Question

The value of $\sqrt{3-2\sqrt{2}}$ is

(a)$\sqrt{3}+\sqrt{2}$ (b)$\sqrt{3}-\sqrt{2}$ (c)$\sqrt{2}+1$ (d)$\sqrt{2}-1$

Answer 1

$3-2\sqrt{2}=2+1-2\times \sqrt{2}\times 1$

This is of the form of $a^2+b^2-2ab=(a-b{)}^2$

${\sqrt{2}}^2+1^2-2\times \sqrt{2}\times 1=(\sqrt{2}-1{)}^2$

So, $\sqrt{3-2\sqrt{2}}=\sqrt{(\sqrt{2}-1{)}^2}=\sqrt{2}-1$

Hence, the correct answer is the option (d)