## 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.

Answer:

6 – x = √x

(6 – x)^{2} = √x^{2}

36 – 12x + x^{2} = x

x^{2} – 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.

Answer:

\(\sqrt{x+3}\) = 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

Answer:

The only solution is 1.

Note that \(\frac{1}{9}\) is an extraneous solution.

Exercise 2.

3 = 4√x – x

Answer:

The two solutions are 9 and 1.

Exercise 3.

\(\sqrt{x+5}\) = x – 1

Answer:

The only solution is 4.

Note that – 1 is an extraneous solution.

Exercise 4.

\(\sqrt{3 x+7}\) + \(2 \sqrt{x-8}\) = 0

Answer:

There are no solutions.

Exercises 5 – 6

Solve the following equations.

Exercise 5.

\(\sqrt{x-3}\) + \(\sqrt{x+5}\) = 4

Answer:

4

Exercise 6.

3 + √x = \(\sqrt{x+81}\)

Answer:

144

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

Solve.

Question 1.

\(\sqrt{2 x-5}-\sqrt{x+6}\) = 0

Answer:

11

Question 2.

\(\sqrt{2 x-5}+\sqrt{x+6}\) = 0

Answer:

No solution

Question 3.

\(\sqrt{x-5}-\sqrt{x+6}\) = 2

Answer:

No solution

Question 4.

\(\sqrt{2 x-5}-\sqrt{x+6}\) = 2

Answer:

43

Question 5.

\(\sqrt{x+4}\) = 3 – \(\sqrt{x}\)

Answer:

\(\frac{25}{36}\)

Question 6.

\(\sqrt{x+4}\) = 3 + \(\sqrt{x}\)

Answer:

No solution

Question 7.

\(\sqrt{x+3}\) = \(\sqrt{5 x+6}\) – 3

Answer:

6

Question 8.

\(\sqrt{2 x+1}\) = x – 1

Answer:

4

Question 9.

\(\sqrt{x+12}+\sqrt{x}\) = 6

Answer:

4

Question 10.

2√x = 1 – \(\sqrt{4 x-1}\)

Answer:

\(\frac{1}{4}\)

Question 11.

2x = \(\sqrt{4 x-1}\)

Answer:

\(\frac{1}{2}\)

Question 12.

\(\sqrt{4 x-1}\) = 2 – 2x

Answer:

\(\frac{1}{2}\)

Question 13.

x + 2 = 4\(\sqrt{x-2}\)

Answer:

6

Question 14.

\(\sqrt{2 x-8}+\sqrt{3 x-12}\) = 0

Answer:

4

Question 15.

x = 2\(\sqrt{x-4}\) + 4

Answer:

4, 8

Question 16.

x – 2 = \(\sqrt{9x – 36}\)

Answer:

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.

Answer:

AC = \(\sqrt{64+x^{2}}\)

b. Find the value of x for which AC – AB = 9.

Answer:

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.

Answer:

AB = BC = \(\sqrt{81+x^{2}}\) cm

b. Write an expression for the perimeter of ∆ABC in terms of x.

Answer:

(2\(\sqrt{81+x^{2}}\) + 18)cm

c. Find the value of x for which the perimeter of ∆ABC is equal to 38 cm.

Answer:

\(\sqrt{19}\)cm

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

Question 1.

Solve \(\) = x + 6. Verify the solution(s).

Answer:

2x+ 15 = x^{2} + 12x + 36

0 = x^{2} + 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.

Answer:

Raising both sides of an equation to a power can produce an equation whose solution set is not equivalent to that of the original equation. In the problem above, x = – 7 does not satisfy the equation.