 and c = 56
x = -(-31) ± √(-31)² – 4(1)(56)/2(1)
x = (31 ± √961-224)/2
x = (31 ± √737)/2
x = 31/2 ± √737/2
x = 29.07
x = 1.92
The solution is x = 29.07 and x = 1.92

Question 57.
x² = 1 − x
Answer:

Question 58.
9x² + 36x + 72 = 0
Answer:
Given,
9x² + 36x + 72 = 0
Take 9 as a common factor.
9(x² + 4x + 8) = 0
x² + 4x + 8 = 0
x = -b ± √b² – 4ac/2a
Use the quadratic equation formula to find the solution.
x = -(4) ± √(4)² – 4(1)(8)/2(1)
x = (-4 ± √16-32)/2
x = (-4 ± √-16)/2
x = -4/2 ± 4i/2
x = -2 + 2i
x = -2 – 2i
The solution is x = -2 + 2i and x = -2 – 2i

MATHEMATICAL CONNECTIONS In Exercises 59 and 60, find the value for x.
Question 59.
Area of the rectangle = 24 m²

Answer:

Question 60.
Area of the triangle = 8ft²

Answer:
Given,
Area of the triangle = 8ft²
We know that,
Area of the triangle = 1/2 × bh
b = (3x – 7)
h = (x + 1)
8 = 1/2 × (3x – 7) (x + 1)
16 = 3x(x + 1) -7(x + 1)
16 = 3x² + 3x – 7x – 7
16 + 7 = 3x² – 4x
3x² – 4x – 23 = 0
x = -b ± √b² – 4ac/2a
x = -(-4) ± √(-4)² – 4(3)(23)/2(3)
x = (4 ± √-260)/6
x = (4 ± √260i)/6
x = 4/6 ± 2√65i/6
x = 2/3 ± √65i/3
x = 0.66 + 2.68i
x = 0.66 – 2.68i
The solution is x = 0.66 + 2.68i and x = 0.66 – 2.68i

Question 61.
MODELING WITH MATHEMATICS
A lacrosse player throws a ball in the air from an initial height of 7 feet. The ball has an initial vertical velocity of 90 feet per second. Another player catches the ball when it is 3 feet above the ground. How long is the ball in the air?

Answer:

Question 62.
NUMBER SENSE
Suppose the quadratic equation ax² + 5x + c = 0 has one real solution. Is it possible for a and c to be integers? rational numbers? Explain your reasoning. Then describe the possible values of a and c.
Answer:

Question 63.
MODELING WITH MATHEMATICS
In a volleyball game, a player on one team spikes the ball over the net when the ball is 10 feet above the court. The spike drives the ball downward with an initial vertical velocity of 55 feet per second. How much time does the opposing team have to return the ball before it touches the court?
Answer:

Question 64.
MODELING WITH MATHEMATICS
An archer is shooting at targets. The height of the arrow is 5 feet above the ground. Due to safety rules, the archer must aim the arrow parallel to the ground.

a. How long does it take for the arrow to hit a target that is 3 feet above the ground?
b. What method did you use to solve the quadratic equation? Explain.
Answer:
An archer is shooting at targets.
The height of the arrow is 5 feet above the ground.
h = 1/2 gt²
h = 5 – 3 = 2
h = 2 ft
g = 32.2 ft/s²
2 = 32.2 × t²
4 – 32.2 t²
0.124 = t²
t = 0.35s
So, the time to hit the target is 0.35 seconds

Question 65.
PROBLEM SOLVING
A rocketry club is launching model rockets. The launching pad is 30 feet above the ground. Your model rocket has an initial vertical velocity of 105 feet per second. Your friend’s model rocket has an initial vertical velocity of 100 feet per second.
a. Use a graphing calculator to graph the equations of both model rockets. Compare the paths.
b. After how many seconds is your rocket 119 feet above the ground? Explain the reasonableness of your answer(s).
Answer:

Question 66.
PROBLEM SOLVING
The number A of tablet computers sold (in millions) can be modeled by the function A = 4.5t² + 43.5t + 17, where t represents the year after 2010.

a. In what year did the tablet computer sales reach 65 million?
b. Find the average rate of change from 2010 to 2012 and interpret the meaning in the context of the situation.
c. Do you think this model will be accurate after a new, innovative computer is developed? Explain.
Answer:
The number A of tablet computers sold (in millions) can be modeled by the function A = 4.5t² + 43.5t + 17, where t represents the year after 2010.
A = 4.5t² + 43.5t + 17
where,
A = 65
65 = 4.5t² + 43.5t + 17
t = 1 and t = -10.67
Therefore, the answer is 2011.

Question 67.
MODELING WITH MATHEMATICS
A gannet is a bird that feeds on fish by diving into the water. A gannet spots a fish on the surface of the water and dives 100 feet to catch it. The bird plunges toward the water with an initial vertical velocity of −88 feet per second.

a. How much time does the fish have to swim away?
b. Another gannet spots the same fish, and it is only 84 feet above the water and has an initial vertical velocity of −70 feet per second. Which bird will reach the fish first? Justify your answer.
Answer:

Question 68.
USING TOOLS
You are asked to find a possible pair of integer values for a and c so that the equation ax² − 3x + c = 0 has two real solutions. When you solve the inequality for the discriminant, you obtain ac < 2.25. So, you choose the values a = 2 and c = 1. Your graphing calculator displays the graph of your equation in a standard viewing window. Is your solution correct? Explain.

Answer:

Question 69.
PROBLEM SOLVING
Your family has a rectangular pool that measures 18 feet by 9 feet. Your family wants to put a deck around the pool but is not sure how wide to make the deck. Determine how wide the deck should be when the total area of the pool and deck is 400 square feet. What is the width of the deck?

Answer:

Question 70.
HOW DO YOU SEE IT?
The graph of a quadratic function y = ax² + bx + c is shown. Determine whether each discriminant of ax² + bx + c = 0 is positive, negative, or zero. Then state the number and type of solutions for each graph. Explain your reasoning.

