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Question

(A) Given the equation $lx^2 + mx + n = 0$,the sum of roots is $\alpha + \beta = -\frac{m}{l}$ and the product of roots is $\alpha\beta = \frac{n}{l}$

Vedclass · Accepted Answer

$l^4x^2 - nl(m^2 - 2nl)x + n^4 = 0$

Vedclass · Answer

(A) Given the equation $lx^2 + mx + n = 0$,the sum of roots is $\alpha + \beta = -\frac{m}{l}$ and the product of roots is $\alpha\beta = \frac{n}{l}$.Let the new roots be $S_1 = \alpha^3\beta$ and $S_2 = \alpha\beta^3$.The sum of the new roots is $S_1 + S_2 = \alpha\beta(\alpha^2 + \beta^2) = \alpha\beta((\alpha + \beta)^2 - 2\alpha\beta) = \frac{n}{l} \left( \frac{m^2}{l^2} - \frac{2n}{l} ight) = \frac{n}{l} \left( \frac{m^2 - 2nl}{l^2} ight) = \frac{n(m^2 - 2nl)}{l^3}$.The product of the new roots is $S_1 S_2 = \alpha^4\beta^4 = (\alpha\beta)^4 = \left( \frac{n}{l} ight)^4 = \frac{n^4}{l^4}$.The required equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.$x^2 - \frac{n(m^2 - 2nl)}{l^3}x + \frac{n^4}{l^4} = 0$.Multiplying by $l^4$,we get $l^4x^2 - nl(m^2 - 2nl)x + n^4 = 0$.

If $\alpha, \beta$ are the roots of the equation $lx^2 + mx + n = 0$,then the equation whose roots are $\alpha^3\beta$ and $\alpha\beta^3$ is

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