If alpha and beta are roots of the equation A | vedclass

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

Question

(A) Given the quadratic equation $Ax^2 + Bx + C = 0$.From the relationship between roots and coefficients,we have:Sum of roots: $\alpha + \beta = -\fr

Vedclass · Accepted Answer

$\frac{3ABC - B^3}{A^3}$

Vedclass · Answer

(A) Given the quadratic equation $Ax^2 + Bx + C = 0$.From the relationship between roots and coefficients,we have:Sum of roots: $\alpha + \beta = -\frac{B}{A}$Product of roots: $\alpha \beta = \frac{C}{A}$We use the algebraic identity: $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta)$.Substituting the values:$\alpha^3 + \beta^3 = \left(-\frac{B}{A}\right)^3 - 3\left(\frac{C}{A}\right)\left(-\frac{B}{A}\right)$$= -\frac{B^3}{A^3} + \frac{3BC}{A^2}$To combine these,find a common denominator $A^3$:$= \frac{-B^3 + 3ABC}{A^3} = \frac{3ABC - B^3}{A^3}$.

If $\alpha$ and $\beta$ are roots of the equation $Ax^2 + Bx + C = 0$,then the value of $\alpha^3 + \beta^3$ is

Similar Questions

The number of real roots of the equation $e^{4x} + e^{3x} - 4e^{2x} + e^x + 1 = 0$ is:

If $x+y+z=0,$ then find the value of $\frac{(x+y)(y+z)(z+x)}{x y z}$

The sum of the solutions of the equation $|\sqrt{x} - 2| + \sqrt{x}(\sqrt{x} - 4) + 2 = 0$ $(x > 0)$ is equal to

Solve the given two equations and select the correct option.
$I.$ $36 x^{2}+47 \sqrt{7} x+105=0$
$II.$ $35 y^{2}+20 \sqrt{3} y+63 \sqrt{2} y+36 \sqrt{6}=0$

Divide $16$ into $2$ parts such that twice the square of the larger part exceeds the square of the smaller part by $164$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module