Question

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

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 \[f\left( x \right)=6{{x}^{2}}+x-2\]. We have to find the value of $\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }$.
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 \[f\left( x \right)=6{{x}^{2}}+x-2\]
$\begin{align}
  & \alpha +\beta =\text{sum of roots }=\dfrac{-1}{6}..........\left( 1 \right) \\
 & \alpha \beta \ \text{=}\ \text{product of roots }=\dfrac{-2}{6}=\dfrac{-1}{3}...........\left( 2 \right) \\
\end{align}$
Now, we have to find the value of$\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }$.
On taking LCM we have,
$\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }=\dfrac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }...........\left( 3 \right)$
Now, we have the value of $\alpha \beta $but we don’t have the value of ${{\alpha }^{2}}+{{\beta }^{2}}$which we will find by using ${{\left( \alpha +\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta $.
So, by substituting the value of ${{\alpha }^{2}}+{{\beta }^{2}}$ in (3) we have,
$\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }=\dfrac{{{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta }{\alpha \beta }$
Now, we will substitute the value of $\left( \alpha +\beta \right)\ and\ \alpha \beta $from (1) and (2). So,
$\begin{align}
  & \dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }=\dfrac{{{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta }{\alpha \beta } \\
 & =\dfrac{{{\left( \dfrac{1}{6} \right)}^{2}}-2\left( \dfrac{-1}{3} \right)}{\dfrac{-1}{3}} \\
 & =\dfrac{\dfrac{1}{36}+\dfrac{2}{3}}{\dfrac{-1}{3}} \\
 & =\dfrac{\dfrac{1+24}{36}}{\dfrac{-1}{3}} \\
 & =\dfrac{\dfrac{25}{36}}{\dfrac{-1}{3}} \\
 & =\dfrac{25\times 3}{-36} \\
 & \dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }=\dfrac{-25}{12} \\
\end{align}$
Therefore, the value of $\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }$is $\dfrac{-25}{12}$.

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.
$\begin{align}
  & \text{sum of roots }=\dfrac{-b}{a} \\
 & \text{product of roots }=\dfrac{c}{a} \\
\end{align}$