Square Root

multiply the top and bottom by sqrt(3)-1

kill the denominator by multiplying with 1 and remembering the binomial formula

>>8472506
[eqn]\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{3 + 1 - 2 \sqrt{3}}{2} = 2 - \sqrt{3}[/eqn]

>>8472512
but it is fuckface

http://www.wolframalpha.com/input/?i=((sqrt3)-1)%2F((sqrt3)%2B1)

>>8472530
>binomial formula
no need

>>8472531
Thank you

Multiply by the conjugate of the bottom

[math] \dfrac {a- \sqrt{b} } {a+ \sqrt{b}} = \dfrac {a- \sqrt{b} } {a+ \sqrt{b}} \dfrac {a- \sqrt{b} } {a- \sqrt{b}} = \dfrac {a^2 +b - 2a\sqrt{b} } {a^2 -b } [/math]

>>8472538
That's not a conjugate you idiot.

>>8472549
https://en.wikipedia.org/wiki/Conjugate_(algebra)

>>8472552
>https://en.wikipedia.org/wiki/Conjugate_(algebra)
Granted, didn't know that.

>>8472556
you do know, smartass

>>8472506
rationalize the denominator

>>8472597
now*

File: qedfgtgetrektlmaobbq.png (7KB, 268x180px) Image search: [Google]

7KB, 268x180px

>>8472506
it equals -1 you retards

>>8472963

2=sqr(4)=sqr(-1^2*4)=-2

2=-2

Fucking Gobshite

>>8472963
>second step
what the fuck? at least make your bait half-decent

>>8473125
-(a-b) = (b-a)
are you stupid or something