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Examveda

Given that $$\sqrt 3 = 1.732{\text{,}}$$   the value of $$\frac{{3 + \sqrt 6 }}{{5\sqrt 3 - 2\sqrt {12} - \sqrt {32} + \sqrt {50} }}$$      is ?

A. 1.414

B. 1.732

C. 2.551

D. 4.899

Answer: Option B

Solution(By Examveda Team)

$$\eqalign{ & {\text{Given expression,}} \cr & = \frac{{3 + \sqrt 6 }}{{5\sqrt 3 - 2\sqrt {12} - \sqrt {32} + \sqrt {50} }} \cr & = \frac{{3 + \sqrt 6 }}{{5\sqrt 3 - 4\sqrt 3 - 4\sqrt 2 + 5\sqrt 2 }} \cr & = \frac{{\left( {3 + \sqrt 6 } \right)}}{{\left( {\sqrt 3 + \sqrt 2 } \right)}} \cr & = \frac{{\left( {3 + \sqrt 6 } \right)}}{{\left( {\sqrt 3 + \sqrt 2 } \right)}} \times \frac{{\left( {\sqrt 3 - \sqrt 2 } \right)}}{{\left( {\sqrt 3 - \sqrt 2 } \right)}} \cr & = \frac{{3\sqrt 3 - 3\sqrt 2 + 3\sqrt 2 - 2\sqrt 3 }}{{\left( {3 - 2} \right)}} \cr & = \sqrt 3 \cr & = 1.732 \cr} $$

This Question Belongs to Arithmetic Ability >> Square Root And Cube Root

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