Solve Radical Equation: (2 root 3 + 3 root 4)/-5

[wScott]

This isn't exactly homework, it was a question we had on a test and I had no idea at all of how to do it, although I think I did good on the other questions.

The result that it came to, I can't remember exactly what it was, looked kind of like this:

(2 root 3 + 3 root 4) over (-5)

My teacher won't let us have any question with a negative denominator. My question is how do you remove the negative from the bottom.

EDIT: Sorry I couldn't figure out how to LaTeX this problem.

 

[PPonte]

Your LaTeX cannot be seen. Could you write the fraction normally?

 

[wScott]

Sorry about that, it's fixed.

 

[Hootenanny]

You can take it outside the fraction or put it to the top. The fraction still has the same value;

\frac{a}{-b} \equiv \frac{-a}{b} \equiv - \frac{a}{b}

They are all identical, but it is usually more acceptable to have the negative 'outside' the fraction as in the third example.

 

[wScott]

Ahh makes sense to me, thanks for teh help. I definitely screwed that up. For my answetr I came out with fractional radicals under 1 :p

 

[PPonte]

Hootenanny explained it very well. But I could add a tip.

Your fraction is:
(2 root 3 + 3 root 4) over (-5)

If you divide all the terms of the fraction by the same number you mantain it equal to the initial fraction.
In this case you could divide by -1. So, we would have:

(-2 root 3 - 3 root 4) over (5)

 

[wScott]

Good point PPonte