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  • 1by sqroot9- sqroot8 is equal to:(A)1by2(3-2 sqroot2)(B)1by3+2 sqroot2(c)3-2 sqroot2(D)3+2 sqroot2

\frac{1}{\sqrt{9}-\sqrt{8}} is equal to :

(A) \frac{1}{2}\left ( 3-2\sqrt{2} \right )
(B) \frac{1}{3+2\sqrt{2}}

(C)3-2\sqrt{2}
(D) 3+2\sqrt{2}

Answers (1)

Answer.[D]
Solution.        
We have, \frac{1}{\sqrt{9}-\sqrt{8}}
We have to rationalize it
\frac{1}{\sqrt{9}-\sqrt{8}}\times \frac{\sqrt{9}+\sqrt{8}}{\sqrt{9}+\sqrt{8}}            [Multiplying and dividing by \sqrt{9}+\sqrt{8}]

= \frac{\sqrt{9}+\sqrt{8}}{\left ( \sqrt{9} \right )^{2}-\left ( \sqrt{8} \right )^{2}}      [\because (a – b) (a + b) = a2 – b2]

=\frac{3+2\sqrt{2}}{9-8}
= 3+2\sqrt{2}

Hence option D is correct.

Posted by

infoexpert27

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