# 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 !

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As shown by @PraphullaKoushik , your 4th line should be re-written as such. In that way, 0 is also a root. –  Mick Feb 19 '14 at 4:54

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