No need for the quadratic formula, just note that $p^2-2p+1=(p-1)(p-1)=(p-1)^2$.

I guess, I am just not seeing it. To me $p^2-2p+1=p(p-2)+1=0$

That's true too, but this trinomial can also be factored into $(p-1)(p-1)$. This is called the product sum method. If you don't see this, feel free to use the quadratic formula, which gives the same answers, but is a bit more difficult in this case.

Right, simple method...$(p-1)(p-1)=0 \implies (p-1)^2=0 \implies (x^2-1)^2=0\\$ Then $(x^2-1)=0 \implies x^2=1 \implies \sqrt{x^2}=\sqrt{1} \implies x=\pm \sqrt{1} = \pm 1$ Then repeat for the other term. Giving 4 roots of $\pm 1$

Your method works fine, but I don't understand what you say after the math. $x^4-2x^2+1$ only has 2 roots, which are $x=\pm1$.

It must have 4 roots though if it is of degree 4, no? The solution to the problem states: $x=\pm 1, \pm 1$ Implying 4 real roots. So if we have from above $(x^2-1)^2=(x^2-1)(x^2-1)$ and thus we have two terms in which to solve the roots for. No? $(x^2-1)=0 \implies x=\pm 1$ and then again $(x^2-1)=0 \implies x=\pm 1$

Good idea to continue here.

So does that make sense?

Let me explain to you. Geometrically, when I plot $f(x)=x^4-2x^2+1$, the graph intersects (actually, it touches) the $x$-axis 2 times. Therefore, we say that it has 2 roots. Those roots are $x=1$ and $x=-1$.

I get your logic. Although, I am not sure if this makes a difference, but this problem is for a differential equations problem in which I need to find the general solution to a 4th order linear homogeneous DE.

To be honest, I don't know. Maybe you need to quote the whole problem in a new question?

Are you using a book or ..?

Gotcha. Yeah, maybe I will do that. But thank you!!!

No problem! ;-) Don't forget to click the checkbox if my answer helped you .. at least a bit! :)

Oh absolutely!

Have a nice day :)

You too.