Question

$\frac{3+\sqrt{12}-\sqrt{27}}{7+\sqrt{192}-\sqrt{48}}=a+b\sqrt{3},$then find a and b.

Answer 1

We have :
$\frac{3+\sqrt{12}-\sqrt{27}}{7+\sqrt{192}-\sqrt{48}}=\frac{3+2\sqrt{3}-3\sqrt{3}}{7+8\sqrt{3}-4\sqrt{3}}=\frac{3-\sqrt{3}}{7+4\sqrt{3}}$

Now, rationalising the denominator of the fraction by multiplying both the numerator and denominator with $7-4\sqrt{3}$, we get:
$$\begin{gathered} \frac{3-\sqrt{3}}{7+4\sqrt{3}}=\frac{3-\sqrt{3}}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}} \\ =\frac{\left(3-\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{{\left(7\right)}^2-{\left(4\sqrt{3}\right)}^2} \\ =\frac{3\left(7-4\sqrt{3}\right)-\sqrt{3}\left(7-4\sqrt{3}\right)}{49-48} \\ =\frac{21-12\sqrt{3}-7\sqrt{3}+12}{1} \\ =33-19\sqrt{3} \end{gathered}$$

∴ $33-19\sqrt{3}$ = a + b$\sqrt{3}$
So, on comparing the LHS and RHS, we get:
a = 33 and b = $-19$