Q. If $\sin\alpha$ and $\cos\alpha$ are the roots of the equation $px^{2} + qx + r = 0$, then
A) $p^{2} + q^{2} - 2pr = 0$
B) $p^{2} - q^{2} + 2pr = 0$
C) $p^{2} - q^{2} - 2pr = 0$
D) $p^{2} + q^{2} + 2qr = 0$
Explanation
(b)
Here, $\sin{\alpha + \cos{\alpha = - \frac{q}{p}}}$ and $\sin{\alpha.\cos{\alpha = \frac{r}{p}}}$
$\therefore\left( \sin{\alpha + \cos\alpha} \right)^{2} = \left( - \frac{q}{p} \right)^{2}$
$\Rightarrow \sin^{2}{\alpha + \cos^{2}{\alpha + 2\sin{\alpha\cos{\alpha = \frac{q^{2}}{p^{2}}}}}}$
$\Rightarrow 1 + 2.\frac{r}{p} = \frac{q^{2}}{p^{2}}$
$\Rightarrow p(p + 2r) = q^{2}$
$\Rightarrow p^{2} - q^{2} + 2rp = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.