If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ –\ x\ –\ 4$, find the value of $\frac{1}{α}\ +\ \frac{1}{β}\ –\ αβ$.
Given:
$α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ –\ x\ -\ 4$.
To do:
Here, we have to find the value of $\frac{1}{α} + \frac{1}{β}–αβ$.
Solution:  
We know that,
The standard form of a quadratic equation is $ax^2+bx+c=0$, where a, b and c are constants and $a≠0$.
Comparing the given equation with the standard form of a quadratic equation,
$a=1$, $b=-1$ and $c=-4$
Sum of the roots $= α+β = \frac{-b}{a} = \frac{– (-1)}{1} = 1$.
Product of the roots $= αβ = \frac{c}{a} = \frac{-4}{1} = -4$.
Therefore,
$\frac{1}{α} +\frac{1}{β}–αβ=\frac{(α +β)}{αβ}- αβ$
$=\frac{1}{-4}– (-4) = -\frac{1}{4}+ 4= \frac{-1+4\times4}{4}=\frac{-1+16}{4}=\frac{15}{4}$.
The value of $\frac{1}{α}+\frac{1}{β}–αβ$ is $\frac{15}{4}$.
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