Rationalise root 6/root 2+ root 3

1.	Rationalise root 6/root 2+ root 3
Answer» To rationalise the denominator of: $\longrightarrow \sf { \dfrac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} }$ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ 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. ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ 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. ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\longrightarrow \sf { \dfrac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} \times \dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}} } \\ \\$ $\longrightarrow \sf { \dfrac{\sqrt{6} (\sqrt{2} - \sqrt{3}) }{(\sqrt{2} + \sqrt{3})(\sqrt{2} + \sqrt{3})} } \\ \\$ By USING identity, (a + b)(a - b) = a² - b² ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\longrightarrow \sf { \dfrac{\sqrt{6} (\sqrt{2}) - \sqrt{6}( \sqrt{3}) }{(\sqrt{2})^2 -(\sqrt{3})^2 } } \\ \\$ $\longrightarrow \sf { \dfrac{\sqrt{12} - \sqrt{18}}{2 -3 } } \\ \\$ $\longrightarrow \sf { \dfrac{\sqrt{12} - \sqrt{18}}{-1 } } \\ \\$ $\longrightarrow \sf { \dfrac{\sqrt{12} - \sqrt{18}}{-1 } \times \dfrac{-1}{-1} } \\ \\$ $\longrightarrow \sf { \dfrac{-1(\sqrt{12} - \sqrt{18})}{(-1)^2 } } \\ \\$ $\longrightarrow \sf { \dfrac{ -\sqrt{12} + \sqrt{18}}{1 } } \\ \\$ $\longrightarrow \sf { -\sqrt{12} + \sqrt{18} } \\ \\$ $\longrightarrow \boxed{\sf { -2\sqrt{3} +3 \sqrt{2} }} \\ \\$ Hence, rationalised. ⠀⠀⠀⠀___________________________ 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

Answer»

To rationalise the denominator of:

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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

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.