What is (x^2-x-12)/(x^2+x-6)>0 ?

The solution to (x^2-x-12)/(x^2+x-6)>0 is x 4

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(x^2-x-12)/(x^2+x-6)>0

\frac{x ^{2} - x - 12}{x ^{2} + x - 6} > 0

x < - 3 or - 3 < x < 2 or x > 4

Interval Notation

(- \infty, - 3) \cup (- 3, 2) \cup (4, \infty)

Solution steps

\frac{x ^{2} - x - 12}{x ^{2} + x - 6} > 0

Factor

\frac{x ^{2} - x - 12}{x ^{2} + x - 6} : \frac{( x + 3 ) ( x - 4 )}{( x - 2 ) ( x + 3 )}

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

Identify the intervals

x < - 3 or - 3 < x < 2 or x > 4

8 x^{2} + 16 x

3 m - 12 \geq 30

\frac{21 x ^{3} y}{14 x ^{2} y ^{2}}

5 x^{2} - 55 x + 150 = 0

4 x^{2} + 22 x + 10

What is (x^2-x-12)/(x^2+x-6)>0 ?
The solution to (x^2-x-12)/(x^2+x-6)>0 is x<-3\lor-3<x<2\lor x>4