# How do you solve sqrt(x-3) = x - 5 and find any extraneous solutions?

Aug 29, 2016

$x = 7 \mathmr{and} x = 4$

Both solutions are valid.

#### Explanation:

The value under the root may not be negative. $x \ge 3$

Square both sides of the equation.
${\sqrt{x - 3}}^{2} = {\left(x - 5\right)}^{2}$

A common error is to square each of the terms on the right side separately, instead of as a square of a binomial.

$x - 3 = {x}^{2} - 10 x + 25$

${x}^{2} - 10 x + 25 - x + 3 = 0$

${x}^{2} - 11 x + 28 = 0$

Find factors of 28 which add to give 11. ($7 \times 4$ works)
The signs will be the same, both negative.

$\left(x - 7\right) \left(x - 4\right) = 0$

Putting each factor equal to 0 gives:

$x = 7 \mathmr{and} x = 4$

Both solutions are valid.