If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ –\ 5x\ +\ 4$, find the value of $\frac{1}{α}\ +\ \frac{1}{β}\ –\ 2αβ$.
Given:
$α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ –\ 5x\ +\ 4$.
To do:
Here, we have to find the value of $\frac{1}{α}\ +\ \frac{1}{β}\ –\ 2αβ$.
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,
$a=1$, $b=-5$ and $c=4$
Sum of the roots $= α+β = \frac{-b}{a} = \frac{– (-5)}{1} = 5$.
Product of the roots $= αβ = \frac{c}{a} = \frac{4}{1} = 4$
Therefore,
$\frac{1}{α} +\frac{1}{β}– 2αβ=\frac{(α +β)}{αβ}- 2αβ$
$=\frac{5}{4}– 2(4) = \frac{5}{4}– 8 = \frac{5-4\times8}{4}=\frac{5-32}{4}=\frac{-27}{4}$.
The value of $\frac{1}{α}\ +\ \frac{1}{β}\ –\ 2αβ$ is $\frac{-27}{4}$.
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