 QUESTION

# If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $p(s)=3{{s}^{2}}-6s+4$, find the value of $\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }+2\left( \dfrac{1}{\alpha }+\dfrac{1}{\beta } \right)+3\alpha \beta$.

Hint: We will use the formula of summation of two roots that is $\alpha +\beta =\dfrac{-b}{a}$ and product of two roots that is $\alpha \beta =\dfrac{c}{a}$ to solve this question. Then using these two we find ${{\alpha }^{2}}+{{\beta }^{2}}$ and then substitute the values of all these terms in the given expression.

Complete step-by-step solution -
Quadratic polynomial given in the question is $p(s)=3{{s}^{2}}-6s+4......(1)$ has two roots(zeros), namely $\alpha$ and $\beta$ from coefficient and zeros relation.
We know that the summation of the two zeros is $\alpha +\beta =\dfrac{-b}{a}........(2)$
Also the product of the two zeros is $\alpha \beta =\dfrac{c}{a}........(3)$
Now comparing equation (1) with the general form of quadratic polynomial $p(s)=a{{s}^{2}}+bs+c$ we get a as 3, b as -6 and c as 4. Now substituting these values of coefficients in equation (2) and equation (3) we get,
$\alpha +\beta =\dfrac{-(-6)}{3}=\dfrac{6}{3}=2........(4)$ and $\alpha \beta =\dfrac{4}{3}........(5)$
Now we will solve this expression $\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }+2\left( \dfrac{1}{\alpha }+\dfrac{1}{\beta } \right)+3\alpha \beta ........(6)$
Rearranging and simplifying equation (6) we get,
$\,\Rightarrow \dfrac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }+2\left( \dfrac{\alpha +\beta }{\alpha \beta } \right)+3\alpha \beta ........(7)$
In equation (7) the value of the term ${{\alpha }^{2}}+{{\beta }^{2}}$ is not known and all other terms value is known. So now finding the value of ${{\alpha }^{2}}+{{\beta }^{2}}$.
From equation (4) we get,
$\Rightarrow \alpha +\beta =2$
Now squaring both sides of this equation we get,
$\,\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta =4......(8)$
Substituting the value of $\alpha \beta$ from equation (5) in equation (8) we get,
\begin{align} & \,\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}+2\times \dfrac{4}{3}=4 \\ & \,\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}=4-\dfrac{8}{3} \\ & \,\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}=\dfrac{4}{3}........(9) \\ \end{align}
Now substituting all the values from equation (4), equation (5) and equation (9) in equation (7) we get,
\begin{align} & \,\Rightarrow \dfrac{\dfrac{4}{3}}{\dfrac{4}{3}}+2\left( \dfrac{2}{\dfrac{4}{3}} \right)+3\times \dfrac{4}{3} \\ & \,\Rightarrow 1+3+4=8 \\ \end{align}
Hence the answer to this expression $\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }+2\left( \dfrac{1}{\alpha }+\dfrac{1}{\beta } \right)+3\alpha \beta$ is 8.

Note: We have to remember both the formulas of sum of roots and product of the roots because these are the key to solve this problem. We may get confused in equation (7) about how to find ${{\alpha }^{2}}+{{\beta }^{2}}$ but here we have can find it by squaring both sides of $\Rightarrow \alpha +\beta =2$. In a hurry we may make simple calculation mistakes, so we need to be very careful with every step.