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If $α$ and $β$ are the zeros of the quadratic polynomial $p(s)\ =\ 3s^2\ -\ 6s\ +\ 4$, find the value of $\frac{α}{β}\ +\ \frac{β}{α}\ +\ 2\left(\frac{1}{α}\ +\ \frac{1}{β}\right)\ +\ 3αβ$.
Given:
$α$ and $β$ are the zeros of the quadratic polynomial $p(s)\ =\ 3s^2\ -\ 6s\ +\ 4$.
To do:
Here, we have to find the value of $\frac{α}{β}\ +\ \frac{β}{α}\ +\ 2\left(\frac{1}{α}\ +\ \frac{1}{β}\right)\ +\ 3αβ$.
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,
$a=3$, $b=-6$ and $c=4$
Sum of the roots $= α+β = \frac{-b}{a} = \frac{-(-6)}{3} = -\frac{6}{3}=2$.
Product of the roots $= αβ = \frac{c}{a} = \frac{4}{3}$.
Therefore,
$ \begin{array}{l}
\frac{\alpha }{\beta } +\frac{\beta }{\alpha } +2\left(\frac{1}{\alpha } +\frac{1}{\beta }\right) +3\alpha \beta =\frac{\alpha ^{2} +\beta ^{2}}{\alpha \beta } +2\left(\frac{\alpha +\beta }{\alpha \beta }\right) +3\alpha \beta \\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{( \alpha +\beta )^{2} -2\alpha \beta }{\alpha \beta } +2\left(\frac{\alpha +\beta }{\alpha \beta }\right) +3\alpha \beta \\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{( 2)^{2} -2\left(\frac{4}{3}\right)}{\frac{4}{3}} +2\left(\frac{2}{\frac{4}{3}}\right) +3\left(\frac{4}{3}\right)\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{4-\frac{8}{3}}{\frac{4}{3}} +4\left(\frac{3}{4}\right) +4\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{4\times 3-8}{3} \times \frac{3}{4} +3+4\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{12-8}{4} +7\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{4}{4} +7\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1+7\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =8
\end{array}$
The value of $\frac{α}{β}\ +\ \frac{β}{α}\ +\ 2\left(\frac{1}{α}\ +\ \frac{1}{β}\right)\ +\ 3αβ$ is $8$.
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