## Engage NY Eureka Math Precalculus Module 1 Lesson 4 Answer Key

### Eureka Math Precalculus Module 1 Lesson 4 Exercise Answer Key

Opening Exercise

Is R(x) = \(\frac{1}{x}\) a linear transformation? Explain how you know.

Answer:

R(2+3) = R(5) = \(\frac{1}{5}\), but R(2)+R(3) = \(\frac{1}{2}\)+\(\frac{1}{3}\) = \(\frac{5}{6}\), which is not the same as \(\frac{1}{5}\). This means that the reciprocal function does not preserve addition, and so it is not a linear transformation.

Exercises

Exercise 1.

Solve 5x^{2} – 3x + 17 = 9.

Answer:

5x^{2} – 3x+17 = 9

5x^{2} – 3x = 9 – 17 = – 8

20(5x^{2} – 3x) = 20( – 8)

100x^{2} – 60x = – 160

100x^{2} – 60x+9 = – 160+9

(10x – 3)^{2} = – 151

x = \(\frac{3 \pm i \sqrt{151}}{10}\) = \(\frac{3}{10}\) ± i\(\frac{\sqrt{151}}{10}\)

Exercise 2.

Use the fact that i^{2} = – 1 to show that i^{3} = – i. Interpret this statement geometrically.

Answer:

We have i^{3} = i^{2}∙i = ( – 1)∙i = – i. Multiplying by i rotates a point through 90 degrees, and multiplying by – 1 rotates it 180 degrees farther. This makes sense with our earlier conjecture that multiplying by i^{3} would induce a 270 – degree rotation.

Exercise 3.

Calculate i^{6}

Answer:

i^{6} = i^{2}∙i^{2}∙i^{2} = ( – 1)( – 1)( – 1) = (1)( – 1) = – 1

Exercise 4.

Calculate i^{5}

Answer:

i^{5} = i^{2}∙i^{2}∙i = ( – 1)( – 1)(i) = (1)(i) = i

### Eureka Math Precalculus Module 1 Lesson 4 Problem Set Answer Key

Question 1.

Solve the equation below.

5x^{2} – 7x+8 = 2

Answer:

5x^{2} – 7x = – 6

100x^{2} – 140x = – 120

100x^{2} – 140x+49 = – 120+49

(10x – 7)^{2} = – 71

x = \(\frac{7 \pm i \sqrt{71}}{10}\) = \(\frac{7}{10}\) ±i\(\frac{\sqrt{71}}{10}\)

Question 2.

Consider the equation x^{3} = 8.

a. What is the first solution that comes to mind?

Answer:

It is easy to see that 2^{3} = 8, so 2 is a solution.

b. It may not be easy to tell at first, but this equation actually has three solutions. To find all three solutions, it is helpful to consider x^{3} – 8 = 0, which can be rewritten as (x – 2)(x^{2}+2x+4) = 0 (check this for yourself). Find all of the solutions to this equation.

Answer:

x^{2}+2x+4 = 0

x^{2}+2x = – 4

x^{2}+2x+1 = – 4+1

(x+1)^{2} = – 3

x = – 1±i\(\sqrt{3}\)

The solutions to x^{3} – 8 = 0 are 2, – 1+i\(\sqrt{3}\), and – 1 – i\(\sqrt{3}\).

Question 3.

Make a drawing that shows the first 5 powers of i (i.e., i^{1},i^{2},…,i^{5}), and then confirm your results algebraically.

Answer:

i^{1} = i

i^{2} = – 1

i^{3} = i^{2}∙i = – 1∙i = – i

i^{4} = i^{2}∙i^{2} = – 1∙ – 1 = 1

i^{5} = i^{4}∙i = 1∙i = i

Question 4.

What is the value of i^{99}? Explain your answer using words or drawings.

Answer:

Multiplying by i four times is equivalent to rotating through 4∙90 = 360 degrees, which is a complete rotation. Since 99 = 4∙24+3, multiplying by i for 99 times is equivalent to performing 24 complete rotations, followed by three 90 – degree rotations. Thus, i^{99} = – i.

Question 5.

What is the geometric effect of multiplying a number by – i? Does your answer make sense to you? Give an explanation using words or drawings.

Answer:

If we multiply a number by i and then by – 1, we get a quarter turn followed by a half turn. This is equivalent to a three – quarters turn in the counterclockwise direction, which is the same as a quarter turn in the clockwise direction. This makes sense because we would expect multiplication by – i to have the opposite effect as multiplication by i, and so it feels right to say that multiplying by – i rotates a point in the opposite direction by the same amount.

### Eureka Math Precalculus Module 1 Lesson 4 Exit Ticket Answer Key

Question 1.

Solve the equation below.

2x^{2} – 3x+9 = 4

Answer:

2x^{2} – 3x = – 5

16x^{2} – 24x = – 40

16x^{2} – 24x+9 = – 40+9

(4x – 3)^{2} = – 31

4x – 3 = ±i\(\sqrt{31}\)

x = \(\frac{3 \pm i \sqrt{31}}{4}\) = \(\frac{3}{4}\)±i \(\frac{\sqrt{31}}{4}\)

Question 2.

What is the geometric effect of multiplying a number by i^{4}? Explain your answer using words or pictures, and then confirm your answer algebraically.

Answer:

If you multiply a number by i four times, you would expect to see four 90 – degree rotations. This amounts to a

360 – degree rotation. In other words, each point is mapped back to itself. This makes sense algebraically as well since the work below shows that i^{4} = 1.

i^{4} = i^{2}∙i^{2} = – 1∙ – 1 = 1