Question

Find the values of a and b in each of the following.
$\frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b\sqrt{2}$

Answer 1

We have:
$\frac{3+\sqrt{2}}{3-\sqrt{2}}$

Now, rationalising the denominator of the given number by multiplying both the numerator and denominator with $3+\sqrt{2}$, we get:
$$\begin{gathered} \frac{3+\sqrt{2}}{3-\sqrt{2}}=\frac{3+\sqrt{2}}{3-\sqrt{2}}\times \frac{3+\sqrt{2}}{3+\sqrt{2}} \\ =\frac{{\left(3+\sqrt{2}\right)}^2}{{\left(3\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{9+2+6\sqrt{2}}{9-2} \\ =\frac{11+6\sqrt{2}}{7} \\ =\frac{11}{7}+\frac{6}{7}\sqrt{2} \end{gathered}$$

∴ $\frac{11}{7}+\frac{6}{7}\sqrt{2}$= a + b$\sqrt{2}$
So, on comparing the LHS and the RHS, we get:
a = $\frac{11}{7}$ and b = $\frac{6}{7}$