Question

http://prntscr.com/930xfa

Answer 1

@amistre64 @ganeshie8 @phi

Answer 2

whats the conjugate of \(1+\sqrt{2}\) ?

Answer 3

do you know what conjugate means ? )

Answer 4

\[1-\sqrt{2}\]

Answer 5

good, multiply that top and bottom

Answer 6

\[\dfrac{1-\sqrt{2}}{1+\sqrt{2}}\]

\[\dfrac{(1-\sqrt{2})\color{red}{(1-\sqrt{2})}}{(1+\sqrt{2})\color{red}{(1-\sqrt{2})}}\]

Answer 7

I did and got \[\frac{-1-\sqrt{2}}{1+\sqrt{2}-\sqrt{2+2}}\]...wait, I might've done something wrong --\[\frac{3+\sqrt{2}}{2}?\]

Answer 8

nope, you may use the identities :
\[(a+b)(a-b)=a^2-b^2\]

and

\[(a-b)^2 = a^2-2ab+b^2\]

Answer 9

wha

Answer 10

brb one sec then lol

Answer 11

if you're not familiar with those identities, you may simply multiply the binomials and simplify

Answer 12

you should get :

\[\dfrac{1-\sqrt{2}}{1+\sqrt{2}}\]

\[\dfrac{(1-\sqrt{2})\color{red}{(1-\sqrt{2})}}{(1+\sqrt{2})\color{red}{(1-\sqrt{2})}}\]

\[\dfrac{1-2\sqrt{2}+2}{1-2}\]

\[\dfrac{3-2\sqrt{2}}{-1}\]

\[-3+2\sqrt{2}\]

Answer 13

okay so I got \[\frac{1-2}{1^{2}-2(1)(\sqrt{2})+\sqrt{2}^{2}}=\frac{-1}{1-2\sqrt{2}+2}=\frac{-1}{3-\sqrt{2}}\]

Answer 14

I mean on the bottom of the last fraction:\[3+2\sqrt{2}\]

Answer 15

what I got it upside down 0-0??

Answer 16

oh I mixed up the formulas whoops

Answer 17

sorry!

Answer 18

yeah looks you have used the wrong identities, thats okay.. try again :)

Answer 19

lol okay, I got this one -- what about the other one o-o

Answer 20

@ganeshie8
I got \[\frac{1^{2}-2(1)(\sin{x})+sin^{2}{x}}{1^{2}-sin^{2}{x}}=\frac{1-2\sin{2}+\sin^{2}{x}}{1-sin^{2}{x}}\]

Answer 21

sinx in second equation*

Answer 22

and the bottom part translates to \[cos^{2}{x}\]

Answer 23

looks good to me !

Answer 24

alright thanks :)