What is (x^2+9x-22)/(x+4)>= 0 ?

The solution to (x^2+9x-22)/(x+4)>= 0 is -11 = 2

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(x^2+9x-22)/(x+4)>= 0

\frac{x ^{2} + 9 x - 22}{x + 4} \geq 0

- 11 \leq x < - 4 or x \geq 2

Interval Notation

[- 11, - 4) \cup [2, \infty)

Solution steps

\frac{x ^{2} + 9 x - 22}{x + 4} \geq 0

Factor

\frac{x ^{2} + 9 x - 22}{x + 4} : \frac{( x - 2 ) ( x + 11 )}{x + 4}

\frac{( x - 2 ) ( x + 11 )}{x + 4} \geq 0

Identify the intervals

- 11 \leq x < - 4 or x \geq 2

2 - (- 8)

\frac{R}{10} \geq - 1

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What is (x^2+9x-22)/(x+4)>= 0 ?
The solution to (x^2+9x-22)/(x+4)>= 0 is -11<= x<-4\lor x>= 2