Let and be the roots of the equation p x 2 + q x - r = 0 , where p ? 0 . If p , q and r be the consecutive terms of a non constant G.P. and 1 + 1 = 3 4 , then the value of is: [2024]

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Solution

Q.
Let $α$ and $β$ be the roots of the equation $p x^{2} + q x - r = 0,$ where $p \neq 0 .$ If $p, q$ and $r$ be the consecutive terms of a non constant G.P. and $\frac{1}{α} + \frac{1}{β} = \frac{3}{4},$ then the value of ${(α - β)}^{2}$ is: [2024]

1 $8$

2 $9$

3 $\frac{20}{3}$

4 $\frac{80}{9}$

Ans.
(4)

We have,

$α + β = - \frac{q}{p}$ ...(i) and $α β = - \frac{r}{p}$ ...(ii)

Now, $\frac{1}{α} + \frac{1}{β} = \frac{3}{4} \Rightarrow \frac{α + β}{α β} = \frac{3}{4} \Rightarrow \frac{q}{r} = \frac{3}{4} \Rightarrow \frac{r}{q} = \frac{4}{3}$

$∵$ $p, q$ and $r$ are in G.P $∴$ $\frac{q}{p} = \frac{r}{q} = \frac{4}{3}$

So, $α + β = - \frac{4}{3}$ and $α β = - {(\frac{r}{q})}^{2} = - \frac{16}{9}$

Now, ${(α - β)}^{2} = {(α + β)}^{2} - 4 α β = \frac{16}{9} + \frac{64}{9} = \frac{80}{9}$

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