Villainous Vile Variegated Vieta Values

Well, it's just $a^2 + b^2 + c^2 + d^2 = (a+b+c+d)^2 - 2(ab + ac + ad + bc + bd + cd) = 12^2-2\cdot46= \boxed{52} .$

But wait, we're not done. We need to show that a quadruplet $(a,b,c,d)$ that satisfy the constraints stated too.

Here's the quadruplet:

$(a,b,c,d) = \left( 3-\sqrt7, 2,4,3 + \sqrt7 \right)$

How do I get it? From the previous problem , we gather that $9 <abcd< 25$ . A simple graphing shows that $a,b,c,d$ must all be in the open interval $(0,6)$ .

So all the possible values of $(b,c)$ must be $(1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), (4,5).$

So at least one of these pairs must satisfy $f(b) = f(c) = 0$ for $f(x) := x^4 - 12x^3 + 46x^2 - 60x + u$ with some constant $u$ .

Trial and error show that only one of these pairs satisfy the 3 given equations. Namely, $(b,c) = (2,4)$ .

And we're done!