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.

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

1. $\color{blue}{x=4}$
2. $\color{blue}{x=121}$
3. $\color{blue}{x=98}$
4. $\color{blue}{x=58}$
5. $\color{blue}{x=4}$
6. $\color{blue}{x=7}$