What is (x^2+x-12)/(x^2-4x+4)>0 ?

The solution to (x^2+x-12)/(x^2-4x+4)>0 is x 3

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(x^2+x-12)/(x^2-4x+4)>0

\frac{x ^{2} + x - 12}{x ^{2} - 4 x + 4} > 0

x < - 4 or x > 3

Solution steps

\frac{x ^{2} + x - 12}{x ^{2} - 4 x + 4} > 0

Factor

\frac{x ^{2} + x - 12}{x ^{2} - 4 x + 4} : \frac{( x - 3 ) ( x + 4 )}{( x - 2 ) ^{2}}

\frac{( x - 3 ) ( x + 4 )}{( x - 2 ) ^{2}} > 0

Identify the intervals

x < - 4 or x > 3

lo g_{2} (\frac{1}{50})

\frac{5}{6} h - \frac{7}{12} = - \frac{3}{4} h - \frac{13}{6}

16 + 8 x - 25 = - 7 (3 - x) + 3 x

(\frac{x ( x + 1 )}{2})^{2}

(\frac{π}{6})^{3}

What is (x^2+x-12)/(x^2-4x+4)>0 ?
The solution to (x^2+x-12)/(x^2-4x+4)>0 is x<-4\lor x>3