# How do you solve x - 9 = sqrt(x - 3)?

Jul 1, 2016

$x = 12$

#### Explanation:

First you fix the domain of the solutions:

$x - 3 \ge 0 \mathmr{and} x - 9 \ge 0$

that's $x \ge 9$

then you square both members:

${\left(x - 9\right)}^{2} = {\left(\sqrt{x - 3}\right)}^{2}$

that's

${x}^{2} - 18 x + 81 = x - 3$

${x}^{2} - 18 x + 81 - x + 3 = 0$

${x}^{2} - 19 x + 84 = 0$

$x = \frac{19 \pm \sqrt{{19}^{2} - 4 \cdot 84}}{2}$

$x = \frac{19 \pm 5}{2}$

$x = 7 \mathmr{and} x = 12$

but only x=12 belongs to the calculated domain