# How do you solve sqrt( x-5) = x-7?

Apr 18, 2016

$x = \left\{9\right\}$

#### Explanation:

square both sides.
x-5=(x-7)².
x-5=x²-14x+49.
0=x²-15
x+54.

$\Delta = \sqrt{{\left(- 15\right)}^{2} - 4 \cdot 1 \cdot 54}$
$\Delta = \sqrt{225 - 216}$
$\Delta = \sqrt{9}$
Delta=±3

${x}_{1} = \frac{15 - 3}{2}$

${x}_{1} = 6$

${x}_{2} = \frac{15 + 3}{2}$

${x}_{2} = 9$
$x = \left\{9\right\}$

Apr 18, 2016

$x = \left\{9\right\}$

#### Explanation:

$\sqrt{x - 5} = x - 7$
${\left(\sqrt{x - 5}\right)}^{2} = {\left(x - 7\right)}^{2}$

$x - 5 = {\left(x - 7\right)}^{2}$

$x - 5 = {x}^{2} - 14 x + 49$

${x}^{2} - 14 x - x + 49 + 5 = 0$

${x}^{2} - 15 x + 54 = 0$

$\left(x - 6\right) \left(x - 9\right) = 0$

$\mathmr{if} \left(x - 6\right) = 0 \text{ "then" } x = 6$

$\mathmr{if} \left(x - 9\right) = 0 \text{ "then" } x = 9$

Watch out for extraneous solutions! 6 does not work in the original equation, however, 9 does. Thus, $\left\{x = 9\right\}$