Engage NY Eureka Math Precalculus Module 3 Lesson 2 Answer Key
Eureka Math Precalculus Module 3 Lesson 2 Example Answer Key
Example: Find the Square Roots of 119 + 120i
Let w = p + qi be the square root of 119 + 120i. Then
w2 = 119 + 120i
and
(p + qi)2 = 119 + 120i.
a. Expand the left side of this equation.
Answer:
p2 – q2 + 2pqi = 119 + 120i
b. Equate the real and imaginary parts, and solve for p and q.
Answer:
p2 – q2 = 119 and 2pq = 120. Solving for q and substituting gives
p2 – (\(\frac{60}{p}\))2 = 119.
Multiplying by p2 gives the equation
p4 – 119p2 – 3600 = 0.
And this equation can be solved by factoring.
(p2 + 25)(p2 – 144) = 0
We now have two polynomial expressions that we know how to factor: the sum and difference of squares.
(p + 5i)(p – 5i)(p – 12)(p + 12) = 0
The solutions are 5i, – 5i, 12, and – 12. Since p must be a real number by the definition of complex number, we can choose 12 or – 12 for p. Using the equation 2pq = 120, when p = 12, q = 5, and when p = – 12, q = – 5.
c. What are the square roots of 119 + 120i?
Answer:
Thus, the square roots of 119 + 120i are 12 + 5i and – 12 – 5i.
Eureka Math Precalculus Module 3 Lesson 2 Exercise Answer Key
Exercises 1–6
Exercise 1.
Use the geometric effect of complex multiplication to describe how to calculate a square root of z = 119 + 120i.
Answer:
The square root of this number would have a modulus equal to the square root of |119 + 120i| and an argument equal to \(\frac{\arg (119 + 120 i)}{2}\).
Exercise 2.
Calculate an estimate of a square root of 119 + 120i.
Answer:
|z| = \(\sqrt{119^{2} + 120^{2}}\) = 169
arg(z) = arctan(\(\frac{120}{119}\))
The square root’s modulus is \(\sqrt{169}\) or 13, and the square root’s argument is \(\frac{1}{2}\) arctan(\(\frac{120}{119}\)). The square root is close to 13(cos(22.6°) + i sin(22.6°) ) or 12 + 4.99i.
Exercise 3.
Every real number has two square roots. Explain why.
Answer:
The fundamental theorem of algebra guarantees that a second – degree polynomial equation has two solutions. To find the square roots of a real number b, we need to solve the equation z2 = b, which is a second – degree polynomial equation. Thus, the two solutions of z2 = b are the two square roots of b. If a is one of the square roots, then –a is the other.
Exercise 4.
Provide a convincing argument that every complex number must also have two square roots.
Answer:
By the same reasoning, if w is a complex number, then the polynomial equation z2 = w has two solutions. The two solutions to this quadratic equation are the square roots of w. If a + bi is one square root, then a – bi is the other.
Exercise 5.
Explain how the polynomial identity x2 – b = (x – \(\sqrt{b}\))(x + \(\sqrt{b}\)) relates to the argument that every number has two square roots.
Answer:
To solve x2 = b, we can solve x2 – b = 0. Since this quadratic equation has two distinct solutions, we can find two square roots of b. The two square roots are opposites of each other.
Exercise 6.
What is the other square root of 119 + 120i?
ans;
It would be the opposite of 12 + 5i, which is the complex number, – 12 – 5i.
Exercises 7–9
Exercise 7.
Use the method in the Example to find the square roots of 1 + \(\sqrt{3}\) i.
Answer:
p2 – q2 = 1 and 2pq = \(\sqrt{3}\)
Substituting and solving for p,
p2 – (\(\frac{\sqrt{3}}{2 p}\))2 = 1
4p4 – 3 = 4p2
4p4 – 4p2 – 3 = 0
(2p2 – 3)(2p2 + 1) = 0
gives the real solutions p = \(\frac{\sqrt{3}}{2}\) or p = – \(\frac{\sqrt{3}}{2}\). The values of q would then be
q = \(\frac{\sqrt{3}}{2 \sqrt{\frac{3}{2}}}\) = \(\frac{\sqrt{2}}{2}\) and q = – \(\frac{\sqrt{2}}{2}\).
