If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ ax^2\ +\ bx\ +\ c$, then evaluate:
$\frac{1}{aα\ +\ b}\ +\ \frac{1}{aβ\ +\ b}$

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 $\frac{1}{aα\ +\ b}\ +\ \frac{1}{aβ\ +\ b}$.

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,

$ \begin{array}{l}
\frac{1}{a\alpha +b} +\frac{1}{a\beta +b} =\frac{a\beta +b+a\alpha +b}{( a\alpha +b)( a\beta +b)}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{a( \alpha +\beta ) +2b}{a^{2} \alpha \beta +ab\alpha +ab\beta +b^{2}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{a( \alpha +\beta ) +2b}{a^{2} \alpha \beta +ab( \alpha +\beta ) +b^{2}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{a\left( -\frac{b}{a}\right) +2b}{a^{2}\left(\frac{c}{a}\right) +ab\left( -\frac{b}{a}\right) +b^{2}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{-b+2b}{ac-b^{2} +b^{2}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{b}{ac}
\end{array}$

The value of $\frac{1}{aα\ +\ b}\ +\ \frac{1}{aβ\ +\ b}$ is $\frac{b}{ac}$.

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Updated on: 10-Oct-2022

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