Question

If $\frac{\sqrt{2}+1}{\sqrt{2}-1}=a+b\sqrt{2}$ then find the values of $a$ and $b$

Answer 1

given, $\frac{\sqrt{2}+1}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}=a+b\sqrt{2}$

$\Rightarrow \frac{2+1+2\sqrt{2}}{2-1}=a+b\sqrt{2}$

$\Rightarrow \frac{3+2\sqrt{2}}{1}=a+b\sqrt{2}$

$\Rightarrow 3+2\sqrt{2}=a+b\sqrt{2}$

Comparing, we get $a=3,b=2$