What is ((x^2-9))/((3x-x^2-24))<0 ?

The solution to ((x^2-9))/((3x-x^2-24)) 3

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((x^2-9))/((3x-x^2-24))<0

\frac{( x ^{2} - 9 )}{( 3 x - x ^{2} - 24 )} < 0

x < - 3 or x > 3

Interval Notation

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

Solution steps

\frac{x ^{2} - 9}{3 x - x ^{2} - 24} < 0

Factor

\frac{x ^{2} - 9}{3 x - x ^{2} - 24} : \frac{( x + 3 ) ( x - 3 )}{3 x - x ^{2} - 24}

\frac{( x + 3 ) ( x - 3 )}{- ( - 3 x + x ^{2} + 24 )} < 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( x + 3 ) ( x - 3 ) ( - 1 )}{- ( - 3 x + x ^{2} + 24 )} > 0 \cdot (- 1)

Simplify

\frac{( x + 3 ) ( x - 3 )}{- 3 x + x ^{2} + 24} > 0

Identify the intervals

x < - 3 or x > 3

2 x^{2} - 2 x - 1 < 0

\frac{( 2 x )}{( 2 x ^{2} + 5 x + 2 )} > \frac{1}{( x + 1 )}

\frac{1}{2 ( 3 - 8 x )} > 20 (\frac{1 - 1}{5 x})

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

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

What is ((x^2-9))/((3x-x^2-24))<0 ?
The solution to ((x^2-9))/((3x-x^2-24))<0 is x<-3\lor x>3