Question

If $\alpha \ and\ \beta $are the zeros of the quadratic polynomial \[f\left( x \right)={{x}^{2}}-x-4\], find the value of $\dfrac{1}{\beta }+\dfrac{1}{\alpha }-\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 \[f\left( x \right)={{x}^{2}}-x-4\]. We have to find the value of $\dfrac{1}{\beta }+\dfrac{1}{\alpha }-\alpha \beta $.
Now, to find this value we will be using 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 \[f\left( x \right)={{x}^{2}}-x-4\]
$\begin{align}
  & \alpha +\beta =\text{sum of roots }=1..........\left( 1 \right) \\
 & \alpha \beta \ \text{=}\ \text{product of roots }=-4...........\left( 2 \right) \\
\end{align}$
Now, we have to find the value of,
$\begin{align}
  & \dfrac{1}{\alpha }+\dfrac{1}{\beta }-\alpha \beta \\
 & =\dfrac{\beta +\alpha }{\alpha \beta }-\alpha \beta \\
\end{align}$
Now, we will substitute the value of $\alpha +\beta \ and\ \alpha \beta $ from (1) and (2).
$\begin{align}
  & =\dfrac{1}{-4}-\left( -4 \right) \\
 & =\dfrac{-1}{4}+4 \\
 & =4-\dfrac{1}{4} \\
 & =\dfrac{15}{4} \\
\end{align}$
Therefore, the value of $\dfrac{1}{\beta }+\dfrac{1}{\alpha }-\alpha \beta =\dfrac{15}{4}$.

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