# How do you solve sqrt(x + 6) + sqrt(2-x) = 4?

Oct 7, 2015

I found $x = - 2$

#### Explanation:

Let us try squaring both sides:

${\left[\sqrt{x + 6} + \sqrt{2 - x}\right]}^{2} = {4}^{2}$

$\cancel{x} + 6 + 2 \sqrt{x + 6} \sqrt{2 - x} + 2 \cancel{- x} = 16$

$8 + 2 \sqrt{\left(x + 6\right) \left(2 - x\right)} = 16$ taking $8$ to the right:

$2 \sqrt{\left(x + 6\right) \left(2 - x\right)} = 8$ taking $2$ to the right (dividing):

$\sqrt{\left(x + 6\right) \left(2 - x\right)} = 4$

let us square again:

$\left(x + 6\right) \left(2 - x\right) = 16$

$2 x - {x}^{2} + 12 - 6 x - 16 = 0$

$- {x}^{2} - 4 x - 4 = 0$ using the Quadratic Formula we get:

${x}_{1 , 2} = \frac{4 \pm \sqrt{16 - 4 \left(- 4 \cdot - 1\right)}}{-} 2 = \frac{4 \pm \sqrt{0}}{-} 2 = - 2$

Let us try this solution into our original equation as $x$:
$\sqrt{- 2 + 6} + \sqrt{2 + 2} = 2 + 2 = 4$ Yes.