Question

2
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2-2√3
class 9
topic : rationalise the denominator​

Answer 1

Question :-

$\longrightarrow \sf Rationalize\: \dfrac{2}{2-\sqrt{3} }$

Answer :-

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

• Denominator's value = 2 - √3
• Inverse = 2 + √3

$\sf Multiplying\:the\:number\:with\:\dfrac{2+\sqrt{3} }{2+\sqrt{3} } ,$

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

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

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

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

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

$\implies \sf \dfrac{4 + 2\sqrt{3} }{1}$

$\implies \sf 4 + 2\sqrt{3}$

$\sf Therefore, the\: rationalized\:form \:of \:\dfrac{2}{2-\sqrt{3}}\:\: is\: 4 + 2\sqrt{3}$

Identities :-

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• a² - b² = (a - b)(a + b)
• (x + a)(x + b) = x² + x(a + b) + ab
• (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
• (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)