SOLUTION: Suppose at^2+ 5t + 4 > 0 for every real number t. Show that a >25/16.

Question 942945: Suppose
at^2+ 5t + 4 > 0
for every real number t. Show that a >25/16.
Found 2 solutions by richard1234, josgarithmetic:
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
The given quadratic never intersects the x-axis so it has two non-real roots. Also, it is above the x-axis so a > 0.

Therefore the discriminant must be < 0.

Rearranging, this is equivalent to .
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
Critical values are the roots.

roots:
and you want the discriminant to be real, not complex with any imaginaries.

, the requirement for the discriminant.
;
Your really want

different from what you expected.
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