What is ((x^2+4x+4))/((2x^2-x-1))>0 ?

The solution to ((x^2+4x+4))/((2x^2-x-1))>0 is x 1

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((x^2+4x+4))/((2x^2-x-1))>0

\frac{( x ^{2} + 4 x + 4 )}{( 2 x ^{2} - x - 1 )} > 0

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

Interval Notation

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

Decimal Notation

x < - 2 or - 2 < x < - 0.5 or x > 1

Solution steps

\frac{x ^{2} + 4 x + 4}{2 x ^{2} - x - 1} > 0

Factor

\frac{x ^{2} + 4 x + 4}{2 x ^{2} - x - 1} : \frac{( x + 2 ) ^{2}}{( 2 x + 1 ) ( x - 1 )}

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

Identify the intervals

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

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

\frac{( 2 x + 3 )}{( x ^{2} + x - 12 )} \leq \frac{1}{2}

\frac{3}{2 x - 1} \geq \frac{5}{x - 1}

\frac{1}{x} - 2 - \frac{1}{x} \leq \frac{2}{x} + 2

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

What is ((x^2+4x+4))/((2x^2-x-1))>0 ?
The solution to ((x^2+4x+4))/((2x^2-x-1))>0 is x<-2\lor-2<x<-1/2 \lor x>1