I'm fairly new to writing proofs so I'd appreciate it if anyone could point out amy holes in this proof, and if there's any comments so I could improve my proof writing!
Question: "Let $a ≠ 0$. If there exists one $x_1$ such that $ax_1^2 + bx_1 + c = 0$ and no other. Without using the quadratic formula, show that $b^2 - 4ac = 0$."
I'm also given the proposition that if $a ≠ 0$, $ax_1^2 + bx_1 + c = 0$ and $ax_2^2 + bx_2 + c = 0$ then $x_2 = -(\frac{b}{a} + x_1)$.
If it's given that only one such $x_1$ exists for the first equation, I assume I may let $x_2 = x_1$, so then I end up with $x_1 = -(\frac{b}{2a})$. Then, by substituting into the first equation, I arrive at
$a(\frac{b^2}{4a^2}) - b(\frac{b}{2a}) + c = 0$
We have $a ≠ 0$, so simplifying we get
$\frac{-(b^2 - 4ac)}{4a} = 0$
Implying $b^2 - 4ac$ is equal to zero.
$$ax^2 +bx +c =a(x^2+\frac{b}{a}x )+c =$$
$$a(x+\frac {b}{2a})^2 +c -\frac {b^2}{4a} = a(x+\frac{b}{2a})^2 -\frac {b^2-4ac}{4a} $$ in order for the equation to have double root we need $$\frac {b^2-4ac}{4a}=0 $$