SOLUTION: Find an interval for b that equation has at least 1 real solution x^2 + bx +64 = 0

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Question 1184481: Find an interval for b that equation has at least 1 real solution
x^2 + bx +64 = 0

Found 3 solutions by josgarithmetic, ikleyn, Edwin McCravy:
Answer by josgarithmetic(39613) About Me  (Show Source):
You can put this solution on YOUR website!
One or two real solutions for x%5E2%2Bbx%2B64=0


b%5E2-256%3E=0-------discriminant must be nonzero.

b%5E2%3E=256or%28b-16%29%28b%2B16%29%3E=0


Answer by ikleyn(52747) About Me  (Show Source):
You can put this solution on YOUR website!
.

For it, the discriminant of the quadratic function must be greater than or equal to zero,

which leads to inequality  


    b^2 - 4*64 >= 0.


It gives


    b^2 >= 256,   


with the solutions  b >= 16  OR  b <= -16.    ANSWER

Solved.



Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
It will have at least 1 real solution if its discriminant is non-negative.

x%5E2+%2B+bx+%2B64+=+0

has discriminant

b%5E2-4ac%3E=0}

b%5E2-4%281%29%2864%29%3E=0

b%5E2-256%3E=0

%28b-16%29%28b%2B16%29%3E=0

Solution: 

Edwin