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

How to solve radical equations, step by step: the rule in plain language, solved examples worked one line at a time, and practice questions so you can try it yourself before moving on.

Radical equations are equations that contain a variable inside a radical sign. Solving them requires isolating the radical and then squaring both sides to eliminate it, but this step can sometimes introduce extraneous solutions that do not satisfy the original equation. This guide walks through the complete solving process, including how to check your answers, with four worked examples and two video lessons.

What Is a Radical Equation?

A radical equation is any equation in which the variable appears inside a radical (square root, cube root, etc.). Examples: (color{blue}{sqrt{x} = 5}), (color{blue}{sqrt{(2x + 1)} = 3}), (color{blue}{sqrt{(x – 3)} + 4 = 8}). The main strategy is to isolate the radical, then raise both sides to a power that eliminates it.

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How to Solve Radical Equations

1. Isolate the Radical

Get the radical expression by itself on one side of the equation. Move all other terms to the opposite side first.

• (color{blue}{sqrt{x} + 2 = 7 rightarrow sqrt{x} = 5})

2. Square Both Sides

Squaring eliminates a square root: (color{blue}{(sqrt{x})^{2} = x}). Apply this to both sides of the equation.

• (color{blue}{(sqrt{x})^{2} = 5^{2} rightarrow x = 25})

3. Solve the Resulting Equation

After squaring, you will have a linear or quadratic equation. Solve it using standard methods.

4. Check for Extraneous Solutions

Squaring both sides can introduce solutions that do not work in the original equation. Always substitute each solution back into the original equation to verify it is valid. Discard any that do not satisfy the original.

Step-by-Step Summary

1. Isolate the radical on one side.
2. Square both sides to eliminate the radical.
3. Solve the resulting equation.
4. Check every solution in the original equation; discard extraneous ones.

Watch: Solving Radical Equations (Video Lesson)

Khan Academy explains the isolation and squaring process with clear, worked problems:

Radical Equations — Worked Examples

Example 1: Solve (color{blue}{sqrt{x} = 5}).

Square both sides: (color{blue}{x = 25}).
Check: (color{blue}{sqrt{25} = 5}) ✓
Answer: (color{blue}{x = 25})

Example 2: Solve (color{blue}{sqrt{(2x + 1)} = 3}).

Square both sides: (color{blue}{2x + 1 = 9}) → (color{blue}{2x = 8}) → (color{blue}{x = 4}).
Check: (color{blue}{sqrt{(8 + 1)} = sqrt{9} = 3}) ✓
Answer: (color{blue}{x = 4})

Example 3: Solve (color{blue}{sqrt{(x – 3)} = 4}).

Square both sides: (color{blue}{x – 3 = 16}) → (color{blue}{x = 19}).
Check: (color{blue}{sqrt{(19 – 3)} = sqrt{16} = 4}) ✓
Answer: (color{blue}{x = 19})

Example 4: Solve (color{blue}{sqrt{(x + 3)} = x – 3}).

Square both sides: (color{blue}{x + 3 = (x – 3)^{2} = x^{2} – 6x + 9}).
Rearrange: (color{blue}{x^{2} – 7x + 6 = 0}) → (color{blue}{(x – 1)(x – 6) = 0}) → (color{blue}{x = 1}) or (color{blue}{x = 6}).
Check (color{blue}{x = 1}): (color{blue}{sqrt{4} = 2}) but (color{blue}{1 – 3 = -2}). Since (color{blue}{2 ne -2}), (color{blue}{x = 1}) is extraneous.
Check (color{blue}{x = 6}): (color{blue}{sqrt{9} = 3}) and (color{blue}{6 – 3 = 3}) ✓
Answer: (color{blue}{x = 6})

More Practice: Solving Radical Equations Video

This algebra video covers solving radical equations including cases with extraneous solutions:

Exercises for Radical Equations

Solve each equation. Check for extraneous solutions.

1. (color{blue}{sqrt{x} = 7})
2. (color{blue}{sqrt{(x + 4)} = 5})
3. (color{blue}{sqrt{(2x – 3)} = 3})
4. (color{blue}{sqrt{(x – 1)} = x – 3})
5. (color{blue}{sqrt{(3x + 1)} = 4})

Answers

1. (color{blue}{x = 49})
2. (color{blue}{x = 21})  ((color{blue}{x + 4 = 25}))
3. (color{blue}{x = 6})  ((color{blue}{2x – 3 = 9}))
4. (color{blue}{x = 5})  ((color{blue}{x = 2}) is extraneous; check: (color{blue}{sqrt{(5-1)} = 2}), (color{blue}{5-3 = 2}) ✓)
5. (color{blue}{x = 5})  ((color{blue}{3x + 1 = 16}))

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Want More Practice?

We haven’t published a worksheet built specifically for Radical Equations just yet. In the meantime, the free worksheets below cover closely related skills and concepts. If you’d like extra practice, download any that look helpful, complete the problems, and check your work. They’re a great way to reinforce what you learned on this page and strengthen the foundations this topic builds on:

• Download Solving Quadratics by Square Roots Worksheet
• Download Graphing Square Root Functions Worksheet

Frequently Asked Questions

What is a radical equation in math?

A radical equation is an equation with the variable inside a square root or other radical, such as √(x + 3) = 5. You solve it by isolating the radical and raising both sides to a power that removes it. Here, squaring gives x + 3 = 25, so x = 22. An equation like x + √5 = 7 is not a radical equation, because the variable is not under the root.

How do you solve radical equations step by step?

Isolate the radical, square both sides, solve, and check. For √(2x + 1) + 3 = 8, subtract 3 to get √(2x + 1) = 5. Square both sides: 2x + 1 = 25, so 2x = 24 and x = 12. Check by substituting: √25 + 3 = 5 + 3 = 8, so x = 12 works. Skipping the check is how extraneous answers slip through.

What is an extraneous solution in radical equations?

An extraneous solution is a value you get from squaring that does not work in the original equation. Solve √(x + 2) = x. Squaring gives x + 2 = x², so x² − x − 2 = 0 and (x − 2)(x + 1) = 0. Test both: x = 2 gives √4 = 2, which works. x = −1 gives √1 = 1, not −1, so −1 is extraneous.

Can radical equations have no solution?

Yes. A principal square root is never negative, so an equation like √x = −3 has no real solution. Sometimes this hides until you isolate the radical: √(x − 1) + 5 = 2 becomes √(x − 1) = −3, which is impossible. In other cases, every answer you find after squaring turns out to be extraneous when you check, which also means no solution.

How do you solve radical equations with two square roots?

If each side has one square root, square both sides once. For √(3x − 2) = √(x + 6), squaring gives 3x − 2 = x + 6, so 2x = 8 and x = 4. Check: √10 = √10. If both roots are on the same side, move one to the other side first, square, isolate the remaining root, and square again, then check every answer.

Related Topics

• Simplifying Radical Expressions
• Rationalizing Radical Expressions
• Multiplying Radical Expressions
• Domain and Range of Radical Functions
• Adding and Subtracting Radical Expressions