Use quantifiers and logical connectives to express the fact that a quadratic polynomial with real number coefficients has at most two real roots.

Question

"Use quantifiers and logical connectives to express the fact that a quadratic polynomial with real number coefficients has at most two real roots."

My solution:
1. ∀a∀b∀c [
2. ∀x [(ax2 + bx + c = 0)⟶∃w∃z(aw2 + bw + c = 0 ⋀ az2 + bz + c = 0 ⋀ (x = w V x = z))]
3. V
4. ㄱ∃x [ax2 + bx + c = 0]]

My thought process:
1. For all triplet real nums a, b and c,
either:
2. For all real numbers x, if ax2 + bx + c = 0, then there are at least one and utmost two real nums w and z that x is equal to. (hence didn't give w != z)
3. or:
4. there is no real number that satisfies the condition ax2 + bx + c = 0. (handling the zero roots possible condition)
Edit: The domain of discourse of all variables is the set of real numbers.

Please correct me if the answer is wrong. If by any chance it is right, maybe I can shorten it in some way or there is a better solution?

Answer 1

For all real $a,b,c,x,y,z$, if
$$(ax^2 + bx + c)^2 + (ay^2 + by + c)^2 + (az^2 + bz + c)^2 = 0,$$ then $x = y$ or $y = z$ or $z = x$.

Answer 2

You have missed several points in your solution. First of all you did not specify that the coefficients are real. You also assumed that every quadratic equation has at least one real root. Why there is no real number that satisfies the condition ax2 + bx + c = 0 ? Over all your solution looks correct.

Answer 3

$$\forall x \ \forall y\ \forall z\ \forall a\ \forall b\ \forall c\ \bigl({a ≠ 0} \wedge {(ax^2 + bx + c = ay^2 + by + c = az^2 + bz + c = 0)\bigr) \rightarrow \bigl((x = y) \vee (y = z) \vee (z = x)\bigr) }$$