Question

If $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}=a+b\sqrt{3}$ , then the value of a and b.

Answer 1

$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}=a+b\sqrt{3}$

Since rationalization of $\frac{a+\sqrt{b}}{c+\sqrt{d}}=\frac{a+\sqrt{b}}{c+\sqrt{d}}\times \frac{c-\sqrt{d}}{c-\sqrt{d}}$

Taking L.H.S.

By rationalization,

$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}$ = $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}\times$ $\frac{7-4\sqrt{3}}{7-4\sqrt{3}}$

Since, $\left\{\right(a+b\left)\right(a-b)=a^2-b^2\}$

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

= $\frac{35+14\sqrt{3}-20\sqrt{3}-8\sqrt{3}\sqrt{3}}{49-16\sqrt{3}\sqrt{3}}$

= $\frac{35-\sqrt{3}(14-8)-24}{49-48}$

= $\frac{35-\sqrt{3}\left(6\right)-24}{1}$

=$11-6\sqrt{3}$

Hence, by comparing a = 11 $\&$ b = -6.