Question

Help Please! Rationalize the denominator:

Answer 1

\[\frac{ \sqrt{-36} }{ (2-3i)+(3+2i) }\]

Answer 2

How do I start this? :/

Answer 3

@starlord6200

Answer 4

you would first do whats in the parenthesis so its -i and 5i then you would add thm and make it sqr 36 /4i

Answer 5

ok, so after i have this, how do i get id of the -36 under the radical?

Answer 6

so what times it self is 36

Answer 7

I thought it would be 5-i as the denominator D: im so lost

Answer 8

^You are correct

@starlord6200 there is no 'i' term regarding the 2 and the 3...they are just constants

Your denominator IS 5 - i

Answer 9

Oh ok, at least i understand that part lol :P. So, how do i get rid of the -36 under the radical?

Answer 10

So what you have is

\[\large \frac{\sqrt{-36}}{5 - i}\]

Now let me ask you ...what is the square root of 36? "forget the negative sign for the moment*

Answer 11

6

Answer 12

Right...and now also

Cant we just write

\(\large \sqrt{-36}\) as \(\large \sqrt{36 \times -1}\) ?

So if we know that the square root of 36 is 6..we have

\[\large 6\sqrt{-1}\]

And we all know the square root of -1 is i right? so the top turns into

\[\large 6i\]

So NOWWW we have

\[\large \frac{6i}{5 - i}\]

making sense so far?

Answer 13

Yep that makes sense

Answer 14

Okay...so..."rationalizing the dnominator" here means we dont want an "i" there...so we need to find a way to get the "i" out of the denominator...do you know how to do that?

Answer 15

Should I multiply by the conjugate? Im not sure.

Answer 16

YES! perfect :) never doubt yourself :P

So what do we have when you do that?

Answer 17

\[\frac{ 6i }{5-i} x \frac{ 5+i}{ 5+i } \] is this the right setup?

Answer 18

*Minus the 'x' but I know you meant *times* so yes that is correct...so go ahead and carry out the multiplication :)

Answer 19

\[\frac{ 30i+6i^2 }{ 25-i^2}\]

Answer 20

Great!

Now we just need to simplify a bit...like those i^2 what is i^2 ?

Answer 21

i think i^2 is (-1)

Answer 22

And you are correct...so go ahead and simplify the expression a little more :)

Answer 23

\[\frac{ 30i-6 }{ 26 } \] :) ?

Answer 24

I think i can simplify further by dividing all terms by 2, right?

Answer 25

oh then i have to rearrange the complex number also, right?

Answer 26

Great!

Now just a little more...notice how both terms in the numerator are divisible by 6? why dont we go ahead and factor out a 6

\[\large \frac{6(5i - 1)}{26}\]

And why dont we just go ahead and divide both the 6 and 26 by 2...

nvm you already got it haha

Answer 27

What do you mean by "rearrange the complex number*?

Answer 28

like, should i make it 6-30i instead of 30i-6?

Answer 29

I should say that rearranging *I'm stressing CORRECTLY* makes no difference

however what you just wrote is very different from one another

\[\large 6 - 30i \cancel{=} 30 - 6i\]

Answer 30

But regardless, dont worry about "which comes first" the order doesnt matter :)

Answer 31

oh ok :) sorry im a pain, but, should I have (15i-3)/13 as my final answer?

Answer 32

Not a pain at all :D

And well...not quite...you still have something you can do...that's why I originally just factored out a 6 of the numerator

But...you now have a 15...and a 3 on the top....anything you can do? both seem to be dividable by 3 right?

so...

Answer 33

5i - 1 as the numerator?

Answer 34

Perfect :)

and the 3 out in front

so finally we would have

\[\large \frac{3}{13}(5i - 1)\]

Answer 35

Ohh ok :) Thank youuu!!! i really needed to see all these steps, thanks for that! c:

Answer 36

Of course :) if you need any more help just let me know :)

Answer 37

ok thanks c: