Sunday, April 1, 2018

algebra precalculus - How to simplify this expression with radicals? $3sqrt2 - sqrt{32} + sqrt{frac{80}{16}}$



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 $\dfrac{9-4\sqrt2}{4}$.


Answer



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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