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

Oct 17, 2015

$x = 9$

#### Explanation:

The first thing to do is figure out which values of $x$ cannot be a solution to the equation.

For real numbers, you need to take the square root of positive numbers only, which means that you will need

$x - 5 \ge 0 \implies x \ge 5$

Next, square both sides of the equation to reduce the number of radical terms from two to one.

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

${\left(\sqrt{x}\right)}^{2} + 2 \sqrt{x \left(x - 5\right)} + {\left(\sqrt{x - 5}\right)}^{2} = 25$

$x + 2 \sqrt{x \left(x - 5\right)} + x - 5 = 25$

Rearrange to get the radical term alone on one side of the equation

$2 \sqrt{x \left(x - 5\right)} = 30 - 2 x$

$\sqrt{x \left(x - 5\right)} = 15 - x$

Square both sides of the equation to get rid of the square root

${\left(\sqrt{x \left(x - 5\right)}\right)}^{2} = {\left(15 - x\right)}^{2}$

$x \left(x - 5\right) = 225 - 30 x + {x}^{2}$

$\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} - 5 x = 225 - 30 x + \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}}$

$25 x = 225 \implies x = \textcolor{g r e e n}{9}$

Since $x = 9$ satisfies the condition $x \ge 5$, this will be a valid solution to the original equation.

Do a quick check to make sure that the calculations came out right

$\sqrt{9} + \sqrt{9 - 5} = 5$

$3 + 2 = 5 \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$