# How do rationalise {(plusminus)2(sqrt)2/(sqrt)3}

how do rationalise this?

{(plusminus)2(sqrt)2/(sqrt)3} ALL MINUS 2

Related Introductory Physics Homework Help News on Phys.org
TD
Homework Helper
What do you mean rationalize? You can't just get rid off the square root, you can simplify though.

$$\pm \frac{{\frac{{2\sqrt 2 }} {{\sqrt 3 }}}} {{ - 2}} = \pm \frac{{2\sqrt 2 }} {{ - 2\sqrt 3 }} = \mp \frac{{\sqrt 2 }} {{\sqrt 3 }} = \mp \frac{{\sqrt 2 \sqrt 3 }} {{\sqrt 3 \sqrt 3 }} = \mp \frac{{\sqrt 6 }} {3}$$

Fermat
Homework Helper
Isn't it supposed to be

$$\pm 2\frac{\sqrt{2}}{\sqrt{3}} - 2$$ ?

fermat is correct
the plusminus 2 belongs to the root 2

Fermat
Homework Helper
I get,

$$\frac{2(\sqrt{2} - \sqrt{3})}{\sqrt{3}}$$

disregarding the $$\pm$$ at the beginning.

TD
Homework Helper
Oh yes, I misread. Though all 'over' "-2".. You still won't be able to just lose the root, so what do you have to do exactly?.

your not supposed to have a square root on the bottom are you?

Here is a letter I wrote to Nexus magazine and a few theories I threw in the letter:

editor@nexusmagazine.com

Sincerely yours,

Paul Dale Roberts
5606 Moonlight Way
Elk Grove, CA 95758
JazmaPika@cs.com
(916) 203-7503

Fermat
Homework Helper
aricho said:
your not supposed to have a square root on the bottom are you?
No. That was just a simplification.
You can now multiply top and bottom by √3 to get,

$$\frac{2(\sqrt{6} - 3)}{3}$$