Question

$\sqrt{3 - 2 \sqrt{2} }$
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Answer 1

Answer:

$We \: \: can \: \: write \: \: 3 \: \: as \: \: 2 + 1 \: \: \: \: and \: \: write \: \: \sqrt{2} \: \: as \: \: \sqrt{2} \times \sqrt{1}$

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

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

${x}^{2} + {y}^{2} - 2xy \: = \: {(x - y)}^{2}$

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

We can cancel square and square root.

$Therefore \: \: the \: \: final \: \: anwer \: \: is \: \sqrt{2} - \sqrt{1}$