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

Apr 7, 2016

x = 4.

#### Explanation:

I have deleted my inclusion of x=0 as a solution. Thanks to Veddesh, for duly pointing out the mistake..
$x \ge 0$.
${\left(\sqrt{5 + x}\right)}^{2} = {\left(5 - \sqrt{x}\right)}^{2}$
This simplifies to $\sqrt{x} = 2$. So, x = 4.

Apr 7, 2016

$x = 4$

#### Explanation:

color(blue)(sqrt(5+x)+sqrtx=5

Subtract $\sqrt{x}$ both sides

$\rightarrow \sqrt{5 + x} + \cancel{\sqrt{x} - \sqrt{x}} = 5 - \sqrt{x}$

$\rightarrow \sqrt{5 + x} = 5 - \sqrt{x}$

Square both sides to get rid of the radical sign

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

In the right hand side use the formula

color(brown)((a-b)^2=a^2-2ab+b^2

So,

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

Subtract $x$ and $5$ both sides

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

$\rightarrow 0 = 20 - 10 \sqrt{x}$

Rewrite the equation in standard form

$\rightarrow 20 - 10 \sqrt{x} = 0$

Subtract $20$ both sides

$\rightarrow 20 - 10 \sqrt{x} - 20 = - 20$

$\rightarrow - 10 \sqrt{x} = - 20$

Divide both sides by $- 10$

$\rightarrow \frac{\cancel{- 10} \sqrt{x}}{\cancel{- 10}} = \frac{- 20}{-} 10$

$\rightarrow \sqrt{x} = 2$

Square both sides

$\rightarrow x = {2}^{2}$

color(green)(rArrx=4