If one root of the equation $lx^{2} + mx + n = 0$ is $\frac{9}{2}$ ($l,\ m\$and$\ n$ are positive integers) and $\frac{m}{4n} = \frac{1}{m}$, then $I + n$ is equal to

Question

Dr. Abubacker · Accepted Answer

(b)
Given,$lx^{2} + mx + n = 0$ ...(i)
Now,
$D = m^{2} - 4ln = 0\ \ \ \ \ \ (\because m^{2} = 4\ln{given)}$
It means roots of given equation are equal
$	herefore\ \left( x - \frac{9}{2} ight)^{2} = 0$
$\Rightarrow 4x^{2} + 81 - 36x = 0$ ...(ii)
On comparing Eqs. (i) and (ii), we get
$l = 4,\ m = - 36,\ n = 81$
$	herefore l + n = 4 + 81 = 85$

Q. If one root of the equation $lx^{2} + mx + n = 0$ is $\frac{9}{2}$ ($l,\ m\$and$\ n$ are positive integers) and $\frac{m}{4n} = \frac{1}{m}$, then $I + n$ is equal to

Explanation

Reviewed by Dr. Abubacker

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