# How do you find extraneous solutions when solving radical equations?

Jan 15, 2015

$x - 6 = \sqrt{x}$

This can be solved easily by:

${x}^{2} - 12 x + 36 = x$

${x}^{2} - 13 x + 36 = 0$

$\left(x - 9\right) \left(x - 4\right) = 0$

$x = 9 , x = 4$

but wait! we have to be careful whenever we square both sides of an equation.

Re-substitute the two answers into the original equation.

$9 - 6 = \sqrt{9}$
$3 = 3$

That's fine, one more.

$4 - 6 = \sqrt{4}$
$- 2 \ne 2$

Our extraneous answer is $x = 4$

To find the extraneous answer of any equation, just find the solutions and then re-substitute the solutions into the original equation. The answer that results in an illogical equation is the extraneous answer.