Rationalizing Denominators of Radical Expressions

Question

The task is to get rid of square root in the denominator in the following equation: $\frac{2\sqrt{7} + \sqrt{14}} {\sqrt{7}} $. To do so I multiplied both denominator and nominator by $\sqrt{7}$ and my result is as follows:

$$\frac{2\sqrt{7} + \sqrt{14}} {\sqrt{7}} = \frac{(2\sqrt{7} + \sqrt{14})\sqrt{7}} {\sqrt{7}\sqrt{7}} = \frac{2\sqrt{7*7} + \sqrt{14* 7}} {7} = \frac {2*7 + \sqrt{98}} {7} = \frac {14 + 7\sqrt{2}} {7} = 2 + 7\sqrt{2}$$

however the correct answer is = $2 + \sqrt{2}$

When I calculate $\frac{2\sqrt{7} + \sqrt{14}} {\sqrt{7}} $ = ~ 3.4142 which is indeed = $2 + \sqrt{2}$

Could you please point out a mistake in my solution?

I really would like to understand why I am not able to get the correct answer.

Answer 1

Isolate $\sqrt2$ from numerator and use the fact that $\sqrt{14} = \sqrt{2}\sqrt{7}$

$$\frac{2\sqrt7+\sqrt{14}}{\sqrt7} = \frac{\sqrt{7}}{\sqrt{7}}(2+\sqrt2) = 2+\sqrt2$$

Answer 2

Multiplying numerator and denominator by $$\sqrt{7}$$ we get $$\frac{2\sqrt{7}\cdot \sqrt{7}+\sqrt{2}\sqrt{7}\cdot\sqrt{7}}{7}=\frac{14+7\cdot\sqrt{2}}{7}=2+\sqrt{2}$$