Determine all possible real values of h so that the equation has 4 real roots

Question

The equation is as following: $$x(x+1)(x+h)(x+1+h)=h^2$$ I did the multiplication and got the expression: $$x^4+2(h+1)x^3+(h+1)^2x^2+hx+h=0$$ From now on I do not know how to proceed.

Answer 1

At $h = 0$ you have double roots at $x=-1$ and $x = 0$

$x(x+1)(x+h)(x+1+h)=h^2\\ u = x+ \frac h2$

$(u-\frac h2)(u+\frac h2)(u+1-\frac h2)(u+1+\frac h2)=h^2\\ (u^2-\frac {h^2}4)((u+1)^2-\frac {h^2}{4})=h^2$

$(u^2-\frac {h^2}4)((u+1)^2-\frac {h^2}{4})$ has its maximum at $u = -\frac 12$

If the max $(u^2-\frac {h^2}4)((u+1)^2-\frac {h^2}{4})< h^2$ then there are complex roots for that value of $h.$

$(\frac 14 - \frac {h^2}{4})^2 > h^2\\ |1 - h^2|> 4|h|$

$(h^2 + 4h - 1)(h^2 - 4h - 1) < 0$

$h = \pm 2\pm\sqrt 5)$

$h\in(-\sqrt 5-2,2-\sqrt 5)\cup 0\cup(\sqrt 5-2,\sqrt 5+2)$ there are only 2 real roots.

Outside of these intervals there are 4