If alpha and beta are roots of x^2 - 3x + 1 = | 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 the equation $x^2 - 3x + 1 = 0$.From the properties of quadratic equations,the sum of roots $\alp

Vedclass · Accepted Answer

$x^2 - x - 1 = 0$

Vedclass · Answer

(C) Given that $\alpha$ and $\beta$ are the roots of the equation $x^2 - 3x + 1 = 0$.From the properties of quadratic equations,the sum of roots $\alpha + \beta = 3$ and the product of roots $\alpha \beta = 1$.Let the new roots be $x_1 = \frac{1}{\alpha - 2}$ and $x_2 = \frac{1}{\beta - 2}$.The sum of the new roots $S = x_1 + x_2 = \frac{1}{\alpha - 2} + \frac{1}{\beta - 2} = \frac{\beta - 2 + \alpha - 2}{(\alpha - 2)(\beta - 2)} = \frac{(\alpha + \beta) - 4}{\alpha \beta - 2(\alpha + \beta) + 4}$.Substituting the values: $S = \frac{3 - 4}{1 - 2(3) + 4} = \frac{-1}{1 - 6 + 4} = \frac{-1}{-1} = 1$.The product of the new roots $P = x_1 \cdot x_2 = \frac{1}{(\alpha - 2)(\beta - 2)} = \frac{1}{\alpha \beta - 2(\alpha + \beta) + 4}$.Substituting the values: $P = \frac{1}{1 - 2(3) + 4} = \frac{1}{1 - 6 + 4} = \frac{1}{-1} = -1$.The required quadratic equation is $x^2 - Sx + P = 0$,which gives $x^2 - (1)x + (-1) = 0$,or $x^2 - x - 1 = 0$.

If $\alpha$ and $\beta$ are roots of $x^2 - 3x + 1 = 0$,then the equation whose roots are $\frac{1}{\alpha - 2}$ and $\frac{1}{\beta - 2}$ is

Similar Questions

If $3p^2 = 5p + 2$ and $3q^2 = 5q + 2$ where $p \ne q$,then the equation whose roots are $3p - 2q$ and $3q - 2p$ is

If $x^{4} + \frac{1}{x^{4}} = 119$,then the value of $x^{3} - \frac{1}{x^{3}}$ is

Let $p, q \in Q$. If $2 - \sqrt{3}$ is a root of the quadratic equation $x^2 + px + q = 0$,then:

Solve the given two equations and select the correct answer from the given options.
$I.$ $x = \sqrt[3]{4913}$
$II.$ $13y + 3x = 246$

Vedclass Test Series

Exam Paper Generator

Online Exam Module