[SOLVED] question about writing roots in fractions?

• Apr 24th 2010, 11:03 PM
dorkymichelle
[SOLVED] question about writing roots in fractions?
I'm doing a calculus problem and i found that $\displaystyle x =\frac{\sqrt3}{3}$
in the answer in the back, it says $\displaystyle \frac{1}{\sqrt3}$
I put them both in the calculator and it comes out to be about 0.57....
why did the book put the root in the denominator? i thought it was standard form to have it in the numerator?

not sure where this topic belongs, sorry if in wrong category!
• Apr 24th 2010, 11:10 PM
Prove It
Quote:

Originally Posted by dorkymichelle
I'm doing a calculus problem and i found that x =\frac{\sqrt3}{3}
in the answer in the back, it says \frac{1}{\sqrt3}
I put them both in the calculator and it comes out to be about 0.57....
why did the book put the root in the denominator? i thought it was standard form to have it in the numerator?

not sure where this topic belongs, sorry if in wrong category!

Note that

$\displaystyle \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$.

It is considered to be more elegant to rationalise denominators, because it better represents what a fraction is - namely, dividing a length into a COUNTABLE number of pieces.

However, rationalising the denominator is not usually done until the very end, as it could result in more work with whatever you happen to be doing. Perhaps your book intends you to use this answer for further calculations...
• Apr 24th 2010, 11:19 PM
dorkymichelle
I found it odd because i didnt encounter 1/root 3 in my calculations at all.
is there a way to solve for x in the equation $\displaystyle -12x^2+4$
• Apr 24th 2010, 11:25 PM
harish21
Quote:

Originally Posted by dorkymichelle
I found it odd because i didnt encounter 1/root 3 in my calculations at all.
is there a way to solve for x in the equation $\displaystyle -12x^2+4$

is your equation $\displaystyle -12x^2+4=0$??
• Apr 24th 2010, 11:27 PM
dorkymichelle
yes
• Apr 24th 2010, 11:31 PM
harish21
Quote:

Originally Posted by dorkymichelle
yes

if $\displaystyle -12x^2+4=0$, then

$\displaystyle 4 = 12x^2$

so,

$\displaystyle x^2 = \frac{4}{12} = \frac{1}{3}$

$\displaystyle \therefore x = \pm \frac{1}{\sqrt 3}$
• Apr 24th 2010, 11:34 PM
dorkymichelle
wow... i can't believe i didn't see that!!!