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

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Answer 1 · 2018-05-12T23:15:52

The answer
$x<[5-sqrt(61)]/(3)$
$x<[5+sqrt(61)]/(3)$
$x<0$

show below

$(3x^2)/(x+3)>(4x)/(x-2)$

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

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

$x[3x^2-10x-12]<0$

$x<0$

now we will solve $[3x^2-10x-12]=0$

$x=ax^2+bx+c$

$a=3, b=-10 and c=-12$

$x=[-b+-sqrt(b^2-4ac)]/(2a)$

$x=[10+-sqrt(100+144)]/(6)$

$x=[5+-sqrt(61)]/(3)$

$x<[5-sqrt(61)]/(3)$

$x<[5+sqrt(61)]/(3)$

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