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) $\sqrt{2}-1$
Solution:
$3-2 \sqrt{2}=2+1-2 \times \sqrt{2} \times 1$
$=(\sqrt{2})^{2}+1^{2}-2 \times \sqrt{2} \times 1$
This is of the form
$a^{2}+b^{2}-2 a b=(a-b)^{2}$
$(\sqrt{2})^{2}+1^{2}-2 \times \sqrt{2} \times 1=((\sqrt{2})-1)^{2}$
So,
$\sqrt{3-2 \sqrt{2}}=\sqrt{((\sqrt{2})-1)^{2}}=\sqrt{2}-1$
Hence, the correct answer is option (d).