If alpha and beta are the roots of the equati | vedclass

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

Question

(B) Given $\alpha + \beta = -b/a$ and $\alpha \beta = c/a$.Expanding the expression $(1 + \alpha + \alpha^2)(1 + \beta + \beta^2)$:$= 1 + (\alpha + \b

Vedclass · Accepted Answer

Positive

Vedclass · Answer

(B) Given $\alpha + \beta = -b/a$ and $\alpha \beta = c/a$.Expanding the expression $(1 + \alpha + \alpha^2)(1 + \beta + \beta^2)$:$= 1 + (\alpha + \beta) + (\alpha^2 + \beta^2) + \alpha \beta + \alpha \beta (\alpha + \beta) + \alpha^2 \beta^2$$= 1 + (\alpha + \beta) + ((\alpha + \beta)^2 - 2\alpha \beta) + \alpha \beta + \alpha \beta (\alpha + \beta) + (\alpha \beta)^2$$= 1 + (\alpha + \beta) + (\alpha + \beta)^2 - \alpha \beta + \alpha \beta (\alpha + \beta) + (\alpha \beta)^2$Substituting the values:$= 1 - \frac{b}{a} + \frac{b^2}{a^2} - \frac{c}{a} + \left(\frac{c}{a}\right)\left(-\frac{b}{a}\right) + \frac{c^2}{a^2}$$= \frac{a^2 - ab + b^2 - ac - bc + c^2}{a^2} = \frac{a^2 + b^2 + c^2 - ab - bc - ca}{a^2}$$= \frac{1}{2a^2} [(a - b)^2 + (b - c)^2 + (c - a)^2]$Since $a, b, c$ are distinct,the sum of squares is always positive. Thus,the expression is always positive.

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) = $

Similar Questions

Solve the given equations and select the correct option.
$I.$ $8x + 7y = 135$
$II.$ $5x + 6y = 99$
$III.$ $9y + 8z = 121$

If the roots of the given equation $(2k + 1)x^2 - (7k + 3)x + k + 2 = 0$ are reciprocal to each other,then the value of $k$ will be

If $\alpha, \beta$ are the roots of $x^2 - px + q = 0$ and $\alpha', \beta'$ are the roots of $x^2 - p'x + q' = 0$,then the value of $(\alpha - \alpha')^2 + (\beta - \alpha')^2 + (\alpha - \beta')^2 + (\beta - \beta')^2$ is

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2} = 64$
$II.$ $2y^{2} + 25y + 72 = 0$

Solve the given two equations and select the correct answer from the given options.
$I.$ $3x^2 - 7x - 20 = 0$
$II.$ $y^2 - 8y + 16 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module