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I don't understand how I could calculate this: $3\sqrt2 - \sqrt{32} + \sqrt{\dfrac{80}{16}}$

My answer is $-\sqrt2 + \sqrt5$, but the real answer should be $\dfrac{9-4\sqrt2}{4}$.

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    $\begingroup$ Maybe it was $\sqrt{\frac{81}{16}}$? $\endgroup$ – Daniel Fischer Dec 27 '14 at 16:31
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    $\begingroup$ Your answer is perfectly correct for what you posted ! $\endgroup$ – Claude Leibovici Dec 27 '14 at 16:31
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    $\begingroup$ Did you try to check the question again? Answer seems right .@DanielFischer it is very likely $\sqrt{\frac{81}{16}}$ $\endgroup$ – Devarsh Ruparelia Dec 27 '14 at 16:32
  • $\begingroup$ Oh... I am so blind. Yes, it's 81/16. Haha... :| Thank you guys for pointing it out. $\endgroup$ – Commander Shepard Dec 27 '14 at 16:58
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As commenters pointed out, $80$ should be $81$. The radicals simplify as follows: $$ 3\sqrt2 - \sqrt{32} + \sqrt{\dfrac{81}{16}} = 3\sqrt{2}- \sqrt{16} \sqrt{2} +\frac94 = (3-4)\sqrt{2}+\frac94 = \frac94-\sqrt{2} $$

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Assuming you meant $81$: $$ \begin {align*} 3\sqrt2 - \sqrt{32} + \sqrt{\dfrac{81}{16}} &= 3\sqrt{2}- \sqrt{16} \sqrt{2} +\frac94 \\&= (3-4)\sqrt{2}+\frac94 \\&= \frac94-\sqrt{2}. \end {align*} $$

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