# How do you solve \sqrt { ( x - 2) ( x + 4) } - x - 2= 0?

May 24, 2018

No solutions

#### Explanation:

writing $\sqrt{\left(x - 2\right) \left(x + 4\right)} = x + 2$
by squaring we get
${x}^{2} + 2 x - 8 = {x}^{2} + 4 x + 4$
or
$- 12 = 2 x$ so $x = - 6$
This fulfilles not our equation since
$\sqrt{\left(- 8\right) \left(- 2\right)} \ne - 4$

May 24, 2018

Move values that are not under the radical to the other side and square both sides. After Rearranging, you will find $x = - 6$

#### Explanation:

First, move the values that aren't under the radical to the Right Hand Side (RHS):

$\sqrt{\left(x - 2\right) \left(x + 4\right)} \cancel{- x - 2} \textcolor{red}{\cancel{+ x + 2}} = 0 \textcolor{red}{+ x + 2}$

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

Next, we'll FOIL the two expressions that are under the radical:

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

Next, we'll square both sides to get rid of the radical:

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

${x}^{2} + 2 x - 8 = {x}^{2} + 4 x + 4$

We can eliminate the ${x}^{2}$ term since its the same for both sides. We will also begin to rearrange the equation so all of the $x$-terms are on one side and all of the constants are on the other:

$\cancel{2 x} - 8 \textcolor{b l u e}{\cancel{- 2 x}} \textcolor{red}{- 4} = 4 x \cancel{+ 4} \textcolor{b l u e}{- 2 x} \textcolor{red}{\cancel{- 4}}$

$- 8 \textcolor{red}{- 4} = 4 x \textcolor{b l u e}{- 2 x}$

$- 12 = 2 x$

Finally, divide both sides by $x$'s coefficient:

$- \frac{12}{\textcolor{red}{2}} = \frac{\cancel{2} x}{\textcolor{red}{\cancel{2}}}$

$\textcolor{g r e e n}{x = - 6}$

Let's apply this back in the (somewhat) original equation:

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

$\sqrt{\left(\textcolor{g r e e n}{\left(- 6\right)} - 2\right) \left(\textcolor{g r e e n}{\left(- 6\right)} + 4\right)} = \textcolor{g r e e n}{\left(- 6\right)} + 2$

$\sqrt{\left(- 8\right) \left(- 2\right)} = - 4$

$\sqrt{16} = - 4$

Now, here's a tricky part. The square root of a number can be either positive or negative, since any real number squared is ALWAYS positive.

$\pm 4 = - 4$

$- 4 = - 4$