What is ((x^2-5x))/((x+2))>= 0 ?

The solution to ((x^2-5x))/((x+2))>= 0 is -2 = 5

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((x^2-5x))/((x+2))>= 0

\frac{( x ^{2} - 5 x )}{( x + 2 )} \geq 0

- 2 < x \leq 0 or x \geq 5

Interval Notation

(- 2, 0] \cup [5, \infty)

Solution steps

\frac{x ^{2} - 5 x}{x + 2} \geq 0

Factor

\frac{x ^{2} - 5 x}{x + 2} : \frac{x ( x - 5 )}{x + 2}

\frac{x ( x - 5 )}{x + 2} \geq 0

Identify the intervals

- 2 < x \leq 0 or x \geq 5

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

(x - 2)^{4} \cdot (x - 3)^{3} \cdot (x - 4)^{2} \cdot (1 - x) \leq 0

\frac{( ( x - 1 ) ( x - 3 ) )}{( ( x + 1 ) ( x - 5 ) )} \geq 0

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

x^{2} - 17 < 0

What is ((x^2-5x))/((x+2))>= 0 ?
The solution to ((x^2-5x))/((x+2))>= 0 is -2<x<= 0\lor x>= 5