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
( sqrt(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.
Example 1 : Find all real solutions to the equation
sqrt ( x + 1) = 4
Solution to Example 1:
 Given
sqrt ( x + 1) = 4
 We raise both sides to power 2 in order to clear the square root.
[ sqrt ( 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 = sqrt (x + 1) = sqrt (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
sqrt ( 3 x + 1) = x  3
Solution to Example 2:
 Given
sqrt ( 3 x + 1) = x  3
 We raise both sides to power 2.
[ sqrt ( 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 = sqrt ( 3 x + 1) = sqrt (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 = sqrt ( 3 x + 1) = sqrt (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:(answers further down the page)
Solve the following equations
1. sqrt(2 x + 15) = 5
2. sqrt(4 x  3) = x  2
Solutions to above exercises
1. x = 5
2. x = 7
More references and links on how to Solve Equations, Systems of Equations and Inequalities.
