What are the solutions for the intersection of y=abs(x) and y=(x^2)-6?

[pb23me]

Homework Statement

The graphs y=abs(x) and y=(x^2)-6 intersect at x=3 and x= -3
What is confusing me is when I set them equal to each other and solve (x^2)-x-6=0 and (x^2)+x-6=0 I get -3,+3,-2,+2
What is the deal with the negative 2 and pos 2?

Homework Equations

The Attempt at a Solution

 

[wukunlin]

abs(x) = x only if x is positive, so when you solve x^2-x-6=0, you need to disregard negative answers, and vice versa

 

[rock.freak667]

Homework Statement

The graphs y=abs(x) and y=(x^2)-6 intersect at x=3 and x= -3
What is confusing me is when I set them equal to each other and solve (x^2)-x-6=0 and (x^2)+x-6=0 I get -3,+3,-2,+2
What is the deal with the negative 2 and pos 2?

y= |x| is the graph of y=x with the bottom half reflected in the x-axis (so any value of x gives a positive value of y). It looks like a v basically.

Since both y=|x| and y=x²-6 are symmetrical about the y-axis, you will get 'mirrored' answers, so if you get x=2, you will get x=-2 on the next side of the graphs.

Draw them out and you will see the symmetry I am talking about.

 

[pb23me]

Thanks, I did draw them out and the graphs intersect at x=3 and x=-3
They do not intersect at x=2 and x=-2 even though 2 and -2 solve the equation when I set the two functions equal to each other.

 

[rock.freak667]

Thanks, I did draw them out and the graphs intersect at x=3 and x=-3
They do not intersect at x=2 and x=-2 even though 2 and -2 solve the equation when I set the two functions equal to each other.

Yeah drawing you would see +3 and -3 but it's like wukunlin said.

abs(x) = x only if x is positive, so when you solve x^2-x-6=0, you need to disregard negative answers, and vice versa

When you solved x²-x-6=0 you had the constraint of x>0, you would ignore the x=-2.

 

[pb23me]

ohhhhhhhh ok I get it now. Thanks guys