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I am trying to find the value of: $$\frac {\sqrt5}{\sqrt3+1} - \sqrt\frac{30}{8} + \frac {\sqrt {45}}{2}$$

I have the key with the answer $\sqrt 5$ but am wondering how I can easily get to that answer?

I realize this is a very basic question on radicals and I know that square roots can be separated into parts, for instance $\sqrt {45} = \sqrt 9 \cdot \sqrt 5$ and I can see how that would be useful here in canceling out radicals but I have not yet been able to reach a solution using this method.

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  • $\begingroup$ You can square that number and try to prove the square equals $25$, then you just need to prove that the given number is positive to conclude. $\endgroup$
    – Git Gud
    Jul 24 '14 at 11:21
  • $\begingroup$ To begin, write all the terms as multiples of $\sqrt 5$ (for example, from what you noted, the third term can be written as ${\sqrt 9\over 2}\sqrt5={3\over2}\sqrt 5$). For the first term, multiply top and bottom by $\sqrt3-1$ to start. $\endgroup$ Jul 24 '14 at 11:21
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$$\begin{align} \frac{\sqrt{5}}{\sqrt{3}+1} - \sqrt{\frac{30}{8}} + \frac{\sqrt{45}}{2} &= \frac{\sqrt{5}(\sqrt{3}-1)}{(\sqrt{3}+1)((\sqrt{3}-1))} - \sqrt{\frac{15}{4}} + \frac{\sqrt{9·5}}{2}\\ &= \frac{\sqrt{5}(\sqrt{3}-1)}{2} - \frac{\sqrt{15}}{2} + \frac{3\sqrt{5}}{2}\\ &= \frac{\sqrt{15}-\sqrt{5} -\sqrt{15} + 3\sqrt{5}}{2}\\ &= \sqrt{5} \end{align}$$

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