Question

find the value of (6-4√3)/(6+√3) by rationalistising the denominator.

Answer 1

$\bf\large\underline\blue{Question:-}$

• Find the value of $\frac{6-4\sqrt{3}}{6+\sqrt{3}}$ by rationalizing the denominator.

$\bf\large\underline\blue{Solution:-}$

$\frac{6 - 4 \sqrt{3} }{6 + \sqrt{3} }$

$= \frac{6 - 4 \sqrt{3} }{6 + \sqrt{3} } \times \frac{6 - \sqrt{3} }{6 - \sqrt{3} }$

$= \frac{(6 - 4 \sqrt{3})(6 - \sqrt{3} ) }{36 - 3}$

$= \frac{36 - 6 \sqrt{3} - 24 \sqrt{3} + 12 }{33}$

$= \frac{48 -30 \sqrt{3} }{33}$

$= \frac{ \cancel{3}(16 - 10 \sqrt{3}) }{ \cancel{3} \times 11}$

$= \frac{16 - 10 \sqrt{3} }{11}$

$\bf\large\underline\blue{Answer:-}$

• $\frac{16 - 10 \sqrt{3} }{11}$

Answer 2

$\huge\sf\pink{Answer}$

☞ $\sf \dfrac{16-10\sqrt{3}}{11}$

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$\huge\sf\blue{Given}$

✭ $\sf \dfrac{6-4\sqrt{3}}{6+\sqrt{3}}$

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$\huge\sf\gray{To \:Find}$

◈ It's rationalised form?

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$\huge\sf\purple{Steps}$

$\large\underline{\underline{\sf Concept}}$

So here we shall simply multiply the given fraction by the conjugate of its denominator, a conjugation is a number with the opposite side. We do so just to eliminate the radical and thus making our calculation much easier!!

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➝ $\sf \dfrac{6-4\sqrt{3}}{6+\sqrt{3}}$

➝ $\sf \dfrac{6-4\sqrt{3}}{6+\sqrt{3}} \times \dfrac{6-\sqrt{3}}{6-sqrt{3}}$

➝ $\sf \dfrac{(6)(6) +(6)(-\sqrt{3})-(4\sqrt{3})(6)-(4\sqrt{3})(-\sqrt{3})}{(6)(6)+(6)(-\sqrt{3})+(\sqrt{3})(6)+(\sqrt{3})(-\sqrt{3})}$

➝ $\sf \dfrac{36-6\sqrt{3}-24\sqrt{3}+12}{36-6\sqrt{3}+6\sqrt{3} -3}$

➝ $\sf \dfrac{48-30\sqrt{3}}{33}$

➝ $\sf \dfrac{3(16-10\sqrt{3}}{3\times 11}$

➝ $\sf \orange{\dfrac{16-10\sqrt{3}}{11}}$

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