What is the square root of $3 - 2 \sqrt{2}$ $3-2sqrt2$ ?

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Answer 1 · 2017-01-18T09:15:10

$\sqrt{3 - 2 \sqrt{2}} = \sqrt{2} - 1$ $sqrt(3-2sqrt2)=sqrt2-1$

$3 - 2 \sqrt{2}$ $3-2sqrt2$

= $2 - 2 \sqrt{2} + 1$ $2-2sqrt2+1$

= ${(\sqrt{2})}^{2} - 2 \times \sqrt{2} \times 1 + 1^{2}$ $(sqrt2)^2-2xxsqrt2xx1+1^2$

= ${(\sqrt{2} - 1)}^{2}$ $(sqrt2-1)^2$

Hence $\sqrt{3 - 2 \sqrt{2}} = \sqrt{2} - 1$ $sqrt(3-2sqrt2)=sqrt2-1$

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