Let α and β be the roots of equation px2 + qx + r = 0, p \neq 0. If p, q, r are in A.P. and \frac{\alpha + \beta}{αβ}= 4, then the value of | α – β

Engineering

Mathematics

Quadratic Equation and Nature of Roots

Question

Let α and β be the roots of equation px² + qx + r = 0, p $\neq$ 0. If p, q, r are in A.P. and $\frac{α + β}{αβ}$ = 4, then the value of | α – β | is

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

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

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

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

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, p $\neq$ 0

Given, p, q, r are in A.P.

So, 2q = p + r ……. (i)

$and \frac{α + β}{αβ} = 4 \Rightarrow \frac{- q}{p} = \frac{4 r}{p} \Rightarrow q + 4 r = 0$

Also, – 8r = p + r ⇒ p = – 9r {from (i)}

Now, | α – β |2 = (α + β)2 – 4αβ

$= \frac{q^{2}}{p^{2}} - \frac{4 r}{p} = \frac{q^{2} - 4 pr}{p^{2}} = \frac{16 r^{2} - 4 pr}{p^{2}}$

$= \frac{16 \times p^{2}}{81 p^{2}} - \frac{4}{p} (\frac{- p}{9}) = \frac{16}{81} + \frac{4}{9}$

$= \frac{16 + 36}{81} = \frac{52}{81}$

Hence, $| α - β | = \sqrt{\frac{52}{81}} = \frac{2 \sqrt{13}}{9}$ .