Question

Rationalize $\sf \dfrac{22}{2+\sqrt{3} +\sqrt{5} }$

Answer 1

Provided Question :

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Rationalize : $\boxed{\boxed{\sf{\dfrac{22}{ 2 + \sqrt{3} + \sqrt{5} }}}}$

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Required Answer :

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$\sf{ \implies \dfrac{22}{ ( 2 + \sqrt{3} ) + \sqrt{5} } \times \dfrac{( 2 + \sqrt{3} ) - \sqrt{5} }{ ( 2 + \sqrt{3} ) - \sqrt{5} } }$

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$\sf{ \implies \dfrac{ 22 ( 2 + \sqrt{3} - \sqrt{5} }{ \{ ( 2 + \sqrt{3} ) - \sqrt{5} \} \{ ( 2 + \sqrt{3} ) - \sqrt{5} \} } }$

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$\sf{ \implies \dfrac{ 22 ( 2 + \sqrt{3} - \sqrt{5} ) }{ \{ ( 2 + \sqrt{3} ) - \sqrt{5} \} \{ ( 2 + \sqrt{3} ) - \sqrt{5} \} } }$

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$\sf{ \implies \dfrac{ 22 ( 2 + \sqrt{3} - \sqrt{5} ) }{ ( 2 + \sqrt{3} )² - ( \sqrt{5} )² } }$

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$\sf{ \implies \dfrac{ 44 + 22\sqrt{3} - 22\sqrt{5} ) }{ 4 + ( \sqrt{3} )² + 4 \sqrt{3} - 5 } }$

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$\sf{ \implies \dfrac{ \cancel{44} + \cancel{22} \sqrt{3} - \cancel{22} \sqrt{5} }{ \cancel{2} + \cancel{4} \sqrt{3} } \:\:\:\:\:\:\:-\:-\:-\:-\: cancellation\:by\:2 }$

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$\sf{ \implies \dfrac{ 22 + 11 \sqrt{3} - 11 \sqrt{5} }{ 1 + 2 \sqrt{3} } }$

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$\sf{\implies \dfrac{ 22 + 11 \sqrt{3} - 11 \sqrt{5} }{ 1 + 2 \sqrt{3} } \times \dfrac{ 1 - 2 \sqrt{3} }{ 1 - 2 \sqrt{3} } }$

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$\sf{\implies \dfrac{ ( 22 + 11 \sqrt{3} - 11 \sqrt{5} ) ( 1 - 2\sqrt{3} ) }{ ( 1 + 2 \sqrt{3} ) ( 1 - 2 \sqrt{3} ) } }$

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$\sf{ \implies \dfrac{ 22 \sqrt{15} - 11 \sqrt{5} - 33 \sqrt{3} - 44 }{ 1 - ( 2 \sqrt{3} )² } }$

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$\sf{ \implies \dfrac{ \cancel{22} \sqrt{15} \bcancel{-} \cancel{11} \sqrt{5} \bcancel{-} \cancel{33} \sqrt{3} \bcancel{-} \cancel{44} }{ \bcancel{-} \cancel{11} } }$

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$\sf{\implies \orange{ 4 - 2\sqrt{15} + \sqrt{5} + 3\sqrt{3} } }$

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Learn Extras ::

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$\boxed{\begin{array}{| c |}\qquad\tt\large\textsf{ ! Formulas ! }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ ( a + b )² = a² + 2ab + b² }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ ( a - b )² = a² - 2ab + b² }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ a² - b² = ( a + b ) ( a - b ) }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ ( a + b )³ = a³ + 3a²b + 3ab² }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ ( a - b )³ = a³ - 3a²b + 3ab² - b³ }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ a³ - b³ = ( a - b ) ( a² + ab + b² ) }\end{array}}$

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Answer 2

Given :-

Rationalize :

$\longrightarrow \sf \dfrac{22}{2+\sqrt{3} +\sqrt{5} }$

Answer :-

In order to rationalize the denominator, we have to multiply the denominator with it's inverse such that (a+b)(a-b) is formed.

