Examveda
Examveda

The value of $$\frac{1}{{\sqrt {\left( {12 - \sqrt {140} } \right)} }}$$   $$ -\, \frac{1}{{\sqrt {\left( {8 - \sqrt {60} } \right)} }}$$   $$ -\, \frac{2}{{\sqrt {\left( {10 + \sqrt {84} } \right)} }}$$    = is ?

A. 0

B. 1

C. 2

D. 3

Answer: Option A

Solution(By Examveda Team)

$$\frac{1}{{\sqrt {\left( {12 - \sqrt {140} } \right)} }} - \frac{1}{{\sqrt {\left( {8 - \sqrt {60} } \right)} }} - \frac{2}{{\sqrt {\left( {10 + \sqrt {84} } \right)} }}$$
$$ = \frac{1}{{\sqrt {\left( {12 - \sqrt {4 \times 35} } \right)} }} - \frac{1}{{\sqrt {\left( {8 - \sqrt {4 \times 15} } \right)} }} - \frac{2}{{\sqrt {\left( {10 + \sqrt {4 \times 21} } \right)} }}$$
$$ = \frac{1}{{\sqrt {\left( {12 - 2\sqrt {35} } \right)} }} - \frac{1}{{\sqrt {\left( {8 - 2\sqrt {15} } \right)} }} - \frac{2}{{\sqrt {\left( {10 + 2\sqrt {21} } \right)} }}$$
$$ = \frac{1}{{\sqrt {{{\left( {\sqrt 7 } \right)}^2} + {{\left( {\sqrt 5 } \right)}^2} - 2.\sqrt 7 .\sqrt 5 } }} - \frac{1}{{\sqrt {{{\left( {\sqrt 5 } \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} - 2.\sqrt 5 .\sqrt 3 } }} - \frac{2}{{\sqrt {{{\left( {\sqrt 7 } \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} + 2.\sqrt 7 .\sqrt 3 } }}$$
$$ = \frac{1}{{\sqrt {{{\left( {\sqrt 7 - \sqrt 5 } \right)}^2}} }} - \frac{1}{{\sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} }} - \frac{2}{{\sqrt {{{\left( {\sqrt 7 + \sqrt 3 } \right)}^2}} }}$$
$$ = \frac{1}{{\sqrt 7 - \sqrt 5 }} - \frac{1}{{\sqrt 5 - \sqrt 3 }} - \frac{2}{{\sqrt 7 + \sqrt 3 }}$$
Rationalizing in above equation,
$$ = \frac{1}{{\sqrt 7 - \sqrt 5 }} \times \frac{{\sqrt 7 + \sqrt 5 }}{{\sqrt 7 + \sqrt 5 }} - \frac{1}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} - \frac{2}{{\sqrt 7 + \sqrt 3 }} \times \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 7 - \sqrt 3 }}$$
$$ = \frac{{\sqrt 7 + \sqrt 5 }}{2} - \frac{{\sqrt 5 + \sqrt 3 }}{2} - \frac{{\sqrt 7 - \sqrt 3 }}{2}$$
$$ = \frac{{\sqrt 7 + \sqrt 5 - \sqrt 5 - \sqrt 3 - \sqrt 7 + \sqrt 3 }}{2}$$
$$ = 0$$

This Question Belongs to Arithmetic Ability >> Surds And Indices

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