Let y = f(x) = ax^2 + 2bx + c (where a, b, c | vedclass

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

Question

(B) Given the quadratic function $f(x) = ax^2 + 2bx + c$.We are given that $f(x) = 0$ has imaginary roots,which implies the discriminant $D = (2b)^2 -

Vedclass · Accepted Answer

$c < 0$

Vedclass · Answer

(B) Given the quadratic function $f(x) = ax^2 + 2bx + c$.We are given that $f(x) = 0$ has imaginary roots,which implies the discriminant $D = (2b)^2 - 4ac We are also given that $4a + 4b + c Since $f(x)$ has imaginary roots,the parabola $y = f(x)$ does not intersect the $x$-axis. This means $f(x)$ is either always positive (if $a > 0$) or always negative (if $a If $a > 0$,then $f(x) > 0$ for all $x$,which contradicts $f(2) Since $a < 0$ and $f(x)$ is always negative,the entire parabola lies below the $x$-axis. The $y$-intercept is $f(0) = c$. Since the entire graph is below the $x$-axis,the $y$-intercept must be negative. Therefore,$c < 0$.

Let $y = f(x) = ax^2 + 2bx + c$ (where $a, b, c \in R$ and $a \neq 0$). If $f(x) = 0$ has imaginary roots and $4a + 4b + c < 0$,then which of the following is true?

Similar Questions

The common root of the equations $x^{2}-7x+10=0$ and $x^{2}-10x+16=0$ is:

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 - 6x - 2 = 0$. If $a_n = \alpha^n - \beta^n$ for $n \ge 1$,then the value of $\frac{a_{10} - 2a_8}{2a_9}$ is equal to:

If one root of the equation ${x^2} + px + q = 0$ is $2 + \sqrt{3}$,then the values of $p$ and $q$ are:

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-11x+24=0$
$II.$ $2y^{2}-9y+9=0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module