Question

rationalise the denominator√3 -1 / √3+1​

Answer 1

$\maltese$Given to Rationalize the denominator :-

$\dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1 }$

$\maltese$SOLUTION:-

$\maltese$What is Rationalization?

Rationalization is the process to remove the surds or radicals in the denominator

$\maltese$Process to do the Rationalization:-

• Firstly find the conjuagate of the denominator i.e Conjugate means in order to change the sign

• Now with the conjuagate of denominator multiply with the numerator and denominator

• Since the denominator forms the (a+b)(a-b) Algebraic identity

• By simplying this, the denominator can be rationalized

The conjugate of √3+1 is √3 -1

$\dfrac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 } \times \dfrac{ \sqrt{3} - 1 }{ \sqrt{3} - 1}$

$= \dfrac{( \sqrt{3} - 1) {}^{2} }{( \sqrt{3} + 1)( \sqrt{3} - 1) }$

$\maltese$Simplifying the numerator by Algebraic Identitiy (a-b)²

$\maltese$Simplifying the denominator by Algebraic Identitiy (a+b)(a-b)

(a-b)² = a²-2ab + b²

(a+b)(a-b) = a²-b²

$= \dfrac{( \sqrt{3} ) {}^{2} + (1) {}^{2} - 2(1)( \sqrt{3} ) }{( \sqrt{3} ) {}^{2} - (1) {}^{2} }$

$= \dfrac{3 + 1 - 2 \sqrt{3} }{3 - 1}$

$= \dfrac{4 - 2 \sqrt{3} }{2}$

Since the denominator rationalized i.e The surds or radicals removed in denominator

$\maltese$Know more Algebraic Identities:-

(a+ b)² = a² + b² + 2ab

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( a + b )² - ( a - b)² = 4ab

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

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc

Answer 2

$\clubsuit \sf \frac{ \sqrt{3 - 1} }{ \sqrt{3 - 1} } :-$

$\dashrightarrow\sf \frac{( \sqrt{3 + 1)( \sqrt{3 - 1)} } }{ (\sqrt{3 + 1})( \sqrt{3 + 1)} }$

$\dashrightarrow\sf \frac{3 - \sqrt{3 + \sqrt{3 - 1} } }{3 + \sqrt{3 + \sqrt{3 + 1} } }$

$\dashrightarrow\sf \frac{3 - 1}{2 \sqrt{3 + 4} } = \frac{2}{ \sqrt{3 + 4} }$

$\dashrightarrow\sf \frac{2}{ \sqrt[2]{3} } + \frac{2}{4}$

$\dashrightarrow\sf \frac{2}{12} + \frac{2}{4}$

$\dashrightarrow\sf \frac{1}{6} + \frac{1}{2} = \frac{2 + 6}{12}$

$\dashrightarrow\sf \frac{8}{4}$

$\dashrightarrow\sf \frac{4}{2}$

$\dashrightarrow\sf \frac{2}{3}$