Everything is possible | A tricky math question

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@higher_mathematics

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[snigdhasingh]

This problem is complex only in the sense that it uses the concept of "complex" numbers.

[Segalmed]

y = 1-x;
insert in 2nd equation;
x(1-x)=1 => x² - x + 1 = 0 ;
standard formula yields x1, 2 = 0.5 +/- 0.5 * SQRT(-3) = 0.5 +/- 0.5*i*SQRT(3)
=> y1, 2 = 0, 5 -/+ 0, 5*I*SQRT(3)

[therongjr]

Everything is possible if you don't limit yourself to reality.

[UserSams-vemj]

There's a much faster way to get x² -x + 1 =0
The sum S and the product P of x and y are directly given;
S = 1 and P = 1.
You can apply the formula
x² - Sx + P = 0
x² - x + 1 = 0

[warrengibson]

This dude’s singsong delivery has me LMAO

[matthewalan]

Wow, what a huge number of comments from people who have an extremely limited background in mathematics. Just because we have used the term "imaginary" does not mean that such numbers are any less "real" than any other kind of number. I recall grade eight students trying to argue that you could not have less than zero and that negative numbers were not real.

Many years ago, as an exercise, I used the general solution for a cubic equation to solve a specific cubic equation. What was fascinating to me is that the three roots were all real but the details of the general solution involved complex numbers. In the end the imaginary parts all canceled out. If, during the solution, you were to conclude that there were no real solutions because square roots of negative numbers were appearing, you would be completely mistaken.

[wes]

The simple approach. Substitute x=1/2+z and y=1/2–z into the second equation: (1/2+z)(1/2–z)=1 or z^2=–3/4. Thus, z=±i√3/2, x=(1±i√3)/2, and y=(1∓i√3)/2

[GiovanniMascellaro]

The square of (-1) is (1), because ((-1) \times (-1) = 1). However, when we talk about (i), the imaginary unit, the situation is different. By definition, (i) is such that (i^2 = -1). By definition i^2 = −1

[octobrot]

I just got here, gotta say I love the handwriting

[RikiFaridoke]

You can using complex analysis method sir.

[Danielle_The_Endearing]

nice to know I am smart enough for that :3

[quintopia]

Not that tricky. Since the curves don't meet, the solution is complex. Since it simplifies to a quadratic, they must be conjugate of each other. Thus, the first equation implies the real part is 1/2. The second constraint implies they are unit length. That pins it down to exp(±pi/3).

[BKNeifert]

What's interesting, is it spits out imaginary sine for an equilateral triangle with side length one, and then the cosine. Don't know why. But it does.

[AMINE-ddqy]

i remember when the professor showed us this equation as introduction to complex numbers

[istvangeai]

I just want to ask what kind of pen you use. Because your writing style looks very nice.

[matthewlloyd]

Why can't you just graph y = 1 - x and y = 1 /x and find the intereception points between the two curves on the graph. It might not be all the solutions but it will be some of them.

[anatolykhmelnitsky]

Let's substitute x for x₁ and y for x₂. Now we have x₁ + x₂ = 1 and x₁ * x₂ = 1. For x² + bx + c = 0 Vieta's formula is x₁ + x₂ = -b and x₁ * x₂ = c. All we need now is to solve quadratic equation where b is -1 and c is 1. It's just x² - x + 1 = 0. Quadratic formula will do the rest of the job. That's all folks!

[remy]

i achieved the same solution setting xy=x+y, it just took more steps! :)

[TransportGeekery]

Love it! Especially the additional steps to work out 1/4 + 3/4 😂

[rainerzufall]

Another Vieta problem => Immediately clear, that x, y are the two z solutions of z^2 - bz + c [-b = 1, c = 1] = z^2 - z + 1 = 0
=> z_1/2 = b/2 +/- sqrt(b^2/4 - c) = 1/2 +/- sqrt(1/4 - 1) = 1/2 +/- 1/2 sqrt(3) i => x = z1, y = z2 OR x = z2, y = z1.
Use the symmetry for God's sake! x and y are totally exchangeable!