Question

rationalise the denominator 2+√3/2-√3

Answer 1

Step-by-step explanation:

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Answer 2

Question

Rationalise the denominator → $\dfrac{2+\sqrt{3}}{2-\sqrt{3}}$

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Answer

So to rationalize the given number, we must multiply the numerator and denominator with the conjugate of the denominator. So the conjugate of 2 - √3 would be 2 + √3

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

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

Now we have to apply two algebraic identities for the numerator and denominator, respectively.

• (a + b)² = a² + 2ab + b²
• (a + b)(a - b) = a² - b²

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

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

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

$\sf \implies \dfrac{7+4\sqrt{3} }{1 }$

$\sf \implies 7+4\sqrt{3}$

∴ Upon rationalizing the denominator of $\bf \dfrac{2+\sqrt{3}}{2-\sqrt{3}}$ we get 7 + 4√3.

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