# How do you solve 4-sqrt(x-3)=sqrt(x+5)?

Apr 10, 2015

$4 - \sqrt{x - 3} = \sqrt{x + 5}$

Square both sides giving
$16 - 8 \sqrt{x - 3} + \left(x - 3\right) = x + 5$

$13 - 8 \sqrt{x - 3} + \cancel{x} = \cancel{x} + 5$

$- 8 \sqrt{x - 3} = - 8$

$\sqrt{x - 3} = 1$

Square both sides again
$x - 3 = 1$

$x = 4$

Verify by substituting $x = 4$ back into the original equation.

Apr 10, 2015

$x = 4$

$4 - \sqrt{x - 3} = \sqrt{x + 5}$

Squaring both sides

${\left[4 - \sqrt{x - 3}\right]}^{2} = {\left[\sqrt{x + 5}\right]}^{2}$

As ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$

So,

${4}^{2} - 8 \sqrt{x - 3} + {\left[\sqrt{x - 3}\right]}^{2} = {\left[\sqrt{x + 5}\right]}^{2}$

$16 - 8 \sqrt{x - 3} + \left(x - 3\right) = x + 5$

$16 + x - 3 - x - 5 - 8 \sqrt{x - 3} = 0$

$16 - 8 - 8 \sqrt{x - 3} = 0$

$8 - 8 \sqrt{x - 3} = 0$

$8 \left[1 - \sqrt{x - 3}\right] = 0$

$1 - \sqrt{x - 3} = 0$

$\sqrt{x - 3} = 1$

Again Squaring both sides

${\left[\sqrt{x - 3}\right]}^{2} = {1}^{2}$

$x - 3 = 1$

$x = 1 + 3$

$x = 4$