# How do you solve sqrt(x+4) = sqrt(x) + sqrt(2)?

Oct 14, 2015

$x = \frac{1}{2}$

#### Explanation:

Right from the start, you know that, for real numbers, you can't take the square root of a negative value, which means that you need to have

$x \ge 0$

Next, square both sides of the equation to get

${\left(\sqrt{x + 4}\right)}^{2} = {\left(\sqrt{x} + \sqrt{2}\right)}^{2}$

$x + 4 = {\left(\sqrt{x}\right)}^{2} + 2 \sqrt{2 x} + {\left(\sqrt{2}\right)}^{2}$

$\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} + 4 = \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} + 2 \sqrt{2 x} + 2$

This is equivalent to

$\sqrt{2 x} = 1$

Once again, you have

${\left(\sqrt{2 x}\right)}^{2} = {1}^{2}$

$2 x = 1 \implies x = \frac{1}{2}$

Since this value of $x$ satisfies the condition $x \ge 0$, it will be a valid solution to the original equation.

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

$\sqrt{\frac{1}{2} + 4} = \sqrt{\frac{1}{2}} + \sqrt{2}$

$\frac{\sqrt{9}}{\sqrt{2}} = \frac{1 + \sqrt{2} \cdot \sqrt{2}}{\sqrt{2}}$

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