Question

If $\alpha$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$1/\alpha +1/\beta =4$, then the value of $|\alpha -\beta |$is

• A. $\frac{\sqrt{61}}{9}$
• B. $\frac{2\sqrt{17}}{9}$
• C. $\frac{\sqrt{34}}{9}$
• D. $\frac{2\sqrt{13}}{9}$

Answer 1

The correct option is D

$\frac{2\sqrt{13}}{9}$

Explanation for the correct option:

Step 1. Apply the given conditions in the questions :

Given that $\alpha$and $\beta$are the roots of the equation$p{x}^2+qx+r=0,p\ne 0$.

$\Rightarrow ɑ+\beta =–q/pandɑ\beta =r/p$

and also given $1/\alpha +1/\beta =4$

$$\begin{gathered} \Rightarrow \frac{\alpha +\beta }{\alpha \beta }=4 \\ \Rightarrow \alpha +\beta =4\alpha \beta ........\left(i\right) \end{gathered}$$

Since $p,q$ and $r$ are in AP .

$$\begin{gathered} \Rightarrow 2q=p+r \\ \Rightarrow 2\left(\frac{q}{p}\right)=1+\left(\frac{r}{p}\right) \\ \Rightarrow -2\alpha -2\beta =1+\alpha \beta \\ \Rightarrow -2(\alpha +\beta )=1+\alpha \beta \\ \Rightarrow -2\left(4\alpha \beta \right)=1+\alpha \beta \mathbf{}\mathbf{\left(}\mathit{f}\mathit{r}\mathit{o}\mathit{m}\mathbf{\right(}\mathit{i}\mathbf{\left)}\mathbf{\right)} \\ \Rightarrow -8\left(\alpha \beta \right)=1+\alpha \beta \\ \Rightarrow \left(\alpha \beta \right)=1/9 \\ \Rightarrow ɑ+\beta =–4/9 \end{gathered}$$

Step2. Find the value of $\mathbf{|}\mathit{\alpha }\mathbf{}\mathbf{-}\mathit{\beta }\mathbf{|}$:

$$\begin{gathered} |ɑ–\beta |=\sqrt{{(ɑ+\beta )}^2–4(ɑ\beta )} \\ \Rightarrow |ɑ–\beta |=\sqrt{{\left(\frac{-4}{9}\right)}^2-4\left(\frac{-1}{9}\right)} \\ \Rightarrow |ɑ–\beta |=\sqrt{\left(\frac{16}{81}\right)+\left(\frac{4}{9}\right)} \\ \Rightarrow |ɑ–\beta |=\sqrt{\frac{(16+36)}{81}} \\ \Rightarrow |ɑ–\beta |=\frac{\sqrt{52}}{9} \\ \Rightarrow |ɑ–\beta |=\frac{2\left(\sqrt{13}\right)}{9} \end{gathered}$$

Hence, the correct option is (D).