If alpha, beta are the roots of the quadratic | vedclass

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

Question

(D) Given the quadratic equation $x^{2}+p x+q=0$, the sum of roots $\alpha+\beta = -p$ and the product of roots $\alpha \beta = q$. First, calculate $

Vedclass · Accepted Answer

$3 p q-p^{3}$ and $(p^{2}-q)(p^{2}-3 q)$

Vedclass · Answer

(D) Given the quadratic equation $x^{2}+p x+q=0$, the sum of roots $\alpha+\beta = -p$ and the product of roots $\alpha \beta = q$. First, calculate $\alpha^{3}+\beta^{3}$: $\alpha^{3}+\beta^{3} = (\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta) = (-p)^{3}-3 q(-p) = -p^{3}+3 p q = 3 p q-p^{3}$. Next, calculate $\alpha^{4}+\alpha^{2} \beta^{2}+\beta^{4}$: $\alpha^{4}+\alpha^{2} \beta^{2}+\beta^{4} = (\alpha^{2}+\beta^{2})^{2} - \alpha^{2} \beta^{2} = [(\alpha+\beta)^{2}-2 \alpha \beta]^{2} - (\alpha \beta)^{2} = [(-p)^{2}-2 q]^{2} - q^{2} = (p^{2}-2 q)^{2} - q^{2} = p^{4}-4 p^{2} q+4 q^{2}-q^{2} = p^{4}-4 p^{2} q+3 q^{2} = p^{2}(p^{2}-3 q)-q(p^{2}-3 q) = (p^{2}-q)(p^{2}-3 q)$. Thus, the values are $3 p q-p^{3}$ and $(p^{2}-q)(p^{2}-3 q)$.

If $\alpha, \beta$ are the roots of the quadratic equation $x^{2}+p x+q=0,$ then the values of $\alpha^{3}+\beta^{3}$ and $\alpha^{4}+\alpha^{2} \beta^{2}+\beta^{4}$ are respectively

Similar Questions

If the product of roots of the equation $mx^2 + 6x + (2m - 1) = 0$ is $-1$,then the value of $m$ will be

For the equation $3x^2 + px + 3 = 0, p > 0$,if one of the roots is the square of the other,then $p$ is equal to:

If $\alpha$ and $\beta$ are the roots of $9x^2 + 6x + 1 = 0$,then the equation with the roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is

If the sum of the squares of the roots of the equation $x^2 - (\sin \alpha - 2)x - (1 + \sin \alpha) = 0$ is least,then $\alpha$ is

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + x + 1 = 0$,then the value of $\alpha^3 \beta^3 \gamma^3$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module