# Fractions and roots I have this problem: \frac{\sqrt{18}+\sqrt{98}+\sqrt{50}+4}{2\sqrt{2}

Fractions and roots
I have this problem:
$\frac{\sqrt{18}+\sqrt{98}+\sqrt{50}+4}{2\sqrt{2}}$
I'm able to get to this part by myself:
$\frac{15\sqrt{2}+2}{2\sqrt{2}}$
But that's when I get stuck. The book says that the next step is:
$\frac{15\sqrt{2}}{2\sqrt{2}}+\frac{2}{2\sqrt{2}}$
But I don't understand why you can take the 2 out of the original fraction, make it the numerator of its own fraction and having root of 2 as the denominator of said fraction.
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alastrimsmnr

$\frac{\sqrt{18}+\sqrt{98}+\sqrt{50}+4}{2\sqrt{2}}=\frac{3\sqrt{2}+7\sqrt{2}+5\sqrt{2}+4}{2\sqrt{2}}=\frac{15\sqrt{2}+4}{2\sqrt{2}}$
Now the standard procedure is to remove the radical in the denominator:
$\frac{15\sqrt{2}+4}{2\sqrt{2}}=\frac{15\sqrt{2}+4}{2\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}=\frac{15\sqrt{2}·\sqrt{2}+4\sqrt{2}}{2\sqrt{2}·\sqrt{2}}=\frac{30+4\sqrt{2}}{4}=\frac{15+2\sqrt{2}}{2}$
One can do it differently: set $a=\sqrt{2}$, so you can write
$\frac{15\sqrt{2}+4}{2\sqrt{2}}=\frac{15a+{a}^{4}}{{a}^{3}}=\frac{15+{a}^{3}}{{a}^{2}}=\frac{15+2\sqrt{2}}{2}$
The final result can also be written by using $\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$ so
$\frac{15+2\sqrt{2}}{2}=\frac{15}{2}+\sqrt{2}$
Whether you want to do this last transformation depends on what you have to do with this number.

ubafumene42h
To me the next obvious step is to remove the radical in the denominator so
$\frac{15\sqrt{2}+2}{2\sqrt{2}}=\frac{15+\sqrt{2}}{2}.$
Personally I would stop there, but you could separate this into $\frac{15}{2}+\frac{\sqrt{2}}{2}$ or into $\frac{15}{2}+\frac{1}{\sqrt{2}}$ if you really wanted to.