Solve Equations With Square Root \( \sqrt{ } \)

Tutorial on how to solve equations containing square roots. Detailed solutions, explanations, and exercises are included.

The main idea in solving equations containing square roots is to square both sides in order to clear the square root using the property

\((\sqrt{x})^{2} = x\)

This holds only for \(x \ge 0\). Because squaring both sides may introduce extraneous solutions, we must always check the answers in the original equation.

Examples with Solutions

Example 1

Find all real solutions to the equation:

\(\sqrt{x + 1} = 4\)

Solution to Example 1

Given: \[ \sqrt{x + 1} = 4 \]
Square both sides to eliminate the square root. \[ (\sqrt{x + 1})^2 = 4^2 \]
Simplify: \[ x + 1 = 16 \]
Solve for \(x\): \[ x = 15 \]
Check the solution (important because we squared both sides).
Left side: \[ \sqrt{15 + 1} = 4 \] Right side: \[ 4 \]

Since LS = RS, \(x = 15\) is a valid solution.

Example 2

Find all real solutions to the equation:

\(\sqrt{3x + 1} = x - 3\)

Solution to Example 2

Given: \[ \sqrt{3x + 1} = x - 3 \]
Square both sides: \[ (\sqrt{3x + 1})^2 = (x - 3)^2 \]
Simplify: \[ 3x + 1 = x^2 - 6x + 9 \]
Rewrite with right-hand side = 0: \[ x^2 - 9x + 8 = 0 \]
Solve the quadratic: \[ x = 8,\quad x = 1 \]
Check both solutions.
Check \(x = 8\):
\[ \sqrt{3(8)+1} = \sqrt{25} = 5 \] \[ x - 3 = 8 - 3 = 5 \] Valid.
Check \(x = 1\):
\[ \sqrt{3(1)+1} = \sqrt{4} = 2 \] \[ x - 3 = 1 - 3 = -2 \] Not valid.
Therefore, the only solution is \(x = 8\). The value \(x = 1\) is an extraneous solution introduced during squaring.

Exercises

Solve the following equations:

\(\sqrt{2x + 15} = 5\)
\(\sqrt{4x - 3} = x - 2\)

Solutions to Above Exercises

\(x = 5\)
\(x = 7\)