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

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

• Maybe it was $\sqrt{\frac{81}{16}}$? – Daniel Fischer Dec 27 '14 at 16:31
• Your answer is perfectly correct for what you posted ! – Claude Leibovici Dec 27 '14 at 16:31
• Did you try to check the question again? Answer seems right .@DanielFischer it is very likely $\sqrt{\frac{81}{16}}$ – Devarsh Ruparelia Dec 27 '14 at 16:32
• Oh... I am so blind. Yes, it's 81/16. Haha... :| Thank you guys for pointing it out. – Commander Shepard Dec 27 '14 at 16:58

As commenters pointed out, $80$ should be $81$. The radicals simplify as follows: $$3\sqrt2 - \sqrt{32} + \sqrt{\dfrac{81}{16}} = 3\sqrt{2}- \sqrt{16} \sqrt{2} +\frac94 = (3-4)\sqrt{2}+\frac94 = \frac94-\sqrt{2}$$
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*}