Question

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

Answer 1

Hint: We will be using the concept of quadratic equations to solve the problem. We will be using sum of roots and product of roots to further simplify the problem.

Complete Step-by-Step solution:
Now, we have been given $\alpha \ and\ \beta $ are the zeros of the quadratic polynomial \[p\left( s \right)=3{{s}^{2}}-6s+4\]. We have to find the value of $\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }+2\left( \dfrac{1}{\alpha }+\dfrac{1}{\beta } \right)+3\alpha \beta $.
Now, to find this value we will be using the sum of roots and product of roots. We know that in a quadratic equation $a{{x}^{2}}+bx+c=0$.
$\begin{align}
  & \text{sum of roots }=\dfrac{-b}{a} \\
 & \text{product of roots }=\dfrac{c}{a} \\
\end{align}$
Therefore, in \[p\left( s \right)=3{{s}^{2}}-6s+4\].
$\begin{align}
  & \alpha +\beta =\text{sum of roots }=\dfrac{-\left( -6 \right)}{3}=2..........\left( 1 \right) \\
 & \alpha \beta \ \text{=}\ \text{product of roots }=\dfrac{4}{3}...........\left( 2 \right) \\
\end{align}$
Now, we have to find the value of$\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }+2\left( \dfrac{1}{\alpha }+\dfrac{1}{\beta } \right)+3\alpha \beta $.
Now, we will simplify the expression by taking LCM,
$\dfrac{{{\alpha }^{2}}+{{\beta }^{2}}+2\left( \beta +\alpha \right)+3{{\alpha }^{2}}{{\beta }^{2}}}{\alpha \beta }$
Now, we have the value of $\left( \alpha +\beta \right)\ and\ \alpha \beta $ but we don’t have the value of ${{\alpha }^{2}}+{{\beta }^{2}}$which we will find by using ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
Now, substituting $a=\alpha \ and\ b=\beta $ we have,
$\begin{align}
  & {{\left( \alpha +\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta \\
 & {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta ={{\alpha }^{2}}+{{\beta }^{2}} \\
\end{align}$
$\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }=\dfrac{{{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta }{\alpha \beta }$
Now, we will substitute the value of $\left( \alpha +\beta \right)\ and\ \alpha \beta $ from (1) and (2). So,
\[\begin{align}
  & {{\alpha }^{2}}+{{\beta }^{2}}={{\left( 2 \right)}^{2}}-2\left( \dfrac{4}{3} \right) \\
 & =4-\dfrac{8}{3} \\
 & =\dfrac{12-8}{3} \\
 & {{\alpha }^{2}}+{{\beta }^{2}}=\dfrac{4}{3}..............\left( 3 \right) \\
\end{align}\]
Now, we have the expression to find as,
$\dfrac{{{\alpha }^{2}}+{{\beta }^{2}}+2\left( \beta +\alpha \right)+3{{\alpha }^{2}}{{\beta }^{2}}}{\alpha \beta }$
Now, we will substitute the values from (1), (2) and (3),
\[\begin{align}
  & =\dfrac{\dfrac{4}{3}+2\left( 2 \right)+3{{\left( \dfrac{4}{3} \right)}^{2}}}{\dfrac{4}{3}} \\
 & =\dfrac{\dfrac{4}{3}+4+3\times \dfrac{16}{9}}{\dfrac{4}{3}} \\
 & =\dfrac{\dfrac{4}{3}+4+\dfrac{16}{3}}{\dfrac{4}{3}} \\
 & =\dfrac{4+12+16}{4} \\
 & =1+3+4 \\
 & =8 \\
\end{align}\]
Therefore, the value of $\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }+2\left( \dfrac{1}{\alpha }+\dfrac{1}{\beta } \right)+3\alpha \beta $ is 8.

Note: To solve these types of questions one must know how to find the relation between sum of roots, product of roots and coefficient of quadratic equation.
$\begin{align}
  & \text{sum of roots }=\dfrac{-b}{a} \\
 & \text{product of roots }=\dfrac{c}{a} \\
\end{align}$