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Can anyone explain how I would simplify this dividing surd:

$$\frac{3\sqrt{14}}{\sqrt{42}}$$

As far as I can see $\sqrt{14}$ and $\sqrt{42}$ can't be simplified, right? So how does the division and the multiplication of 3 come into it? Thanks for the help!

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$$\frac{3\sqrt{14}}{\sqrt{42}}=\frac{3\sqrt{14}}{\sqrt{3}\sqrt{14}}=\frac{3}{\sqrt{3}}=\sqrt{3}$$

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Hint: The key is that $42=3\cdot 14$.

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$$ \frac{3\sqrt{14}}{\sqrt{42}}= \sqrt{\frac{9 \cdot 14}{42}} = \sqrt{3} $$

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And yet another way to do it. The key is that you can reduce out of the roots, for example $\sqrt{4}/\sqrt{8} = \sqrt{1}/\sqrt{2}$ (because you can extend the root over the whole fraction).

First, you can see that 14 and 42 are both divisible by 2: $$\frac{3\sqrt{14}}{\sqrt{42}} = 3\frac{\sqrt{7}}{\sqrt{21}}$$ Now, 21 is 3 × 7, so $$\ldots = 3\frac{\sqrt{1}}{\sqrt{3}}, $$ and then we just bring the three into the root, and we're done: $$\ldots = \frac{\sqrt{9}}{\sqrt{3}} = \sqrt{3} $$

I think this is an easy way because you don't have to 'recognize' prime factors like 42=7×2×3.

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We have $42=7\times2\times3$ and $14=7\times 2$, therefore:

$$\sqrt{42}=\sqrt{7}\sqrt{2}\sqrt{3}=\sqrt{14}\times\sqrt{3}$$

Hence:

$$\frac{3\sqrt{14}}{\sqrt{42}}=\frac{3}{\sqrt{3}}=\sqrt{3}$$

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$$ \frac{\sqrt{14}}{\sqrt{42}} = \sqrt{\frac{14}{42}} = \sqrt{\frac{14}{3\cdot14}} =\sqrt{\frac13}\text{ etc.} $$

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In general, to simplify a surd, you want to "rationalize the denominator".

In this case, that would involve multiplying by $\sqrt{42}$ to get $\displaystyle\frac{3\sqrt{588}}{42}$, factoring out the $196$ from the $588$ to get $\displaystyle\frac{42\sqrt{3}}{42}$, and reducing the fraction to obtain $\sqrt{3}$.

However, one property of square roots is that they can be divided. So, you can simply divide $\sqrt{42}$ in the denominator by $\sqrt{14}$ in the numerator to get $\displaystyle \frac{3}{\sqrt{3}}$, which is simply $\sqrt{3}$.

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