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

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

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

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

- 2 < x \leq 1

Interval Notation

(- 2, 1]

Solution steps

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

Factor

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

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

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

\frac{x - 1}{x + 2} \leq 0

Identify the intervals

- 2 < x \leq 1

\frac{( 2 x ^{2} + 3 x + 5 )}{( 2 x ^{2} + 8 x + 1 )} \leq 0

\frac{( x ^{2} - 3 x + 7 )}{( x ^{2} + 1 )} < 1

\frac{13}{( x - 3 )} < 0

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

(t + 2)^{2} \cdot t (t + 1) > t^{2} \cdot (t + 1)^{2}

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