If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ ax^2\ +\ bx\ +\ c$, then evaluate:
$\frac{1}{α}\ +\ \frac{1}{β}\ -\ 2αβ$
Given:
$α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ ax^2\ +\ bx\ +\ c$.
To do:
We have to find the value of $\frac{1}{α}\ +\ \frac{1}{β}\ -\ 2αβ$.
Solution:
The given quadratic equation is $ax^2+bx+c=0$, where a, b and c are constants and $a≠0$.
Sum of the roots $= α+β = \frac{-b}{a}$.
Product of the roots $= αβ = \frac{c}{a}$.
We know that,
$\frac{1}{\alpha } +\frac{1}{\beta } -2\alpha \beta =\frac{\alpha +\beta }{\alpha \beta } -2\alpha \beta$
$=\frac{-\frac{b}{a}}{\frac{c}{a}} -2\left(\frac{c}{a}\right)$
$=-\frac{b}{c} -\frac{2c}{a}$
$=-\left(\frac{b}{c} +\frac{2c}{a}\right)$
The value of $\frac{1}{α}\ +\ \frac{1}{β}\ -\ 2αβ$ is $-(\frac{b}{c}+\frac{2c}{a})$.
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