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Q.

If ${ax}^{2} + bx + c = 0$ does not have real and distinct roots, then

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a

4a-2b +c > 0

b

4a-2b +c < 0

c

c(4a - 2b +c)>0

d

b(4a - 2b +c)<0

answer is D.

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Detailed Solution

If ax² + b x +c= 0 does not have real and distinct root, then

$\begin{array}{l} f (0) f (- 2) > 0 \\ \Rightarrow c (4 a - 2 b + c) > 0 \end{array}$

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