What is (9-x^2)/(x^2-x-2)>= 0 ?

The solution to (9-x^2)/(x^2-x-2)>= 0 is -3<= x<-1\lor 2<x<= 3

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(9-x^2)/(x^2-x-2)>= 0

\frac{9 - x ^{2}}{x ^{2} - x - 2} \geq 0

- 3 \leq x < - 1 or 2 < x \leq 3

Solution steps

\frac{9 - x ^{2}}{x ^{2} - x - 2} \geq 0

Factor

\frac{9 - x ^{2}}{x ^{2} - x - 2} : \frac{- ( x + 3 ) ( x - 3 )}{( x + 1 ) ( x - 2 )}

\frac{- ( x + 3 ) ( x - 3 )}{( x + 1 ) ( x - 2 )} \geq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x + 3 ) ( x - 3 ) ) ( - 1 )}{( x + 1 ) ( x - 2 )} \leq 0 \cdot (- 1)

Simplify

\frac{( x + 3 ) ( x - 3 )}{( x + 1 ) ( x - 2 )} \leq 0

Identify the intervals

- 3 \leq x < - 1 or 2 < x \leq 3

e^{2 iπ}

3 (- 2 x - 7) > 9

6 x - \frac{2}{3} x^{2} = 3

(e^{2})^{3}

What is (9-x^2)/(x^2-x-2)>= 0 ?
The solution to (9-x^2)/(x^2-x-2)>= 0 is -3<= x<-1\lor 2<x<= 3