Question

rationalise root 6/root 2+ root 3​

Answer 1

Answer:

3√2-2√3 this is the required answer

Answer 2

To rationalise the denominator of:

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

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To rationalise the denominator of the given fraction, we need to multiply the the rationalising factor of the denominator with both the numerator and the denominator of the given fraction.

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Here,

$\longrightarrow \sf {Denominator = (\sqrt{2} + \sqrt{3} )}$

Rationalising factor of $\sf { (\sqrt{2} + \sqrt{3} )}$ is $\sf { (\sqrt{2} - \sqrt{3} )}$ . So, we'll multiply $\sf { (\sqrt{2} - \sqrt{3} )}$ with both the numerator and the denominator.

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

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

By using identity,

• (a + b)(a - b) = a² - b²

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

$\longrightarrow \sf { \dfrac{\sqrt{12} - \sqrt{18}}{2 -3 } }$

$\longrightarrow \sf { \dfrac{\sqrt{12} - \sqrt{18}}{-1 } }$

$\longrightarrow \sf { \dfrac{\sqrt{12} - \sqrt{18}}{-1 } \times \dfrac{-1}{-1} }$

$\longrightarrow \sf { \dfrac{-1(\sqrt{12} - \sqrt{18})}{(-1)^2 } }$

$\longrightarrow \sf { \dfrac{ -\sqrt{12} + \sqrt{18}}{1 } }$

$\longrightarrow \sf { -\sqrt{12} + \sqrt{18} }$

$\longrightarrow \boxed{\sf { -2\sqrt{3} +3 \sqrt{2} }}$

Hence, rationalised.

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More Identities :

• (√a)² = a

• √a√b = √ab

• √a/√b = √a/b

• (√a + √b)(√a - √b) = a - b

• (a + √b)(a - √b) = a² - b

• (√a ± √b)² = a ± 2√ab + b

• (√a + √b)(√c + √d) = √ac + √ad + √bc + √bd