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

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

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

\frac{1 - x ^{2}}{x + 2} \geq 0

x < - 2 or - 1 \leq x \leq 1

Interval Notation

(- \infty, - 2) \cup [- 1, 1]

Solution steps

\frac{1 - x ^{2}}{x + 2} \geq 0

Factor

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

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

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

Identify the intervals

x < - 2 or - 1 \leq x \leq 1

3 x^{2} + 8 x = 0

16 x - 7 \leq - 71

3 e^{z^{2}} = 9

(x - 2) (x + 8) \geq 0

\frac{12}{- 4}

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