If p and q are non-zero real numbers and alph | vedclass

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

Question

(B) Given $\alpha^3 + \beta^3 = -p$ and $\alpha \beta = q$. Let the roots of the required quadratic equation be $x_1 = \frac{\alpha^2}{\beta}$ and $x_

Vedclass · Accepted Answer

$qx^2 + px + q^2 = 0$

Vedclass · Answer

(B) Given $\alpha^3 + \beta^3 = -p$ and $\alpha \beta = q$. Let the roots of the required quadratic equation be $x_1 = \frac{\alpha^2}{\beta}$ and $x_2 = \frac{\beta^2}{\alpha}$. The sum of the roots is $x_1 + x_2 = \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha \beta} = \frac{-p}{q}$. The product of the roots is $x_1 	imes x_2 = \frac{\alpha^2}{\beta} 	imes \frac{\beta^2}{\alpha} = \alpha \beta = q$. The required quadratic equation is given by $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$. Substituting the values,we get $x^2 - (\frac{-p}{q})x + q = 0$. $x^2 + \frac{p}{q}x + q = 0$. Multiplying by $q$,we get $qx^2 + px + q^2 = 0$.

If $p$ and $q$ are non-zero real numbers and $\alpha^3 + \beta^3 = -p$,$\alpha \beta = q$,then a quadratic equation whose roots are $\frac{\alpha^2}{\beta}$ and $\frac{\beta^2}{\alpha}$ is

Similar Questions

The $G.M.$ of the roots of the equation $x^2 - 18x + 9 = 0$ is

If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$,then the value of $\alpha^6+\beta^6$ is

Let $p$ and $q$ be two positive numbers such that $p + q = 2$ and $p^{4} + q^{4} = 272$. Then $p$ and $q$ are roots of the equation:

If the sum of two roots of the equation $x^4-2x^3+x^2+4x-6=0$ is zero,then the sum of the squares of the other two roots is

If the sum of the roots of the equation $ax^2 + bx + c = 0$ is equal to the sum of the squares of their roots,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module