Find the value of p when the equation has 3 real roots( Derived from a symmetric line question) 
Question statement: Two points M and N are on a parabola $y^2=2px$ ($p\neq0$) such that they are symmetric points about a straight line $x+y=1$. Determine the range of $p$.

What I did was I reflected that parabola over $x+y=1$, obtaining the equation: 
$$(1-x)^2=2p(1-y)$$
So now we got 2 equations.
$$(1-x)^2=2p(1-y)\tag{1}$$
$$y^2=2px \tag{2}$$
I solved for $y^2$ from the first equation then substitute it into the second equation. After simplification I end up with the following:
$$0=x^4 - 4 x^3 +(6-4p)x^2 +(8p-4-8p^3) x +1$$
According to the question I have to find the range where the equation has 4 roots, but then I realized I can just find the two points where the function has only 3 roots, and the range between them would be the answer. 
That's how far I've gone through. Thank you.
 A: Let $M(a,b),N(c,d)$ where $a\not=c$ or $b\not=d$.
These points are on the parabola $y^2=2px$, so
$$b^2=2pa\iff a=\frac{b^2}{2p}\tag1$$
$$d^2=2pc\iff c=\frac{d^2}{2p}\tag2$$
Also, the two points are symmetric points about a straight line $x+y=1$, so the midpoint of the line segemnt $MN$ is on the line $x+y=1$, and the line $MN$ is perpendicular to $y=-x+1$, so
$$\frac{a+c}{2}+\frac{b+d}{2}=1\iff a+b+c+d=2\tag3$$
$$\frac{b-d}{a-c}\times (-1)=-1\iff b-d=a-c\tag4$$
From $(1),(2),(3),(4)$, we have
$$\frac{b^2}{2p}+b+\frac{d^2}{2p}+d=2\tag5$$
$$b-d=\frac{b^2}{2p}-\frac{d^2}{2p}\tag6$$
From $(6)$, we get
$$(b-d)(2p-b-d)=0$$
Supposing that $b=d$ gives $a=c$, so $b\not=d$.
So, we have $2p-b-d=0$,
i.e.$$d=2p-b\tag7$$
From $(5),(7)$, we get
$$\frac{b^2}{2p}+b+\frac{(2p-b)^2}{2p}+2p-b=2,$$
i.e.
$$b^2-2pb+4p^2-2p=0\tag8$$
What we are seeking is the condition on $p$ that $(8)$, a quadratic equation on $b$, has at least one real solution and $b=p$ is not a solution of $(8)$ (because of $b\not=d$) .
So, the answer is
$$(-2p)^2-4\cdot 1\cdot (4p^2-2p)\ge 0\quad\text{and}\quad p^2-2p^2+4p^2-2p\not=0,$$
i.e.
$$\color{red}{0\lt p\lt\frac 23}$$
