SOLUTION: For what value of k does x^2 + kx + 1 = 0 have two different real roots? This is what I have done so far: D > 0 k^2 - 4(1)(1) > 0 k^2 > 4 sqrt(k) > + or - sqrt(4)

Algebra ->  Equations -> SOLUTION: For what value of k does x^2 + kx + 1 = 0 have two different real roots? This is what I have done so far: D > 0 k^2 - 4(1)(1) > 0 k^2 > 4 sqrt(k) > + or - sqrt(4)       Log On


   

Question 1054306: For what value of k does x^2 + kx + 1 = 0 have two different real roots?
This is what I have done so far:
D > 0
k^2 - 4(1)(1) > 0
k^2 > 4
sqrt(k) > + or - sqrt(4)
k > + or - 2
My answer sheet says k < -2 or k > 2, but I don't understand how to get to that answer.
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
the discriminant, b^2-4ac=k^2-4>0
k^2>4
abs(k)>2
that means k>2 OR k<-2