1. Given quadratic in $x$

We are given

where $a,b,c$ are side lengths of a triangle.

We need the set of all possible real values of $x$ for which this equation can hold for some triangle $(a,b,c)$.

2. Condition for real roots in $x$

For some real $x$ to be possible, the quadratic equation must be satisfied by positive $a,b,c$ forming a triangle.

Treating it as a quadratic in $x$, its discriminant must be nonnegative:

So,

Expand:

Cancelling $a^2b^2$ and $b^2c^2$,

This is

Hence necessarily

Thus the discriminant is zero, so the quadratic has a repeated root.

3. Find the root $x$

When $b^2=ac$, the root is

Using $b^2=ac$,

Therefore,

Also,

Let

Then

So the three sides are proportional to

4. Apply triangle inequalities

Since $a,b,c$ are sides of a triangle, with proportional sides $1,x,x^2$, we need:

• $1+x>x^2$
• $x+x^2>1$
• $1+x^2>x$

Now analyze each:

1. $1+x>x^2 \Rightarrow x^2-x-1<0.$ The roots are $\frac{1\pm\sqrt5}{2}.$ Since $x>0$ (side lengths are positive), this gives $0<x<\frac{1+\sqrt5}{2}.$

2. $x+x^2>1 \Rightarrow x^2+x-1>0.$ The roots are $\frac{-1\pm\sqrt5}{2}.$ Since $x>0$, this gives $x>\frac{\sqrt5-1}{2}.$

3. $1+x^2>x$ This is always true because $x^2-x+1=\left(x-\frac12\right)^2+\frac34>0.$

Combining,

Hence

5. Compute $12(\alpha^2+\beta^2)$

Let

Now,

Therefore,

Hence,

6. Final answer

The derived answer matches the stored correct answer.