Question

rationalize the denominater

6/2√3 + √6​

Answer 1

Answer:

Solution →

$\frac{ \sqrt{6} }{ \sqrt{2} } + \sqrt{3 } \\ \frac{ \sqrt{6} }{ \sqrt{2} } + \sqrt{3} \times \sqrt{2} - \frac{ \sqrt{3} } { \sqrt{2} } - \sqrt{3} \\ \sqrt{6} ( \frac{ \sqrt{2 - \sqrt{3} } }{2 - 3} ) \\ - \sqrt{6} ( \sqrt{2 - \sqrt{3} } ) \\ - \sqrt{12} + \sqrt{18}$

Answer 2

$\large{ \sf \underline{Solution - }}$

$\bf 1^{st} \: Method : [ Without \: using \: identity]$

To rationalise :

$\sf \implies\dfrac{6 }{2 \sqrt{3} + \sqrt{6} }$

➢ Here the denominator is in the form of (a+b). Rationalising factor of (a+b) is (a-b). So, rationalising factor of (2√3+√6) is (2√3-√6). We will multiply (2√3-√6) with both the numerator and denominator to rationalise the denominator.

$\sf \implies\dfrac{6 }{2 \sqrt{3} + \sqrt{6} } \times \dfrac{2 \sqrt{3} - \sqrt{6}}{2 \sqrt{3} - \sqrt{6}}$

Combine fractions

$\sf \implies \dfrac{6(2 \sqrt{3} - \sqrt{6})}{(2 \sqrt{3} + \sqrt{6})(2 \sqrt{3} - \sqrt{6})}$

$\sf \implies \dfrac{6(2 \sqrt{3} - \sqrt{6})}{2 \sqrt{3}(2 \sqrt{3} - \sqrt{6}) + \sqrt{6}(2 \sqrt{3} - \sqrt{6}) }$

$\sf \implies \dfrac{6(2 \sqrt{3} - \sqrt{6})}{12 - 6 \sqrt{2} + 6 \sqrt{2} - 6 }$

$\sf \implies \dfrac{6(2 \sqrt{3} - \sqrt{6})}{6 - 6 \sqrt{2} + 6 \sqrt{2} }$

$\sf \implies \dfrac{6(2 \sqrt{3} - \sqrt{6})}{6 - 0 }$

$\sf \implies \dfrac{ \cancel{6}(2 \sqrt{3} - \sqrt{6})}{ \cancel{6} }$

Cancel the common factor of 6,

$\sf \implies 2 \sqrt{3} - \sqrt{6}$

Hence,

On rationalising we got,

$\bf \implies 2 \sqrt{3}- \sqrt{6}$

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$\bf 2^{nd} \: Method : [Using \: identity]$

To rationalise :

$\sf \implies\dfrac{6 }{2 \sqrt{3} + \sqrt{6} }$

$\sf \implies\dfrac{6 }{2 \sqrt{3} + \sqrt{6} } \times \dfrac{2 \sqrt{3} - \sqrt{6}}{2 \sqrt{3} - \sqrt{6}}$

Combine fractions

$\sf \implies \dfrac{6(2 \sqrt{3} - \sqrt{6})}{(2 \sqrt{3} + \sqrt{6})(2 \sqrt{3} - \sqrt{6})}$

We know that,

$\sf \implies (a - b)(a + b) = a^{2} - b ^{2}$

So,

$\sf \implies \dfrac{6(2 \sqrt{3} - \sqrt{6})}{(2 \sqrt{3} )^{2}-(\sqrt{6})^{2}}$

$\sf \implies \dfrac{6(2 \sqrt{3} - \sqrt{6})}{12-6}$

$\sf \implies \dfrac{\cancel{6}(2 \sqrt{3} - \sqrt{6})}{\cancel{6}}$

Cancel the common factor of 6,

$\sf \implies 2 \sqrt{3} - \sqrt{6}$

Hence,

On rationalising we got,

$\bf \implies 2 \sqrt{3}- \sqrt{6}$