# What is the solution set for 8/(x+2)=(x+4)/(x-6)?

Aug 12, 2015

There are no real solutions and two complex solutions $x = 1 \setminus \pm i \sqrt{55}$

#### Explanation:

First, cross multiply to get $8 \left(x - 6\right) = \left(x + 2\right) \left(x + 4\right)$. Next, expand to get $8 x - 48 = {x}^{2} + 6 x + 8$. Now rearrange to obtain ${x}^{2} - 2 x + 56 = 0$.

The quadratic formula now gives solutions

$x = \frac{2 \setminus \pm \sqrt{4 - 224}}{2} = 1 \setminus \pm \frac{1}{2} \sqrt{- 220}$

$= 1 \setminus \pm \frac{1}{2} i \sqrt{4} \sqrt{55} = 1 \setminus \pm i \sqrt{55}$

These are definitely worth checking in the original equation. I'll check the first and you can check the second.

The left-hand-side of the original equation, upon substitution of $x = 1 + i \sqrt{55}$ becomes:

$\frac{8}{3 + i \sqrt{55}} = \frac{8 \left(3 - i \sqrt{55}\right)}{9 + 55} = \frac{3}{8} - i \frac{\sqrt{55}}{8}$

Now do the same substitution on the right-hand-side of the original equation:

$\frac{5 + i \sqrt{55}}{- 5 + i \sqrt{55}} = \frac{\left(5 + i \sqrt{55}\right) \cdot \left(- 5 - i \sqrt{55}\right)}{25 + 55}$

$= \frac{- 25 - 10 i \sqrt{55} + 55}{80} = \frac{3}{8} - i \frac{\sqrt{55}}{8}$

It works! :-)