If alpha and beta are the roots of x^{2}-px+1 | vedclass

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

Question

(A) Given that $\alpha$ and $\beta$ are the roots of $x^{2}-px+1=0$. By the relation between roots and coefficients, $\alpha+\beta=p$ and $\alpha\beta

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given that $\alpha$ and $\beta$ are the roots of $x^{2}-px+1=0$. By the relation between roots and coefficients, $\alpha+\beta=p$ and $\alpha\beta=1$. Also, $\gamma$ is a root of $x^{2}+px+1=0$, so $\gamma^{2}+p\gamma+1=0$, which implies $\gamma^{2}=-p\gamma-1$. Now, consider the expression $(\alpha+\gamma)(\beta+\gamma) = \alpha\beta + \gamma(\alpha+\beta) + \gamma^{2}$. Substituting the known values: $(\alpha+\gamma)(\beta+\gamma) = 1 + \gamma(p) + (-p\gamma-1)$. $= 1 + p\gamma - p\gamma - 1 = 0$.

If $\alpha$ and $\beta$ are the roots of $x^{2}-px+1=0$ and $\gamma$ is a root of $x^{2}+px+1=0,$ then $(\alpha+\gamma)(\beta+\gamma)$ is

Similar Questions

The cubic equation whose roots are the squares of the roots of $x^3-2x^2+10x-8=0$ is

Let $\alpha, \beta$ be the roots of the equation $ax^2 + 2bx + c = 0$ and $\gamma, \delta$ be the roots of the equation $px^2 + 2qx + r = 0$. If $\alpha, \beta, \gamma, \delta$ are in $G.P.$,then:

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

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$ ($a \ne 0$; $a, b, c$ being distinct),then $(1 + \alpha + \alpha^2)(1 + \beta + \beta^2) = $

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$ ($a \neq 0$; $a, b, c \in \mathbb{R}$),then $(1 + \alpha + \alpha^2)(1 + \beta + \beta^2)$ is . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module