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

Mar 24, 2017

$x = 4$ is applicable for +ve value of both sides
$x = 1$ is applicable for -ve value of both sides

#### Explanation:

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

square both sides
${\left(\left(\sqrt{x}\right) - 1\right)}^{2} = {\left(\sqrt{5 - x}\right)}^{2}$

$x - 2 \sqrt{x} + 1 = 5 - x$

$x + x - 2 \sqrt{x} + 1 - 5 = 0$

$2 x - 2 \sqrt{x} - 4 = 0$

Let say $\sqrt{x} = y$, then $x = {\left(\sqrt{x}\right)}^{2} = {y}^{2}$

$2 {y}^{2} - 2 y - 4 = 0$
$2 \left({y}^{2} - y - 2\right) = 0$
$2 \left(y - 2\right) \left(y + 1\right) = 0$
$y = 2 , y = - 1$

When $y = 2$, $x = {2}^{2} = 4$
When $y = - 1$, $x = {\left(- 1\right)}^{2} = 1$