Question

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

The correct option is C

$x^3+4x–1=0$

Explanation for the correct option:

Step 1. Find the required quadratic equation:

Given, $\alpha ,\beta$ and $\gamma$ are the roots of $x^3+4x+1=0$

$\Rightarrow$$\sum \alpha =0,\sum \alpha \beta =4$ and $\alpha \beta \gamma =–1$

Step 2. Find the Sum of the roots:

$\begin{array}{rcl}\frac{{\alpha }^2}{\left(\beta +\gamma \right)}+\frac{{\beta }^2}{\left(\gamma +\alpha \right)}+\frac{{\gamma }^2}{\left(\alpha +\beta \right)}& =& \frac{{\alpha }^2}{–\alpha }+\frac{{\beta }^2}{\beta }+\frac{{\gamma }^2}{–\gamma }\\ & =& –(\alpha +\beta +\gamma )\\ & =& 0\end{array}$

Step 3. Find the sum of Product of the roots:

$\begin{array}{rcl}\frac{{\alpha }^2{\beta }^2}{\left(\beta +\gamma \right)\left(\gamma +\alpha \right)}+\frac{{\beta }^2{\gamma }^2}{\left(\gamma +\alpha \right)\left(\alpha +\beta \right)}+\frac{{\gamma }^2{\alpha }^2}{\left(\beta +\gamma \right)\left(\alpha +\beta \right)}& =& \alpha \beta +\beta \gamma +\gamma \alpha \\ & =& 4\end{array}$

Step 4. Find the Product of the roots:

$\begin{array}{rcl}\frac{{\alpha }^2{\beta }^2{\gamma }^2}{\left(\beta +\gamma \right)\left(\gamma +\alpha \right)\left(\alpha +\beta \right)}& =& –\alpha \beta \gamma \\ & =& 1\end{array}$

$\therefore$The required equation is ${\mathit{x}}^{\mathbf{3}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{4}\mathit{x}\mathbf{}\mathbf{–}\mathbf{}\mathbf{1}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$.

Hence, Option ‘C’ is Correct.