System of 4 quartic equations \begin{align*}a &=\sqrt{4+\sqrt{5+a}},\\ b &=\sqrt{4-\sqrt{5+b}},\\ c &=\sqrt{4+\sqrt{5-c}},\\ d &=\sqrt{4-\sqrt{5-d}}.\end{align*} 
Compute $abcd$.
I set up each as a quartic and got $a^4-8a^2-a+11=0$ and similar equations for $b,c,d$, but am unable to figure out how to format the roots as a product of all of them.  Could someone give me a bit of a hint?  Thank you.

Notice that $a, b, c, d$ are the roots of the equation 
$$(x^2 - 4)^2 = 5 + x$$
$$x^2 - 4 = \pm\sqrt{5 + x}$$
$$x^2 = 4 \pm \sqrt{5 + x}$$
$$x = 4 \pm \sqrt{5 + x}, -(4 \pm \sqrt{5 + x})$$
$$x = 4 \pm \sqrt{5 + x}, 4 \pm \sqrt{5 - x}$$
Now, the quartic equation corresponding to the first equation is, as you've expanded,
$$x^4 - 8x^2 - x + 11 = 0$$
By Vieta's Formulas, the product of the roots ($a, b, c, d$) is given by
$$abcd = (-1)^4\frac{11}{1} = 11$$
