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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Let $\alpha$ and $\beta$ be the roots of the equation $p x^{2}+q x-r=0$, where $p\ne 0$.
If $p,q,r$ are consecutive terms of a non-constant G.P. and
$\dfrac1\alpha+\dfrac1\beta=\dfrac34$, then the value of $(\alpha-\beta)^{2}$ is:
Solution
Solution:
$ px^2 + qx - r = 0 $
$ \alpha + \beta = -\frac{q}{p}, \;\; \alpha\beta = -\frac{r}{p} $
Given $ p,q,r $ are consecutive GP ⇒ let ratio = $k$
$ q = pk,\;\; r = pk^2 $
So,
$ \alpha + \beta = -k,\;\; \alpha\beta = -k^2 $
Given:
$ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{-k}{-k^2} = \frac{1}{k} = \frac{3}{4} $
$ k = \frac{4}{3} $
Now,
$ \alpha + \beta = -\frac{4}{3}, \;\; \alpha\beta = -\frac{16}{9} $
$ (\alpha - \beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta $
$ = \frac{16}{9} + \frac{64}{9} = \frac{80}{9} $
$ \boxed{\frac{80}{9}} $
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