+0

0
274
1

2 times the square root of 5

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2 times the square root of 7 + 3 times the square root 3

Guest Apr 2, 2017
#1
+7324
+2

We want to get the radicals out of the denominator.

Like this:

$$\frac{2\sqrt5}{2\sqrt7+3\sqrt3}*\frac{2\sqrt7-3\sqrt3}{2\sqrt7-3\sqrt3}=\frac{2*2\sqrt{5*7}-2*3\sqrt{5*3}}{2*2\sqrt{7*7}-3*3\sqrt{3*3}} \\~\\ = \frac{4\sqrt{35}-6\sqrt{15}}{4*7-9*3}= \frac{4\sqrt{35}-6\sqrt{15}}{1}=4\sqrt{35}-6\sqrt{15}$$

If you want an approximate answer, it is $$\approx 0.426$$

hectictar  Apr 2, 2017