# How do you solve 1 + 8/(x - 5) = 3/x and find any extraneous solutions?

Aug 31, 2016

There are no Real solutions, but there are Complex solutions:

$x = \pm \sqrt{15} i$

#### Explanation:

Given:

$1 + \frac{8}{x - 5} = \frac{3}{x}$

Note that neither $0$ nor $5$ can be a solution since they result in division by $0$. So if our derived equations have solutions $0$ or $5$ then they are extraneous.

Multiply both sides by $x$ (possibly introducing an extraneous solution $0$) to get:

$x + \frac{8 x}{x - 5} = 3$

Multiply both sides by $\left(x - 5\right)$ (possibly introducing an extraneous solution $5$) to get:

$x \left(x - 5\right) + 8 x = 3 \left(x - 5\right)$

Expand both sides to get:

${x}^{2} - 5 x + 8 x = 3 x - 15$

which simplifies to:

${x}^{2} + 3 x = 3 x - 15$

Subtract $3 x$ from both sides to get:

${x}^{2} = - 15$

This has no Real solutions since ${x}^{2} \ge 0$ for any Real value of $x$.

If we are interested in Complex solutions, then add $15$ to both sides and transpose to get:

$0 = {x}^{2} + 15 = {x}^{2} - {\left(\sqrt{15} i\right)}^{2} = \left(x - \sqrt{15} i\right) \left(x + \sqrt{15} i\right)$

where $i$ is the imaginary unit, with the property that ${i}^{2} = - 1$.

Hence solutions $x = \pm \sqrt{15} i$