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Question Number 33575 by MURALI last updated on 19/Apr/18

$$\mathrm{simplify}\sqrt{(3+2\sqrt{2)}}$$

Answered by Rasheed.Sindhi last updated on 19/Apr/18

$$\mathrm{Let}\sqrt{(3+2\sqrt{2)}}=\mathrm{a}+\mathrm{b}\sqrt{2}$$$$\mathrm{where}\mathrm{a},\mathrm{b}\in \mathbb{Z}$$$${\left(\sqrt{(3+2\sqrt{2)}}\right)}^2={(\mathrm{a}+\mathrm{b}\sqrt{2})}^2$$$${\mathrm{a}}^2+{2\mathrm{b}}^2+2\mathrm{ab}\sqrt{2}=3+2\sqrt{2}$$$${\mathrm{a}}^2+{2\mathrm{b}}^2=3\wedge \mathrm{ab}=1$$$$\mathrm{ab}=1\Rightarrow \mathrm{b}=1/\mathrm{a}$$$${\mathrm{a}}^2+{2\mathrm{b}}^2=3\Rightarrow {\mathrm{a}}^2+\frac{2}{{\mathrm{a}}^2}=3$$$${\mathrm{a}}^4-{3\mathrm{a}}^2+2=0$$$$({\mathrm{a}}^2-1)({\mathrm{a}}^2-2)=0$$$$\mathrm{a}=\pm 1,\mathrm{a}=\pm \sqrt{2}\notin \mathbb{Z}$$$$\mathrm{Hence}\mathrm{a}=\pm 1$$$$\mathrm{b}=1/(\pm 1)=\pm 1$$$$\therefore \sqrt{(3+2\sqrt{2)}}=1+\sqrt{2},-1-\sqrt{2}$$

Answered by MURALI last updated on 19/Apr/18

$${(3+2\sqrt{2})}^2={\left(\sqrt{2}\right)}^2$$$$$$

Commented by Rasheed.Sindhi last updated on 21/Apr/18

$$\mathrm{Perhaps}\mathrm{you}\mathrm{mean}$$$${(3+2\sqrt{2})}^2={(1+\sqrt{2})}^2$$$$$$

Answered by MURALI last updated on 19/Apr/18

$$(3+2\sqrt{2})={\left(\sqrt{2}\right)}^2+2\sqrt{2}\times 1+{\left(1\right)}^2$$$$(3+2\sqrt{2})={(\sqrt{2}+1)}^2$$$$\therefore \sqrt{3+2\sqrt{2)}}=\sqrt{2}+1$$$$$$

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