First slide

Theory of equations

Question

If ${(b^{2} - 4 ac)}^{2} (1 + 4 a^{2}) < 64 a^{2}, a < 0$ , then the maximum value of quadratic expression ${ax}^{2} + bx + c$ is always less than

Easy

A

0
B

2
C

-1
D

-2

Solution

$\frac{{(b^{2} - 4 ac)}^{2}}{16 a^{2}} < \frac{4}{1 + 4 a^{2}} (1)$
Now, $max ({ax}^{2} + bx + c) = - \frac{b^{2} - 4 ac}{4 a}$
Also, $\frac{- 2}{\sqrt{1 + 4 a^{2}}} < - \frac{b^{2} - 4 ac}{4 a} < \frac{2}{\sqrt{1 + 4 a^{2}}}$ [From (1)]
So, maximum value is always less than 2 (when $a \to 0$ )

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