Question

If $\alpha ,\beta $ and $\gamma $ are the roots of the equation $x^{3}-13x^{2}+15x+189=0$ and one root exceeds the other by $2,$ then the value of $\left|\alpha \right|+\left|\beta \right|+\left|\gamma \right|$ is equal to

• A. $23$
• B. $17$
• C. $13$
• D. $19$

Answer 1

Answer: (D)

Let, the roots be $\alpha ,\beta ,\alpha +2.$
$S_{1}=\alpha +\beta +\alpha +2=2\alpha +\beta +2=13\Rightarrow 2\alpha +\beta =11\Rightarrow \beta =11-2\alpha $
$S_{2}=\alpha \beta +\beta \left(\alpha + 2\right)+\left(\alpha + 2\right)\alpha =15$
$\Rightarrow \beta \left(\alpha + \alpha + 2\right)+\alpha \left(\alpha + 2\right)=15$
$\Rightarrow \left(11 - 2 \alpha \right)\left(2 \alpha + 2\right)+\alpha \left(\alpha + 2\right)=15$
$\Rightarrow 22\alpha +22-4\alpha ^{2}-4\alpha +\alpha ^{2}+2\alpha =15$
$\Rightarrow 3\left(\alpha \right)^{2}-20\alpha -7=0\Rightarrow \left(\alpha - 7\right)\left(3 \alpha + 1\right)=0$
$\Rightarrow \alpha =7$ or $-\frac{1}{3}.$
$\alpha =7,\beta =11-2\alpha =11-14=-3,\gamma =\alpha +2=9$
$\alpha =-\frac{1}{3},\beta =11-2\alpha =11+\frac{2}{3}=\frac{35}{3},\gamma =\alpha +2=\frac{5}{3}.$
Since, $\alpha \beta \gamma =-189$ , hence we will take the first case.
$\left|\alpha \right|+\left|\beta \right|+\left|\gamma \right|=\left|7\right|+\left|- 3\right|+\left|9\right|=19$