Solve this

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). 

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