# How do you solve x^2 / (x+3) - 5 / (x+3) = 0 and find any extraneous solutions?

Sep 27, 2016

$x = \pm \sqrt{5}$

#### Explanation:

As there are only two fractions - let's move one to the other side.
Then we can cross-multiply..

$\textcolor{b l u e}{{x}^{2} / \left(x + 3\right) = \frac{5}{x + 3} \text{ } \leftarrow \left(x + 3 \ne 0\right)}$

${x}^{2} \left(x + 3\right) = 5 \left(x + 3\right)$

${x}^{3} + 3 {x}^{2} = 5 x + 15 \text{ } \leftarrow$ no fractions! Make equal to 0

${x}^{3} + 3 {x}^{2} - 5 x - 15 = 0 \text{ } \leftarrow$ find factors by grouping

${x}^{2} \left(x + 3\right) - 5 \left(x + 3\right) = 0 \text{ } \leftarrow$ common bracket

$\left(x + 3\right) \left({x}^{2} - 5\right) = 0$

If $x + 3 = 0 \text{ } \rightarrow x = - 3$ (extraneous solution x-3 !=0)

If ${x}^{2} - 5 = 0 \text{ "rarr x = +-sqrt5 " } \leftarrow$ the only solutions.

However if we consider the fraction right at the beginning, we could have arrived at the answer in a much quicker way!

$\textcolor{b l u e}{{x}^{2} / \left(x + 3\right) = \frac{5}{x + 3} \text{ } \leftarrow \left(x + 3 \ne 0\right)}$

The denominators are equal, so the numerators are equal..

${x}^{2} = 5$

$x = \pm \sqrt{5}$

$x + 3 \ne 0 , \text{ so } x \ne - 3$