First slide

Theory of equations

Question

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

Moderate

A

$4 ac < b^{2}$
B

$4 ac > b^{2}$
C

$ac < b^{2}$
D

$ac > b^{2}$

Solution

Let $f (x) = {ax}^{2} + bx + c .$ 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. $f (x) \neq 0$ for any real x.
Hence $f (x) = 0$ i.e., ${ax}^{2} + bx + c = 0$ has imaginary roots and so $b^{2} < 4 ac$

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