What is ((2x+5))/(x^2+2x-35)>= 0 ?

The solution to ((2x+5))/(x^2+2x-35)>= 0 is -7 5

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((2x+5))/(x^2+2x-35)>= 0

\frac{( 2 x + 5 )}{x ^{2} + 2 x - 35} \geq 0

- 7 < x \leq - \frac{5}{2} or x > 5

Interval Notation

(- 7, - \frac{5}{2}] \cup (5, \infty)

Decimal Notation

- 7 < x \leq - 2.5 or x > 5

Solution steps

\frac{2 x + 5}{x ^{2} + 2 x - 35} \geq 0

Factor

\frac{2 x + 5}{x ^{2} + 2 x - 35} : \frac{2 x + 5}{( x - 5 ) ( x + 7 )}

\frac{2 x + 5}{( x - 5 ) ( x + 7 )} \geq 0

Identify the intervals

- 7 < x \leq - \frac{5}{2} or x > 5

\frac{a ^{m}}{a ^{n}}

9 = - x^{2} - 6 x

x^{3} - 12 x^{2} + 48 x - 64

8 x^{3} + 27 g^{3}

0 = - x^{2} + 9

What is ((2x+5))/(x^2+2x-35)>= 0 ?
The solution to ((2x+5))/(x^2+2x-35)>= 0 is -7<x<=-5/2 \lor x>5