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Step-by-step Solution
Step 1: Express the condition that p, q, r are in Arithmetic Progression (A.P.)
Since p, q, r are in A.P., it follows that
$2q = p + r.$
This will be used later to relate the coefficients p, q, and r.
Step 2: Use the relation for sum and product of roots
Let the roots of $p x^2 + q x + r = 0$ be $ \alpha $ and $ \beta $. Then,
$ \alpha + \beta = -\frac{q}{p} $ and $ \alpha \beta = \frac{r}{p}. $
Step 3: Apply the given condition on the roots
We are given
$ \frac{1}{\alpha} + \frac{1}{\beta} = 4.$
This can be rewritten using sum and product of roots as
$ \frac{\alpha + \beta}{\alpha \beta} = 4.$
Substituting
$ \alpha + \beta = -\frac{q}{p} $
and
$ \alpha \beta = \frac{r}{p}, $
we get
$ \frac{-\frac{q}{p}}{\frac{r}{p}} = 4 \implies \frac{-q}{r} = 4 \implies q = -4r. $
Step 4: Relate p, q, and r using the A.P. condition
From
$2q = p + r$
and
$q = -4r,$
substitute $q$ into the A.P. relation:
$ 2(-4r) = p + r \implies -8r = p + r \implies p = -9r.$
Thus, we have
$ p = -9r, \quad q = -4r. $
Step 5: Use the difference of roots formula
The difference of the roots is given by
$ |\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4 \alpha \beta}. $
Substituting $\alpha + \beta = -\frac{q}{p}$ and $\alpha\beta = \frac{r}{p},$ we obtain:
$
|\alpha - \beta|
= \sqrt{\Bigl(-\frac{q}{p}\Bigr)^2 - 4 \cdot \frac{r}{p}}
= \sqrt{\frac{q^2}{p^2} - \frac{4r}{p}}
= \sqrt{\frac{q^2 - 4pr}{p^2}}
= \frac{\sqrt{q^2 - 4 p r}}{|p|}.
$
Step 6: Substitute the values of p and q in terms of r
From the previous steps,
$
p = -9r,\quad q = -4r.
$
Then,
$
q^2 - 4pr
= (-4r)^2 - 4(-9r)(r)
= 16r^2 + 36r^2
= 52r^2.
$
Hence,
$
|\alpha - \beta|
= \frac{\sqrt{52\,r^2}}{|-9r|}
= \frac{\sqrt{52}\,\lvert r \rvert}{9 \lvert r \rvert}
= \frac{2\sqrt{13}}{9}.
$
Final Answer
$|\alpha - \beta| = \frac{2\sqrt{13}}{9}.$
