How is $\sqrt{9+2\sqrt6+2\sqrt8+2\sqrt{12}}$ simplified into $\sqrt2+\sqrt3+\sqrt4$ ?
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1$$9=2+3+4,\sqrt{ab}=\sqrt a\sqrt b$$ for $a,b\ge0$ – lab bhattacharjee Apr 25 '17 at 05:49
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2I guess they expect you to recognize and thus use backwards the formula of the square of a trinomial. – Apr 25 '17 at 05:50
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One does not have to recognize it, it could be derived $(a+b+c)^2 = \cdots$ – mathreadler Apr 25 '17 at 10:47
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It's just the observation that $$(\sqrt2+\sqrt3+\sqrt4)^2=2+3+4+2\sqrt2\sqrt3+2\sqrt2\sqrt4+2\sqrt3\sqrt4=9+2\sqrt6+2\sqrt8+2\sqrt{12}.$$
Angina Seng
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