Question

Rationalize the denominator of -

equation will be in comments

Answer 1

\[\frac{ \sqrt{-4}\ }{ (7-3i)+(2+5i) }\]

Answer 2

anyone help? I have no idea how to do this and just need some help.

Answer 3

@ybarrap

Answer 4

@mathstudent55

Answer 5

thats not one of my answers, the closest is \[\frac{ -4+18i }{ 85 }\] and how did you get this answer @Love_333

Answer 6

hey @Nnesha any help?

Answer 7

@paki

@dan815

@Jaynator495

Answer 8

first combine like terms (7-3i)+(2+5i)
to rationalize the denominator multiply top and bottom with the conjugate of the denominator :)

Answer 9

do I fully do (7-3i)+(2+5i) then combine the like terms? yes right

Answer 10

hmm what ? sorry i don't know what you mean

Answer 11

ok so I turn (7-3i)+(2+5i) into 9+12i-1i+2i^2

correct, then I have to go further

Answer 12

there is a plus sign so can remove the parentheses \[\large\rm \frac{ \sqrt{-4}\ }{ 7-3i + 2+5i }\]
so first step would be combine like terms
(remember if there would be negative sign then we should distribute first )

Answer 13

ohhh ok so I dont do that I just add the terms like they are,

9+2i^2 <-- this right

Answer 14

hmm how did you get i^2?
remember : when we `multiply` same bases THEN we should add their exponents

x times x =x^2
x+x = 2x :=))

Answer 15

5i - 3i isn't 2i^2
\[\large\rm x+x \cancel{= }x^2\]
\[\rm x \times x=x^2\]
when we combine like terms the exponent of the variable would stay the same so
5i-3i is just 2i

Answer 16

ok, so 30i-40i would be -10 i not -10i^2, ok

Answer 17

right!
when we multiply like terms then we should add their exponents
when we combine them we just add their coefficient exponent would stay the same alright \[\large\rm \frac{ \sqrt{-4} }{ 9-2i }\]
now multiply top and bottom of the fraction by the conjugate of the denominator

Answer 18

do you know what conjugate is ?

Answer 19

ummm no, I remember the word conjugate from learning, but I cant remember what it means...

Answer 20

alright it's simple
`a-bi` conjugate would be `a+bi` so just change the sign of `imaginary term`
9+2i conjugate would be ?

Answer 21

9+2i** sorry

Answer 22

ohhhhh ok , 9-2i would be 9+2i, 9+2i would then be 9-2i right?

Answer 23

\[\frac{ \sqrt{-4} }{ 9+2i }\]

Answer 24

right \[\frac{\sqrt{-4}}{9+2i} * \frac{ 9-2i}{9-2i}\]
first deal with sqrt{-4}

Answer 25

we can rewrite \[\sqrt{-4} \rightarrow \sqrt{-1 \times 4}\]
use the fact sqrt{-1} = i

Answer 26

i have to go sorry

Answer 27

\[\sqrt{-4}\] =1?

Answer 28

ok thanks man, @ybarrap

Answer 29

can you help?

Answer 30

You are definitely on the right track
$$
\frac{ \sqrt{-4}\ }{ (7-3i)+(2+5i) }
$$
Let me summarize what you've done right so far

Answer 31

ok

Answer 32

$$
\frac{ \sqrt{-4}\ }{ (7-3i)+(2+5i) }\\
=\frac {\sqrt {-1}\sqrt 4}{9+2i}=\frac {i\sqrt 4}{9+2i}
$$
Next, we need to make the denominator real so we multiply it by the conjugate:

$$
=\frac {i\sqrt 4}{9+2i}=\cfrac{2i}{(9+2i)}\cfrac{(9-2i)}{(9-2i)}=\cfrac{2i(9-2i)}{81+4}
$$
Make sense so far?

Answer 33

how exactly did you get 81+4 on the bottom at the last? what did you multiply?

Answer 34

wait no I got it now

Answer 35

(9+2i) times the (9-2i) right

Answer 36

$$
(9+2i)(9-2i)=9\times9-9\times 2i+2i\times9-2i\times 2i=81-4i^2
$$
See that?
But \(i^2=-1\). Then we get
$$
81-4i^2=81+4
$$

Answer 37

ok yea I got that, then the 2i you put on the outside of the (9-2i) to get the 2i(9-2i)

Answer 38

$$
\cfrac{2i(9-2i)}{81+4}=\cfrac{-4+18i}{85}=-\cfrac{4}{85}+\cfrac{18}{85}i
$$
That's your answer :)

Answer 39

dude, are you a teacher? Im understanding this better from you than her :-:.... so my answer is \[\frac{ 4+18i }{ 85 }\]

Answer 40

yes, you got my error in the minus sign :)

Answer 41

ok thanks, I got a 90, and it was on a question that I did myself without posting for a check xD, thanks man if I ever need more help ill tag you :D

Answer 42

ok

Answer 43

but im done for toniight, doing part two of the test tomorrow, gnight man :)