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Solving Radical Equations Solutions

Solve equations with radicals using inverse operations .
Recall that radicals can be written as: $\sqrt[d]{x^{n}} = x^{\frac{n}{d}}$
To find the inverse of a radical, raise both sides of the equation by the index d.
If you raise both sides of an equation to an even power, it is important to check for extraneous solutions.
Raising to an even power can produce extraneous solutions because it cannot be undone.
- If $c = d$ , then $c^{2} = d^{2}$ is always true. But if $c^{2} = d^{2}$ , then sometimes $c = d$ . For example:
  $x = 5 x^{2} = 25 \sqrt{x^{2}} = \sqrt{25} x = \pm 5 - 5 is extraneous$
Raising to an odd power does not result in extraneous solutions.
- If $c = d$ , then $c^{3} = d^{3}$ , and if $c^{3} = d^{3}$ , then $c = d$ . For example:
  $x = - 5 x^{3} = - 125 \sqrt[3]{x^{3}} = \sqrt[3]{- 125} x = - 5$
Extraneous solutions can occur when:
- a solved value results in a negative radicand.
- there are radicals on both sides of the equation.
To check for extraneous solutions, use substitution .

Example 1

Solve.
$\sqrt{3 x - 5} = \sqrt{8 - x}$

Implement

${(\sqrt{3 x - 5})}^{2} = {(\sqrt{8 - x})}^{2}$

$\begin{array}{rcl} 3 x - 5 & = & 8 - x \\ 4 x & = & 13 \\ x & = & \frac{13}{4} = 3.25 \end{array}$

Check

$\begin{array}{rcl} \sqrt{3 (3.25) - 5} & = & \sqrt{8 - (3.25)} \\ \sqrt{9.75 - 5} & = & \sqrt{4.75} \\ \sqrt{4.75} & = & \sqrt{4.75} ✓ \end{array}$

Explain

Square both sides
Isolate x
Using decimal values may be more efficient because a common denominator is not needed. Recommended: Use a calculator to check radical equations

Example 2

Solve.
$2 + \sqrt{x} = \sqrt{x - 12}$

Implement

${(2 + \sqrt{x})}^{2} = {(\sqrt{x - 12})}^{2}$
$(2 + \sqrt{x}) (2 + \sqrt{x}) = x - 12$
$\begin{array}{rcl} 4 + 4 \sqrt{x} + x & = & x - 12 \\ 4 \sqrt{x} & = & - 16 \\ \sqrt{x} & = & - 4 \end{array}$

no solution

Explain

Square both sides
Distribute on the left side
Isolate x
The principal square root cannot equal a negative number

Example 3

Solve.

$\sqrt{x + 8} - \sqrt{x - 5} = 3$

Implement

$\sqrt{x + 8} = \sqrt{x - 5} + 3$

$\begin{array}{rcl} {(\sqrt{x + 8})}^{2} & = & {(\sqrt{x - 5} + 3)}^{2} \end{array}$
$\begin{array}{rcl} x + 8 & = & x - 5 + 6 \sqrt{x - 5} + 9 \\ x + 8 & = & x + 4 + 6 \sqrt{x - 5} \\ 4 & = & 6 \sqrt{x - 5} \\ {(\frac{2}{3})}^{2} & = & {(\sqrt{x - 5})}^{2} \\ \frac{4}{9} & = & x - 5 \\ x & = & 5 \frac{4}{9} \end{array}$

Check

$\begin{array}{rcl} \sqrt{5 \frac{4}{9} + 8} - \sqrt{5 \frac{4}{9} - 5} & = & 3 \\ \sqrt{13 \frac{4}{9}} - \sqrt{\frac{4}{9}} & = & 3 ✓ \end{array}$

Explain

Rewrite with one radical expression on each side of the equation
Square both sides

Note

The distributive property is used here to multiply all terms on the right side of the equation.

Isolate the radical to solve for x
Isolate the radical on the right side to solve for the remaining variable in the equation
Remember to use a simplified fraction