Question

If $\alpha ,\beta ,\gamma$ are the roots of $x^3+4x+1=0$, then the equation whose roots are $\frac{{\alpha }^2}{\beta +\gamma },\frac{{\beta }^2}{\gamma +\alpha },\frac{{\gamma }^2}{\alpha +\beta }$ is

• A. $x^3-4x-1=0$
• B. $x^3-4x+1=0$
• C. $x^3+4x-1=0$
• D. $x^3+4x+1=0$

Answer 1

Answer: (C)

Given, $\alpha ,\beta$ and $\gamma$ are the roots of
$x^3+4x+1=0$
$\alpha +\beta +\gamma =0,\alpha \beta +\beta \gamma +\gamma \alpha =4,\alpha \beta \gamma =-1$
Now, $\frac{{\alpha }^2}{\beta +\gamma }+\frac{{\beta }^2}{\gamma +\alpha }+\frac{{\gamma }^2}{\alpha +\beta }=\frac{{\alpha }^2}{-\alpha }+\frac{{\beta }^2}{-\beta }+\frac{{\gamma }^2}{-\gamma }$
$=-(\alpha +\beta +\gamma )=0$
$\frac{{\alpha }^2{\beta }^2}{(\beta +\gamma )(\gamma +\alpha )}+\frac{{\beta }^2{\gamma }^2}{(\gamma +\alpha )(\alpha +\beta )}+\frac{{\gamma }^2{\alpha }^2}{(\beta +\gamma )(\alpha +\beta )}$
$=\alpha \beta +\beta \gamma +\gamma \alpha =4$
and $\frac{{\alpha }^2{\beta }^2{\gamma }^2}{(\beta +\gamma )(\gamma +\alpha )(\alpha +\beta )}=-\alpha \beta \gamma =1$
$(\because \alpha +\beta +\gamma =0)$
$\therefore$ Required equation is
$x^3+4x-1=0$