Question 41234: Please help with this.
Decide all values of b that will give one or more real number solutions.
5x^2 + bx + 1 = 0
Thank You! Answer by fractalier(6550) (Show Source):
You can put this solution on YOUR website! We look at the discriminant...its value must be ≥ 0 to be real...so
b^2 - 4ac ≥ 0
b^2 - 4(5)(1) ≥ 0
b^2 - 20 ≥ 0
b^2 ≥ 20
|b| ≥ sqrt(20)
|b| ≥ 2*sqrt(5)
