SOLUTION: Find the solution set of x^2+2x-8>0 a. {x: -4<x<2} b. {x: x<-4 or x>2} c. {x: -2<x<4} d. {x: x<-2 or x>4} I know that answer is b. But I don't get how to get that. Pleas

Algebra ->  Graphs -> SOLUTION: Find the solution set of x^2+2x-8>0 a. {x: -4<x<2} b. {x: x<-4 or x>2} c. {x: -2<x<4} d. {x: x<-2 or x>4} I know that answer is b. But I don't get how to get that. Pleas      Log On


   

Question 389988: Find the solution set of x^2+2x-8>0
a. {x: -4 b. {x: x<-4 or x>2}
c. {x: -2 d. {x: x<-2 or x>4}
I know that answer is b.
But I don't get how to get that.
Please solve it.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi,        



Note: x^2+2x-8 = 0 when x = 2, x = -4.  Note: Factors are (x+4)(x-2)

for x^2+2x-8 > 0 , then {x: x<-4 or x>2}