Question

Find the values of $a$ and $b,$ if $\frac{\sqrt{2}+1}{\sqrt{2}-1}-\frac{\sqrt{2}-1}{\sqrt{2}+1}=a+\sqrt{2}b$

Answer 1

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

$⟹\frac{(\sqrt{2}+1{)}^2}{(\sqrt{2}-1)(\sqrt{2}+1)}-\frac{(\sqrt{2}-1{)}^2}{(\sqrt{2}+1)(\sqrt{2}-1)}=a+\sqrt{2}b$

$⟹\frac{(2+1+2\sqrt{2})}{2-1}-\frac{(2+1-2\sqrt{2})}{2-1}=a+\sqrt{2}b$

$⟹3+2\sqrt{2}-3+2\sqrt{2}=a+b\sqrt{2}$

$⟹a+b\sqrt{2}=0+4\sqrt{2}$

$⟹a=0,b=4$