Answer:
The graph shows the parabola with branches going down in a negative y-direction.
This gives you a < 0.
Since the parabola has no x-intercepts, then the equation ax² + bx + c = 0 has no solutions and the discriminant is less than 0.
d < 0

Question 71.
CRITICAL THINKING
Solve each absolute value equation.
a. |x² – 3x – 14| = 4
b. x² = |x| + 6
Answer:

Question 72.
MAKING AN ARGUMENT
The class is asked to solve the equation 4x² + 14x + 11 = 0. You decide to solve the equation by completing the square. Your friend decides to use the Quadratic Formula. Whose method is more efficient? Explain your reasoning.
Answer:
Given,
The class is asked to solve the equation 4x² + 14x + 11 = 0.
4x² + 14x + 11 = 0
x = -b ± √b² – 4ac/2a
x = -(14) ± √(14)² – 4(4)(11)/2(4)
a = 4, b = 14 and c = 11
x = (-14 ± √196 – 176)/8
x = (-14 ± √20)/8
x = -14/8 ± 2√5/8
x = -7/4 ± √5/4
x = -1.19
x = -2.3
The solution is x = -1.19 and x = -2.3

Question 73.
ABSTRACT REASONING
For a quadratic equation ax² + bx + c = 0 with two real solutions, show that the mean of the solutions is \(\frac{b}{2a}\). How is this fact related to the symmetry of the graph of y = ax² + bx + c?
Answer:

Question 74.
THOUGHT PROVOKING
Describe a real-life story that could be modeled by h = −16t² + v₀t + h₀ . Write the height model for your story and determine how long your object is in the air.
Answer:
h = −16t² + v₀t + h₀ .
Consider the real-life example as, if a stone is launched straight up from the top of a 48 meter tall building and it has an initial speed of 32 meter per second.
h = −16t² + 32t + 48
let h = 0
−16t² + 32t + 48 = 0
Take -16 as a common factor.
-16(t² + 2t – 3) = 0
t² + 2t – 3 = 0
t² – 3t + 1t – 3 = 0
t(t – 3) + 1(t – 3) = 0
(t-3) (t + 1) = 0
t – 3 = 0
t = 3
t + 1 = 0
t = -1
Thus the time the stone stays in the air is 3 seconds.

Question 75.
REASONING
Show there is no quadratic equation ax²+bx+c= 0 such that a, b, and c are real numbers and 3i and −2i are solutions.
Answer:

Question 76.
MODELING WITH MATHEMATICS
The Stratosphere Tower in Las Vegas is 921 feet tall and has a “needle” at its top that extends even higher into the air. A thrill ride called Big Shot catapults riders 160 feet up the needle and then lets them fall back to the launching pad.

a. The height h (in feet) of a rider on the Big Shot can be modeled by h = −16t² + v₀ t + 921, where t is the elapsed time (in seconds) after launch and v₀ is the initial vertical velocity (in feet per second). Find v0 using the fact that the maximum value of h is 921 + 160 = 1081 feet.
b. A brochure for the Big Shot states that the ride up the needle takes 2 seconds. Compare this time to the time given by the model h = −16t² + v₀ t + 921, where v0 is the value you found in part (a). Discuss the accuracy of the model.
Answer:

Maintaining Mathematical Proficiency

Solve the system of linear equations by graphing.
Question 77.
−x + 2y = 6
x + 4y = 24
Answer:

Question 78.
y = 2x − 1
y = x + 1
Answer:
Given two equations
y = 2x − 1
y = x + 1

The equation of the line meet at 2. It has one solution.
The graph intersects at point (2,3)

Question 79.
3x + y = 4
6x + 2y = −4
Answer:

Question 80.
y = −x + 2
−5x + 5y = 10
Answer:
Given two linear equations
y = −x + 2
−5x + 5y = 10

The graph intersects at point 2, so the solution is (0,2)

Graph the quadratic equation. Label the vertex and axis of symmetry.
Question 81.
y = −x² + 2x + 1
Answer:

Question 82.
y = 2x² − x + 3
Answer:
Given quadratic equation is y = 2x² − x + 3
2x² − x + 3
2x² + 2x – 3x + 3 = 0

Question 83.
y = 0.5x² + 2x + 5
Answer:

Question 84.
y = −3x² − 2
Answer:
Given equation is y = −3x² − 2

Lesson 3.5 Solving Nonlinear Systems

Essential Question How can you solve a nonlinear system of equations?

EXPLORATION 1

Solving Nonlinear Systems of Equations
Work with a partner. Match each system with its graph. Explain your reasoning. Then solve each system using the graph.

EXPLORATION 2

Solving Nonlinear Systems of Equations
Work with a partner. Look back at the nonlinear system in Exploration 1(f). Suppose you want a more accurate way to solve the system than using a graphical approach.
a. Show how you could use a numerical approach by creating a table. For instance, you might use a spreadsheet to solve the system.

b. Show how you could use an analytical approach. For instance, you might try solving the system by substitution or elimination.

Communicate Your Answer

Question 3.
How can you solve a nonlinear system of equations?
Answer: We can solve a nonlinear system of equations with two variables by graphing, substitution, and elimination.

• Identify the graph of each equation.
• Graph the first equation.
• Solve one of the equations for either variable.
• Substitute the expression from Step 2 into the other equation.
• Check whether the graph intersects or not.
• Identify the point of intersection.
• Solve the resulting equation.

Question 4.
Would you prefer to use a graphical, numerical, or analytical approach to solve the given nonlinear system of equations? Explain your reasoning.
Answer:
The numerical methods approximate solutions to equations when exact solutions can’t be determined by algebraic methods. Graphical methods are used for solving non-linear equations by seeing graphs we have an idea for solving systems. We can use the analytical method to find the nonlinear system of equations.

