Please, all you people who are just roaming the boards, use your math skills to help me out!
If you have 2 roots of a quadratic equation as (1+3i) and (1-3i) how do you write the quadratic equation in the normal ax^2 +bx+c form?
please and thank you!
Also, I know there is some FOILing involved, could you be ever so kind as to type out those steps? My brian isnt' working and I can't remember this for my final tomorrow!
...Please? I'll love you forever and I'll share my 100 thanks to the answering of this question from my final with you ;D!
I'm sorry, I wish I could help. Even if I could help you, you wouldn't want my help seeing as how I failed my math final this year.
Broadway Legend Joined: 12/31/69
DO you have the answer in the back of the book? If what I did is correct, I can explain how to get the answer.
Broadway Legend Joined: 8/16/05
(1 + 3i)(1 - 3i)
F: 1*1=1
O: 1*-3i=-3i
I: 1*3i=3i
L: 3i*-3i=-9i^2
= -9i^2 + 1 (the 3i's cancel eachother out as one is pos. and one is neg.)
I don't know. I could be totally wrong. It's been probably 10 years since I've done one of those.
Broadway Legend Joined: 12/31/69
I just pm-ed you the same thing.
You can simplify further b/c i^2= -1
so it is -9(-1) + 1 =10
Right?
well the answer is x^2 -2x + 10 = O somehow..
I think the second step is supposed to be
(x-(1+3i)) (x-(1-3i)) but I have no clue how to FOIL that thing.
Thanks for your replies though!! Anyone know where to go from there?
double post
Broadway Legend Joined: 12/31/69
Oh- I understand what the question is now.. let me try again. I didn't read that it was the two roots of the quad eq'n. I'm tired... will reply in a few.
Don't worry about it :)
I've been staring at this question all day. ANY help is appreciated :)
Broadway Legend Joined: 12/31/69
First, distribute the (-):
(x-1-3i)(x-1+3i)
then foil:
x^2-x+3xi-x+1-3i-3ix+3i-9i^2
cancel out 3xi-3xi, -3i+3i
that leaves you with:
x^2-2x-9i^2+1
x^2-2x-9(-1)+1
x^2-2x+10
Broadway Legend Joined: 9/4/05
Oops
Broadway Legend Joined: 9/4/05
Shameful triple post...
Broadway Legend Joined: 9/4/05
(1+3i)(1-3i)
1*1=1
1*-3i= -3i
3i*1= 3i
3i*-3i= -9i^2= 9 because i^2= -1
I think at least...
THANK YOU!!!
!
Broadway Legend Joined: 9/4/05
Wow... I've never double posted like that before...
Broadway Legend Joined: 8/16/05
Oops. I misread the question. Sorry!
Broadway Legend Joined: 12/31/69
Good luck tomorrow! Remember, when you see a problem like that, don't get overwhelmed. Just take it one step at a time and you will be fine!
(1+3i)(1-3i)
Oh geez those are nasty...precalculus from last term that I vaguely remember.
(x-(1+3i)) (x-(1-3i))
So basically FOIL:
F: (x)(x) = x^2
O: (x)(-1+3i) = -x+3xi
I: (x)(-1-3i) = -x-3xi
L: (-1-3i)(-1+3i) = ?
***remember to distribute the negative in parts O & I
You basically have to foil the L part of the original FOIL, so it's a foil inside a FOIL.
(-1-3i)(-1+3i)
F: (-1)(-1) = 1
O: (-1)(3i) = -3i
I: (-3i)(-1) = 3i
L: (-3i)(3i) = -9i^2 = (-9)(-1) = 9
So you get: 1-3i+3i+9 which equals 10
So going back to the first part of the FOIL you know know the product of (-1-3i)(-1+3i), which is 10.
F: (x)(x) = x^2
O: (x)(-1+3i) = -x+3xi
I: (x)(-1-3i) = -x-3xi
L: (-1-3i)(-1+3i) = 10
So then you have: x^2-x+3xi-x-3xi+10, combine like terms, the 3xi's cancle, and you have x^2-2x+10!
[edit] b/c some of the formating became
Updated On: 6/12/06 at 08:21 PM
THank you everyone :)
Geez, I'll have to split my grade amongst many-a-smart-people :)hehe
Thanks again :)
Oh god math! Bad flashback!
I couldn't understand a thing of that!!! I just had my math final today, and let me tell you: Bea isn't the smartest person in the world!!!
Broadway Legend Joined: 6/12/05
*blinks*
*Stares*
*leaves*
***Offers cookies to the fellow math-a-phobes***
Let's take the cookies and go huddle in another room with a good novel. All these numbers are making me nervous.
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