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

Oct 16, 2015

$x = 4$

#### Explanation:

Before doing any calculations, make a note of the fact that any possible solution to this equation must satisfy

$x > = 0$

because, for real numbers, you can only take the square root of a positive number.

The first thing to do is square both sides of the equation to reduce the number of radical terms from two to one

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

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

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

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

$2 \sqrt{x \left(x + 5\right)} = 20 - 2 x$

$\sqrt{x \left(x + 5\right)} = 10 - x$

Once again, square both sides of the equation to get rid of the square root

${\left(\sqrt{x \left(x + 5\right)}\right)}^{2} = {\left(10 - x\right)}^{2}$

$x \left(x + 5\right) = 100 - 20 x + {x}^{2}$

This is equivalent to

$\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} + 5 x = 100 - 20 x + \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}}$

$25 x = 100 \implies x = \frac{100}{25} = \textcolor{g r e e n}{4}$

Since $x = 4 \ge 0$, this will be a valid solution to the original equation.

Do a quick check to make sure that the calculations are correct

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

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