# Factoring

• May 19th 2010, 10:04 PM
hydride
Factoring
$\displaystyle 3x - \frac{7}{x} = 5$

I dont like fractions.-_-"
• May 19th 2010, 10:31 PM
Prove It
Quote:

Originally Posted by hydride
$\displaystyle 3x - \frac{7}{x} = 5$

I dont like fractions.-_-"

$\displaystyle 3x - \frac{7}{x} = 5$

$\displaystyle x\left(3x - \frac{7}{x}\right) = 5x$

$\displaystyle 3x^2 - 7 = 5x$

$\displaystyle 3x^2 - 5x - 7 = 0$

$\displaystyle 3\left(x^2 - \frac{5}{3} - \frac{7}{3}\right) = 0$

$\displaystyle 3\left[x^2 - \frac{5}{3} + \left(-\frac{5}{6}\right)^2 - \left(-\frac{5}{6}\right)^2 - \frac{7}{3}\right] = 0$

$\displaystyle 3\left[\left(x - \frac{5}{6}\right)^2 - \frac{25}{36} - \frac{84}{36} \right] = 0$

$\displaystyle 3\left[\left(x - \frac{5}{6}\right)^2 - \frac{109}{36}\right] = 0$

$\displaystyle 3\left[\left(x - \frac{5}{6}\right)^2 - \left(\frac{\sqrt{109}}{6}\right)^2\right] = 0$

$\displaystyle 3\left(x - \frac{5}{6} + \frac{\sqrt{109}}{6}\right)\left(x - \frac{5}{6} - \frac{\sqrt{109}}{6}\right) = 0$