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.

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

Problem 2

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

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

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

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

  1. Check Tyler’s answer: Is the equation true if \(x=22\)? Explain or show your reasoning.
  2. 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 \(\frac14\) \(\frac{1}{16}\)
(From Unit 3, Lesson 3.)

Problem 6

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

A:

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

B:

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

C:

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

D:

The equation has no solutions.

(From Unit 3, Lesson 8.)