Question

If $\sin \alpha$ and $\cos \alpha$ are the roots of the equation $l{x}^2+mx+2ln=0$, then

• A. $l^2-m^2+2ln=0$
• B. $l^2+m^2+2ln=0$
• C. $l^2+m^2-2ln=0$
• D. $l^2+m^2+ln=0$

Answer 1

Answer: (B)

$l^2+m^2+2ln=0$

Given that $\sin \alpha$ and $\cos \alpha$ are the roots of the equation $l{x}^2+mx+2ln=0$

Thus $\sin \alpha +\cos \alpha =-\frac{m}{l}$ ....(1)
$\sin \alpha \cos \alpha =\frac{n}{l}$ ....(2)
Squaring (1), we get
${\sin }^2\alpha +{\cos }^2\alpha +2\sin \alpha \cos \alpha =\frac{m^2}{l^2}$
$\Rightarrow 1+\frac{2n}{l}=\frac{m^2}{l^2}$

$\Rightarrow l^2-m^2+2ln=0$