Question

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

Answer 1

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

Now, rationalising the denominator of the left side of the given equation
by multiplying both the numerator and denominator with $7-4\sqrt{3}$, we get:
$$\begin{gathered} \frac{5+2\sqrt{3}}{7+4\sqrt{3}}=\frac{5+2\sqrt{3}}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}} \\ =\frac{\left(5+2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{{\left(7\right)}^2-{\left(4\sqrt{3}\right)}^2} \\ =\frac{5\left(7-4\sqrt{3}\right)+2\sqrt{3}\left(7-4\sqrt{3}\right)}{49-48} \\ =\frac{35-20\sqrt{3}+14\sqrt{3}-24}{1} \\ =11-6\sqrt{3} \end{gathered}$$

∴ $11-6\sqrt{3}$ = a + b$\sqrt{3}$
So, on comparing the LHS and RHS, we get:
a = 11 and b = $-6$