Finding intersection points of 2 functions. My method is incomplete.

Question

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 !

Answer 1

$x^2(x^2-1)=0\Rightarrow \text{either } x^2=0 \text{ or } x^2-1=0$

Answer 2

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)