# How do you solve the quadratic 4-1/x=3/x^2 using any method?

Dec 27, 2016

Get rid of the denominators by multiplying the whole equation by $\textcolor{b l u e}{{x}^{2}}$

$4 \textcolor{b l u e}{\times {x}^{2}} - \frac{1}{x} \textcolor{b l u e}{\times {x}^{2}} = \frac{3}{x} ^ 2 \textcolor{b l u e}{\times {x}^{2}}$

$4 {x}^{2} - x = 3 \text{ } \leftarrow$ make equal to 0

$4 {x}^{2} - x - 3 = 0$

Find factors of 4 and 3 which subtract to give 1.

$\left(4 x + 3\right) \left(x - 1\right) = 0 \text{ } \leftarrow$ either factor could be =0

If $4 x + 3 = 0 \text{ "rarr 4x=-3" } \rightarrow x = - \frac{3}{4}$

If$x - 1 = 0 \text{ } x = 1$