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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Sequence and Series

Harmonic mean of roots of equation $(5+\sqrt{2})\;x^2$ - $(4+\sqrt{2})\;x$ + $8$ + $2\;\sqrt{2}$ will be

$(a)\;6\qquad(b)\;4\qquad(c)\;2\qquad(d)\;8$

1 Answer

Answer : (b) 4
Explanation : $\alpha+\beta=\frac{4+\sqrt{2}}{5+\sqrt{2}}$
$\alpha\;\beta=\frac{8+2\sqrt{2}}{5+\sqrt{2}}$
$H.M=\frac{2\alpha\beta}{\alpha+\beta}=4.$
answered Jan 21, 2014 by yamini.v
 

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