Q. If $\alpha\ and\ \beta$ be the roots of the equation $x^{2} + x\sqrt{\alpha} + \beta = 0$, then
A) $\alpha = 1$ and $\beta = - 1$
B) $\alpha = 1$ and $\beta = - 2$
C) $\alpha = 2$ and $\beta = 1$
D) $\alpha = 2$ and $\beta = - 2$
Explanation
(b)
Since, $\alpha\ and\ \beta$ be the roots of the equation $x^{2} + \sqrt{\alpha}x + \beta = 0$, therefore
$\alpha + \beta = - \sqrt{\alpha}$ and $\alpha\beta = \beta$
From second relation $\beta \neq 0$
$\therefore\ \ \alpha = 1$
$\therefore\ 1 + \beta = - 1\ \Rightarrow \ \beta = - 2$
Hence, $\alpha = 1$ and $\beta = - 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.