Let α and β be the roots of the equation px² + qx + r = 0, p ≠ 0. If p, q, r are in A.P. and $\frac{1}{α}$ + $\frac{1}{β}$ = 4, then the value of |α − β| is:
$\frac{2\sqrt{13}}{9}$
Since \(\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = 4\), and from the quadratic \(\alpha+\beta = -q/p\), \(\alpha\beta = r/p\), we get \(-q/r = 4\), so \(q = -4r\).
Since \(p, q, r\) are in A.P., \(2q = p+r\), so \(p = 2q-r = 2(-4r)-r = -9r\).
Taking \(r = t\) (with \(t \neq 0\)), we have \(p=-9t\), \(q=-4t\), \(r=t\).
So \(\alpha+\beta = -q/p = -\frac{-4t}{-9t} = -\frac{4}{9}\), and \(\alpha\beta = r/p = \frac{t}{-9t} = -\frac{1}{9}\).
Then \(|\alpha-\beta|^2 = (\alpha+\beta)^2-4\alpha\beta = \frac{16}{81}-4\left(-\frac{1}{9}\right) = \frac{16}{81}+\frac{36}{81} = \frac{52}{81}\).
So \(|\alpha-\beta| = \dfrac{\sqrt{52}}{9} = \dfrac{2\sqrt{13}}{9}\).
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