# Solve Equations With Square Root (√)

Tutorial on how to solve equations containing square roots. Detailed solutions to examples, explanations and exercises are included.
The main idea behind solving equations containing square roots is to raise to power 2 in order to clear the square root using the property

( √x ) 2 = x.

The above holds only for x greater than or equal to 0. In solving equations we square both sides of the equations and instead of putting similar conditions we check the solutions obtained.

## Examples with Solutions

### Example 1

Find all real solutions to the equation
√( x + 1) = 4

#### Solution to Example 1:

• Given
√( x + 1) = 4
• We raise both sides to power 2 in order to clear the square root.
[ √( x + 1) ] 2 = 4 2
• and simplify
x + 1 = 16
• Solve for x.
x = 15
• NOTE: Since we squared both sides without putting any conditions, extraneous solutions may be introduced, checking the solutions is necessary.
Left side (LS) of the given equation when x = 15
LS = √(x + 1) = √(15 + 1) = 4
Right Side (RS) of the given equation when x = 15
RS = 4
• For x = 15, the left and the rigth sides of the given equation are equal: x = 15 is a solution to the given equation.

### Example 2

Find all real solutions to the equation

√( 3 x + 1) = x - 3

#### Solution to Example 2:

• Given
√( 3 x + 1) = x - 3
• We raise both sides to power 2.
[ √( 3 x + 1) ] 2 = (x - 3) 2
• and simplify.
3 x + 1 = x 2 - 6 x + 9
• Write the equation with right side equation to 0.
x 2 - 9 x + 8 = 0
• It is a quadratic equation with 2 solutions
x = 8 and x = 1
• NOTE: Since we squared both sides , extraneous solutions may be introduced, checking the solutions in the original equation is necessary.
1. check equation for x = 8.
Left side (LS) of the given equation when x = 8
LS = √( 3 x + 1) = √(3 * 8 + 1) = 5
Right Side (RS) of the given equation when x = 8
RS = x - 3 = 8 - 3 = 5
2. check equation for x = 1.
Left side (LS) of the given equation when x = 8
LS = √( 3 x + 1) = √(3 * 1 + 1) = 2
Right Side (RS) of the given equation when x = 8
RS = x - 3 = 1 - 3 = -2
• For x = 8 the left and right sides of the equation are equal and x = 8 is a solution to the given equation. x = 1 is not a solution to the given equation; it is an extraneous solution introduced because of the raising to power 2.

## Exercises

Solve the following equations
1.   sqrt(2 x + 15) = 5
2.   sqrt(4 x - 3) = x - 2

1.   x = 5
2.   x = 7