Question

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

• A. $x^3-2x-1=0$
• B. $x^3+2x^2+1=0$
• C. $x^3+2x-1=0$
• D. $x^3-2x^2-1=0$

Answer 1

Answer: (C)

$x^3+2x-1=0$

Let $\alpha ,\beta ,\gamma$ be roots of the given equation. $s_1=\alpha +\beta +\gamma$. It is equal to $-1\times$ the coefficient of $x^2$, which is $0$. Hence, $\alpha +\beta +\gamma =0$. So we have,

$-\alpha =\beta +\gamma ⟹-\frac{\alpha }{\beta +\gamma }=1⟹-\frac{{\alpha }^2}{\beta +\gamma }=\alpha$. Similarly, all the three given terms can be shown to be equal to the negative of the roots of the given equation. Hence, our required equation is the same as the given one except for the signs of the coefficient of $x^2$ and the constant term. So, by changing them we obtain,

$x^3+2x-1$.