# How do you solve the quadratic 3+5/(2x)=1/x^2 using any method?

Sep 21, 2016

$x = 0.295 \text{ or } x = - 1.129$

#### Explanation:

If you have an equation which has fractions, you can get rid of them by multiplying each term by the LCM of the denominators.

In the equation $\text{ "3+5/(2x)=1/x^2" the LCD is } \textcolor{b l u e}{2 {x}^{2}}$

$\textcolor{w h i t e}{x} \textcolor{b l u e}{2 {x}^{2}} \times 3 + \textcolor{b l u e}{{\cancel{2 {x}^{2}}}^{x}} \times \frac{5}{\cancel{2 x}} = \textcolor{b l u e}{2 {\cancel{x}}^{2}} \times \frac{1}{\cancel{x}} ^ 2$

$\textcolor{w h i t e}{\times \times \times \times \times x} 6 {x}^{2} + 5 x = 2 \text{ } \leftarrow$ no fractions!

$6 {x}^{2} + 5 x - 2 = 0 \text{ } \rightarrow$ the options are: $\text{find factors}$
$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times x . \times \times \times \times x} \text{complete the square}$
$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times x . \times \times \times \times x} \text{quadratic formula}$

This quadratic expression does not factor, use the formula.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \left(6\right) \left(- 2\right)}}{2 \left(6\right)}$

$x = \frac{- 5 \pm \sqrt{25 + 48}}{12}$

$x = \frac{- 5 + \sqrt{73}}{12} \text{ or } x = \frac{- 5 + \sqrt{73}}{12}$

$x = 0.295 \text{ or } x = - 1.129$