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

$$\begin{gathered} 3-2\sqrt{2}=2+1-2\times \sqrt{2}\times 1 \\ ={\left(\sqrt{2}\right)}^2+1^2-2\times \sqrt{2}\times 1 \end{gathered}$$
This is of the form
$a^2+b^2-2ab={\left(a-b\right)}^2$
$$\begin{gathered} {\left(\sqrt{2}\right)}^2+1^2-2\times \sqrt{2}\times 1={\left(\left(\sqrt{2}\right)-1\right)}^2 \\ \mathrm{So}, \\ \sqrt{3-2\sqrt{2}}=\sqrt{{\left(\left(\sqrt{2}\right)-1\right)}^2}=\sqrt{2}-1 \end{gathered}$$
Hence, the correct answer is option (d).