# For what value of x is (3x^2)/(x+3)>(4x)/(x-2)?

May 12, 2018

$x < \frac{5 - \sqrt{61}}{3}$
$x < \frac{5 + \sqrt{61}}{3}$
$x < 0$

#### Explanation:

show below

$\frac{3 {x}^{2}}{x + 3} > \frac{4 x}{x - 2}$

$4 {x}^{2} + 12 x > 3 {x}^{3} - 6 {x}^{2}$

$3 {x}^{3} - 10 {x}^{2} - 12 x < 0$

$x \left[3 {x}^{2} - 10 x - 12\right] < 0$

$x < 0$

now we will solve $\left[3 {x}^{2} - 10 x - 12\right] = 0$

$x = a {x}^{2} + b x + c$

$a = 3 , b = - 10 \mathmr{and} c = - 12$

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

$x = \frac{10 \pm \sqrt{100 + 144}}{6}$

$x = \frac{5 \pm \sqrt{61}}{3}$

$x < \frac{5 - \sqrt{61}}{3}$

$x < \frac{5 + \sqrt{61}}{3}$