Training Radicals & Rational Exponents Solving Radical Equations
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Solving Radical Equations

20 min Radicals & Rational Exponents

Solving Radical Equations

Solving radical equations requires isolating the radical on one side and then raising both sides to the appropriate power to eliminate it. This process can introduce extraneous solutions — values that satisfy the transformed equation but not the original — so checking your answers is mandatory.

When an equation contains two radical terms, you may need to isolate and square twice, leading to a multi-step process that demands careful bookkeeping.

This lesson walks through the solution process for equations with one and two radical terms, with emphasis on the check step that catches extraneous roots.

A radical equation is an equation in which the variable appears inside a radical.

Procedure
  1. Isolate the radical on one side of the equation.
  2. Raise both sides to the power that eliminates the radical (square both sides for square roots, cube both sides for cube roots).
  3. Solve the resulting equation.
  4. Check every solution in the original equation — squaring can introduce extraneous solutions.
Caution

An extraneous solution is a value that satisfies the squared equation but NOT the original. Always verify your answers.

Example 1

Solve $\sqrt{x} = 7$.

Square both sides: $x = 49$.

Check: $\sqrt{49} = 7$ ✓

Example 2

Solve $\sqrt{x + 3} = 5$.

$(\sqrt{x+3})^2 = 5^2 \implies x + 3 = 25 \implies x = 22$.

Check: $\sqrt{22 + 3} = \sqrt{25} = 5$ ✓

Example 3

Solve $\sqrt{2x - 1} = x - 2$.

Square: $2x - 1 = (x-2)^2 = x^2 - 4x + 4$.

$0 = x^2 - 6x + 5 = (x-1)(x-5)$.

$x = 1$ or $x = 5$.

Check $x = 1$: $\sqrt{1} = 1$ but $1 - 2 = -1$. $1 \ne -1$ ✗ (extraneous)

Check $x = 5$: $\sqrt{9} = 3$ and $5 - 2 = 3$. ✓

Solution: $x = 5$.

Example 4

Solve $\sqrt{3x + 4} + 2 = x$.

Isolate: $\sqrt{3x+4} = x - 2$.

Square: $3x + 4 = x^2 - 4x + 4$.

$0 = x^2 - 7x = x(x - 7)$.

$x = 0$ or $x = 7$.

Check $x = 0$: $\sqrt{4} + 2 = 4 \ne 0$ ✗

Check $x = 7$: $\sqrt{25} + 2 = 7$ ✓

Solution: $x = 7$.

Example 5

Solve $\sqrt[3]{x - 1} = 3$.

Cube both sides: $x - 1 = 27 \implies x = 28$.

Check: $\sqrt[3]{27} = 3$ ✓

Practice Problems

1. Solve $\sqrt{x} = 9$
2. Solve $\sqrt{x - 5} = 4$
3. Solve $\sqrt{2x + 3} = 7$
4. Solve $\sqrt{x + 1} = x - 1$
5. Solve $\sqrt{3x - 2} = x - 2$
6. Solve $\sqrt[3]{2x - 5} = 3$
7. Solve $\sqrt{5x + 1} - 3 = x$
8. Solve $\sqrt{x} = -4$
9. Solve $\sqrt{4x + 5} = 2x - 1$
10. Solve $2\sqrt{x} + 1 = 7$
11. Solve $\sqrt{x + 7} = \sqrt{2x - 1}$
12. Solve $\sqrt{3x + 10} = x + 2$
Show Answer Key

1. $x = 81$

2. $x = 21$

3. $x = 23$

4. $x = 3$ ($x = 0$ is extraneous)

5. $x = 6$ ($x = 1$ is extraneous)

6. $x = 16$

7. $x = 1$ ($x = -8$ is extraneous)

8. No solution (square root cannot be negative)

9. $x = 1$ ($x = -1/4$ is extraneous)

10. $x = 9$

11. $x = 8$

12. $x = 2$ ($x = -3$ is extraneous)

Rational Exponents Placement Test Practice — Radicals