Can Lagrange's Undetermined Multiplier Have Imaginary Points?

Main Question or Discussion Point

Hey guys, new here and this is my first post. Wondering if anyone could help me.
So I've encountered a problem on Lagrange's undetermined multiplier. Usually i have no problem with these, but this one caught me off a little.

g(x,y) = x^2 + y^2 - 4xy - 6 = 0
Find the points closest to the origin.
With this in mind:
f(x,y) = x^2 + y^2
Using the formula:
d(f + λg) = 0
Let (f + g) = F
F_x = 2x + 2λx -λ4y
F_y = 2y + 2λy -λ4x
By inspection you can see x = y, so into g(x,y) and...
-2x^2-6=0
x^2 = -3

∴ x = √-3

I haven't encountered an imaginary point yet in this type of question, and since i can't quite make sense of it in my mind, i was wondering if anyone could help me? Have I made an error somewhere, or can you have imaginary points closest to the origin?

Cheers.

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AlephZero
Homework Helper
By inspection you can see x = y
That's your error. Check the signs!

That's your error. Check the signs!
F_x = 2x + 2λx -λ4y
F_y = 2y + 2λy -λ4x
I only say by inspection to save writing out the rest of the calculation but I did try it.

2x + λ(2x - 4y) = 0
2y + λ(2y - 4x) = 0

λ= -2x/(2x - 4y) = -2y/(2y - 4x)
[-2x/(2x - 4y) = -2y/(2y - 4x)] *-1
2x/(2x - 4y) = 2y/(2y - 4x)
[2x/(2x - 4y) = 2y/(2y - 4x)]/2
x/(2x - 4y) = y/(2y - 4x)
rearrange:
x(2y - 4x) = y(2x - 4y)
2xy -4x^2 = 2xy -4y^2
-4x^2=-4y^2
y^2 = x^2
y = x.

Unless I'm making a basic floor I'm not seeing.

AlephZero