# 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.

• $\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$ – egreg Oct 4 '13 at 23:38
• Such an "obvious" solution! Thanks, egreg! – Jose Oct 4 '13 at 23:48
• Important detail: a "problem" really needs an English-language direction or question of what to do, and you didn't include that. You cannot look at a mathematical expressions and infer "what to do" from it, so pay attention to the actual directions. A short example I run in class is to ask: "What's the degree of $3x^2+5x^2$?", and a lot of students answer, $8x^2$, which isn't a degree at all. – Daniel R. Collins Sep 21 '15 at 6:33

$$\frac{\sqrt{18}+\sqrt{98}+\sqrt{50}+4}{2\sqrt{2}}= \frac{3\sqrt{2}+7\sqrt{2}+5\sqrt{2}+4}{2\sqrt{2}}= \frac{15\sqrt{2}+4}{2\sqrt{2}}$$ Now the standard procedure is to remove the radical in the denominator: $$\frac{15\sqrt{2}+4}{2\sqrt{2}}= \frac{15\sqrt{2}+4}{2\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}= \frac{15\sqrt{2}\cdot\sqrt{2}+4\sqrt{2}}{2\sqrt{2}\cdot\sqrt{2}}= \frac{30+4\sqrt{2}}{4}=\frac{15+2\sqrt{2}}{2}$$ 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.

Why shouldn't it be $+4$ at the end of the second expression? That should carry down to the third. From the second to the third you are just splitting one fraction into two.

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 $\dfrac{15}{2}+\dfrac{\sqrt{2}}{2}$ or into $\dfrac{15}{2}+\dfrac{1}{\sqrt{2}}$ if you really wanted to.