Question

If $\alpha ,\beta ,\gamma$ are the roots of the equation $x^3+4x+1=0$, then ${\left(\alpha +\beta \right)}^{-1}+{\left(\beta +\gamma \right)}^{-1}+{\left(\gamma +\alpha \right)}^{-1}$ is equal to

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is C

$4$

Explanation for the correct option:

Find the value of the expression:

Since, $\alpha ,\beta ,\gamma$ are the roots of the equation $x^3+4x+1=0$, So

$S_1=-\frac{\mathrm{coefficient}\mathrm{of}{x}^2}{\mathrm{coefficient}\mathrm{of}{x}^3}$

$\begin{array}{rcl}& \Rightarrow & S_1=\alpha +\beta +\gamma =-\left(\frac{0}{1}\right)=0\end{array}...\left(1\right)$

$S_2=\frac{\mathrm{coefficient}\mathrm{of}x}{\mathrm{coefficient}\mathrm{of}{x}^3}$

$\begin{array}{rcl}& \Rightarrow & S_2=\alpha \beta +\beta \gamma +\gamma \alpha =\frac{4}{1}=4\end{array}...\left(2\right)$

$S_3=-\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{x}^3}$

$\begin{array}{rcl}& \Rightarrow & S_2=\alpha \beta \gamma =-1\end{array}...\left(3\right)$

Here, $S_1,S_2$ and $S_3$ are the coefficient of the cubic equation $x^3+S_1{x}^2+S_{2}x+S_3=0$.

Now,

$\begin{array}{rcl}{\left(\alpha +\beta \right)}^{-1}+{\left(\beta +\gamma \right)}^{-1}+{\left(\gamma +\alpha \right)}^{-1}& =& \frac{1}{\alpha +\beta }+\frac{1}{\beta +\alpha }+\frac{1}{\gamma +\alpha }\\ & =& \left(\frac{1}{-\gamma }\right)+\left(\frac{1}{-\alpha }\right)+\left(\frac{1}{-\beta }\right)\left[\mathrm{from}\left(1\right)\right]\\ & =& -\left[\frac{\alpha \beta +\beta \gamma +\alpha \gamma }{\alpha \beta \gamma }\right]\\ & =& -\left[\frac{4}{-1}\right]\left[\mathrm{from}\left(2\right)\&\left(3\right)\right]\\ & =& 4\end{array}$

Hence, the correct option is C.