Disgustinlgy difficult math problem!!!

[chompchompchomsky]

I challenge all of you to find two numbers that have a sum of 2 and a product of 2. I know it is possible, I know I can't figure it out. :rant:

[Nefret]

The discriminant is imaginary.

[Fuggles]

x=1+i
y=1-i

[chompchompchomsky]

The discriminant is imaginary.

I am more than aware that the discriminant is imaginary, But I was assigned this problem, and there are two numbers (even if they do include i) that satisfy a+b=2 and ab=2.

[Fuggles]

I am more than aware that the discriminant is imaginary, But I was assigned this problem, and there are two numbers (even if they do include i) that satisfy a+b=2 and ab=2.
You mean like the answer I posted?
x=1+i
y=1-i

[chompchompchomsky]

x=1+i
y=1-i

THANKYOUTHANKYOUTHANKYOU!!!!!!!!!!!!!!!! That's it!

[Fuggles]

THANKYOUTHANKYOUTHANKYOU!!!!!!!!!!!!!!!! That's it!
I win. :cool:

[chompchompchomsky]

(I was writing a reply while you posted that)

[Mathematics]

x=1+i
y=1-i

Yes.

Also:
x=1-i
y=1+i

[Nefret]

I am more than aware that the discriminant is imaginary, But I was assigned this problem, and there are two numbers (even if they do include i) that satisfy a+b=2 and ab=2.

Yeah, in which case, the answer is nowhere near difficult in the least.

[Fuggles]

Yes.

Also:
x=1-i
y=1+i
True.

Also:
a=1+i
b=1-i

[Fuggles]

Yeah, in which case, the answer is nowhere near difficult in the least.
Actually, it is "disgustinlgy difficult." :p

[Dog]

Eleventeen and flob. :cool:

[Mathematics]

True.

Also:
a=1+i
b=1-i

This is the same as your first solution.

There are two solutions to equations x+y=2 and x*y=2.

[Fuggles]

This is the same as your first solution.

There are two solutions to equations x+y=2 and x*y=2.
Your post where you simply flipped the x and y was the exact same as my original post; since no variable was set in the original problem, x and y merely were used because they were easier to read than 1+i,1-i. Since the question asked for two numbers, not for x and y, there is only one solution.

[Mathematics]

Your post where you simply flipped the x and y was the exact same as my original post; since no variable was set in the original problem, x and y merely were used because they were easier to read than 1+i,1-i. Since the question asked for two numbers, not for x and y, there is only one solution.

In this case its pedantic, but strictly x=1+i and x=1-i. Take the case of x^2=4, x=2 and x=-2.

[Agent 008]

This problem is easy as fuck.

[Dog]

This problem is easy as fuck.

It's sort of a trick question though. I mean, if you know that you're supposed to be using imaginary numbers, then it's pretty easy, but who regularly thinks in terms of imaginary numbers?

[Agent 008]

It's sort of a trick question though. I mean, if you know that you're supposed to be using imaginary numbers, then it's pretty easy, but who regularly thinks in terms of imaginary numbers?

Meh, you just write a simple system of two equations, substitute the variables, and see instantly that there is no solution in the real numbers.

[Mantikore]

Not having done a scrap of math for about 5 months, i decided to do the working as practise. i dont know how to do that fancy math formatting in UBB code, so bear with me. (the FAQ link is fucked for me http://bbs.zoklet.net/misc.php?do=bbcode )

ok, so lets say our two numbers are x and y. our two equations will therefore be

x+y=2............(i)
xy=2.........(ii)

there are just 2 equations with 2 variables. no need to do anything complicated. just solve simultaneously by substitution

from (i),
x+y=2
x=2-y.......(iii)

substitute this into (ii)
(2-y)y=2...................expanding, we get
2y-y^2=2.................rearranging into something nice looking, we get
-y^2+2y-2=0...........use the quadratic formula (-b+sqrt(b^2-4*a*c))/2a
y=(-2+sqrt(4-4*-1*-2))/2*-1
=(-2+sqrt(-4))/-2
=(-2+sqrt(4)*sqrt(-1))/-2
=(-2+2*sqrt(-1))/-2..............we can simplify by factorising the top (with -2 as the highest common factor), then cancelling. sqrt(-1) becomes imaginary number, i. the although there should be a -i, it doesnt matter, since it would just swap something that was alread plus or minus.
=1+i.....................(iv)

so we have y, now we find x. go back to equation (iii).
x=2-y....................sub in
y=1+i

x=2-1+i
x=1+i or 1-i

so yeah, it didnt really matter if you substituted back into the original equation. the solution will still be 1+i and 1-i

[Dog]

Meh, you just write a simple system of two equations, substitute the variables, and see instantly that there is no solution in the real numbers.

ASSES!