Question

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

• A. $l^4{x}^2-nl(m^2-2nl)x+n^4=0$
• B. $l^4{x}^2+nl(m^2-2nl)x+n^4=0$
• C. $l^4{x}^2+nl(m^2-2nl)x-n^4=0$
• D. $l^4{x}^2-nl(m^2+2nl)x+n^4=0$

Answer 1

The correct option is A $l^4{x}^2-nl(m^2-2nl)x+n^4=0$

$l{x}^2+mx+n=0$ (Given equation).Equation whose roots are ${\alpha }^3\beta ,\alpha {\beta }^3$ will be $x^2-({\alpha }^3\beta +\alpha {\beta }^3)x+{\alpha }^3\beta \left(\alpha {\beta }^3\right)=0$
Now ${\alpha }^3\beta +\alpha {\beta }^3=\alpha \beta ({\alpha }^2+{\beta }^2)$
But $\alpha \beta =+\frac{n}{l}$
$\Rightarrow (\alpha \beta {)}^4=\frac{n^4}{l^4}$
and ${\alpha }^2+\beta =(\alpha +\beta {)}^2-2\alpha \beta$
$={(-\frac{m}{l})}^2-\frac{2n}{l}=\frac{m^2}{l^2}-\frac{2n}{l}$
then, $x^2-\frac{n}{l}(\frac{m^2}{l^2}-\frac{2n}{l})x+\frac{n^4}{l^4}=0$
or, $l^4{x}^2-\frac{n{l}^4}{l}\left(\frac{m^2-2nl}{l^2}\right)x+n^4=0$
or, $l^4{x}^2-nl(m^2-2nl)x+n^4=0$