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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 hypothesis $f(x) > 0$.

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 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,$f(x) = 0$,i.e.,$ax^2 + bx + c = 0$ has imaginary roots,which implies the discriminant $D = b^2 - 4ac < 0$,or $b^2 < 4ac$.

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

Similar Questions

The number of real roots of the equation $\sqrt{x^2-4x+3}+\sqrt{x^2-9}=\sqrt{4x^2-14x+6}$ is:

The solution to the equation $\sqrt{x + 10} + \sqrt{x - 2} = 6$ is:

If the roots of the equation $6x^2 - 7x + k = 0$ are rational,then $k = .......$

The equation $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$ for $x \in R$ has:

If $r$ and $s$ are positive,what is the nature of the roots of the quadratic equation $ax^2 - rx - s = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module