Question

Let $\alpha $ and $\beta $ are the roots of the equation $p{{x}^{2}}+qx+r=0$ , $p\ne 0$. If $p$ , $q$ and $r$ are in AP and $\dfrac{1}{\alpha }+\dfrac{1}{\beta }=4$ , then the value of $\left| \alpha -\beta \right|$ is
1) $\dfrac{\sqrt{61}}{9}$
2) $\dfrac{2\sqrt{17}}{9}$
3) $\dfrac{\sqrt{34}}{9}$
4) $\dfrac{2\sqrt{13}}{9}$

Answer 1

Hint: In this problem we need to find the value of $\left| \alpha -\beta \right|$. For this we will first consider the given quadratic equation $p{{x}^{2}}+qx+r=0$ and the roots $\alpha $ and $\beta $, use the relation between the roots and coefficients of the quadratic equation and write the values of $\alpha +\beta $ and $\alpha \beta $. Now consider the statement that $p$ , $q$ and $r$ are in AP. From this we can establish a relation between $p$ , $q$ and $r$. Now consider the value $\dfrac{1}{\alpha }+\dfrac{1}{\beta }=4$ apply the lcm and use the values of $\alpha +\beta $ and $\alpha \beta $ to get the relation between $p$ , $q$ and $r$ based on these relations we can calculate the required value.

Complete step-by-step solution:
The quadratic equation is $p{{x}^{2}}+qx+r=0$ and the roots of the equation are $\alpha $ and $\beta $.
We can write the values of $\alpha +\beta $ and $\alpha \beta $ as
$\alpha +\beta =-\dfrac{q}{p}$ and $\alpha \beta =\dfrac{r}{p}$ .
Given that $p$ , $q$ and $r$ are in AP, we can write that $2q=p+r$ .
Also we have the value $\dfrac{1}{\alpha }+\dfrac{1}{\beta }=4$. Simplify the above value using LCM, then we will have
$\begin{align}
  & \dfrac{\alpha +\beta }{\alpha \beta }=4 \\
 & \alpha +\beta =4\alpha \beta \\
\end{align}$
Substitute the values $\alpha +\beta =-\dfrac{q}{p}$ and $\alpha \beta =\dfrac{r}{p}$ in the above equation, then we will get
$\begin{align}
  & -\dfrac{q}{p}=\dfrac{4r}{p} \\
 & \Rightarrow q=-4r \\
\end{align}$
We have the relation $2q=p+r$. Substituting the value $q=-4r$ in this relation, then we will have
$\begin{align}
  & 2\left( -4r \right)=p+r \\
 & \Rightarrow -8r=p+r \\
 & \Rightarrow p=-9r \\
\end{align}$
Now consider the value $\alpha +\beta =-\dfrac{q}{p}$. Substitute the value of $q$ and $p$ , then we will get
$\begin{align}
  & \alpha +\beta =-\dfrac{\left( -4r \right)}{\left( -9r \right)} \\
 & \Rightarrow \alpha +\beta =-\dfrac{4}{9} \\
\end{align}$
Consider the value $\alpha \beta =\dfrac{r}{p}$. Substitute the value of $p$, then we will have
$\begin{align}
  & \alpha \beta =\dfrac{r}{-9r} \\
 & \Rightarrow \alpha \beta =-\dfrac{1}{9} \\
\end{align}$
From algebraic formulas ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, we can write the value ${{\left( \alpha -\beta \right)}^{2}}$ as
${{\left( \alpha -\beta \right)}^{2}}={{\left( \alpha +\beta \right)}^{2}}-4\alpha \beta $
Substituting the values of $\alpha +\beta $ and $\alpha \beta $ in the above equation, then we will have
$\begin{align}
  & {{\left( \alpha -\beta \right)}^{2}}={{\left( -\dfrac{4}{9} \right)}^{2}}-4\left( -\dfrac{1}{9} \right) \\
 & \Rightarrow {{\left( \alpha -\beta \right)}^{2}}=\dfrac{16}{81}+\dfrac{4}{9} \\
 & \Rightarrow {{\left( \alpha -\beta \right)}^{2}}=\dfrac{16+9\left( 4 \right)}{81} \\
 & \Rightarrow {{\left( \alpha -\beta \right)}^{2}}=\dfrac{52}{81} \\
\end{align}$
Applying square root on both sides of the above equation, then we will get
$\left| \alpha -\beta \right|=\dfrac{2\sqrt{13}}{9}$
Hence option 4 is the correct answer.

Note: For this problem we can also use different methods. In which we can directly use the algebraic formula ${{\left( \alpha -\beta \right)}^{2}}={{\left( \alpha +\beta \right)}^{2}}-4\alpha \beta $ at the beginning and substitute the values of $\alpha +\beta $ and $\alpha \beta $. After that use the remaining data and simplify the equation to get the direct result. But it will have some complicated calculations. So we have not used that method.