# How do you solve sqrtx- sqrt(x-5)=1?

Oct 13, 2015

$x$ = 9

#### Explanation:

$\sqrt{x} - \sqrt{x - 5} = 1$ equation 1

multiply both sides with $\sqrt{x} + \sqrt{x - 5}$

(sqrt (x) -sqrt(x-5)) (sqrt (x) +sqrt(x-5)) = 1$\left(\sqrt{x} + \sqrt{x - 5}\right)$

L H S is in the form of (a+b)(a-b) =${a}^{2} - {b}^{2}$

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

Solve equation 1 and 2 for $x$

$\sqrt{x} - \sqrt{x - 5} = 1$ equation 1

$\sqrt{x} + \sqrt{x - 5} = 5$ equation 2

Sum of equation 1 & 2

2$\sqrt{x} = 6$
$\sqrt{x} = \frac{6}{2}$ = 3
Squaring on both side to get $x$

$x = 9$