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{\frac{8 x}{x ^{2} - 1}}{\frac{10 x}{x + 1}}

- 16 t^{2} + 36 t

\frac{3 π}{4}

\frac{10}{3 ( 4 ) ^{\frac{7}{2}}}

5 x^{2} - 2 x + 5 = 0

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