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

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

Question

(C) Given that $\alpha$ and $\beta$ are the roots of $x^2 + bx - c = 0$.From the properties of roots,we have:Sum of roots: $\alpha + \beta = -b \impli

Vedclass · Accepted Answer

$x^2 + [(\alpha + \beta) + \alpha \beta]x + \alpha \beta(\alpha + \beta) = 0$

Vedclass · Answer

(C) Given that $\alpha$ and $\beta$ are the roots of $x^2 + bx - c = 0$.From the properties of roots,we have:Sum of roots: $\alpha + \beta = -b \implies b = -(\alpha + \beta)$Product of roots: $\alpha \beta = -c \implies c = -\alpha \beta$We need to find the equation whose roots are $b$ and $c$.Sum of new roots: $b + c = -(\alpha + \beta) - \alpha \beta = -[(\alpha + \beta) + \alpha \beta]$Product of new roots: $bc = [ -(\alpha + \beta) ] 	imes [ -\alpha \beta ] = (\alpha + \beta)(\alpha \beta)$The required quadratic equation is given by $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values: $x^2 - (-[(\alpha + \beta) + \alpha \beta])x + (\alpha + \beta)(\alpha \beta) = 0$$x^2 + [(\alpha + \beta) + \alpha \beta]x + \alpha \beta(\alpha + \beta) = 0$.

If $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 + bx - c = 0$,then the equation whose roots are $b$ and $c$ is

Similar Questions

The equation $\log_e x + \log_e(1 + x) = 0$ can be written as

Solve the given two equations and select the correct answer from the given options.
$I.$ $\sqrt{x} - \frac{(18)^{15/2}}{x^2} = 0$
$II.$ $\sqrt{y} = \frac{(19)^{9/2}}{y}$

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

Let $P(x) = x^3 - ax^2 + bx + c$ where $a, b, c \in R$ has integral roots such that $P(6) = 3$,then '$a$' cannot be equal to

If ${x_1}, {x_2}, {x_3}$ are distinct roots of the equation $ax^2 + bx + c = 0$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module