# How do you solve 2(sqrt x) + x=8?

Mar 3, 2016

You must square both sides to get rid of the radical.

#### Explanation:

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

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

$4 x = 64 - 16 x + {x}^{2}$

$0 = {x}^{2} - 20 x + 64$

$0 = \left(x - 16\right) \left(x - 4\right)$

$x = 16 \mathmr{and} 4$

You must check the solutions inside the original equation to ensure they work. 4 does but 16 doesn't, so the solution is $\left\{4\right\}$

Practice exercises:

1. Solve for x in the following equations. Don't forget to check your solutions in the original equation.

a) $\sqrt{5 x + 6} = x$

b) $\sqrt{4 x + 1} - \sqrt{x + 2} = 1$

c) $\sqrt{2 x + 1} + \sqrt{3 x + 4} = \sqrt{6 x + 1}$

Good luck!