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 $\frac{b}{c}$.