The expression y = ax^2 + bx + c has always t | vedclass

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(B) Let $f(x) = ax^2 + bx + c$. Then $f(0) = c$. Thus,the graph of $y = f(x)$ meets the $y$-axis at $(0, c)$.If $c > 0$,then by the hypothesis,$f(x) >

Vedclass · Accepted Answer

$4ac > b^2$

Vedclass · Answer

(B) Let $f(x) = ax^2 + bx + c$. Then $f(0) = c$. Thus,the graph of $y = f(x)$ meets the $y$-axis at $(0, c)$.If $c > 0$,then by the hypothesis,$f(x) > 0$. This means that the curve $y = f(x)$ does not meet the $x$-axis.If $c Thus,in both cases,$y = f(x)$ does not intersect the $x$-axis,i.e.,$f(x) 
eq 0$ for any real $x$.Hence,the equation $ax^2 + bx + c = 0$ has imaginary roots,which implies that the discriminant $D = b^2 - 4ac Therefore,$b^2  b^2$.

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

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