The square roots of 1 + \(\sqrt{3}\) i are \(\frac{\sqrt{6}}{2}\) + \(\frac{\sqrt{2}}{2}\) i and – \(\frac{\sqrt{6}}{2}\) – \(\frac{\sqrt{2}}{2}\) i.
Exercise 8.
Find the square roots of each complex number.
a. 5 + 12i
Answer:
The square roots of 5 + 12i satisfy the equation (p + qi)2 = 5 + 12i.
Expanding (p + qi)2 and equating the real and imaginary parts gives
p2 – q2 = 5
2pq = 12.”
Substituting q = \(\frac{6}{p}\) into p2 – q2 = 5 gives
p2 – (\(\frac{6}{p}\))2 = 5
p4 – 36 = 5p2
p4 – 5p2 – 36 = 0
(p2 – 9)(p2 + 4) = 0.
The one positive real solution to this equation is 3. Let p = 3. Then q = 2. A square root of 5 + 12i is
3 + 2i.
The other square root is when p = – 3 and q = – 2. Therefore, – 3 – 2i is the other square root of 5 + 12i.
b. 5 – 12i
Answer:
The square roots of 5 – 12i satisfy the equation (p + qi)2 = 5 – 12i.
Expanding (p + qi)2 and equating the real and imaginary parts gives
p2 – q2 = 5
2pq = – 12.
Substituting q = – \(\frac{6}{p}\) into p2 – q2 = 5 gives
p2 – ( – \(\frac{6}{p}\))2 = 5
p4 – 36 = 5p2
p4 – 5p2 – 36 = 0
(p2 – 9)(p2 + 4) = 0.
The one positive real solution to this equation is 3. Let p = 3. Then q = – 2. A square root of 5 – 12i is
3 – 2i.
The other square root is when p = – 3 and q = 2. Therefore, – 3 + 2i is the other square root of 5 – 12i.
Exercise 9.
Show that if p + qi is a square root of z = a + bi, then p – qi is a square root of the conjugate of z, z ̅ = a – bi.
a. Explain why (p + qi)2 = a + bi.
Answer:
If p + qi is a square root of a + bi, then it must satisfy the definition of a square root. The square root of a number raised to the second power should equal the number.
b. What do a and b equal in terms of p and q?
Answer:
Expanding (p + qi)2 = p2 – q2 + 2pqi. Thus, a = p2 – q2 and b = 2pq.
c. Calculate (p – qi)2. What is the real part, and what is the imaginary part?
Answer:
(p – qi)2 = p2 – q2 – 2pqi The real part is p2 – q2, and the imaginary part is – 2pq.
d. Explain why (p – qi)2 = a – bi.
Answer:
From part (c), p2 – q2 = a and 2pq = b. Substituting,
(p – qi)2 = p2 – q2 – 2pqi = a – bi.
Eureka Math Precalculus Module 3 Lesson 2 Problem Set Answer Key
Find the two square roots of each complex number by creating and solving polynomial equations.
Question 1.
z = 15 – 8i
Answer:
(p + qi)2 = 15 – 8i, p2 – q2 + 2pqi = 15 – 8i,
p2 – q2 = 15, pq = – 4, q = – \(\frac{4}{p}\)
p2 – \(\frac{16}{p^{2}}\) – 15 = 0, p4 – 15p2 – 16 = 0, (p2 + 1)(p2 – 16) = 0, p = ±4,
p = 4, q = – 1; p = – 4, q = 1
Therefore, the square roots of 15 – 8i are 4 – i and – 4 + i.
Question 2.
z = 8 – 6i
Answer:
(p + qi)2 = 8 – 6i, p2 – q2 + 2pqi = 8 – 6i,
p2 – q2 = 8, pq = – 3, q = – \(\frac{3}{p}\)
p2 – \(\frac{9}{p^{2}}\) – 8 = 0, p4 – 8p2 – 9 = 0, (p2 + 1)(p2 – 9) = 0, p = ±3,
p = 3, q = – 1; p = – 3, q = 1
Therefore, the square roots of 8 – 6i are 3 – i and – 3 + i.
Question 3.
z = – 3 + 4i
Answer:
(p + qi)2 = – 3 + 4i, p2 – q2 + 2pqi = – 3 + 4i,
p2 – q2 = – 3, pq = 2, q = \(\frac{2}{p}\)
p2 – \(\frac{4}{p^{2}}\) + 3 = 0, p4 + 3p2 – 4 = 0, (p2 + 4)(p2 – 1) = 0, p = ±1,
p = 1, q = 2; p = – 1, q = – 2
Therefore, the square roots of – 3 + 4i are 1 + 2i and – 1 – 2i.
