Help me please

**Quick** · August 1st 2006, 06:25 PM

Originally Posted by Chester

I am a little confused (not that that’s hard to do at this point.) But the problem is

$<br /> \frac{\sqrt3}{\sqrt{6-1}}-\frac{\sqrt3}{\sqrt{6+1}}<br />$

Not

$<br /> \frac{\sqrt3}{\sqrt{6-1}}=\frac{\sqrt3}{\sqrt{6+1}}$

Do I solve this the same way?

No you definitely do not. let me explain why I wrote $<br /> \frac{\sqrt3}{\sqrt{6-1}}=\frac{\sqrt3}{\sqrt{6-1}}$

everyone knows that $x-x=0$

now look at your paper and you realize that you wrote [tex]
$\frac{\sqrt3}{\sqrt{6-1}}-\frac{\sqrt3}{\sqrt{6-1}}$
so I was just pointing out the obvious by saying they were zero.

but now you write $\frac{\sqrt3}{\sqrt{6-1}}-\frac{\sqrt3}{\sqrt{6+1}}$ which is COMPLETELY different, in fact right now I'm drawing a blank on this problem (I feel like the answers easy to get, but I just can't think of it)

**Chester** · August 1st 2006, 06:39 PM

The problem says to "simplify"

I am at a loss also. Buti will keep looking.

Thanks again,
Chester

**topsquark** · August 1st 2006, 06:44 PM

Originally Posted by Chester

The problem says to "simplify"

I am at a loss also. Buti will keep looking.

Thanks again,
Chester

My only thought is that you might want to rationalize the denominator in each term. This has the result that both terms will have the same denominator and then can be added.

-Dan

**Chester** · August 1st 2006, 06:51 PM

I will see what i come up with

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Thanks Dan