First slide

Theory of expressions

Question

Let $α$ and $β$ be the roots of equation px²+ qx+ r = 0 $p \neq 0$ If p ,q and r, n AP and $\frac{1}{α} + \frac{1}{β} = 4$ then the value of $| α - β |$ is

Easy

A

$\frac{\sqrt{61}}{9}$
B

$\frac{2 \sqrt{17}}{9}$
C

$\frac{\sqrt{34}}{9}$
D

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

Solution

Given $α$ and $β$ are roots of px² + qx + r =0, $p \neq 0$
$∴ α + β = \frac{- q}{p} and αβ = \frac{r}{p}$ …………………….(i)
Since, p, q and r are in AP
$∴ 2 q = p + r$ ……………………………………….(ii)
$\begin{array}{ll} Also, \frac{1}{α} + \frac{1}{β} = 4 \\ \Rightarrow \frac{α + β}{αβ} = 4 [given] \end{array}$
$\begin{aligned} \Rightarrow α + β = 4 αβ \\ \Rightarrow \frac{- q}{p} = \frac{4 r}{p} \\ \Rightarrow q = - 4 r \end{aligned} . . . . . . . . . . . . . . . . . . . . . . . . [from Eq . (i)]$
On putting the value of qin Eq. (ii), we get
$\begin{aligned} 2 (- 4 r) = p + r \Rightarrow p = - 9 r \\ Now, α + β = \frac{- q}{p} = \frac{4 r}{p} = \frac{4 r}{- 9 r} = - \frac{4}{9} \\ and αβ = \frac{r}{p} = \frac{r}{- 9 r} = \frac{1}{- 9} \\ ∴ \begin{aligned} (α - β)^{2} & = (α + β)^{2} - 4 αβ \\ = \frac{16}{81} + \frac{4}{9} = \frac{16 \cdot + 36}{81} \\ \Rightarrow & (α - β)^{2} = \frac{52}{81} \Rightarrow | α - β | = \frac{2}{9} \sqrt{13} \end{aligned} \end{aligned}$

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