Solve the system using any method. Explain your choice of method.
Question 1.
y = −x² + 4
y = −4x + 8
Answer:

Question 2.
x² + 3x + y = 0
2x + y = 5
Answer:

Question 3.
2x² + 4x − y =−2
x² + y = 2
Answer:

Solve the system.
Question 4.
x² + y² = 16
y = −x + 4
Answer:

Question 5.
x² + y² = 4
y = x + 4
Answer:

Question 6.
x² + y² = 1
y = \(\frac{1}{2}\)x + \(\frac{1}{2}\)
Answer:

Solve the equation by graphing.
Question 7.
x² − 6x + 15 = −(x − 3)² + 6
Answer:

Question 8.
(x + 4)(x − 1) = −x² + 3x + 4
Answer:

Solving Nonlinear Systems 3.5 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Describe the possible solutions of a system consisting of two quadratic equations.
Answer: The possible number of solutions of a system of two quadratic equations is two, when they intersect in two points, one when they intersect in one point, or zero when they do not intersect.

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

Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, solve the system by graphing. Check your solution(s).
Question 3.
y = x + 2
y = 0.5(x + 2)²
Answer:

Question 4.
y = (x − 3)² + 5
y = 5
Answer:

Question 5.
y = \(\frac{1}{3}\)x + 2
y = −3x² − 5x − 4
Answer:

Question 6.
y = −3x² − 30x − 71
y = −3x − 17
Answer:

Question 7.
y = x² + 8x + 18
y = −2x² − 16x − 30
Answer:

Question 8.
y = −2x² − 9
y = −4x − 1
Answer:

Question 9.
y = (x − 2)²
y = −x² + 4x − 2
Answer:

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

In Exercises 11–14, solve the system of nonlinear equations using the graph.
Question 11.

Answer:

Question 12.

Answer:
Observe the two functions carefully.
The two graphs represent two parabolas: one opens up and the other opens down.
The functions do not intersect.
So the system has no solution.

Question 13.

Answer:

Question 14.

Answer:

In Exercises 15–24, solve the system by substitution.
Question 15.
y = x + 5
y = x² − x + 2
Answer:

Question 16.
x² + y² = 49
y = 7 − x
Answer:

Question 17.
x² + y² = 64
y = −8
Answer:

Question 18.
x = 3
−3x² + 4x − y = 8
Answer:

Question 19.
2x² + 4x − y = −3
−2x + y = −4
Answer:

Question 20.
2x − 3 = y + 5x²
y = −3x − 3
Answer:

Question 21.
y = x² − 1
−7 = −x² − y
Answer:

Question 22.
y + 16x − 22 = 4x²
4x² − 24x + 26 + y = 0
Answer:

Question 23.
x² + y² = 7
x + 3y = 21
Answer:

Question 24.
x² + y² = 5
−x + y = −1
Answer:

Question 25.
USING EQUATIONS
Which ordered pairs are solutions of the nonlinear system?
y = \(\frac{1}{2}\)x² − 5x + \(\frac{21}{2}\)
y = −\(\frac{1}{2}\)x + \(\frac{13}{2}\)
A. (1, 6)
B. (3, 0)
C. (8, 2.5)
D. (7, 0)
Answer:

Question 26.
USING EQUATIONS
How many solutions does the system have? Explain your reasoning.
y = 7x² − 11x + 9
y = −7x² + 5x − 3
A. 0
B. 1
C. 2
D. 4
Answer:

In Exercises 27–34, solve the system by elimination.
Question 27.
2x² − 3x −y =−5
−x + y = 5
Answer:

Question 28.
−3x² + 2x − 5 = y
−x + 2 = −y
Answer:

Question 29.
−3x² + y = −18x + 29
−3x² − y = 18x − 25
Answer:

Question 30.
y = −x² − 6x 10
y = 3x² + 18x + 22
Answer:

Question 31.
y + 2x = −14
−x² − y − 6x = 11
Answer:

Question 32.
y = x² + 4x + 7
−y = 4x + 7
Answer:
Given,
y = x² + 4x + 7
−y = 4x + 7
Add both the equations
y = x² + 4x + 7
−y = 4x + 7
0 = x² + 8x + 14
x² + 8x + 14 = 0
Use the quadratic function to find the solution of x.
x = -b ± √b² – 4ac/2a
a = 1, b = 8 and c = 14.
x = -(8) ± √(8)² – 4(1)(14)/2(1)
x = (-8 ± √64 – 56)/2
x = (-8 ± √8)/2
x = -8/2 ± √8/2
x = -4 ± √2
x = -2.58
x = -5.41
So, the solution is -2.58 and -5.41

Question 33.
y = −3x² − 30x − 76
y = 2x² + 20x + 44
Answer:

Question 34.
−10x² + y = −80x + 155
5x² + y = 40x − 85
Answer:
Given,
−10x² + y = −80x + 155
5x² + y = 40x − 85
–     –       –       +
-15x² = -120x + 240
-x² = -8x + 16
x² – 8x + 16 = 0
Use the quadratic function to find the solution of x.
x = -b ± √b² – 4ac/2a
a = 1, b = -8 and c = 16
x = -(8) ± √(8)² – 4(1)(16)/2(1)
x = (-8 ± √64 – 64)/2
x = (-8)/2
x = -4
So, the solution is x = -4

Question 35.
ERROR ANALYSIS
Describe and correct the error in using elimination to solve a system.

Answer:

Question 36.
NUMBER SENSE
The table shows the inputs and outputs of two quadratic equations. Identify the solution(s) of the system. Explain your reasoning.

