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.Com
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) (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
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