Debug

How to Solve Radical Equations

Video Tutorial and practice problems

How To solve Radical Equations

  • 1) Isolate radical on one side of the equation
  • 2) Square both sides of the equation to eliminate radical
  • 3) Simplify and solve as you would any equations
  • 4) Substitute answers back into original equation to make sure that your solutions are valid (there could be some extraneous roots that do not satisfy the original equation and that you must throw out)

The video below and our examples explain these steps and you can then try our practice problems below.

Video of How to Solve Radical Equations

Example 1
Example 2

Practice Problems

Problem 1

Solve the radical Equation Below.

equation

Follow the steps for solving radical equations.

Step 1

Isolate the radical.

equation
Step 2

Square both sides.

equation
Step 3

Solve expression.

x = 10
Step 4

Substitute answer into original radical equation to verify that the answer is a real number.

equation
Problem 2

Solve the radical Equation Below.

radical equation

Remember how to solve radical equations.

Step 1

Isolate the radical.

radical equation
Step 2

Square both sides.

radical equation
Step 3

Solve expression.

3x = 23
Step 4

Substitute answer into original radical equation to verify that the answer is a real number.

radical equation
Problem 3

Solve the following radical equation:

$$ \sqrt{3x -11} = 3x -x $$

Remember how to solve radical equations.

Step 1

Isolate the radical.

radical equation
Step 2

Square both sides.

radical equation
Step 3

Solve expression.

radical equation

This quadratic equation can be solved by factoring.

0 =(x - 4)(x - 5)
x = 4, x = 5
Step 4

Substitute answer into original radical equation to verify that the answer is a real number.

$$ \sqrt{3x -11} = 3x -x \\ \sqrt{3 (\color{Red}{4}) -11} = 3 \cdot (\color{Red}{4}) -\color{Red}{4} \\ \sqrt{1} = 8 \\ 1 \color{red}{ \ne } 8 $$

Therefore, reject 4 as a solution, check 5.

$$ \sqrt{3x -11} = 3x -x \\ \sqrt{3 (\color{Red}{5}) -11} = 3(\color{Red}{5}) - \color{Red}{5} \\ \sqrt{15 -11} = 15 - 5 \\ \sqrt{15 -11} = 15 - 5 \\ \sqrt{4} = 10 \\ 2 = 10 \\ \color{red}{ \ne } 10 $$

Therefore, reject 5 as a solution.

Since both our solutions were rejected, there are no real solutions to this equation.


Back to Algebra Next to Radical Equation Solver