A great math question for math lovers

[@JSSTyger]

I've learned that any time I see a problem set up like this (x+y = a and xy = b), the answer is always of the form x = a/2±z. So I already know x = 1±z. And if I choose x = 1+z, then y = 1-z. That allows a fast solution of xy = (1+z)(1-z) = 4 and we get z = isqrt(3). So your solution sets will be (1+isqrt(3), 1-isqrt(3)) and (1-isqrt(3), 1+isqrt(3)).

[@MarlowePeter-p5f]

Coty Forks

[@DriftinVr]

Y=2-x X(2-x)=4 -x^2+2x=4 X^2-2x+4 2+-sqrt(4-16)/2 2+-isqrt(12)/2 2+-2isqrt(3)/2 1+-isqrt(3) X=1+isqrt(3),1-isqrt(3) 1+isqrt(3)+y=2 Y= 1-isqrt(3) 1-isqrt(3)+y=2 Y=1+isqrt(3) Answers is (1+isqrt(3),1-isqrt(3)) and (1-isqrt(3),1+isqrt(3))

[@Godzilla_28]

Fun equation :D x+y = 2 xy = 4 x = 2-y (2-y)y = 4 2y-y² = 4 -y²+2y-4 = 0 y²-2y+4 = 0 D = (-2)²-4×1×4 = 4-16 = -12 y1;2 = (2±√ -12)/2 √ D = √ -|12| = i×√ 12 y1 = (2+i√ 12)/2 y2 = (2-i√ 12)/2 y = 1±i√ 3 x1+(1+i√ 3) = 2 x1 = 2-1-i√ 3 x1 = 1-i√ 3 x2+(1-i√ 3) = 2 x2 = 2-1+i√ 3 x2 = 1+i√ 3 x = 1±i√ 3

[@peterjansen4826]

This time an easy one which I could solve, all that you need is the ability to solve a second grade equation (root formula), basic knowledge of complex numbers (i² = -1) and the knowledge that sqrt(a*b) = sqrt(a)*sqrt(b). So basically 5th/6th grade high school level if the mathematics teachers do their job, if not then don't blame the students, blame the teachers or educational system. 😉 I regret to say that I am one of those pupils, I had to learn complex numbers at the university. Shame for the educational system in my country.

[@bimbitashaikh8331]

cant we just use whole square formula of (x+y)^2 - 4xy = (x+y)^2 and then plus minus. thats like. goddamn easy mates

[@meurdesoifphilippe5405]

Another method : (x+y)^2 =4 => x^2+y^2+2xy=4 => x^2+y^2-2xy = 4-4xy =4-4*4=-12 => (x-y)^2 =-12 => x-y=+/- i sqrt(12) then you know x+y and x-y so you are done since x = 1/2( (x+y) + (x-y)) and so on

[@Segalmed]

x + y = 2 => x = 2- y. Insert in equation 2: (2-y)y = 4 => y² - 2y + 4 = 0 Standard formula for quadratic equations yields 2 complex solutions: y1,2 = 1 +/- i*SQRT(3) The solution for x are therefore x1,2 = 1-/+ i*SQRT(3), i.e. the conjugated complex numbers.

[@jcinthemountains]

That was easy to follow, tho I'd have lost my way on my own long before you finished. Thanks for helping dust off some brain cells that haven't thought through this level of thinking for a long time. I'll have to confess I think it'll be a while before I'm ready to wake up to differential equations, but I'm game to try.

[@GeometriCuber]

maths is fun

[@RealQinnMalloryu4]

(1)+(1) =2 (y ➖ 1x+1). 2^2 (xy ➖ 2xy+2).

[@jpptubie]

Oh dear. 10 mins to solve simple equations that can easily be solved in a minute.