What is ((x+1))/((x^2-x-12))>2 ?

The solution to ((x+1))/((x^2-x-12))>2 is -3<x<(3-sqrt(209))/4 \lor 4<x<(3+sqrt(209))/4

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((x+1))/((x^2-x-12))>2

\frac{( x + 1 )}{( x ^{2} - x - 12 )} > 2

- 3 < x < \frac{3 - 209}{4} or 4 < x < \frac{3 + 209}{4}

Interval Notation

(- 3, \frac{3 - 209}{4}) \cup (4, \frac{3 + 209}{4})

Decimal Notation

- 3 < x < - 2.86420 \dots or 4 < x < 4.36420 \dots

Solution steps

\frac{x + 1}{x ^{2} - x - 12} > 2

Rewrite in standard form

\frac{- 2 x ^{2} + 3 x + 25}{x ^{2} - x - 12} > 0

Factor

\frac{- 2 x ^{2} + 3 x + 25}{x ^{2} - x - 12} : \frac{- ( x - \frac{3 + 209}{4} ) ( x - \frac{3 - 209}{4} )}{( x + 3 ) ( x - 4 )}

\frac{- ( x - \frac{3 + 209}{4} ) ( x - \frac{3 - 209}{4} )}{( x + 3 ) ( x - 4 )} > 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x - \frac{3 + 209}{4} ) ( x - \frac{3 - 209}{4} ) ) ( - 1 )}{( x + 3 ) ( x - 4 )} < 0 \cdot (- 1)

Simplify

\frac{( x - \frac{3 + 209}{4} ) ( x - \frac{3 - 209}{4} )}{( x + 3 ) ( x - 4 )} < 0

Identify the intervals

- 3 < x < \frac{3 - 209}{4} or 4 < x < \frac{3 + 209}{4}

\frac{( 4 x - 1 )}{( 3 x + 1 )} > 0

\frac{( x ^{2} )}{( 2 x ^{2} + 4 x + 5 )} > 0

\frac{( ( x + 3 ) ( x - 1 ) )}{( x ^{2} \cdot ( x - 20 ) ^{3} )} \leq 0

(x + 1) (x - 1) (x - 3) > 0

(x^{2}) - 4 x + 3 < 0

What is ((x+1))/((x^2-x-12))>2 ?
The solution to ((x+1))/((x^2-x-12))>2 is -3<x<(3-sqrt(209))/4 \lor 4<x<(3+sqrt(209))/4