Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 !

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

1 Answer 1

up vote 4 down vote accepted

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.