# Eureka Math Algebra 2 Module 1 Lesson 29 Answer Key

## Engage NY Eureka Math Algebra 2 Module 1 Lesson 29 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 29 Example Answer Key

Example 1.
Solve the equation 6 = x + √x.
6 – x = √x
(6 – x)2 = √x2
36 – 12x + x2 = x
x2 – 13x + 36 = 0
(x – 9) (x – 4) = 0
The solutions are 9 and 4.
Check x = 9:
9 + √9 = 9 + 3 = 12
6 ≠ 12

Check x = 4:
4 + √4 = 4 + 2 = 6
So, 9 is an extraneous solution.
The only valid solution is 4.

Example 2.
Solve the equation $$\sqrt{x}+\sqrt{x+3}$$ = 3.
$$\sqrt{x+3}$$ = 3 – √x
$$(\sqrt{x+3})^{2}$$ = $$(3-\sqrt{x})^{2}$$
x + 3 = 9 – 6√x + x
1 = √x
1 = x

Check:
√1 + $$\sqrt{1+3}$$ = 1 + 2 = 3
So the solution is 1.

### Eureka Math Algebra 2 Module 1 Lesson 29 Exercise Answer Key

Solve:
Exercise 1.
3x = 1 + 2√x
The only solution is 1.
Note that $$\frac{1}{9}$$ is an extraneous solution.

Exercise 2.
3 = 4√x – x
The two solutions are 9 and 1.

Exercise 3.
$$\sqrt{x+5}$$ = x – 1
The only solution is 4.
Note that – 1 is an extraneous solution.

Exercise 4.
$$\sqrt{3 x+7}$$ + $$2 \sqrt{x-8}$$ = 0
There are no solutions.

Exercises 5 – 6

Solve the following equations.

Exercise 5.
$$\sqrt{x-3}$$ + $$\sqrt{x+5}$$ = 4
4

Exercise 6.
3 + √x = $$\sqrt{x+81}$$
144

### Eureka Math Algebra 2 Module 1 Lesson 29 Problem Set Answer Key

Solve.

Question 1.
$$\sqrt{2 x-5}-\sqrt{x+6}$$ = 0
11

Question 2.
$$\sqrt{2 x-5}+\sqrt{x+6}$$ = 0
No solution

Question 3.
$$\sqrt{x-5}-\sqrt{x+6}$$ = 2
No solution

Question 4.
$$\sqrt{2 x-5}-\sqrt{x+6}$$ = 2
43

Question 5.
$$\sqrt{x+4}$$ = 3 – $$\sqrt{x}$$
$$\frac{25}{36}$$

Question 6.
$$\sqrt{x+4}$$ = 3 + $$\sqrt{x}$$
No solution

Question 7.
$$\sqrt{x+3}$$ = $$\sqrt{5 x+6}$$ – 3
6

Question 8.
$$\sqrt{2 x+1}$$ = x – 1
4

Question 9.
$$\sqrt{x+12}+\sqrt{x}$$ = 6
4

Question 10.
2√x = 1 – $$\sqrt{4 x-1}$$
$$\frac{1}{4}$$

Question 11.
2x = $$\sqrt{4 x-1}$$
$$\frac{1}{2}$$

Question 12.
$$\sqrt{4 x-1}$$ = 2 – 2x
$$\frac{1}{2}$$

Question 13.
x + 2 = 4$$\sqrt{x-2}$$
6

Question 14.
$$\sqrt{2 x-8}+\sqrt{3 x-12}$$ = 0
4

Question 15.
x = 2$$\sqrt{x-4}$$ + 4
4, 8

Question 16.
x – 2 = $$\sqrt{9x – 36}$$
5, 8

Question 17.
Consider the right triangle ABC shown to the right, with AB = 8 and BC = x. a. Write an expression for the length of the hypotenuse in terms of x.
AC = $$\sqrt{64+x^{2}}$$

b. Find the value of x for which AC – AB = 9.
The solutions to the mathematical equation $$\sqrt{64+x^{2}}$$ – 8 = 9 are – 15 B and 15. Since lengths must be positive, – 15 is an extraneous solution, and x = 15.

Question 18.
Consider the triangle ABC shown to the right where AD = DC, and $$\overline{B D}$$ is the altitude of the triangle. a. If the length of is x cm, and the length of $$\overline{A C}$$ is 18 cm, write an expression for the lengths of $$\overline{A B}$$ and $$\overline{B C}$$ in terms of x.
AB = BC = $$\sqrt{81+x^{2}}$$ cm

b. Write an expression for the perimeter of ∆ABC in terms of x.
(2$$\sqrt{81+x^{2}}$$ + 18)cm

c. Find the value of x for which the perimeter of ∆ABC is equal to 38 cm.
$$\sqrt{19}$$cm

### Eureka Math Algebra 2 Module 1 Lesson 29 Exit Ticket Answer Key

Question 1.
Solve  = x + 6. Verify the solution(s).
2x+ 15 = x2 + 12x + 36
0 = x2 + 10x + 21
0 = (x + 3) (x + 7)
The solutions are – 3 and – 7.

Check x = – 3:
$$\sqrt{2(-3)+15}$$ = $$\sqrt{9}$$ = 3
– 3 + 6 = 3
So, – 3 is a valid solution.
Therefore, the only solution to the original equation is – 3.

Check x = – 7:
$$\sqrt{2(-7)+15}$$ = $$\sqrt{1}$$ = 1
– 7 + 6 = – 1
Since – 1 ≠ 1, we see that – 1 is an extraneous solution.

Question 2.
Explain why it is necessary to check the solutions to a radical equation.