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

Apr 6, 2017

$\left\{4\right\}$

#### Explanation:

We start by squaring both sides.

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

$x + 5 = 25 - 10 \sqrt{x} + x$

$10 \sqrt{x} = 20$

$\sqrt{x} = \frac{20}{10}$

$\sqrt{x} = 2$

We square both sides one more time.

$x = 4$

$\sqrt{4 + 5} = 5 - \sqrt{4}$

3 = 5 - 2 color(green)(√)

Practice exercises

$1$. Solve for $x$. Make sure to verify your solutions because they may or not be extraneous.

a) $\sqrt{2 x - 2} = \sqrt{x} + 1$
b) $\sqrt{3 x - 5} = \sqrt{x + 6} + 1$

Solutions

$1$. a) $x = 9$
b) $x = 10$

Hopefully this helps, and good luck!

Apr 6, 2017

I got $x = 4$

#### Explanation:

We can try write it as:
$\sqrt{x + 5} + \sqrt{x} = 5$
square both sides:
${\left(\sqrt{x + 5} + \sqrt{x}\right)}^{2} = {5}^{2}$
$x + 5 + 2 \sqrt{x + 5} \sqrt{x} + x = 25$
$2 \sqrt{x + 5} \sqrt{x} = 20 - 2 x$
square again:
$4 x \left(x + 5\right) = 400 - 80 x + 4 {x}^{2}$
$\cancel{4 {x}^{2}} + 20 x = 400 - 80 x + \cancel{4 {x}^{2}}$
$100 x = 400$
$x = \frac{400}{100} = 4$