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{3 x - 5}{4 + 1} > \frac{1}{2 x - 1}

(4 x - 6)^{2} \geq (6 x - 4)^{2}

7 x^{9} - 4 > 17

x^{2} + x - 8 \geq x - x^{2} - 10

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

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