# Rationalizing denominator of $\frac{18}{\sqrt{162}}$. Cannot match textbook solution

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 $$\sqrt{2}$$ ?

• Hint: $162=2\cdot 81$. Jan 3 '19 at 4:52
• $162=2*81=2*9^2$ 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 $\sqrt 2$. Jan 3 '19 at 5:49

$$\frac{\sqrt{162}}{9} = \frac{\sqrt{2 \cdot 9^2}}{9} = \frac{9\sqrt{2}}{9} = \sqrt{2}$$.

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

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.

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

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, $$a = \sqrt{2}$$.