Finding the Critical Point(negative square root)

[Luscinia]

Homework Statement

Find the critical points for the function g(x, y, z) = x³+xy²+x²+y²+3z².

Homework Equations

The Attempt at a Solution

I've come up with the following 3 equations (derivatives set so that they are equal to 0)
(1) 3x²+y²+2x=0
(2) 2xy+2y=0
(3) 6z=0

From (3),

z=0​

From (2),

2y(x)+2y(1)=0
x+1=0
x=-1​

From (1) using what I have obtained from (2),

3(-1)²+y²+2(-1)=0
3-2+y²=0
1=-y²
y²=-1​

I do not know how I am supposed to isolate y in this case since square roots are supposed to be positive. Do I need to get i involved? If so, how?

 

[pasmith]

From (2) you have two possibilities: $0 = 2xy + 2y = 2y(x + 1)$ so either $y = 0$ or $x = -1$ or both.

You must also satisfy (1). You've shown that if $x = -1$ then (1) requires that $y^2 = -1$, so that doesn't give you a critical point. There remains the $y = 0$ possibility.

 

[Ray Vickson]

Homework Statement

Find the critical points for the function g(x, y, z) = x³+xy²+x²+y²+3z².

Homework Equations

The Attempt at a Solution

I've come up with the following 3 equations (derivatives set so that they are equal to 0)
(1) 3x²+y²+2x=0
(2) 2xy+2y=0
(3) 6z=0

From (3),

z=0​

From (2),

2y(x)+2y(1)=0
x+1=0
x=-1​

From (1) using what I have obtained from (2),

3(-1)²+y²+2(-1)=0
3-2+y²=0
1=-y²
y²=-1​

I do not know how I am supposed to isolate y in this case since square roots are supposed to be positive. Do I need to get i involved? If so, how?

$(2) \Longrightarrow 2y(x+1) = 0 \Longrightarrow y = 0 \; \text{or }\; x+1 = 0$.