Q. If $\alpha,\beta$ are the roots of the equation $x^{2} - \left( 1 + n^{2} \right)x + \frac{1}{2}\left( 1 + n^{2} + n^{4} \right) = 0$, then $\alpha^{2} + \beta^{2}$ is
A) $n^{2}$
B) $- n^{2}$
C) $n^{4}$
D) ${- n}^{4}$
Explanation
(a)
Here, $\alpha + \beta = 1 + n^{2}$ and $\alpha\beta = \frac{1 + n^{2} + n^{4}}{2}$
Now, $\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta$
$= \left( 1 + n^{2} \right)^{2} - \left( 1 + n^{2} + n^{4} \right) = n^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.