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

May 18, 2016

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

$\sqrt{x} = 1 + \sqrt{x - 1}$

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

$x = 1 + x - 1 + 2 \sqrt{x - 1}$

$0 = 2 \sqrt{x - 1}$

${\left(0\right)}^{2} = {\left(2 \sqrt{x - 1}\right)}^{2}$

$0 = 4 \left(x - 1\right)$

$0 = 4 x - 4$

$4 = 4 x$

$x = 1$

Therefore, $\left\{x = 1\right\}$. Checking the solution back in the original equation we find it works.

Hopefully this helps!