Question

Rationalise the denominator of5/ roots 3 +root2​

Answer 1

Question :-

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

Answer :-

• In order to rationalize the denominator, we have to multiply the number with the denominator's inverse.

• Multiplying by  $\sf\dfrac{\sqrt{3} -\sqrt{2} }{\sqrt{3} -\sqrt{2} }$  

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

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

By using the identity a² - b² = (a+b)(a-b) in the denominator,

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

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

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

Any number with the denominator 1, can be written as the number itself.

Therefore,

$\implies \sf Answer = 5\sqrt{3} - 5\sqrt{2}$

The rationalized form of $\implies \sf \dfrac{5}{\sqrt{3} -\sqrt{2} }$ is $\sf 5\sqrt{3} - 5\sqrt{2}$

Identities :-

• (a+b)² = a² + 2ab + b²

• (a-b)² = a² - 2ab + b²

• (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca

• (x+a)(x+b) = x² + x(a+b) + ab

• a²-b² = (a+b)(a-b)

• (a+b)³ = a³ + 3a²b + 3ab² + b³

• (a-b)³ = a³ - 3a²b + 3ab² - b³

• 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)