A) \[\frac{\sqrt{61}}{9}\]
B) \[\frac{2\sqrt{17}}{9}\]
C) \[\frac{\sqrt{34}}{9}\]
D) \[\frac{2\sqrt{13}}{9}\]
Correct Answer: D
Solution :
\[\alpha ,\beta \]are roots of \[p{{x}^{2}}+qx+r=0\] \[\alpha +\beta =-\frac{q}{p},\alpha \beta =\frac{r}{p}\text{ }\]p, q, r are in A.P. \[2q=p+r\] ? (1) \[\frac{1}{\alpha }+\frac{1}{\beta }=4\] \[\alpha +\beta \] \[=4\alpha \beta \] \[-\frac{q}{p}\] \[=\frac{4r}{p}\] \[-q=4r\] \[\frac{q}{r}=-4\] ?(2) From (1) and (2)\[\frac{p}{r}=-9\] ?.(3) \[|\alpha -\beta |=\sqrt{{{(\alpha +\beta )}^{2}}-4\alpha \beta }\]\[=\sqrt{\frac{{{q}^{2}}}{{{p}^{2}}}+\frac{4r}{p}}\] \[=\sqrt{\frac{16}{81}+\frac{4}{9}}=\frac{2\sqrt{13}}{9}\]
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