Why does Solve produce a result in terms of one variable, but not the other?

Question

I have the following quartic equation:\begin{equation} x^4 + ax^3 + bx^2 + ax + 1 = 0 \end{equation} and I'm trying to find relations for $a$ and $b$ for which we have multiple roots. To this end, I set

x1 = -(a/4)-1/4 Sqrt[8+a^2-4 b]-1/2 Sqrt[-2+a^2/2-b-(-8 a-a^3+4 a b)/(2 Sqrt[8+a^2-4 b])]
x3 = -(a/4)+1/4 Sqrt[8+a^2-4 b]-1/2 Sqrt[-2+a^2/2-b+(-8 a-a^3+4 a b)/(2 Sqrt[8+a^2-4 b])]

and I use Solve to get expressions for $b$ in terms of $a$, which satisfy $x_{1}=x_{3}$:

In[156]:= Solve[x1 == x3, b]
Out[156]= {}

It says that there is no solution. Then, I try to solve for $a$:

In[143]:= Solve[x1 == x3, a]
Out[143]= {{a -> -2 Sqrt[-2 + b]}, {a -> 2 Sqrt[-2 + b]}}

It does produce a solution. Now, I managed to get the solution for $b$ if I allow an extra condition:

In[152]:= Solve[x1 == x3, b, MaxExtraConditions -> 1] 
During evaluation of In[152]:=Solve::useq: The answer found by Solve contains equational condition(s) 
0==Indeterminate,0==Indeterminate,0==Indeterminate,0==Indeterminate}. A likely reason for this is that the 
solution set depends on branch cuts of Wolfram Language functions.
Out[152]= {{b -> ConditionalExpression[1/4 (8 + a^2), Indeterminate == 0 || Indeterminate == 0]}}

which, I believe, is because at $b=\frac{1}{4}\left(8+a^{2}\right)$ $$\frac{-8 a - a^3 + 4 a b}{2\sqrt{8 + a^2 - 4 b}} = \frac{0}{0}$$ However, the same holds for $a = \pm2\sqrt{-2+b}$ and yet, no extra conditions are needed to solve for $a$. Could someone explain to me why that is? Thanks.

Answer 1

Try the code

poly = x^4 + a x^3 + b x^2 + a x + 1;
soln = Solve[Discriminant[poly, x] == 0];
Print[soln // InputForm]
Print[poly /. soln // Factor // InputForm]
(* {{b -> -2 - 2*a}, {b -> -2 + 2*a}, {b -> (8 + a^2)/4}} *)
(* {(-1 + x)^2*(1 + 2*x + a*x + x^2),
    (1 + x)^2*(1 - 2*x + a*x + x^2), 
    (2 + a*x + 2*x^2)^2/4} *)

The reason is that setting the discriminant to zero is precisely the condition to have multiple roots.