• Aug 24th 2008, 10:05 AM
Forkmaster
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.

• Aug 24th 2008, 10:08 AM
Moo
Quote:

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 ;)
• Aug 24th 2008, 10:13 AM
Chris L T521
Quote:

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! (Sun)

--Chris

EDIT: Moo beat me! :eek:
• Aug 24th 2008, 10:56 AM
Forkmaster
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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