2
$\begingroup$

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 $\sqrt{2}$ ?

$\endgroup$
  • 4
    $\begingroup$ Hint: $162=2\cdot 81$. $\endgroup$ – Michael Burr Jan 3 '19 at 4:52
  • 1
    $\begingroup$ $162=2*81=2*9^2$ 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
8
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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, $a = \sqrt{2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.