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For what value of x is $\frac{3{x}^{2}}{x+3}>\frac{4x}{x-2}$?
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Step 1
$\frac{3{x}^{2}}{x+3}>\frac{4x}{x-2}$
$4{x}^{2}+12x>3{x}^{3}-6{x}^{2}$
$3{x}^{3}-10{x}^{2}-12x<0$
$x\left[3{x}^{2}-10x-12\right]<0$
$x<0$
Step 2
now we will solve $\left[3{x}^{2}-10x-12\right]=0$
$x=a{x}^{2}+bx+c$
$a=3,b=-10\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}c=-12$
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
$x=\frac{10±\sqrt{100+144}}{6}$
$x=\frac{5±\sqrt{61}}{3}$
$x<\frac{5-\sqrt{61}}{3}$
$x<\frac{5+\sqrt{61}}{3}$
The answer is $x<\frac{5-\sqrt{61}}{3}$
$x<\frac{5+\sqrt{61}}{3}$
$x<0$