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  • The number obtained on rationalizing the denominator of 1 by Sqroot 7-2 is : (A)Sqroot 7+2by 3 (B)Sqroot7-2by3(C) Sqroot7+2by5 (D)Sqroot7+2by45

The number obtained on rationalizing the denominator of \frac{1}{\sqrt{7}-2} is :
(A) \frac{\sqrt{7}+2}{3}
(B)\frac{\sqrt{7}-2}{3}
(C)\frac{\sqrt{7}+2}{5}
(D)\frac{\sqrt{7}+2}{45}

Answers (1)

Answer. [A]
Solution.        

We have, \frac{1}{\sqrt{7}-2}
We have to rationalize it

\frac{1}{\sqrt{7}-2}\times \frac{\sqrt{7}+2}{\sqrt{7}+2}       [Multiplying numerator and denominator by \sqrt{7}+2]

= \frac{\sqrt{7}+2}{\left ( \sqrt{7} \right )^{2}-\left ( 2 \right )^{2}}         [\because (a – b) (a + b) = a2 – b2]

= \frac{\sqrt{7}+2}{7-4}
= \frac{\sqrt{7}+2}{3}

Hence option A is correct.

Posted by

infoexpert27

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