The expression y = a x^2 + bx + c has always the same sign as c i… - Vidyadip

MCQ

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

A
$4ac < {b^2}$
✓
$4ac > {b^2}$
C
$ac < {b^2}$
D
$ac > {b^2}$

✓

Answer

Correct option: B.

$4ac > {b^2}$

b
(b) Let $f(x) = a{x^2} + bx + c$. Then $f(0) = c$. Thus the graph of $y = f(x)$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 cases $y = f(x)$ does not intersect with $x$-axis i.e. $f(x) \ne 0$for any real $x$.

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

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