Question

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

• A. 80
• B. 85
• C. 90
• D. 95

Answer 1

Answer: (B)

85

Answer 2

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