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NIMCET 2020

Roots of equation are $ax^2-2bx+c=0$ are n and m , then the value of $\frac{b}{an^2+c}+\frac{b}{am^2+c}$ is

Solution

Given quadratic equation is $ax^2 - 2bx + c = 0$. 
 Since $n$ and $m$ are the roots of this equation: 
$an^2 - 2bn + c = 0 \implies an^2 + c = 2bn$ 
$am^2 - 2bm + c = 0 \implies am^2 + c = 2bm$ 

 Now, the given expression is: $\frac{b}{an^2+c} + \frac{b}{am^2+c}$ 

 Substitute $an^2 + c = 2bn$ and $am^2 + c = 2bm$: $= \frac{b}{2bn} + \frac{b}{2bm}$ 

 Canceling $b$ from numerator and denominator: $= \frac{1}{2n} + \frac{1}{2m}$ $= \frac{1}{2} \left( \frac{n+m}{nm} \right)$ 

 From Vieta's formulas: 

Sum of roots, $n + m = \frac{-(-2b)}{a} = \frac{2b}{a}$ 

Product of roots, $nm = \frac{c}{a}$ 
Substituting these values: $= \frac{1}{2} \left( \frac{\frac{2b}{a}}{\frac{c}{a}} \right)$ $= \frac{1}{2} \left( \frac{2b}{c} \right)$ $= \frac{b}{c}$ 
 Therefore, the value is $\frac{b}{c}$.
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