Theory of Equations 1 Question 19
20. Let $\alpha$ and $\beta$ be the roots of equation $p x^{2}+q x+r=0$, $p \neq 0$. If $p, q$ and $r$ are in AP and $\frac{1}{\alpha}+\frac{1}{\beta}=4$, then the value of $|\alpha-\beta|$ is
(2014 Main)
(a) $\frac{\sqrt{61}}{9}$
(b) $\frac{2 \sqrt{17}}{9}$
(c) $\frac{\sqrt{34}}{9}$
(d) $\frac{2 \sqrt{13}}{9}$
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Solution:
- PLAN If $a x^{2}+b x+c=0$ has roots $\alpha$ and $\beta$, then $\alpha+\beta=-b / a$ and $\alpha \beta=\frac{c}{a}$. Find the values of $\alpha+\beta$ and $\alpha \beta$ and then put in $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta$ to get required value.
Given, $\alpha$ and $\beta$ are roots of $p x^{2}+q x+r=0, p \neq 0$.
$$ \therefore \quad \alpha+\beta=\frac{-q}{p}, \quad \alpha \beta=\frac{r}{p} $$
Since, $p, q$ and $r$ are in AP.
$$ \therefore \quad 2 q=p+r $$
Also, $\quad \frac{1}{\alpha}+\frac{1}{\beta}=4 \Rightarrow \frac{\alpha+\beta}{\alpha \beta}=4$
$$ \begin{array}{ccrl} & \Rightarrow & \alpha+\beta & =4 \alpha \beta \Rightarrow \\ \Rightarrow & & q & =-4 r \end{array} $$
[from Eq. (i)]
On putting the value of $q$ in Eq. (ii), we get
$$ \Rightarrow \quad 2(-4 r)=p+r \Rightarrow p=-9 r $$
$$ \begin{aligned} & \text { Now, } \quad \alpha+\beta=\frac{-q}{p}=\frac{4 r}{p}=\frac{4 r}{-9 r}=-\frac{4}{9} \\ & \text { and } \quad \alpha \beta=\frac{r}{p}=\frac{r}{-9 r}=\frac{1}{-9} \\ & \therefore \quad(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=\frac{16}{81}+\frac{4}{9}=\frac{16+36}{81} \\ & \Rightarrow \quad(\alpha-\beta)^{2}=\frac{52}{81} \\ & \Rightarrow \quad|\alpha-\beta|=\frac{2}{9} \sqrt{13} \end{aligned} $$
