# How do you solve 1-13/x+36/x^2=0 and check for extraneous solutions?

May 15, 2017

$x = 9$ or $x = 4$ and there are no extraneous solutions.

#### Explanation:

Observe that as in denominator, we just have $x$ or ${x}^{2}$ only restrictions on domain is that $x \ne 0$

Let $\frac{1}{x} = u$ then $1 - \frac{13}{x} + \frac{36}{x} ^ 2 = 0$ can be written as

$36 {u}^{2} - 13 u + 1 = 0$

or $36 {u}^{2} - 9 u - 4 u + 1 = 0$

or $9 u \left(4 u - 1\right) - 1 \left(4 u - 1\right) = 0$

or $\left(9 u - 1\right) \left(4 u - 1\right) = 0$

i.e. $u = \frac{1}{9}$ or $u = \frac{1}{4}$

Hence $x = 9$ or $x = 4$ and there are no extraneous solutions.