Q. If $\alpha,\beta$ and $\gamma$ are the roots of equation $x^{3} - 3x^{2} + x + 5 = 0$, then $y = \sum_{}^{}{\alpha^{2} + \alpha\beta\gamma}$ satisfies the equation
A) $y^{3} + y + 2 = 0$
B) $y^{3} - y^{2} - y - 2 = 0$
C) $y^{3} + 3y^{2} - y - 3 = 0$
D) $y^{2} + 4y^{2} + 5y + 20 = 0$
Explanation
(b)
Here, $\alpha + \beta + \gamma = 3,\ \alpha\beta + \beta\gamma + \gamma\alpha = 1$ and $\alpha\beta\gamma = - 5$
Now, $y = \alpha^{2} + \beta^{2} + \gamma^{2} + \alpha\beta\gamma$
$= (\alpha + \beta + \gamma)^{2} - 2(\alpha\beta + \beta\gamma + \gamma\alpha) + \alpha\beta\gamma$
$= (3)^{2} - 2(1) - 5$
$\Rightarrow y = 2$
So, $y = 2$ satisfies the equation
$y^{3} - y^{2} - y - 2 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.