# Question efd04

Feb 26, 2017

Begin by finding the restrictions on the domain

$5 + x \ge 0 \mathmr{and} 3 - x \ge 0$

Simplify

$x \ge - 5 \mathmr{and} - x \ge - 3$

$x \ge - 5 \mathmr{and} x \le 3$

$- 5 \le x \le 3$

Append the restrictions to the original equation:

sqrt(5+x)-sqrt(3-x)=2;-5<=x<=3

Square both sides:

(sqrt(5+x)-(sqrt(3-x))^2=2^2;-5<=x<=3

Expanding the square of the left side may appear to be a bit daunting but the pattern, ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$, makes it quite simple:

Let $a = \sqrt{5 + x} \mathmr{and} b = \sqrt{3 - x}$, then ${a}^{2} = 5 + x , {b}^{2} = 3 - x , \mathmr{and} - 2 a b = - 2 \sqrt{\left(3 - x\right) \left(5 + x\right)}$

5+x -2sqrt((3-x)(5+x)) + 3-x=4;-5<=x<=3

Collect everything but the radical term to one side:

-2sqrt((3-x)(5+x)) = -4;-5<=x<=3

Divide both sides by -2:

sqrt((3-x)(5+x)) = 2;-5<=x<=3

Square both sides:

(3-x)(5+x)=4;-5<=x<=3

Expand the left side, using the F.O.I.L. method:

15+3x-5x-x^2=4;-5<=x<=3

Combine like terms

11-2x-x^2=0;-5<=x<=3

Multiply both side by -1:

x^2+2x-11=0;-5<=x<=3#

Check the discriminant:

$d = {b}^{2} - 4 \left(a\right) \left(c\right) = {2}^{2} - 4 \left(1\right) \left(- 11\right) = 4 + 44 = 48$

$x = \frac{- b \pm \sqrt{d}}{2 a}$

$x = \frac{- 2 \pm \sqrt{48}}{2}$

$x = - 1 \pm \sqrt{12}$

$x = - 1 \pm 2 \sqrt{3}$

If you substitute the negative root into the original equation you obtain, $- 2 = 2$, which indicates an extraneous root caused by squaring; therefore, it must be discarded:

$x = 2 \sqrt{3} - 1$ is the only solution.