If $α$ and $β$ are the zeros of the quadratic polynomial $p(y)\ =\ 5y^2\ –\ 7y\ +\ 1$, find the value of $\frac{1}{α}\ +\ \frac{1}{β}$.
Given:
$α$ and $β$ are the zeros of the quadratic polynomial $p(y)\ =\ 5y^2\ –\ 7y\ +\ 1$.
To do:
Here, we have to find the value of $\frac{1}{α}+\frac{1}{β}$.
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,
$a=5$, $b=-7$ and $c=1$
Sum of the roots $= α+β = \frac{-b}{a}= \frac{– (-7)}{5}= \frac{7}{5}$.
Product of the roots $= αβ = \frac{c}{a}= \frac{1}{5}$.
Therefore,
$\frac{1}{α}+\frac{1}{β}=\frac{(α +β)}{αβ}$
$=\frac{\frac{7}{5}}{\frac{1}{5}}= 7$.
The value of $\frac{1}{α} + \frac{1}{β}$ is $7$.
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