Question 4.
z = – 5 – 12i
Answer:
(p + qi)2 = – 5 – 12i, p2 – q2 + 2pqi = – 5 – 12i,
p2 – q2 = – 5, pq = – 6, q = – \(\frac{6}{p}\)
p2 – \(\frac{6}{p^{2}}\) + 5 = 0, p4 + 5p2 – 36 = 0, (p2 + 9)(p2 – 4) = 0, p = ±2,
p = 2, q = – 3; p = – 2, q = 3
Therefore, the square roots of – 5 – 12i are 2 – 3i and – 2 + 3i.
Question 5.
z = 21 – 20i
Answer:
(p + qi)2 = 21 – 20i, p2 – q2 + 2pqi = 21 – 20i,
p2 – q2 = 21, pq = – 10, q = – \(\frac{10}{p}\)
p2 – \(\frac{100}{p^{2}}\) – 21 = 0, p4 – 21p2 – 100 = 0, (p2 + 4)(p2 – 25) = 0, p = ±5,
p = 5, q = – 2; p = – 5, q = 2
Therefore, the square roots of 21 – 20i are 5 – 2i and – 5 + 2i.
Question 6.
z = 16 – 30i
Answer:
(p + qi)2 = 16 – 30i, p2 – q2 + 2pqi = 16 – 30i,
p2 – q2 = 16, pq = – 15, q = – \(\frac{15}{p}\)
p2 – \(\frac{225}{p^{2}}\) – 16 = 0, p4 – 16p2 – 225 = 0, (p2 + 9)(p2 – 25) = 0, p = ±5,
p = 5, q = – 3; p = – 5, q = 3
Therefore, the square roots of 16 – 30i are 5 – 3i and – 5 + 3i.
Question 7.
z = i
Answer:
(p + qi)2 = 0 + i, p2 – q2 + 2pqi = 0 + i,
p2 – q2 = 0, pq = \(\frac{1}{2}\), q = \(\frac{1}{2}\)p
p2 – \(\frac{1}{4p^{2}}\) = 0, 4p4 – 1 = 0, (2p2 + 1)(2p2 – 1) = 0, p = ±\(\frac{\sqrt{2}}{2}\)
p = \(\frac{\sqrt{2}}{2}\), q = \(\frac{\sqrt{2}}{2}\); p = – \(\frac{\sqrt{2}}{2}\), q = – \(\frac{\sqrt{2}}{2}\)
Therefore, the square roots of i are \(\frac{\sqrt{2}}{2}\) + \(\frac{\sqrt{2}}{2}\)i and – \(\frac{\sqrt{2}}{2}\) – \(\frac{\sqrt{2}}{2}\)i.
A Pythagorean triple is a set of three positive integers a, b, and c such that a2 + b2 = c2. Thus, these integers can be the lengths of the sides of a right triangle.
Question 8.
Show algebraically that for positive integers p and q, if
a = p2 – q2
b = 2pq
c = p2 + q2
then a2 + b2 = c2.
Answer:
a2 + b2 = (p2 – q2 )2 + (2pq)2
= p4 – 2p2 q2 + q4 + 4p2 q2
= p4 + 2p2 q2 + q4
= p4 + 2p2 q2 + q4
= c2
Question 9.
Select two integers p and q, use the formulas in Problem 8 to find a, b, and c, and then show those numbers satisfy the equation a2 + b2 = c2.
Answer:
Let p = 3 and q = 2. Calculate the values of a, b, and c.
a = 32 – 22 = 5
b = 2(3)(2) = 12
c = 32 + 22 = 13
a2 + b2 = 52 + 122 = 25 + 144 = 169 = 132 = c2
Question 10.
