This is confusing me. The book's method is basically this...
Problem:
4
-

Multiply by
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with (Radical) meaning the checkmark thing, and the hyphens being a division line.
That multiplication gets you...

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7

Which the book marks as correct.

That's all well and good, but I'm confused by the next step up...
What would I do on a problem like this?

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I tried that method, and couldn't do it. The book gives no other methods.

2. Originally Posted by Forkmaster
That's all well and good, but I'm confused by the next step up...
What would I do on a problem like this?

--------------

I tried that method, and couldn't do it. The book gives no other methods.

Hmmm there, you can simplify by $\displaystyle \sqrt{3}$ first, because $\displaystyle \sqrt{6}=\sqrt{3} \cdot \sqrt{2}$.
Then, use the given method

3. Originally Posted by Forkmaster
This is confusing me. The book's method is basically this...
Problem:
4
-

Multiply by
-----------

with (Radical) meaning the checkmark thing, and the hyphens being a division line.
That multiplication gets you...

-------------
7

Which the book marks as correct.

That's all well and good, but I'm confused by the next step up...
What would I do on a problem like this?

--------------

I tried that method, and couldn't do it. The book gives no other methods.

$\displaystyle \frac{2\sqrt{3}}{\sqrt{6}}$

Here is one way of doing this:

since $\displaystyle 2=\sqrt{4}$ we see that our expression becomes $\displaystyle \frac{\sqrt{4}\sqrt{3}}{\sqrt{6}}$

There is a theorem that says $\displaystyle \sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$ given that both $\displaystyle a~and~b>0$

So, we then see that our expression becomes $\displaystyle \frac{\sqrt{12}}{\sqrt{6}}=\sqrt{\frac{12}{6}}=\co lor{red}\boxed{\sqrt{2}}$

Here is another:

Using this idea $\displaystyle \sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$ when $\displaystyle a~and~b>0$ we can maninpulate the demonimator a little bit:

$\displaystyle \frac{2\sqrt{3}}{\sqrt{6}}=\frac{2\sqrt{3}}{\sqrt{ 2\cdot 3}}=\frac{2\sqrt{3}}{\sqrt{2}\sqrt{3}}$

Now, the $\displaystyle \sqrt{3}$ term cancels out.

So we are left with $\displaystyle \frac{2}{\sqrt{2}}$

Now rationalize this expression to get $\displaystyle \frac{2}{\sqrt{2}}\cdot{\color{red}\frac{\sqrt{2}} {\sqrt{2}}}=\frac{2\sqrt{2}}{2}=\color{red}\boxed{ \sqrt{2}}$

I hope this makes sense!

--Chris

EDIT: Moo beat me!

4. Alright, I get it now. Thanks!

Edit: Found more that I can't figure out. It's not my day today...>_>

How would I solve problems like these:

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