# How to Solve Radical Equations? (+FREE Worksheet!)

An equation that contains a radical expression is called a radical equation, and in this blog post, we will teach you how to solve this type of equations.

## Related Topics

- How to Rationalize Radical Expressions
- How to Multiply Radical Expressions
- How to Simplify Radical Expressions
- How to Rationalize Radical Expressions
- How to Find Domain and Range of Radical Functions

## A step-by-step guide to solving Radical Equations

- Isolate the radical on one side of the equation.
- Square both sides of the equation to remove the radical.
- Solve the equation for the variable.
- Plugin the answer (answers) into the original equation to avoid extraneous values.

## Examples

### Radical Equations – Example 1:

Solve \(\sqrt{x}-5=15\)

**Solution**:

Add \(5\) to both sides: \(\sqrt{x}-5+5=15+5 → \) \(\sqrt{x}=20\), Square both sides: \((\sqrt{x})^2=20^2→x=400\), Plugin the value of \(400\) for \(x\) in the original equation and check the answer: \(x=400→\sqrt{x}-5=\sqrt{400}-5=20-5=15\), So, the value of \(400\) for \(x\) is correct.

### Radical Equations – Example 2:

What is the value of \(x\) in this equation? \(2\sqrt{x+1}=4\)

**Solution**:

Divide both sides by \(2\). Then: \(2\sqrt{x+1}=4→\frac{2\sqrt{x+1}}{2}=\frac{4}{2}→\sqrt{x+1}=2\). Square both sides: \((\sqrt{(x+1)})^2=2^2\), Then \(x+1=4→x+1-1=4-1 → x=3\)

Substitute \(x\) by \(3\) in the original equation and check the answer:

\( x=3→2\sqrt{x+1}=2\sqrt{3+1}=2\sqrt{4}=2(2)=4\)

So, the value of \(3\) for \(x\) is correct.

### Radical Equations – Example 3:

Solve \(\sqrt{x}-8=-3\)

**Solution**:

Add \(8\) to both sides: \(\sqrt{x}-8+8=-3+8 → \) \(\sqrt{x}=5\)

Square both sides: \((\sqrt{x})^2=5^2→x=25\)

Substitute \(x\) by \(25\) in the original equation and check the answer:

\(x=25→\sqrt{x}-8=\sqrt{25}-8=5-8=-3\)

So, the value of \(25\) for \(x\) is correct.

### Radical Equations – Example 4:

What is the value of \(x\) in this equation? \(4\sqrt{x+3}=40\)

**Solution**:

Divide both sides by \(4\). Then: \(4\sqrt{x+3}=40→\frac{4\sqrt{x+3}}{4}=\frac{40}{4}→\sqrt{x+3}=10\). Square both sides: \((\sqrt{(x+3)})^2=10^2\), Then \(x+3=100→x+3-3=100-3 → x=97\)

Substitute \(x\) by \(97\) in the original equation and check the answer:

\( x=97→4\sqrt{x+3}=4\sqrt{97+3}=4\sqrt{100}=4(10)=40\)

So, the value of \(97\) for \(x\) is correct.

## Exercises for Radical Equations

### Solve Radical Equations.

- \(\color{blue}{\sqrt{x}+6=8}\)
- \(\color{blue}{\sqrt{x}-7=4}\)
- \(\color{blue}{\sqrt{x+2}=10}\)
- \(\color{blue}{2\sqrt{x-9}=14}\)
- \(\color{blue}{\sqrt{2x-5}=\sqrt{x-1}}\)
- \(\color{blue}{\sqrt{x+8}=\sqrt{2x+1}}\)

- \(\color{blue}{x=4}\)
- \(\color{blue}{x=121}\)
- \(\color{blue}{x=98}\)
- \(\color{blue}{x=58}\)
- \(\color{blue}{x=4}\)
- \(\color{blue}{x=7}\)

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