Question

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

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\[f\left( x \right)={{x}^{2}}+x-2\]. We have to find the value of$\dfrac{1}{\alpha }-\dfrac{1}{\beta }$.
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 \[f\left( x \right)={{x}^{2}}+x-2\]
$\begin{align}
  & \alpha +\beta =\text{sum of roots }=-1..........\left( 1 \right) \\
 & \alpha \beta \ \text{=}\ \text{product of roots }=-2...........\left( 2 \right) \\
\end{align}$
Now, we have to find the value of,
$\begin{align}
  & \dfrac{1}{\alpha }-\dfrac{1}{\beta }=\dfrac{\beta -\alpha }{\alpha \beta } \\
 & =\dfrac{-\left( \alpha -\beta \right)}{\alpha \beta } \\
\end{align}$
Now, we know the value of $\alpha \beta $but we don’t know the value of $\left( \alpha -\beta \right)$. Now, we know that ${{\left( \alpha -\beta \right)}^{2}}={{\left( \alpha +\beta \right)}^{2}}-4\alpha \beta $.
Now, we will substitute the value of $\alpha +\beta \ and\ \alpha \beta $ from (1) and (2) we have,
$\begin{align}
  & {{\left( \alpha -\beta \right)}^{2}}={{\left( -1 \right)}^{2}}-4\left( -2 \right) \\
 & =1+8 \\
 & =9 \\
 & \alpha -\beta =\sqrt{9} \\
 & \alpha -\beta =\pm 3............\left( 3 \right) \\
\end{align}$
Now, we have two values of $\left( \alpha -\beta \right)$ because in subtracting two roots the answer depends from which root you are subtracting to another. Therefore, both are correct. So,
$\dfrac{1}{\alpha }-\dfrac{1}{\beta }=\dfrac{-\left( \alpha -\beta \right)}{\alpha \beta }$
From (2) and (3) we substitute value and get,
$\begin{align}
  & \dfrac{-\left( \alpha -\beta \right)}{\alpha \beta }=\dfrac{-\left( 3 \right)}{-2}=\dfrac{3}{2} \\
 & or \\
 & \dfrac{-\left( \alpha -\beta \right)}{\alpha \beta }=\dfrac{-\left( -3 \right)}{-2}=\dfrac{-3}{2} \\
\end{align}$
Therefore, $\dfrac{1}{\alpha }-\dfrac{1}{\beta }=\dfrac{3}{2}\ or\ \dfrac{-3}{2}$.

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