whenever i do this problem i keep getting 3 - square root of 3

not 1 - square root of 3

idk wtf im doing wrong so easy too lol

when i do this i see the 2's cancel out. and then u have square root of 3 times the other square root of 3, which equals 9. then u got 9/3 - square root of 3 which is what i keep getting. 9/3=3.


wait ops. nvm im pretty sure i messed up when doing root 3 types root 3. its actually that the two radicals cancel and its just 3. so 3/3 = 1. not 9/3. ops.





This post was edited by noob_whacker on Jul 29 2014 03:37pm

since multiplication and division have a higher order than subtraction the first term is

[sqrt(3)/2] * 2 * [sqrt(3)/3] = [sqrt(3) * sqrt (3)] * [2/2] * [1/3] = 1

Here's your problem, root 3 * root 3 = 3 not 9. 3 *3 = 9 which is much different.

(√3*2*√3) / (2 * 3) - √3

(3 * 2) / (2 * 3) - √3

1 - √3

It's easy if you split it all up
root3 * root3 * 2 * 1/2 * 1/3 - root 3
the 2 and the 1/2 go away because they multiply to equal 1
we are left with root3 times root3 times 1/3 - root3
Remembering the order of operations, save the subtraction for last.
root3 times root3 is 3, then multiplying that by the 1/3, we get 1.
1 - root3

Hope this helps in some way.

nevermind, looking at the posted date, I can see this is probably no longer relevant >.<

This post was edited by ringo794 on Aug 14 2014 10:29am

√3 . √3 = 3

so,

= 3/3 . 2/2 - √3
= 1 - √3