Answer:

In Exercises 37–42, solve the system using any method. Explain your choice of method.
Question 37.
y = x² − 1
−y = 2x² + 1
Answer:

Question 38.
y = −4x² − 16x − 13
−3x² + y + 12x = 17
Answer:

Question 39.
−2x + 10 + y = \(\frac{1}{3}\)x²
y = 10
Answer:

Question 40.
y = 0.5x² − 10
y = −x² + 14
Answer:

Question 41.
y = −3(x − 4)² + 6
(x − 4)² + 2 − y = 0
Answer:

Question 42.
−x² + y² = 100
y = −x + 14
Answer:
Given,
−x² + y² = 100
y = −x + 14
x² + y² – 28x = 196

−x² + y² = 100
(-)x² + y² = 28x + 196
-2x² = 28x – 96
-x² = 14x – 48
x² + 14x – 48 = 0
Use the quadratic function to find the solution of x.
x = -b ± √b² – 4ac/2a
a = 1, b = 14 and c = -48
x = -(14) ± √(14)² – 4(1)(-48)/2(1)
x = (-14 ± √196 + 192)/2
x = (-14 ± √388)/2
x = -14/2 ± 2√97/2
x = -7 ± √97
The solution is x = -7 + √97 and x = -7 – √97

USING TOOLS In Exercises 43–48, solve the equation by graphing.
Question 43.
x² + 2x = −\(\frac{1}{2}\)x² + 2x
Answer:

Question 44.
2x² − 12x − 16 = −6x² + 60x − 144
Answer:

Question 45.
(x + 2)(x − 2) = −x² + 6x − 7
Answer:

Question 46.
−2x² − 16x − 25 = 6x² + 48x + 95
Answer:

Question 47.
(x − 2)² − 3 = (x + 3)(−x + 9) − 38
Answer:

Question 48.
(−x + 4)(x + 8) − 42 = (x + 3)(x + 1) − 1
Answer:

Question 49.
REASONING
A nonlinear system contains the equations of a constant function and a quadratic function. The system has one solution. Describe the relationship between the graphs.
Answer: There is only one solution to the system, the graph of the constant function intersects the graph of the quadratic function at the vertex.

Question 50.
PROBLEM SOLVING
The range (in miles) of a broadcast signal from a radio tower is bounded by a circle given by the equation x² + y²= 1620.

A straight highway can be modeled by the equation y = −\(\frac{1}{3}\)x + 30.
For what lengths of the highway are cars able to receive the broadcast signal?
Answer:

Question 51.
PROBLEM SOLVING
A car passes a parked police car and continues at a constant speed r. The police car begins accelerating at a constant rate when it is passed. The diagram indicates the distance d (in miles) the police car travels as a function of time t (in minutes) after being passed. Write and solve a system of equations to find how long it takes the police car to catch up to the other car.

Answer:

Question 52.
THOUGHT PROVOKING
Write a nonlinear system that has two different solutions with the same y-coordinate. Sketch a graph of your system. Then solve the system.
Answer:

Question 53.
OPEN-ENDED
Find three values for m so the system has no solution, one solution, and two solutions. Justify your answer using a graph.
3y = −x² + 8x − 7
y = mx + 3
Answer:

Question 54.
MAKING AN ARGUMENT
You and a friend solve the system shown and determine that x = 3 and x = −3. You use Equation 1 to obtain the solutions (3, -3), (3, −3), (−3, 3), and (−3, −3). Your friend uses Equation 2 to obtain the solutions (3, 3) and (−3, −3). Who is correct? Explain your reasoning.
x² + y² = 18 Equation 1
x − y = 0 Equation 2
Answer:

Question 55.
COMPARING METHODS
Describe two different ways you could solve the quadratic equation. Which way do you prefer? Explain your reasoning.
−2x² + 12x − 17 = 2x² − 16x + 31
Answer:

Question 56.
ANALYZING RELATIONSHIPS
Suppose the graph of a line that passes through the origin intersects the graph of a circle with its center at the origin. When you know one of the points of intersection, explain how you can find the other point of intersection without performing any calculations.
Answer:

Question 57.
WRITING
Describe the possible solutions of a system that contains (a) one quadratic equation and one equation of a circle, and (b) two equations of circles. Sketch graphs to justify your answers.
Answer:

Question 58.
HOW DO YOU SEE IT?
The graph of a nonlinear system is shown. Estimate the solution(s). Then describe the transformation of the graph of the linear function that results in a system with no solution.

Answer:

Question 59.
MODELING WITH MATHEMATICS
To be eligible for a parking pass on a college campus, a student must live at least 1 mile from the campus center.

a. Write equations that represent the circle and Oak Lane.
b. Solve the system that consists of the equations in part (a).
c. For what length of Oak Lane are students not eligible for a parking pass?
Answer:

Question 60.
CRITICAL THINKING
Solve the system of three equations shown.
x² + y² = 4
2y = x² − 2x + 4
y = −x + 2
Answer:

Maintaining Mathematical Proficiency

Solve the inequality. Graph the solution on a number line.
Question 61.
4x − 4 > 8
Answer:

Question 62.
−x + 7 ≤ 4 − 2x
Answer:

Question 63.
−3(x − 4) ≥ 24
Answer:

Write an inequality that represents the graph.
Question 64.

Answer:

Question 65.

Answer:

Question 66.

Answer:

Lesson 3.6 Quadratic Inequalities

Essential Question How can you solve a quadratic inequality?

EXPLORATION 1

Solving a Quadratic Inequality
Work with a partner. The graphing calculator screen shows the graph of f(x) = x² + 2x − 3.

Explain how you can use the graph to solve the inequality x² + 2x − 3 ≤ 0.
Then solve the inequality.

EXPLORATION 2

Solving Quadratic Inequalities
Work with a partner. Match each inequality with the graph of its related quadratic function. Then use the graph to solve the inequality.
a. x² − 3x + 2 > 0
b. x² − 4x + 3 ≤ 0
c. x² − 2x − 3 < 0
d. x² + x − 2 ≥ 0
e. x² − x − 2 < 0
f. x² − 4 > 0

Communicate Your Answer

Question 3.
How can you solve a quadratic inequality?
Answer:
In order to solve a quadratic inequality we have to try to find all possible values of the variables which will make the equality true.
Suppose we have an equation
x² – x – 6 > 0.
x² – 3x + 2x – 6 > 0
x(x – 3) + 2(x – 3) > 0
x = -2 and x = 3
Here you can try to find all the values of x for which the quadratic is greater than zero or positive.
The internals of the equation is (-∞, -2) or (3, ∞).

