Question

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

Answer 1

Step 1: Rationalize
Rationalization is a process by which radicals in the denominator of an algebraic fraction are eliminated.
Multiply numerator & denominator by $\left(3+\sqrt{2}\right)$ on left hand side.

$$\begin{gathered} \frac{3+\sqrt{2}}{3-\sqrt{2}}=\frac{3+\sqrt{2}}{3-\sqrt{2}}\times \frac{(3+\sqrt{2})}{(3+\sqrt{2})} \\ =\frac{(3+\sqrt{2}{)}^2}{(3{)}^2-(\sqrt{2}{)}^2}\left[(a+b)(a-b)=a^2-b^2\right] \\ =\frac{9+6\sqrt{2}+2}{9-2}\left[(a+b{)}^2=a^2+2ab+b^2\right] \\ =\frac{11+6\sqrt{2}}{7} \end{gathered}$$

Step 2: Find value of$a\&b$

By above we get$\frac{3+\sqrt{2}}{3-\sqrt{2}}=\frac{11+6\sqrt{2}}{7}$

So, the given expression can be rewritten as:

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

Compare both sides to get:$a=\frac{11}{7},b=\frac{6}{7}$

Therefore for $\frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b\sqrt{2}$ value of $a\&b$are $a=\frac{11}{7},b=\frac{6}{7}$.