Finding intersection points of 2 functions. My method is incomplete. These are the 2 functions :
$y = x^{4}-2x^{2}+1$
$y = 1-x^{2} $
Here's how I solved It :
$x^{4}-2x^{2}+1 = 1-x^{2}$
$x^{4}-x^{2} = 0$
$x^2(x^2-1)=0$
$x^2-1=0$
$x=\pm \sqrt{1} $
Value of $y$ when $x=1$
$y=1-x^2\\y=1-1\\y=0$
Value of $y$ when $x=(-1)$
$y=1-x^2\\y=1-(-1)^2\\y=1-1\\y=0$
So the intersection points of the 2 functions are $(1,0)$ and $(-1,0)$.
The problem
The problem is when I used a graphing calculator to find the intersection points of the above 2 functions it gave me 3 results instead of 2. They were :
$(-1,0),(0,1),(1,0)$
So what am I missing ? Why don't I get 3 intersection points ?
Best Regards !

Functions:
a.)  y  =  x^4 - 2x^2 + 1
b.)  y  =  1 - x^2
Solving:
x^4 - 2x^2 + 1  =  1 - x^2
x^4 - x^2  =  0
x^2 * ( x^2 - 1)  =  0
Now set each part equal to 0
x^2  =  0 
x  =  0
x^2 - 1  =  0
x  =  +/-1
Now plug in each value of x to get the corresponding value of y
X  =  0
y  =  1 - (0)^2
y  =  1
Intersection at (0,1)
X  =  1 
y  =  1 - (1)^2
y  =  1 - 1
y  =  0
Intersection at (1,0)
X =  -1
y  =  1 - (-1)^2
y  =  1 - 1
y  =  0
Intersection at (-1,0)
Intersections for function a and b are:
(0,1)
(-1,0)
(1,0)
