Question

lf $\alpha ,\beta ,\gamma$ are the roots of $x^3+2x-3=0$, then the transformed equation having the roots $\frac{\alpha }{\beta }+\frac{\beta }{\alpha },\frac{\beta }{\gamma }+\frac{\gamma }{\beta },\frac{\gamma }{\alpha }+\frac{\alpha }{\gamma }$ is obtained by taking $x=$

• A. $\frac{3}{2}(1-y)$
• B. $-\frac{3}{2}(1+y)$
• C. $3(1-y)$
• D. $3(1+y)$

Answer 1

Answer: (B)

$-\frac{3}{2}(1+y)$

As $\alpha ,\beta ,\gamma$ are roots of $x^3+2x-3=0$
$s_1=\alpha +\beta +\gamma =0s_2=\alpha \beta +\beta \gamma +\alpha \gamma =2s_3=\alpha \beta \gamma =3$
$\alpha \beta +\beta \gamma +\alpha \gamma =2\Rightarrow \alpha \beta \gamma +\beta {\gamma }^2+\alpha {\gamma }^2=2\gamma \Rightarrow {\gamma }^2(\alpha +\beta )=2\gamma -\alpha \beta \gamma \Rightarrow -{\gamma }^3=2\gamma -3\Rightarrow {\gamma }^3=3-2\gamma$
Let $y=\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{{(\alpha +\beta )}^2-2\alpha \beta }{\alpha \beta }=\frac{{\gamma }^2-2\alpha \beta }{\alpha \beta }$
$=\frac{{\gamma }^3-2\alpha \beta \gamma }{\alpha \beta \gamma }=\frac{{\gamma }^3-6}{3}=\frac{3-2\gamma -6}{3}$
$\Rightarrow y=-1-\frac{2}{3}x\Rightarrow x=-\frac{3}{2}(y+1)$