Question

How to solve x^2+2x-35 <0

Answer 1

$x^{2}+2x-35 \ \textless \ 0 \\ \\ x^{2}+7x-5x-35=0 \\ \\ x(x+7)-5(x+7)=0 \\ \\ (x-5)(x+7)=0 \\ \\ x-5=0 \ \vee \ x+7=0 \\ \\ x-5 \ \vee \ x=-7 \\ \\ a \ \textgreater \ 0 \\ \\ x\in (-7,5)$

Answer 2

Solve your inequality step-by-step.

x2+2x−35<0

Let's find the critical points of the inequality.

x2+2x−35=0

(x−5)(x+7)=0(Factor left side of equation)

x−5=0 or x+7=0(Set factors equal to 0)

x=5 or x=−7

Check intervals in between critical points. (Test values in the intervals to see if they work.)

x<−7(Doesn't work in original inequality)

−7<x<5(Works in original inequality)

x>5(Doesn't work in original inequality)