Let f(x) be a quadratic expression such that | vedclass

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

Question

(D) Let $f(x) = ax^2 + bx + c$.Given $f(0) + f(1) = 0$,we have $c + (a + b + c) = 0$,which implies $a + b + 2c = 0$ $(i)$.Given $f(-2) = 0$,we have $4

Vedclass · Accepted Answer

$f\left(\frac{3}{5}\right)=0$

Vedclass · Answer

(D) Let $f(x) = ax^2 + bx + c$.Given $f(0) + f(1) = 0$,we have $c + (a + b + c) = 0$,which implies $a + b + 2c = 0$ $(i)$.Given $f(-2) = 0$,we have $4a - 2b + c = 0$ (ii).From $(i)$,$b = -a - 2c$. Substituting into (ii):$4a - 2(-a - 2c) + c = 0$ $\Rightarrow 4a + 2a + 4c + c = 0$ $\Rightarrow 6a + 5c = 0$.Let $a = 5k$,then $c = -6k$.Substituting into $(i)$: $5k + b + 2(-6k) = 0$ $\Rightarrow 5k + b - 12k = 0$ $\Rightarrow b = 7k$.Thus,$f(x) = k(5x^2 + 7x - 6) = k(5x^2 + 10x - 3x - 6) = k(5x(x + 2) - 3(x + 2)) = k(5x - 3)(x + 2)$.The roots are $x = -2$ and $x = \frac{3}{5}$.Therefore,$f\left(\frac{3}{5}\right) = 0$.

Let $f(x)$ be a quadratic expression such that $f(0)+f(1)=0$. If $f(-2)=0$,then

Similar Questions

If the roots of the equation $bx^2 + cx + a = 0$ are imaginary,then for all real values of $x$,the expression $3b^2x^2 + 6bcx + 2c^2$ is:

$\alpha$ is a root of the equation $\frac{x-1}{\sqrt{2x^2-5x+2}} = \frac{41}{60}$. If $-\frac{1}{2} < \alpha < 0$,then $\alpha = $

For how many values of $k$ is the equation $(1 + 2k)x^2 + (1 - 2k)x + (1 - 2k) = 0$ a perfect square?

The condition that ${x^3} - 3px + 2q$ may be divisible by a factor of the form ${x^2} + 2ax + {a^2}$ is

The quadratic equation whose roots are the numbers having arithmetic mean $34$ and geometric mean $16$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module