Question

Algebra ll math question please help!:

Answer 1

It is something about rationalizing the denominator?

Answer 2

@amistre64 Do you know how to solve this?

Answer 3

I kind of know how to solve it but its the negative square root of 36 that I do not know

Answer 4

yes i do add the like terms and use a conjugate to get rid of th ei part in the bottom

Answer 5

sqrt(-36) = sqrt(-1*36) = sqrt(36) sqrt(-1)

Answer 6

Yeah, and then I multiply the conjugate on the top and bottem right?

Answer 7

yep

Answer 8

So (2-3i)+ (2+3i)= 6-i?

Answer 9

And then the conjugate would be 6+i

Answer 10

\[\frac{ci}{a+bi}\]
\[\frac{ci(a-bi)}{(a+bi)(a-bi)}\]
\[\frac{(ac) i -(bc)i^2}{a^2+b^2}\]

Answer 11

So then I would use sqrt(36) sqrt(-1) times 6+i?

Answer 12

adding like terms i would say we have 5 - i, which conjugates to 5+i

Answer 13

Oh ok

Answer 14

So then 5+i/5+i times sqrt(36) sqrt(-1)/ 5-i?

Answer 15

so, a = 5, b=-1, c=6 and recall that i^2 = -1

Answer 16

\[\frac{(ac) i -(bc)i^2}{a^2+b^2}\]
\[\frac{6(5)i -6(-1)(-1)}{(5)^2+(-1)^2}\]

Answer 17

Ok so I solve using that equation?

Answer 18

you dont have to, but if my generality doesnt have any errors in it we can shk

Answer 19

generality?

Answer 20

30i - 6
------
26

we have alot of evens so divide out a 2


15i - 3
------
13

Answer 21

yeah the abc parts are just a generalization which allows us to simply use it as a formula

Answer 22

Ah k

Answer 23

So what next after 15i-3/13?

Answer 24

see which option is the correc tone of course

Answer 25

Oh wait

Answer 26

Thats one of the answers!

Answer 27

So the answer is b?

Answer 28

i see no letters for options, but the second on the list is fine

Answer 29

Ok thanks so much!

Answer 30

a generality allows us to use properties without having to deal with values, and alot of times makes th eprocess simpler since the end result represents a formula

Answer 31

good luck

Answer 32

Ty! You too in anything your doing!