# How do you solve  [sqrt(4 - x)] - [sqrt(x + 6)] = 2 and find any extraneous solutions?

Jul 10, 2017

Since the arguments of square roots may not be negative, we already know that $x \le 4 \mathmr{and} x \ge - 6$

#### Explanation:

Or: $- 6 \le x \le + 4$

And now for the difference of $2$:
It may be clear that $x$ has to be negative, or the difference ($2$) will never be positive.
We try whole values, and $x = - 5$ works out, as
$\sqrt{4 - \left(- 5\right)} - \sqrt{- 5 + 6} = \sqrt{9} - \sqrt{1} = 3 - 1 = 2$
graph{sqrt(4-x)-sqrt(x+6) [-11.25, 11.25, -5.625, 5.625]}

Jul 10, 2017

color(blue)(x=4 or x=-6 extraneous solution

#### Explanation:

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

$\therefore \sqrt{4 - x} = 2 + \sqrt{x + 6}$

square both sides

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

$\sqrt{4 - x} \cdot \sqrt{4 - x} = 4 - x$

$\therefore 4 - x = 10 + 4 \sqrt{x + 6} + x$

$\therefore 4 - 10 - x - x = 4 \sqrt{x + 6}$

$\therefore - 6 - 2 x = 4 \sqrt{x + 6}$

square both sides

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

$\therefore 36 + 24 x + 4 {x}^{2} = 16 {\left(\sqrt{x + 6}\right)}^{2}$

$\therefore 36 + 24 x + 4 {x}^{2} = 16 \left(x + 6\right)$

$\therefore 36 + 24 x + 4 {x}^{2} = 16 x + 96$

$\therefore 36 - 96 + 24 x - 16 x + 4 {x}^{2} = 0$

$\therefore 4 {x}^{2} + 8 x - 96 = 0$

$\therefore 4 \left({x}^{2} + 2 x - 24 = 0\right)$

$\therefore \left(x - 4\right) \left(x + 6\right) = 0$

:.color(blue)(x=4 or x=-6 extraneous solution