Question

The condition that the roots of the equation $l{x}^2+mx+n=0$ may be in the ratio 3 : 4 is

• A. $14n^2=49ml$
• B. $m^2=9nl$
• C. $12m^2=49nl$
• D. $4l^2=49ml$

Answer 1

Answer: (C)

$12m^2=49nl$

Let the two roots be $3\alpha$ and $4\alpha$.

sum of roots$=3\alpha +4\alpha =-\frac{m}{l}$

$\therefore 7\alpha =-\frac{m}{l}$

$\therefore \alpha =-\frac{m}{7l}$ ...(1)

product of roots$=\left(3\alpha \right)\left(4\alpha \right)=\frac{n}{l}$

$\therefore 12{\alpha }^2=\frac{n}{l}$

Substituting the value of $\alpha$ from equation (1), we get

$\therefore 12(-\frac{m}{7l}{)}^2=\frac{n}{l}$

$\therefore \frac{12m^2}{49l^2}=\frac{n}{l}$

$\therefore 12m^2=49nl$

So, the answer is option (C)