Question

help algebra2

Answer 1

Post it chewyti

Answer 2

is there anything with choice\[(\sqrt{3}-1)^{2}/2\] ??

Answer 3

or \[(3-\sqrt{3})^2/6\]???

Answer 4

@chewyti First of all WELCOME TO OPENSTUDY.

Answer 5

let me know that , are you aware with "rationalization" ?

Answer 6

thx

Answer 7

no

Answer 8

wait... for 2 minutes m coming

Answer 9

D is the answer

Answer 10

i attatched the choices

Answer 11

omg thx i really hate algebra i dont get it at all

Answer 12

I am back now... I think I should teach you rationalization now.

Answer 13

ok

Answer 14

are you still gonna teach me

Answer 15

are you still gonna teach me @mathslover

Answer 16

Yep.

Answer 17

So suppose we have a fraction :
\[\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}}\]
We now want that the denominator should be "rational" .

Answer 18

Now. I to make the denominator rational I would multiply the conjugate of denominator to both numerator and denominator :
\[\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} \times \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}}\]

Answer 19

Getting it?

Answer 20

yea kinda keep going i think i remember something like this in one of my lessons

Answer 21

Now we have in denominator : (x-y)(x+y) form... so it will be : x^2 - y^2

Answer 22

oh thx but you kinda lost me

Answer 23

can yu explain it more

Answer 24

do you kno how to do asymptotes

Answer 25

Yes. see let me take easy example :

suppose we have :
\[\large{\frac{1+\sqrt{3}}{1-\sqrt{3}}}\]
Now multiply the opposite of denominator
that is : 1+ sqrt{3}

\[\large{\frac{(1+\sqrt{3})(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}}\]
Now solve this

Answer 26

getting it now?

Answer 27

is it \[(2+\sqrt{9} \over 2 - \sqrt{9}\]

Answer 28

Now : (1^2 - (sqrt(3))^2 ) will be denominator

Answer 29

http://www.mathsisfun.com/algebra/rationalize-denominator.html

Answer 30

mannnn and i was confident on my answer