What is (x^2+4x+10)/(x^2-x-12)>0 ?

The solution to (x^2+4x+10)/(x^2-x-12)>0 is x 4

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(x^2+4x+10)/(x^2-x-12)>0

\frac{x ^{2} + 4 x + 10}{x ^{2} - x - 12} > 0

x < - 3 or x > 4

Solution steps

\frac{x ^{2} + 4 x + 10}{x ^{2} - x - 12} > 0

Factor

\frac{x ^{2} + 4 x + 10}{x ^{2} - x - 12} : \frac{x ^{2} + 4 x + 10}{( x + 3 ) ( x - 4 )}

\frac{x ^{2} + 4 x + 10}{( x + 3 ) ( x - 4 )} > 0

Identify the intervals

x < - 3 or x > 4

\frac{- 2}{4}

4 (x - 4) + 19 = 3 (x + 3)

\frac{x ^{7} y ^{5} z ^{6}}{x ^{5} y ^{4} z ^{2}}

4 x^{2} + 20 x = 0

\frac{10 !}{5}

What is (x^2+4x+10)/(x^2-x-12)>0 ?
The solution to (x^2+4x+10)/(x^2-x-12)>0 is x<-3\lor x>4