What is (8-3x^2)/((x^2)+x-2)>0 ?

The solution to (8-3x^2)/((x^2)+x-2)>0 is -2<x<-(2sqrt(2))/(sqrt(3))\lor 1<x<(2sqrt(2))/(sqrt(3))

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(8-3x^2)/((x^2)+x-2)>0

\frac{8 - 3 x ^{2}}{( x ^{2} ) + x - 2} > 0

- 2 < x < - \frac{2 2}{3} or 1 < x < \frac{2 2}{3}

Interval Notation

(- 2, - \frac{2 2}{3}) \cup (1, \frac{2 2}{3})

Decimal Notation

- 2 < x < - 1.63299 \dots or 1 < x < 1.63299 \dots

Solution steps

\frac{8 - 3 x ^{2}}{x ^{2} + x - 2} > 0

Factor

\frac{8 - 3 x ^{2}}{x ^{2} + x - 2} : \frac{- ( 3 x + 2 2 ) ( 3 x - 2 2 )}{( x - 1 ) ( x + 2 )}

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

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

Identify the intervals

- 2 < x < - \frac{2 2}{3} or 1 < x < \frac{2 2}{3}

(x - 1) (4 - x) (x - 5)^{2} < 0

4 x^{2} - 7 x - 15 > 0

\frac{3}{( x + 1 )} > 0

2 x - x^{2} \geq 1 x

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

What is (8-3x^2)/((x^2)+x-2)>0 ?
The solution to (8-3x^2)/((x^2)+x-2)>0 is -2<x<-(2sqrt(2))/(sqrt(3))\lor 1<x<(2sqrt(2))/(sqrt(3))