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

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

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:

  1. \(\sqrt{2x + 15} = 5\)
  2. \(\sqrt{4x - 3} = x - 2\)
  3. \(\sqrt{x + 1} + \sqrt{x + 6} = 5\) Challenging

Solutions to Above Exercises

  1. \(x = 5\)
  2. \(x = 7\)
  3. \(x = 3\)
Detailed Solution for Exercise 3

Given the equation containing two square roots:

\[ \sqrt{x + 1} + \sqrt{x + 6} = 5 \]

Isolate one radical by moving \(\sqrt{x + 1}\) to the right side:

\[ \sqrt{x + 6} = 5 - \sqrt{x + 1} \]

Square both sides to eliminate the radical on the left:

\[ (\sqrt{x + 6})^2 = (5 - \sqrt{x + 1})^2 \] \[ x + 6 = 25 - 10\sqrt{x + 1} + (x + 1) \]

Simplify and isolate the remaining radical term \(10\sqrt{x + 1}\):

\[ x + 6 = 26 + x - 10\sqrt{x + 1} \] \[ 10\sqrt{x + 1} = 20 \]

Divide both sides by 10:

\[ \sqrt{x + 1} = 2 \]

Square both sides again to solve for \(x\):

\[ x + 1 = 2^2 = 4 \] \[ x = 3 \]

Check the solution:

\[ \sqrt{3 + 1} + \sqrt{3 + 6} = \sqrt{4} + \sqrt{9} = 2 + 3 = 5 \]

Since LS = RS, \(x = 3\) is verified as a valid solution.