Question

hi everyone/plz rationalize the denominator of 1/2+√3​

Answer 1

Answer:

2 - √3

Explanation:

Given that ;

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

To rationalise the denominator, we need to multiply the numerator and denominator with it's rationalising factor, i.e., (2 - √3).

$\implies \sf \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3} } \\ \\ \implies \sf \frac{2 - \sqrt{3} }{(2) {}^{2} - ( \sqrt{3}) {}^{2} } \\ \\ \implies \sf \frac{2 - \sqrt{3} }{4 - 3} \\ \\ \implies \sf \frac{2 - \sqrt{3} }{1} \\ \\ \implies \sf 2 - \sqrt{3}$

Hence, the answer is (2 - √3).

Answer 2

Question :-

Rationalize the denominator $\tt \dfrac{1}{2 + \sqrt{3} }$

Answer :-

2 - √3

Solution :-

$\tt \dfrac{1}{2 + \sqrt{3} }$

The rationalising factor of 2 + √3 is 2 - √3. So multiply both numerator and denominator with rationalising factor.

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

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

[ Since (x + y)(x - y) = x² - y²]

$\tt = \dfrac{2 - \sqrt{3} }{4 - \sqrt{9} }$

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

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

$\tt = 2 - \sqrt{3}$

Extra Information :-

• If the product of two irrational numbers is a rational number then each of the two is rationalising factor of the other.

• To rationalise $\tt \dfrac{1}{ \sqrt{a} + b}$

, we multiply this by $\tt \dfrac{ \sqrt{a} - b}{ \sqrt{a} -b }$,

where a and b are integers.