# Eureka Math Algebra 2 Module 1 Lesson 28 Answer Key

## Engage NY Eureka Math Algebra 2 Module 1 Lesson 28 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 28 Example Answer Key

Example 1.
$$\sqrt{3 x+5}$$ – 2 = – 1
$$\sqrt{3 x+5}$$ = 1
3x + 5 = 1
3x = – 4
x = –$$\frac{4}{3}$$
Check: $$\sqrt{3\left(-\frac{4}{3}\right)+5}$$ – 2 = $$\sqrt{-4+5}$$ – 2 = $$\sqrt{1}$$ – 2 = – 1, so – $$\frac{4}{3}$$ is a valid solution.

Example 2.
Rationalize the denominator in each expression. That is, rewrite the expression so that there is a rational expression in the denominator.
a. $$\frac{x-9}{\sqrt{x-9}}$$

b. $$\frac{x-9}{\sqrt{x}+3}$$

### Eureka Math Algebra 2 Module 1 Lesson 28 Exercise Answer Key

For Exercises 1 – 4, describe each step taken to solve the equation. Then, check the solution to see if it is valid. If it is not a valid solution, explain why.

Exercise 1.
√x – 6 = 4
√x = 10 Add 6 to both sides.
x = 100 Square both sides.
Check √100 – 6 = 10 – 6 = 4
So 100 is a valid a solution .

Exercise 2.
$$\sqrt[3]{x}$$ – 6 = 4
$$\sqrt[3]{x}$$ = 10 Add 6 to both sides.
x = 1000 Cube both sides.
Check: $$\sqrt[3]{1000}$$ – 6 = 10 – 6 = 4
So 1,000 is a valid solution.

Exercise 3.
√x + 6 = 4
√x = – 2
x = 4
Check: √4 + 6 = 2 + 6 = 8, and 8 ≠ 4, s0 4 is not a valid solution.

Exercise 4.
$$\sqrt[3]{x}$$ + 6 = 4
$$\sqrt[3]{x}$$ = – 2
x = – 8
Check: $$\sqrt[3]{-8}$$ + 6 = – 2 + 6 = 4, so, – 8 is a valid solution.

Exercises 5 – 15

Exercise 5.
$$\sqrt{2 x-3}$$ = 11
62

Exercise 6.
$$\sqrt[3]{6-x}$$ = – 3
33

Exercise 7.
$$\sqrt{x+5}$$ – 9 = – 12
No solution

Exercise 8.
$$\sqrt{4 x-7}$$ = $$\sqrt{3 x+9}$$
16

Exercise 9.
-12$$\sqrt{x-6}$$ = 18
No solution

Exercise 10.
$$3 \sqrt[3]{x+2}$$ = 12
62

Exercise 11.
$$\sqrt{x^{2}-5}$$ = 2
3, – 3

Exercise 12.
$$\sqrt{x^{2}+8 x}$$ = 3
– 9, 1

Compute each product, and combine like terms.

Exercise 13.
(√x + 2) (√x – 2)
x – 4

Exercise 14.
(√x + 4) (√x + 4)
x + 8√x + 16

Exercise 15.
$$(\sqrt{x-5})(\sqrt{x-5})$$
x – 5

Exercises 16 – 18

Exercise 16.
Rewrite $$\frac{1}{\sqrt{x}-5}$$ in an equivalent form with a rational expression in the denominator.
$$\frac{\sqrt{x}+5}{x-25}$$

Exercise 17.
Solve the radical equation $$\frac{3}{\sqrt{x+3}}$$ = 1. Be sure to check for extraneous solutions.
x = 6

Exercise 18.
Without solving the radical equation $$\sqrt{x+5}$$ + 9 = 0, how could you tell that it has no real solution?
The value of the radical expression $$\sqrt{x+5}$$ must be positive or zero. In either case, adding 9 to it cannot give zero.

### Eureka Math Algebra 2 Module 1 Lesson 28 Problem Set Answer Key

Question 1.
a. If √x = 9, then what is the value of x?
x = 81

b. If x2 = 9, then what is the value of x?
x = 3 or x = – 3

c. Is there a value of x such that $$\sqrt{x+5}$$ = 0? If yes, what is the value? If no, explain why not.
Yes, x = – 5

d. Is there a value of x such that √x + 5 = 0? If yes, what is the value? If no, explain why not.
No, √x will be a positive value or zero for any value of x, so the sum cannot equal 0. If x = 25, then
√25 + 5 = 10.

