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
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
Updated On: Mar 4, 2024
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The Correct Option is B
Solution and Explanation
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 $
$=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 $
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