Question 4.
Explain how you can use the graph in Exploration 1 to solve each inequality. Then solve each inequality.
Answer:
In exploration 1 the equation of the graph is
x² + 2x – 3 ≤ 0.
Use the quadratic function to find the solution of the graph.
x = -b ± √b² – 4ac/2a
a = 1, b = 2, c = -3.
x = -2 ± √(2² – 4 × 1 × (-3))/2.
x = -2 ± √(6)/2
x = -2 ± 4/2
x = -3 or x = 1.
So, the inequality can be simplified into (x+3) (x-1) ≤ 0.
They’re one of the factors is positive and the other is negative, so the solution will be x = [-3, 1].

Monitoring Progress

Graph the inequality.
Question 1.
y ≥ x² + 2x − 8
Answer:

Question 2.
y ≤ 2x² −x − 1
Answer:

Question 3.
y > −x² + 2x + 4
Answer:

Question 4.
Graph the system of inequalities consisting of y ≤ −x² and y > x² − 3.
Answer:

Solve the inequality.
Question 5.
2x² + 3x ≤ 2
Answer:

Question 6.
−3x² − 4x + 1 < 0
Answer:

Question 7.
2x² + 2 > −5x
Answer:

Question 8.
WHAT IF?
In Example 6, the area must be at least 8500 square feet. Describe the possible lengths of the parking lot.
Answer:

Quadratic Inequalities 3.6 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Compare the graph of a quadratic inequality in one variable to the graph of a quadratic inequality in two variables.
Answer: The graph of a quadratic inequality in one variable is an interval on the real number line, whereas the graph of a quadratic inequality in two variables is a region in the coordinate plane.

Question 2.
WRITING
Explain how to solve x² + 6x − 8 < 0 using algebraic methods and using graphs.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, match the inequality with its graph. Explain your reasoning.
Question 3.
y ≤ x² + 4x + 3
Answer:

Question 4.
y > −x² + 4x − 3
Answer:

Question 5.
y < x² − 4x + 3
Answer:

Question 6.
y ≥ x² + 4x + 3
Answer:

In Exercises 7–14, graph the inequality.
Question 7.
y < −x²
Answer:

Question 8.
y ≥ 4x²
Answer:

Question 9.
y > x² − 9
Answer:

Question 10.
y < x² + 5
Answer:

Question 11.
y ≤ x² + 5x
Answer:

Question 12.
y ≥ −2x² + 9x − 4
Answer:
graph y≥-2x²+9x-4
X= – b/2a
X= – 9/-4 = 2.25
y = -2(2.25)² + 9(2.25) – 4
y = -2(5.0625)+20.25 – 4
y = -10.125 +20.25 – 4
y = 6.125 or 49/8
V(2.25,6.125)

Question 13.

y > 2(x + 3)² − 1
Answer:

Question 14.
y ≤ (x − \(\frac{1}{2}\))² + \(\frac{5}{2}\)
Answer:

ANALYZING RELATIONSHIPS In Exercises 15 and 16, use the graph to write an inequality in terms of f (x) so point P is a solution.
Question 15.

Answer:

Question 16.

Answer:

ERROR ANALYSIS In Exercises 17 and 18, describe and correct the error in graphing y ≥ x² + 2.
Question 17.

Answer:

Question 18.

Answer:

Question 19.
MODELING WITH MATHEMATICS
A hardwood shelf in a wooden bookcase can safely support a weight W (in pounds) provided W ≤ 115x², where x is the thickness (in inches) of the shelf. Graph the inequality and interpret the solution.
Answer:

Question 20.
MODELING WITH MATHEMATICS
A wire rope can safely support a weight W (in pounds) provided W ≤ 8000d², where d is the diameter (in inches) of the rope. Graph the inequality and interpret the solution.
Answer:

In Exercises 21–26, graph the system of quadratic inequalities.
Question 21.
y ≥ 2x²
y < −x² + 1
Answer:

Question 22.
y > −5x²
y > 3x² − 2
Answer:

Question 23.
y ≤ −x² + 4x − 4
y < x² + 2x − 8
Answer:

Question 24.
y ≥ x² − 4
y ≤ −2x² + 7x + 4
Answer:

Question 25.
y ≥ 2x² + x − 5
y < −x² + 5x + 10
Answer:

Question 26.
y ≥ x² − 3x − 6
y ≥ x² + 7x + 6
Answer:

In Exercises 27–34, solve the inequality algebraically.
Question 27.
4x² < 25
Answer:

Question 28.
x² + 10x + 9 < 0
Answer:
Given,
x² + 10x + 9 < 0
Use the factoring method to find the solution of the equation.
x² + 1x + 9x + 9 < 0
x(x + 1) + 9(x + 1) < 0
(x + 9) (x + 1) < 0
x = -9 or x = -1
The solution is x = -9 and x = -1.
-9 < 0 and -1 < 0

Question 29.
x² − 11x ≥ −28
Answer:

Question 30.
3x² − 13x > −10
Answer:
Given,
3x² − 13x > −10
3x² − 13x + 10 > 0
Use the factoring method to find the solution of the equation.
3x² – 3x – 10x + 10 >0
3x(x – 1) -1(x – 1) > 0
(3x – 1) >0 or (x – 1) >0
3x – 1 > 0
x > 1/3 or x > 1
The solution is x = 1/3 and x = 1.
1/3 > x > 1

Question 31.
2x² − 5x − 3 ≤ 0
Answer:

Question 32.
4x² + 8x − 21 ≥ 0
Answer:
Given,
4x² + 8x − 21 ≥ 0
Use the factoring method to find the solution of the equation.
2x(2x + 3) + 1(2x + 3)
(2x + 1)(2x + 3) ≥ 0
x = -1/2 or x = -3/2
-1/2 < 0 and -3/2 < 0

Question 33.
\(\frac{1}{2}\)x² − x > 4
Answer:

