Question

Rationales the denominator and simplify:

${\displaystyle \frac{5+2\sqrt{3}}{7+4\sqrt{3}}}$

Answer 1

We know that rationalization factor for ${\displaystyle 7+4\sqrt{3}}$ is ${\displaystyle 7-4\sqrt{3}}$. We will multiply numerator and  denominator of the given expression ${\displaystyle \frac{5+2\sqrt{3}}{7+4\sqrt{3}}}$ by ${\displaystyle 7-4\sqrt{3}}$ to get

${\displaystyle \frac{5+2\sqrt{3}}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}}=\frac{5\times 7-5\times 4\sqrt{3}+2\times 7\times \sqrt{3}-2\times 4\times {\left(\sqrt{3}\right)}^2}{{\left(7\right)}^2-{\left(4\sqrt{3}\right)}^2}}$

${\displaystyle =\frac{35-20\sqrt{3}+14\sqrt{3}-8\times 3}{49-49}}$

${\displaystyle =\frac{11-6\sqrt{3}}{1}}$

${\displaystyle =11-6\sqrt{3}}$

Hence the given expression is simplified to ${\displaystyle 11-6\sqrt{3}}$