# Lesson 9

Solving Radical Equations

- Let’s practice solving radical equations.

### Problem 1

Find the solution(s) to each of these equations, or explain why there is no solution.

- \(\sqrt{x+5}+7 = 10\)
- \(\sqrt{x-2}+3=\text-2\)

### Problem 2

For each equation, decide how many solutions it has and explain how you know.

- \((x-4)^2= 25\)
- \(\sqrt{x-4} = 5\)
- \(x^3 -7 = \text-20\)
- \(6 \boldcdot \sqrt[3]{x} = 0\)

### Problem 3

Jada was solving the equation \(\sqrt{6-x}=\text-16\). She was about to square each side, but then she realized she could give an answer without doing any algebra. What did she realize?

### Problem 4

Here are the steps Tyler took to solve the equation \(\sqrt{x+3}=\text-5\).

\(\begin{align} \sqrt{x+3} & =\text-5 \\ x+3 &=25 \\ x &=22 \\ \end{align} \)

- Check Tyler’s answer: Is the equation true if \(x=22\)? Explain or show your reasoning.
- What mistake did Tyler make?

### Problem 5

Complete the table. Use powers of 16 in the top row and radicals or rational numbers in the bottom row.

\(16^1\) | \(16^{\frac13}\) | \(16^{\text-1}\) | |||

4 | 1 | \(\frac14\) | \(\frac{1}{16}\) |

### Problem 6

Which are the solutions to the equation \(x^3=35\)?

\(\sqrt[3]{35}\)

\(\text-\sqrt[3]{35}\)

both \(\sqrt[3]{35}\) and \(\text-\sqrt[3]{35}\)

The equation has no solutions.