# How do you solve  (x^2-8x+4)/(x^2-2) = 3?

Jun 6, 2016

First, let's determine the restrictions on x. This can be done by setting the denominator to 0 and solving for x.

${x}^{2} - 2 = 0$

${x}^{2} = 2$

$x = \pm \sqrt{2}$

Therefore, $x \ne \pm \sqrt{2}$

Now we can solve:

${x}^{2} - 8 x + 4 = 3 \left({x}^{2} - 2\right)$

${x}^{2} - 8 x + 4 = 3 {x}^{2} - 6$

$0 = 2 {x}^{2} + 8 x - 10$

$0 = 2 {x}^{2} + 10 x - 2 x - 10$

$0 = 2 x \left(x + 5\right) - 2 \left(x + 5\right)$

$0 = \left(2 x - 2\right) \left(x + 5\right)$

$x = 1 \mathmr{and} - 5$

Checking our solutions in the original equation we find that both work.

Our solution set is therefore {x = 1, -5; x !=+-sqrt(2)}

Hopefully this helps!