Division with Radicals

**Forkmaster** · August 24th 2008, 10:05 AM

This is confusing me. The book's method is basically this...
Problem:
4
-
(Radical) 7

Multiply by
(Radical) 7
-----------
(Radical) 7

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

4 * (Radical) 7
-------------
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?

2 * (Radical) 3
--------------
(Radical) 6

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

Help please...?

**Moo** · August 24th 2008, 10:08 AM

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?

2 * (Radical) 3
--------------
(Radical) 6

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

Help please...?

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

**Chris L T521** · August 24th 2008, 10:13 AM

Originally Posted by Forkmaster

This is confusing me. The book's method is basically this...
Problem:
4
-
(Radical) 7

Multiply by
(Radical) 7
-----------
(Radical) 7

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

4 * (Radical) 7
-------------
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?

2 * (Radical) 3
--------------
(Radical) 6

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

Help please...?

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

Here is one way of doing this:

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

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

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

Here is another:

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

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

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

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

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

I hope this makes sense!

--Chris

EDIT: Moo beat me!

**Forkmaster** · August 24th 2008, 10:56 AM

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:

(Radical) 5
----------
(Radical) 10

(Radical) 27
-----------
(Radical) 45

(Radical) 32
-----------
(Radical) 5

Or

(Radical) 18
-----------
3 * (Radical) 2

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