 # Fractions and roots I have this problem: \frac{\sqrt{18}+\sqrt{98}+\sqrt{50}+4}{2\sqrt{2} Deven Livingston 2022-04-23 Answered
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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$\frac{\sqrt{18}+\sqrt{98}+\sqrt{50}+4}{2\sqrt{2}}=\frac{3\sqrt{2}+7\sqrt{2}+5\sqrt{2}+4}{2\sqrt{2}}=\frac{15\sqrt{2}+4}{2\sqrt{2}}$
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.

We have step-by-step solutions for your answer! 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.

We have step-by-step solutions for your answer!