SOLUTION: I have a couple of questions. The first one I would like you to help me finish. I am supposed to find 2 complex numbers whose sum is 1 and whose product is 5. So far I have: x+y

Question 63540: I have a couple of questions. The first one I would like you to help me finish.
I am supposed to find 2 complex numbers whose sum is 1 and whose product is 5. So far I have:
x+y=1
xy=5
since y=1-x I substitute it for y in xy=5
x(1-x)=5
x-x^2=5
x^2-x=-5 this is where I am stuck. I believe I am supposed to use 1/2b^2 in the equation somewhere, I just don't know how.
The next nonlinear equation I would like help with is:
x^2+xy+y^2=12
x+y=2
Thank you very much.
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
You are right on target!!!! Good work!! I'll pick up where you left off.
x^2-x=-5 Now we are going to make a perfect square out of the left side of the equation. To do this, we take half of the "B" term as you indicated (which is 1 , square it (we'll get (1/2)^2 and add it to both sides. This will give us:
x2-x+1/4=-5+1/4 and we have
(x-1/2)^2=-19/4 Take sqrt of both sides:
x-1/2=+or-(i(sqrt19))/2 and
x=(1+or-i(sqrt19))/2
I'll get you started on this one. I don't think you will need too much help.
(1) x^2+xy+y^2=12
(2) x+y=2
First, lets rearrange (1) and we have:
x^2+y(x+y)=12
We see from (2), x+y=2 so lets plug that in (1)
Now we have:
(1a) x^2+2y=12
but y=2-x. Now lets plug that in (1a)
x^2+2(2-x)=12 or
x^2+4-2x=12 simplifying
x^2-2x-8=0 and we can factor this:
(x-4)(x+2)=0
I think that you can take it from here.
Hope this helps. Have a nice holiday----ptaylor

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