Sketching Graph: Solving (x+y+3)^2 + (x-y-3)^2 = 0

[elitewarr]

Homework Statement

Sketch (x+y+3)^2 + (x-y-3)^2 = 0

Homework Equations

The Attempt at a Solution

How?? I have totally no idea.

 

[Mentallic]

Expand, simplify and complete the square. Now think about what the solutions are for a^2+b^2=0 for all real numbers a and b, and see if you can apply this idea to your problem.

 

[LCKurtz]

Expand, simplify and complete the square. Now think about what the solutions are for a^2+b^2=0 for all real numbers a and b, and see if you can apply this idea to your problem.

For that matter, think about a^2+b^2=0 before you expand the terms.

 

[Mentallic]

For that matter, think about a^2+b^2=0 before you expand the terms.

Haha that's a much more effective method

 

[elitewarr]

I tried using this method.
Let a be x, b be y+3
(a+b)^2 + (a-b)^2 = 0
2a^2 + 2b^2 = 0
(x)^2 + (y+3)^2 = 0
But if I bring x^2 over,
(y+3)^2 = -x^2
I doubt there is such curve and if there is, it is undefined isn't it?

thanks for the reply.

 

[LCKurtz]

You are right, it isn't much of a graph. The left side of your last equation is non-negative and the right side is non-positive. The only way they can be equal is if both sides are zero. Can you find any (x,y) that does that? If so, whatever you find constitutes your graph.

 

[elitewarr]

Ok. So, (0,-3) is the only value? Is the graph just a dot?

Thanks.

 

[LCKurtz]

Ok. So, (0,-3) is the only value? Is the graph just a dot?

Thanks.

Yes. Also note that in your original equation:

(x+y+3)² + (x-y-3)² = 0

both terms must be zero, so the intersection of the two lines:

x+y+3=0 and x - y - 3 = 0

is the only point which satisfies the equation. This agrees with your answer.