The expression $y = a x^{2} + b x + c$ has always the same sign as, c if

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$4 a c > b^{2}$

$a c < b^{2}$

$a c > b^{2}$

$4 a c < b^{2}$

answer is B.

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Detailed Solution

Let $f (x) = a x^{2} + b x + c .$ Then $, f (0) = c$

Thus, the curve $y = f (x)$ meets $y$ $-axis$ at $(0, c)$

If $c > 0,$ then by hypothesis $f (x) > 0 .$ This means that the curve $y = f (x)$ does not meet x-axis.

If $c < 0,$ then by hypothesis, $f (x) < 0,$ which means that the curve $y = f (x)$ is always below x-axis and so it does not intersect with x-axis.

Thus, in both the cases $y = f (x)$ does not intersect with x-axis i.e. $f (x) \neq 0$ for any real $x .$

Hence, $f (x) = 0$ i.e. $a x^{2} + b x + c = 0$ has imaginary roots and so we have $b^{2} < 4 a c$

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