Question 2.
a. Is the statement $$\sqrt{x^{2}}$$ = x true for all x-values? Explain.
No, this statement is only true for x ≥ 0. If x < 0, it is not true. For example, if x = – 5, $$\sqrt{(-5)^{2}}$$ = √25 = 5, then $$\sqrt{(-5)^{2}}[latex] ≠ – 5. b. Is the statement [latex]\sqrt[3]{x^{3}}$$ = x true for all x-values? Explain.
Yes, this statement is true for all x-values. For example, if x = 2, then $$\sqrt[3]{2^{3}}$$ = 2. If x = – 2, then $$\sqrt[3]{(-2)^{3}}$$ = – 2. Since the cube root of a positive number is positive, and the cube root oía negative number is negative, this statement is true for any value of x.

Rationalize the denominator in each expression.

Question 3.
$$\frac{4-x}{2+\sqrt{x}}$$
2 – √x

Question 4.
$$\frac{2}{\sqrt{x-12}}$$
$$\frac{2 \sqrt{x-12}}{x-12}$$

Question 5.
$$\frac{1}{\sqrt{x+3}-\sqrt{x}}$$
$$\frac{\sqrt{x+3}+\sqrt{x}}{3}$$

Solve each equation, and check the solutions.

Question 6.
$$\sqrt{x+6}$$ = 3
x = 3

Question 7.
2$$\sqrt{x+3}$$ = 6
x = 6

Question 8.
$$\sqrt{x+3}$$ + 6 = 3
No solution

Question 9.
$$\sqrt{x+3}$$ – 6 = 3
x = 78

Question 10.
16 = 8 + √x
x = 64

Question 11.
$$\sqrt{3 x-5}$$ = 7
x = 18

Question 12.
$$\sqrt{2 x-3}$$ = $$\sqrt{10-x}$$
x = $$\frac{13}{3}$$

Question 13.
3$$\sqrt{x+2}$$ + $$\sqrt{x-4}$$ = 0
No solution

Question 14.
$$\frac{\sqrt{x+9}}{4}$$ = 3
x = 135

Question 15.
$$\frac{12}{\sqrt{x+9}}$$ = 3
x = 7

Question 16.
$$\sqrt{x^{2}+9}$$ = 5
x = 4 or x = – 4

Question 17.
$$\sqrt{x^{2}-6 x}$$ = 4
x = 8 or x = – 2

Question 18.
$$\frac{5}{\sqrt{x-2}}$$ = 5
x = – 1

Question 19.
$$\frac{5}{\sqrt{x}-2}$$ = 5
x = 9

Question 20.
$$\sqrt[3]{5 x-3}$$ + 8 = 6
x = – 1

Question 21.
$$\sqrt[3]{9-x}$$ = 6
x = – 207

Question 22.
Consider the inequality $$\sqrt{x^{2}+4 x}$$ > 0. Determine whether each x-value is a solution to the inequality.
a. x = – 10
Yes

b. x = – 4
No

c. x = 10
Yes

d. x = 4
Yes

Question 23.
Show that $$\frac{a-b}{\sqrt{a}-\sqrt{b}}$$ = $$\sqrt{a}+\sqrt{b}$$ for all values of a and b such that a > 0 and b > 0 and a B b.
If we multiply the numerator and denominator of $$\frac{a-b}{\sqrt{a}-\sqrt{b}}$$ by $$\sqrt{a}+\sqrt{b}$$ to rationalize the denominator, then we have

Question 24.
Without actually solving the equation, explain why the equation $$\sqrt{x+1}$$ + 2 = 0 has no solution.
The value of $$\sqrt{x+1}$$ must be positive, which is then added to 2. The sum of two positive numbers is positive; therefore, the sum cannot equal 0.

### Eureka Math Algebra 2 Module 1 Lesson 28 Exit Ticket Answer Key

Consider the radical equation 3$$\sqrt{6-x}$$ + 4 = – 8.

Question 1.
Solve the equation. Next to each step, write a description of what is being done.
3$$\sqrt{6-x}$$ = 1 12 Subtract 4 from both sides.
$$\sqrt{6-x}$$ = – 4 Divide both sides by 3 in order to isolate the radical.
6 – x = 16 Square both sides to eliminate the radical.
x = – 10 Subtract 6 from both sides and divide by – 1.

Question 2.
Check the solution.
3$$\sqrt{6-(-10)}$$ + 4 = 3$$\sqrt{16}$$ + 4 = 3(4) + 4 = 16, and B – 8, so -10 is not a valid solution.
Because the square root of a positive number is positive, 3$$\sqrt{6-x}$$ will be positive. A positive number added to 4 cannot be – 8.