# How do you solve sqrt( 3x-3) + sqrt(2x+8) +1=0?

Apr 6, 2016

Isolate one of the radicals on one side of the equation.

#### Explanation:

$\sqrt{3 x - 3} = - 1 - \sqrt{2 x + 8}$

${\left(\sqrt{3 x - 3}\right)}^{2} = {\left(- 1 - \sqrt{2 x + 8}\right)}^{2}$

$3 x - 3 = 1 + 2 x + 8 + 2 \sqrt{2 x + 8}$

$x - 12 = 2 \sqrt{2 x + 8}$

${\left(x - 12\right)}^{2} = {\left(2 \sqrt{2 x + 8}\right)}^{2}$

${x}^{2} - 24 x + 144 = 4 \left(2 x + 8\right)$

${x}^{2} - 24 x + 144 - 8 x - 32 = 0$

${x}^{2} - 32 x + 112 = 0$

$\left(x - 28\right) \left(x - 4\right) = 0$

$x = 28 \mathmr{and} 4$

Checking both these solutions in the original equation we find that neither work. Therefore, there is no solution $\left(\emptyset\right)$

Hopefully this helps!