If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ 6x^2\ +\ x\ –\ 2$, find the value of $\frac{α}{β}\ +\ \frac{β}{α}$.
Given:
$α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ 6x^2\ +\ x\ –\ 2$.
To do:
Here, we have to find the value of $\frac{α}{β}\ +\ \frac{β}{α}$.
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,
$a=6$, $b=1$ and $c=-2$
Sum of the roots $= α+β = \frac{-b}{a} = \frac{-1}{6} = -\frac{1}{6}$.
Product of the roots $= αβ = \frac{c}{a} = \frac{-2}{6} = -\frac{1}{3}$.
Therefore,
$\frac{α}{β} +\frac{β}{α}=\frac{(α^2 +β^2)}{αβ}$
$=\frac{(α+β)^2-2αβ}{αβ}$
$ \begin{array}{l}
=\frac{\left( -\frac{1}{6}\right)^{2} -2\left( -\frac{1}{3}\right)}{-\frac{1}{3}}\\
\\
=\frac{\frac{1}{36} +\frac{2}{3}}{-\frac{1}{3}}\\
\\
=\frac{( 1+2\times 12)}{36} \times \frac{-3}{1}\\
\\
=\frac{-( 1+24)}{12}\\
\\
=\frac{-25}{12}
\end{array}$
The value of $\frac{α}{β}\ +\ \frac{β}{α}$ is $\frac{-25}{12}$.
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