Question 34.
−\(\frac{1}{2}\)x² + 4x ≤ 1
Answer:
Given,
−\(\frac{1}{2}\)x² + 4x ≤ 1
−\(\frac{1}{2}\)x² + 4x – 1 ≤ 0
Use the quadratic function method to find the solution.
x = -b ± √b² – 4ac/2a
a = -1/2, b = 4 and c = -1
x = -(4) ± √(4)² – 4(-0.5)(-1)/2(-0.5)
x = (-14 ± √16 – 2)/-1
x = (-4 ± √14)/-1
x = 4 ± √14
The solution is x = 4 + √14 and x = 4 – √14

In Exercises 35–42, solve the inequality by graphing.
Question 35.
x² − 3x + 1 < 0
Answer:

Question 36.
x² − 4x + 2 > 0
Answer:

Question 37.
x² + 8x > −7
Answer:

Question 38.
x² + 6x < −3
Answer:

Question 39.
3x² − 8 ≤ − 2x
Answer:

Question 40.
3x² + 5x − 3 < 1
Answer:

Question 41.
\(\frac{1}{3}\)x² + 2x ≥ 2
Answer:

Question 42.
\(\frac{3}{4}\)x² + 4x ≥ 3
Answer:

Question 43.
DRAWING CONCLUSIONS
Consider the graph of the function f(x) = ax² + bx + c.

a. What are the solutions of ax² + bx + c < 0?
b. What are the solutions of ax² + bx + c > 0?
c. The graph of g represents a reflection in the x-axis of the graph of f. For which values of x is g(x) positive?
Answer:

Question 44.
MODELING WITH MATHEMATICS
A rectangular fountain display has a perimeter of 400 feet and an area of at least 9100 feet. Describe the possible widths of the fountain.

Answer:

Question 45.
MODELING WITH MATHEMATICS
The arch of the Sydney Harbor Bridge in Sydney, Australia, can be modeled by y = −0.00211x² + 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. For what distances x is the arch above the road?

Answer:

Question 46.
PROBLEM SOLVING
The number T of teams that have participated in a robot-building competition for high-school students over a recent period of time x(in years) can be modeled by T(x) = 17.155x² + 193.68x + 235.81, 0 ≤ x ≤ 6.
After how many years is the number of teams greater than 1000? Justify your answer.

Answer:

Question 47.
PROBLEM SOLVING
A study found that a driver’s reaction time A(x) to audio stimuli and his or her reaction time V(x) to visual stimuli (both in milliseconds) can be modeled by
A(x) = 0.0051x² − 0.319x + 15, 16 ≤ x ≤ 70
V(x) = 0.005x² − 0.23x + 22, 16 ≤ x ≤ 70
where x is the age (in years) of the driver.
a. Write an inequality that you can use to find the x-values for which A(x) is less than V(x).
b. Use a graphing calculator to solve the inequality A(x) < V(x). Describe how you used the domain 16 ≤ x ≤ 70 to determine a reasonable solution. c. Based on your results from parts (a) and (b), do you think a driver would react more quickly to a traffic light changing from green to yellow or to the siren of an approaching ambulance? Explain.
Answer:

Question 48.
HOW DO YOU SEE IT?
The graph shows a system of quadratic inequalities.

a. Identify two solutions of the system.
b. Are the points (1, −2) and (5, 6) solutions of the system? Explain.
c. Is it possible to change the inequality symbol(s) so that one, but not both of the points in part (b), is a solution of the system? Explain.
Answer:

Question 49.
MODELING WITH MATHEMATICS
The length L (in millimeters) of the larvae of the black porgy fish can be modeled by L(x) = 0.00170x² + 0.145x + 2.35, 0 ≤ x ≤ 40 where x is the age (in days) of the larvae. Write and solve an inequality to find at what ages a larva’s length tends to be greater than 10 millimeters. Explain how the given domain affects the solution.

Answer:

Question 50.
MAKING AN ARGUMENT
You claim the system of inequalities below, where a and b are real numbers, has no solution. Your friend claims the system will always have at least one solution. Who is correct? Explain.
y < (x + a)²
y < (x + b)²
Answer:

Question 51.
MATHEMATICAL CONNECTIONS
The area A of the region bounded by a parabola and a horizontal line can be modeled by A= \(\frac{2}{3}\)bh, where b and h are as defined in the diagram. Find the area of the region determined by each pair of inequalities.

Answer:

Question 52.
THOUGHT PROVOKING
Draw a company logo that is created by the intersection of two quadratic inequalities. Justify your answer.
Answer:

Question 53.
REASONING
A truck that is 11 feet tall and 7 feet wide is traveling under an arch. The arch can be modeled by y = −0.0625x² + 1.25x + 5.75, where x and y are measured in feet.

a. Will the truck fit under the arch? Explain.
b. What is the maximum width that a truck 11 feet tall can have and still make it under the arch?
c. What is the maximum height that a truck 7 feet wide can have and still make it under the arch?
Answer:

Maintaining Mathematical Proficiency

Graph the function. Label the x-intercept(s) and the y-intercept.
Question 54.
f(x) = (x + 7)(x − 9)
Answer:
Given
f(x) = (x + 7)(x − 9)
f(x) = 0
x + 7 = 0 or x – 9 = 0
x = -7 or x = 9

Question 55.
g(x) = (x − 2)² − 4
Answer:

Question 56.
h(x) = −x² + 5x − 6
Answer:
Given,
h(x) = −x² + 5x − 6
h(x) = 0
−x² + 5x − 6 = 0
x = -b ± √b² – 4ac/2a
a = -1, b = 5 and c = -6
x = -5 ± √5² – 4(-1)(-6)/2(-1)
x = -5 ± √25 – 24/-2
x = -5 ±-(1/2)
x = -5 – 1/2 = -5 1/2
x = -5 + 1/2 = -4 1/2

Find the minimum value or maximum value of the function. Then describe where the function is increasing and decreasing.
Question 57.
f(x) = −x² − 6x − 10
Answer:

Question 58.
h(x) = \(\frac{1}{2}\)(x + 2)² − 1
Answer:

Question 59.
f(x) = −(x − 3)(x + 7)
Answer:

Question 60.
h(x) = x² + 3x − 18
Answer:

Quadratic Equations and Complex Numbers Performance Task: Algebra in Genetics: The Hardy-Weinberg Law

3.4–3.6 What Did You Learn?