Use the formulas from Problem 8, and find values for p and q that give the following famous triples.
a. (3, 4, 5)
Answer:
a = p2 – q2 = 3, b = 2pq = 4, c = p2 + q2 = 5
2p2 = 8, p = 2, q = 1
(p2 – q2 )2 + (2pq)2 = (p2 – q2 )2
(22 – 12 )2 + (2(2)(1))2 = (22 + 12 )2
(4 – 1)2 + 42 = (4 + 1)2, (3)2 + (4)2 = (5)2
b. (5, 12, 13)
Answer:
a = p2 – q2 = 5, b = 2pq = 12, c = p2 + q2 = 13
2p2 = 18, p = 3, q = 2
(p2 – q2 )2 + (2pq)2 = (p2 – q2 )2
(32 – 22 )2 + (2(3)(2))2 = (32 + 22 )2
(9 – 4)2 + 122 = (9 + 4)2, (5)2 + (12)2 = (13)2
c. (7, 24, 25)
Answer:
a = p2 – q2 = 7, b = 2pq = 24, c = p2 + q2 = 25,
2p2 = 32, p = 4, q = 3
(p2 – q2 )2 + (2pq)2 = (p2 – q2 )2
(42 – 32 )2 + (2(4)(3))2 = (42 + 32 )2
(16 – 9)2 + 242 = (16 + 9)2, (7)2 + (24)2 = (25)2
d. (9, 40, 41)
Answer:
a = p2 – q2 = 9, b = 2pq = 40, c = p2 + q2 = 41,
2p2 = 50, p = 5, q = 4
(p2 – q2 )2 + (2pq)2 = (p2 – q2 )2
(52 – 42 )2 + (2(5)(4))2 = (52 + 42 )2
(25 – 16)2 + 402 = (25 + 16)2, (9)2 + (40)2 = (41)2
Question 11.
Is it possible to write the Pythagorean triple (6, 8, 10) in the form a = p2 – q2, b = 2pq, c = p2 + q2 for some integers p and q? Verify your answer.
Answer:
a = p2 – q2 = 6, b = 2pq = 8, c = p2 + q2 = 10,
2p2 = 16, p2 = 8, p = ±2√2, q = ±√2
The Pythagorean triple (6, 8, 10) cannot be written in the form a = p2 – q2, b = 2pq, c = p2 + q2, for any integers p and q.
Question 12.
Choose your favorite Pythagorean triple (a, b, c) that has a and b sharing only 1 as a common factor, for example (3, 4, 5), (5, 12, 13), or (7, 24, 25), … Find the square of the length of a square root of a + bi; that is, find
|p + qi|2, where p + qi is a square root of a + bi. What do you observe?
Answer:
For (3, 4, 5), a = 3, b = 4, a + bi = (p + qi)2 = ( – p – qi)2,
a + bi = 3 + 4i = (p + qi)2 = p2 – q2 + 2pqi; therefore, p2 – q2 = 3, 2pq = 4, q = \(\frac{2}{p}\)
p2 – 4/p2 = 3, p^4 – 3p2 – 4 = 0, (p2 + 1)(p2 – 4) = 0, p = ±2
p = 2, q = 1; therefore, p + qi = 2 + i.
p = – 2, q = – 1; therefore, p + qi = – 2 – i.
The two square roots of a + bi are 2 + i and –2 – i. Both of these have length √5, so the squared length is 5. This is the third value c in the Pythagorean triple (a, b, c).
Eureka Math Precalculus Module 3 Lesson 2 Exit Ticket Answer Key
Question 1.
Find the two square roots of 5 – 12i.
Answer:
(p + qi)2 = 5 – 12i, p2 – q2 + 2pqi = 5 – 12i,
p2 – q2 = 5, pq = – 6, q = – \(\frac{6}{p}\)
p2 – \(\frac{36}{p^{2}}\) – 5 = 0, p4 – 5p2 – 36 = 0, (p2 + 4)(p2 – 9) = 0, p = ±3,
p = 3, q = – 2; p = – 3, q = 2
Therefore, the square roots are 3 – 2i and – 3 + 2i.
Question 2.
Find the two square roots of 3 – 4i.
Answer:
Let the square roots have the form p + qi. Then (p + qi)2 = 3 – 4i. This gives p2 – q2 = 3 and 2pq = – 4. Then p = – \(\frac{2}{p}\), so p2 – q2 = \(\frac{4}{q^{2}}\) – q2 = 3, and q4 + 3q2 – 4 = 0. This expression factors into
(q + 1)(q – 1)(q + 2i)(q – 2i) = 0, and we see that the only real solutions are q = 1 and q = – 1. If q = 1, then p = – 2, and if q = – 1, then p = 2. Thus, the two square roots of 3 – 4i are – 2 + i and 2 – i.