# How do you solve the rational equation x^2/(x-3) - 5/(x-3) = 0?

Jan 15, 2016

$x = \pm \sqrt{5}$

#### Explanation:

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

First, notice that the denominator is the same for both terms, so we can combine the numerator over the common denominator.

$\frac{{x}^{2} - 5}{x - 3} = 0$

Now we need to contend with the $x$ in the denominator. We can multiply both sides of the equation by $\left(x - 3\right)$. The left hand side cancels the denominator, and the right hand side multiplies by zero.

$\left({x}^{2} - 5\right) {\cancel{\frac{x - 3}{x - 3}}}^{1} = 0 {\cancel{\left(x - 3\right)}}^{0}$

Now we are left with;

${x}^{2} - 5 = 0$

We only have one $x$ term and a constant, so lets add $5$ to both sides.

${x}^{2} = 5$

Now we just need to get rid of the exponent. We can take the square root of both sides to get our $x$ values;

$x = \pm \sqrt{5}$