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I am given this expression and asked to simplify by rationalizing the denominator:

$$\frac{18}{\sqrt{162}}$$

The solution is provided:

$\sqrt{2}$

I arrived at:

$$\frac{\sqrt{162}}{9}$$

Here is my thought process to arrive at this incorrect answer:

$\frac{18}{\sqrt{162}}$

= $\frac{18}{\sqrt{162}}$ * $\frac{\sqrt{162}}{\sqrt{162}}$

= $\frac{18\sqrt{162}}{162}$

= $\frac{\sqrt{162}}{9}$

How can I arrive at $\sqrt{2}$ ?

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    $\begingroup$ Hint: $162=2\cdot 81$. $\endgroup$ – Michael Burr Jan 3 '19 at 4:52
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    $\begingroup$ $162=2*81=2*9^2$ so $\sqrt {162}=\sqrt {2*9^2}=9\sqrt 2$. If your hadn't "deradicalized" the denominator you would have ended up with $\frac 2 {\sqrt 2} $ which is also deradicalized as $\sqrt 2$. $\endgroup$ – fleablood Jan 3 '19 at 5:49
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$\frac{\sqrt{162}}{9} = \frac{\sqrt{2 \cdot 9^2}}{9} = \frac{9\sqrt{2}}{9} = \sqrt{2}$.

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$$\require{cancel}\frac{18}{\sqrt{162}}=\frac{2\cdot3^2}{\sqrt{2\cdot3^4}}=\frac{2\cdot\cancel{3^2}}{\sqrt{2}\cdot\cancel{3^2}}=\frac{2}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{\cancel2\sqrt{2}}{\cancel{2}}=\sqrt{2}$$

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Your thought process is good. But just continue with factorizing $162=2*81=2*3^4$.

So $\sqrt {162}=\sqrt {2*3^4}=\sqrt {2}\sqrt {3^4}=\sqrt 2*3^2=9\sqrt 2$ and from there.... it's just mechanics.

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$ \sqrt{162} $ needs to be simplified further, as $ 162 $ is the product containing a perfect square (i.e. $ 81 $). Thus $$ \sqrt{162} = \sqrt{2 \cdot 81} = \sqrt{2}\sqrt{81} = 9\sqrt{2}$$

and hence $$ \frac{\sqrt{162}}{9} = \frac{9\sqrt{2}}{9} = \sqrt{2} $$

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Alternatively if $a = \frac{18}{\sqrt{162}}$, $a^2 = \frac{18^2}{\sqrt{162}^2} = \frac{324}{162} = 2$. Since $a$ is a positive number, $a = \sqrt{2}$.

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