Question

Rationalise the denominator of 5+2√3 / 7+4√3​

Answer 1

$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \\ \\ = \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{(7 - 4 \sqrt{3} )}{(7 - 4 \sqrt{3}) } \\ \\ = \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24}{49 - 48} \\ \\ = \frac{11 - 6 \sqrt{3} }{1}$

Here I used an identity that is :

(A+B) (A-B) = A²-B²

Answer 2

Answer: 11 - 6√3

Step-by-step explanation:

Given,

$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}\\\;\\\text{Multiplying numerator and denominator by } 7-4\sqrt{3}\\\;\\=\frac{(5+2\sqrt{3})(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3})}\\\;\\=\frac{(5\times7)-(5\times4\sqrt{3})+(7\times2\sqrt{3})-(2\sqrt{3}\times4\sqrt{3})}{7^2-(4\sqrt{3})^2}\;\;\;\;\;\;\;\because(a+b)(a-b)=a^2-b^2\\\;\\=\frac{35-20\sqrt{3}+14\sqrt{3}-24}{49-48}\\\;\\=\frac{11-6\sqrt{3}}{1}\\\;\\=11-6\sqrt{3}$