Question

Rationalize the denominator √3-1/√3+1

Answer 1

HEYA!!
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$\frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 } \times \frac{ \sqrt{3} - 1}{ \sqrt{3} - 1} \\ \\ \\ \frac{( \sqrt{3} - 1) {}^{2} }{ \sqrt{3} {}^{2} - 1 {}^{2} } \\ \\ = \frac{3 + 1 - 2 \sqrt{3} }{2} \\ \\ = \frac{4 - 2 \sqrt{3} }{2} \\ \\ = \frac{2(2 - \sqrt{3}) }{2} \\ \\ = 2 - \sqrt{3}$
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Answer 2

Hey mate!

Thank you for asking this question! ❤

Answer:

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Identities used:

(a + b)(a - b) = a² - b²

(a + b)² = a² + 2ab + b²

Given,

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

On rationalizing the denominator, we get:

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

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

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

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

$= \frac{2(2 - \sqrt{3}) }{2}$

$= 2 - \sqrt{3}$

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