Solution :-

$\implies \sf \dfrac{22}{(2+\sqrt{3}) + \sqrt{5} } \times \dfrac{(2 + \sqrt{3} ) - \sqrt{5} }{(2 + \sqrt{3}) - \sqrt{5} }$

$\implies \sf \dfrac{22\times \{(2+\sqrt{3}) -\sqrt{5}\}}{\{(2+\sqrt{3} )+\sqrt{5}\} -\{(2+\sqrt{3})-\sqrt{5}\} }$

• a = (2+√3)
• b =√5

$\implies \sf \dfrac{22 \times \{(2 + \sqrt{3} ) - \sqrt{5}\} }{(2+\sqrt{3})^{2} - ( \sqrt{5} )^{2} }$

Applying identity (a+b)² = a² + 2ab + b² in the denominator,

$\implies \sf \dfrac{22 \times \{(2 + \sqrt{3} ) - \sqrt{5}\} }{\{ {2}^{2} + 2(2)( \sqrt{3}) + ( \sqrt{ 3})^{2} \} - 5}$

$\implies \sf \dfrac{22\{(2 + \sqrt{3} ) - \sqrt{5}\}}{(4 + 4 \sqrt{3} + 3) - 5 }$

$\implies \sf \dfrac{22\{(2 + \sqrt{3}) - \sqrt{5} \}}{(7 + 4 \sqrt{3}) - 5 }$

$\implies \sf \dfrac{22\{(2 + \sqrt{3} ) - \sqrt{5}\}}{7 + 4 \sqrt{3} - 5 }$

$\implies \sf \dfrac{22\{(2 + \sqrt{3}) - \sqrt{5}\}}{2 + 4 \sqrt{3} }$

$\implies \sf \dfrac{22\{(2 \sqrt{ 3} ) - \sqrt{5} \}}{2(1 + 2 \sqrt{3} )}$

Cancelling 22 and 2, we get :-

$\implies \sf \dfrac{11\{(2 + \sqrt{3}) - \sqrt{5} \}}{1 + 2 \sqrt{3} }$

We have to rationalize this denominator now.

By multiplying by it's inverse,

$\implies \sf \dfrac{11\{(2+\sqrt{3}) - \sqrt{5}\}}{1+2\sqrt{3}} \times \dfrac{1- 2\sqrt{3}}{1- 2\sqrt{3}}$

$\implies \sf \dfrac{11\{(2+\sqrt{3}) - \sqrt{5}\} \times (1-\sqrt{3})}{(1+2\sqrt{3})\times(1- 2\sqrt{3})}$

$\implies \sf \dfrac{11\{1(2+\sqrt{3} - \sqrt{5}) - 2\sqrt{3} (2+ \sqrt{3} - \sqrt{5})\}}{(1)^{2} - (2\sqrt{3})^{2}}$

$\implies \sf \dfrac{11(2+\sqrt{3} - \sqrt{5}) + (-4\sqrt{3} - 6 + 2\sqrt{15})}{1 - 12}$

$\implies \sf \dfrac {11(2 + \sqrt{3} - \sqrt{5}) - 4\sqrt{3} - 6 + 2\sqrt{15})}{-11}$

Cancelling (-11) and (11) respectively, we get :-

$\implies \sf -1(2 + \sqrt{3} - \sqrt{5}) - 4\sqrt{3} - 6 + 2\sqrt{15})$

Adding and subtracting all like terms,

$\implies \sf -1(-4 - 3\sqrt{3} - \sqrt{5} + 2\sqrt{15})$

Required answer :-

$\implies \sf 4 + 3\sqrt{3} + \sqrt{5} - 2\sqrt{15}$

More Formulas :-

• (a+b)² = a² + 2ab b²
• (a-b)² = a² - 2ab + b²
• a² - b² = (a+b)(a-b)
• (x + a)(x + b) = x² + (a + b) x + ab
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• (a + b)³ = a³ + b³ + 3ab (a + b)
• (a – b)³ = a³ – b³ – 3ab (a – b)
• a³+b³ = (a+b)(a² - ab + b²)
• a³-b³ = (a-b)(a² + ab + b²)
• a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)

Conditional identity :-

If a+b+c = 0,

• a³ + b³ + c³ = 3abc