In each of the following determine rational numbers $a$ and $b$:$ \frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b \sqrt{2} $

Given:

\( \frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b \sqrt{2} \)

To do: 

We have to determine rational numbers $a$ and $b$.

Solution:

We know that,

Rationalising factor of a fraction with denominator ${\sqrt{a}}$ is ${\sqrt{a}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}-\sqrt{b}}$ is ${\sqrt{a}+\sqrt{b}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}+\sqrt{b}}$ is ${\sqrt{a}-\sqrt{b}}$.

LHS $=\frac{3+\sqrt{2}}{3-\sqrt{2}}=\frac{(3+\sqrt{2})(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$

$=\frac{(3+\sqrt{2})^{2}}{(3)^{2}-(\sqrt{2})^{2}}$

$=\frac{9+2+2 \times 3 \sqrt{2}}{9-2}$

$=\frac{11+6 \sqrt{2}}{7}$

$=\frac{11}{7}+\frac{6}{7} \sqrt{2}$

Therefore,

$a+b \sqrt{2}=\frac{11}{7}+\frac{6}{7} \sqrt{2}$

Comparing both sides, we get,

$a=\frac{11}{7}$ and $b=\frac{6}{7}$

Hence, $a=\frac{11}{7}$ and $b=\frac{6}{7}$.