If the roots of the equation ax^2 + bx + c = | vedclass

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

Question

(A) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$,we have:$\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{

Vedclass · Accepted Answer

$\frac{c(a - b)}{a^2}$

Vedclass · Answer

(A) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$,we have:$\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.We need to find the value of $\alpha\beta^2 + \alpha^2\beta + \alpha\beta$.Factoring out $\alpha\beta$,we get:$\alpha\beta(\beta + \alpha) + \alpha\beta = \alpha\beta(\alpha + \beta + 1)$.Substituting the values of $\alpha + \beta$ and $\alpha\beta$:$= \frac{c}{a} \left( -\frac{b}{a} + 1 \right)$$= \frac{c}{a} \left( \frac{a - b}{a} \right)$$= \frac{c(a - b)}{a^2}$.

If the roots of the equation $ax^2 + bx + c = 0$ are $\alpha$ and $\beta$,then the value of $\alpha\beta^2 + \alpha^2\beta + \alpha\beta$ will be

Similar Questions

If $\frac{x-4}{x^2-5x-2k} = \frac{2}{x-2} - \frac{1}{x+k}$,then $k$ is equal to

If the roots of $ax^2 + bx + c = 0$ are $\alpha, \beta$ and the roots of $Ax^2 + Bx + C = 0$ are $\alpha - k, \beta - k$,then $\frac{B^2 - 4AC}{b^2 - 4ac}$ is equal to

If $\alpha, \beta, \gamma$ are the roots of $x^3-2x^2+4x-1=0$,then the equation having the roots $\beta\gamma+\frac{1}{\alpha}, \alpha\beta+\frac{1}{\gamma}, \gamma\alpha+\frac{1}{\beta}$ is

If $x^2-3x+2$ is a factor of $x^4-ax^2+b$,then the equation whose roots are $a$ and $b$ is

If one root of the equation $x^2 + px + q = 0$ is the square of the other,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module