Question

rationalise denominator of 6-4√3 / 6+4√3​

Answer 1

$\frac{6 - 4 \sqrt{3} }{6 + 4 \sqrt{3} } \times \frac{6 - 4 \sqrt{3} }{6 - 4 \sqrt{3} } \\ \\ = \frac{ ({6 - 4 \sqrt{3} )}^{2} }{ {(6)}^{2} - {(4 \sqrt{3} )}^{2} } \\ \\ = \frac{ {6}^{2} + {(4 \sqrt{3} )}^{2} - 2 \times 6 \times 4 \sqrt{3} }{36 - (16 \times 3)} \\ \\ = \frac{36 + (16 \times 3) - 12 \times 4 \sqrt{3} }{36 - 48} \\ \\ = \frac{36 + 48 - 12 \times 4 \sqrt{3} }{ - 12} \\ \\ = \frac{72 \times 4 \sqrt{3} }{ - 12} \\ \\ = - 6 \times 4 \sqrt{3} \\ \\ = 2( - 3 \times 2 \sqrt{3} )$

Answer 2

Given:

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

Solution:

$\frac{6 - 4 \sqrt{3} }{6 + 4 \sqrt{3} } \times \frac{6 - 4 \sqrt{3} }{6 - 4 \sqrt{3} } \\ = > \frac{ {(6 - 4 \sqrt{3} )}^{2} }{(6 + 4 \sqrt{3} )(6 - 4 \sqrt{3}) } \\ = > \frac{36 - 48 \sqrt{3} + 48 }{36 - 48} \\ = > \frac{84 - 48 \sqrt{3} }{ - 12} \\ = > - (7 - 4 \sqrt{3} ) \\ = > 4 \sqrt{3} - 7$

Therefore:

$4 \sqrt{3} - 7$

Ans