Core Vocabulary

Core Concepts
Section 3.4
Solving Equations Using the Quadratic Formula, p. 122
Analyzing the Discriminant of ax²+bx+c= 0, p. 124
Methods for Solving Quadratic Equations, p. 125
Modeling Launched Objects, p. 126

Section 3.5
Solving Systems of Nonlinear Equations, p. 132
Solving Equations by Graphing, p. 135

Section 3.6
Graphing a Quadratic Inequality in Two Variables, p. 140
Solving Quadratic Inequalities in One Variable, p. 142

Mathematical Practices
Question 1.
How can you use technology to determine whose rocket lands first in part (b) of Exercise 65 on page 129?
Answer:

Question 2.
What question can you ask to help the person avoid making the error in Exercise 54 on page 138?
Answer:

Question 3.
Explain your plan to find the possible widths of the fountain in Exercise 44 on page 145.
Answer:

Performance Task: Algebra in Genetics: The Hardy-Weinberg Law
Some people have attached earlobes, the recessive trait. Some people have free earlobes, the dominant trait. What percent of people carry both traits?
To explore the answers to this question and more, go to BigIdeasMath.com.

Quadratic Equations and Complex Numbers Chapter Review

3.1 Solving Quadratic Equations (pp. 93–102)

Question 1.
Solve x² − 2x − 8 = 0 by graphing.
Answer:

Solve the equation using square roots or by factoring.
Question 2.
3x² − 4 = 8
Answer:

Question 3.
x² + 6x − 16 = 0
Answer:

Question 4.
2x² − 17x = −30
Answer:

Question 5.
A rectangular enclosure at the zoo is 35 feet long by 18 feet wide. The zoo wants to double the area of the enclosure by adding the same distance x to the length and width. Write and solve an equation to find the value of x. What are the dimensions of the enclosure?
Answer:

3.2 Complex Numbers (pp. 103–110)

Question 6.
Find the values x and y that satisfy the equation 36 − yi = 4x + 3i.
Answer:

Perform the operation. Write the answer in standard form.
Question 7.
(−2 + 3i ) + (7 − 6i )
Answer:

Question 8.
(9 + 3i ) − (−2 − 7i )
Answer:

Question 9.
(5 + 6i )(−4 + 7i )
Answer:

Question 10.
Solve 7x² + 21 = 0.
Answer:

Question 11.
Find the zeros of f(x) = 2x² + 32.
Answer:

3.3 Completing the Square (pp. 111–118)

Question 12.
An employee at a local stadium is launching T-shirts from a T-shirt cannon into the crowd during an intermission of a football game. The height h (in feet) of the T-shirt after t seconds can be modeled by h = −16t² + 96t + 4. Find the maximum height of the T-shirt.
Answer:
Given,
h = −16t² + 96t + 4.
t = 3
h(t) = −16t² + 96t + 4
h(t) = 0
−16t² + 96t + 4 = 0
Put t = 3 in the equation.
h = -16(3)² + 96 (3) + 4
h = 148 ft
Thus the height of the T-shirt is 148 ft.

Solve the equation by completing the square.
Question 13.
x² + 16x + 17 = 0
Answer:

Question 14.
4x² + 16x + 25 = 0
Answer:

Question 15.
9x(x − 6) = 81
Answer:

Question 16.
Write y = x² − 2x + 20 in vertex form. Then identify the vertex.
Answer:

3.4 Using the Quadratic Formula (pp. 121–130)

Solve the equation using the Quadratic Formula.
Question 17.
−x² + 5x = 2
Answer:
Given,
−x² + 5x = 2
x² – 5x + 2 = 0
Use the quadratic formula to solve the equation.
x = -b ± √b² – 4ac/2a
x = -(-5) ± √(-5)² – 4(1)(2)/2(1)
x = (5 ± √25 – 8)/2
x = (5± √17)/2
x = 5/2 ± √17/2
x = 4.56
x = 0.43
The solution is x = 4.56 and 0.43

Question 18.
2x² + 5x = 3
Answer:
Given,
2x² + 5x = 3
2x² + 5x – 3 = 0
Use the quadratic formula to solve the equation.
x = -b ± √b² – 4ac/2a
x = -(5) ± √(5)² – 4(2)(-3)/2(2)
x = (-5 ± √25 +24)/4
x = (-5± √49)/4
x = 2/4 ± 12/4
x = 1/2 ± 3
The solution is x = 1/2 + 3 = 3 1/2 and x = 1/2 – 3 = -2 1/2

Question 19.
3x² − 12x + 13 = 0
Answer:

Find the discriminant of the quadratic equation and describe the number and type of solutions of the equation.
Question 20.
−x² − 6x − 9 = 0
Answer:

Question 21.
x² − 2x − 9 = 0
Answer:
Given,
x² − 2x − 9 = 0
Use the quadratic formula to solve the equation.
x = -b ± √b² – 4ac/2a
a = 1, b = -2 and c = -9
x = -(-2) ± √(-2)² – 4(1)(-9)/2(1)
x = (2 ± √4 +36)/2
x = (2 ± √40)/2
x = 1 ± √10
The solution is x = 1 + √10 and x = 1 – √10

Question 22.
x² + 6x + 5 = 0
Answer:

3.5 Solving Nonlinear Systems (pp. 131–138)

Solve the system by any method. Explain your choice of method.
Question 23.
2x² − 2 = y
−2x + 2 = y
Answer:

Question 24.
x² − 6x + 13 = y
−y = −2x + 3
Answer:

Question 25.
x² + y² = 4
−15x + 5 = 5y
Answer:

Question 26.
Solve −3x² + 5x − 1 = 5x² − 8x − 3 by graphing.
Answer:

3.6 Quadratic Inequalities (pp. 139–146)

Graph the inequality.
Question 27.
y > x² + 8x + 16
Answer:

Question 28.
y ≥ x² + 6x + 8
Answer:

Question 29.
x² + y ≤ 7x − 12
Answer:

Graph the system of quadratic inequalities.
Question 30.
x² − 4x + 8 > y
−x² + 4x + 2 ≤ y
Answer:

Question 31.
2x² − x ≥ y − 5
0.5x²> y − 2x− 1
Answer:

Question 32.
−3x² − 2x ≤ y + 1
−2x² + x − 5 > −y
Answer:

Solve the inequality.
Question 33.
3x² + 3x − 60 ≥0
Answer:

Question 34.
−x² − 10x < 21
Answer:

Question 35.
3x² + 2 ≤ 5x
Answer:

Quadratic Equations and Complex Numbers Chapter Test

Solve the equation using any method. Provide a reason for your choice.
Question 1.
0 = x² + 2x + 3
Answer:
Given equation is
0 = x² + 2x + 3
x² + 2x + 3 = 0
Use the quadratic function method to solve the equation
x = -b ± √b² – 4ac/2a
a = 1, b = 2 and c = 3
x = -(2) ± √(2)² – 4(1)(3)/2(1)
x = (-2 ± √4 – 12)/2
x = (-2± √-8)/2
x = (-2± 2√2i)/2
x = -2/2 ± 2√2i/2
x = -1 ± √2i
The solution is x = -1 + √2i or x = -1 – √2i

Question 2.
6x = x² + 7
Answer:
Given equation is
6x = x² + 7
x² – 6x + 7 = 0
Use the quadratic function method to solve the equation
x = -b ± √b² – 4ac/2a
a = 1, b = -6 and c = 7
x = -(-6) ± √(-6)² – 4(1)(7)/2(1)
x = (6 ± √36 – 28)/2
x = (6± √8)/2
x = (6± 2√2)/2
x = 6/2 ± 2√2/2
x = 3 ± √2
The solution is x = 3 + √2 or 3 – √2

Question 3.
x² + 49 = 85
Answer:
Given equation is
x² + 49 = 85
x² + 49 – 85 = 0
x² – 36 = 0
x² = 36
x = 6
The solution is x = 6

Question 4.
(x + 4)(x − 1) = −x² + 3x + 4
Answer:

Explain how to use the graph to find the number and type of solutions of the quadratic equation. Justify your answer by using the discriminant.
Question 5.
\(\frac{1}{2}\)x² + 3x + \(\frac{9}{2}\) = 0

Answer:

Question 6.
4x² + 16x + 18 = 0

Answer:

Question 7.
−x² + \(\frac{1}{2}\)x + \(\frac{3}{2}\) = 0

Answer:

Solve the system of equations or inequalities.
Question 8.
x² + 66 = 16x − y
2x − y = 18
Answer:

Question 9.
y ≥ \(\frac{1}{4}\)x² − 2
y < −(x + 3)2x − y = 18 + 4
Answer:

Question 10.
0 = x² + y² − 40
y = x + 4
Answer:

Question 11.
Write (3 + 4i )(4 − 6i ) as a complex number in standard form.
Answer:

Question 12.
The aspect ratio of a widescreen TV is the ratio of the screen’s width to its height, or 16 : 9. What are the width and the height of a 32-inch widescreen TV? Justify your answer. (Hint: Use the Pythagorean Theorem and the fact that TV sizes refer to the diagonal length of the screen.)

Answer:

Question 13.
The shape of the Gateway Arch in St. Louis, Missouri, can be modeled by y = −0.0063x² + 4x, where x is the distance (in feet) from the left foot of the arch and y is the height (in feet) of the arch above the ground. For what distances x is the arch more than 200 feet above the ground? Justify your answer.
Answer:

Question 14.
You are playing a game of horseshoes. One of your tosses is modeled in the diagram, where x is the horseshoe’s horizontal position (in feet) and y is the corresponding height (in feet). Find the maximum height of the horseshoe. Then find the distance the horseshoe travels. Justify your answer.

Answer:

Quadratic Equations and Complex Numbers Cumulative Assessment

Question 1.
The graph of which inequality is shown?

A. y² > x² + x – 6
B. y ≥ x² + x – 6
C. y > x² – x – 6
D. y ≥ x² – x – 6
Answer:

Question 2.
Classify each function by its function family. Then describe the transformation of the parent function.
a. g(x) = x + 4
b. h(x) = 5
c. h(x) = x² − 7
d. g(x) = −∣x + 3∣− 9
e. g(x) = \(\frac{1}{4}\)(x − 2)² − 1
f. h(x) = 6x+ 11
Answer:

Question 3.
Two baseball players hit back-to-back home runs. The path of each home run is modeled by the parabolas below, where x is the horizontal distance (in feet) from home plate and y is the height (in feet) above the ground. Choose the correct symbol for each inequality to model the possible locations of the top of the outfield wall.(HSA-CED.A.3)

Answer:

Question 4.
You claim it is possible to make a function from the given values that has an axis of symmetry at x = 2. Your friend claims it is possible to make a function that has an axis of symmetry at x = −2. What values can you use to support your claim? What values support your friend’s claim?(HSF-IF.B.4)

Answer:

Question 5.
Which of the following values are x-coordinates of the solutions of the system?
y = x² – 6x + 14
y = 2x + 7

Answer:

Question 6.
The table shows the altitudes of a hang glider that descends at a constant rate. How long will it take for the hang glider to descend to an altitude of 100 feet? Justify your answer.

A. 25 seconds
B. 35 seconds
C. 45 seconds
D. 55 seconds
Answer:

Question 7.
Use the numbers and symbols to write the expression x² + 16 as the product of two binomials. Some may be used more than once. Justify your answer.

Answer:

Question 8.
Choose values for the constants h and k in the equation x = \(\frac{1}{4}\)( y − k)² + h so that each statement is true.(HSA-CED.A.2)

Answer:

Question 9.
Which of the following graphs represents a perfect square trinomial? Write each function in vertex form by completing the square.

Answer: