Question

Rationalise the denominator 5+2 root 3 by 7 +4 root3

Answer 1

$ANSWER:\\ \\ \frac{5+2\sqrt{3} }{7+4\sqrt{3} } = \frac{5+2\sqrt{3} }{7+4\sqrt{3} }\times\frac{7-4\sqrt{3} }{7-4\sqrt{3} }\\ \\ \\  =>\frac{11-6\sqrt{3} }{49-48} =11-6\sqrt{3}$

Answer 2

Hello,

Solution:

$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \\ \\ \frac{5+ 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ = \frac{(5 + 2 \sqrt{3})(7 - 4 \sqrt{3} ) }{ {7}^{2} - ( {4 \sqrt{3}) }^{2} } \\ \\ = \frac{(5 + 2 \sqrt{3})(7 - 4 \sqrt{3} ) }{ 49 - 48} \\ \\ = \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24}{1} \\ \\ = 11 - 6 \sqrt{3}$
is the final expression.
Hope it helps you