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

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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_2

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

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