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Let's say I have a cubic equation $(x-a)(x+b)(x-c) = 0$, and I want to represent the solutions to this equation, what is the formal/conventional way that one would arrive and state the solution to the equation?

The following two conventions are the ones I tend to use, as I've picked them up along the way from textbooks and such, but I'm not sure how formal or conventional they would be considered.


Convention 1

$$\begin{equation} \begin{split} & (x-a)(x+b)(x-c) = 0 \\ \\ \implies \ & (x = a) \lor (x = -b) \lor (x= c) \\ \\ \implies \ & x \in \{a, -b, \ c\} \\ \\ \therefore \ &(x-b)(x+b)(x-c) = 0 &\forall \ x \in \{a,- b,c\} \end{split} \end{equation}$$


Convention 2

$$\begin{equation} \begin{split} & (x-a)(x+b)(x-c) = 0 \\ \\ \implies \ & (x = a) \lor (x = -b) \lor (x= c) \\ \\ \implies \ & S = \{a, -b, \ c\} & \text{The solution set}\\ \\ \therefore \ &(x-b)(x+b)(x-c) = 0 &\forall \ x \in S \end{split} \end{equation}$$


How close to standard convention are the above conventions I've shown? Are they correct? Furthermore how would you go about arriving and representing the solutions to equations in the most formal way possible?

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  • $\begingroup$ The statement $(x-a)(x+b)(x-c)\Rightarrow x\in \{a,-b,c\}$ definitely does not imply $(x-a)(x+b)(x-c) = 0$ for all $x\in \{a,-b,c\}$. Disregarding notation, there's a structural issue in not distinguishing between these two cases. $\endgroup$ – Milo Brandt Jun 3 '16 at 0:32
  • $\begingroup$ @MiloBrandt, why does $(x-a)(x+b)(x-c) = 0 \implies x \in \{a,-b,c\} \not\implies (x-a)(x+b)(x-c) = 0 \ \ \forall x \in \{a,-b,c\}$ $\endgroup$ – Perturbative Jun 3 '16 at 19:08
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    $\begingroup$ It is true that $x=0\Rightarrow x\in \{0,1\}$, but it is not true that $x\in \{0,1\}\Rightarrow x=0$. $\endgroup$ – Milo Brandt Jun 3 '16 at 20:37
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It sounds like what you want to learn is logic (see https://math.stackexchange.com/a/1684208). After that, all such questions should disappear.

Anyway in this case:

Given any reals $x,a,b,c$, we have $(x-a)(x+b)(x-c) = 0$ if and only if $x = a$ or $x = -b$ or $x = c$.

which in set-theoretic notation can be expressed as:

Given any reals $x,a,b,c$, we have $(x-a)(x+b)(x-c) = 0$ if and only if $x \in \{a,-b,c\}$.

This is what "solving" an equation really means, because we want all and only the solutions. So we need both directions:

(A) Given any reals $x,a,b,c$, if $(x-a)(x+b)(x-c) = 0$ then $x = a$ or $x = -b$ or $x = c$.

(B) Given any reals $x,a,b,c$, if $x = a$ or $x = -b$ or $x = c$ then $(x-a)(x+b)(x-c) = 0$.

Your "convention 1" is more or less correct up to the third line, but only proves (A). If you don't prove (B) then you may 'have extra solutions'. For example, given any real $x \ge -2$, it is true that if $\sqrt{x+2} = x$ then $x+2 = x^2$, but the latter equation is equivalent to ( $x = -1$ or $x = 2$ ), but the former equation is not satisfied by $x = -1$. Why? Because each step you did ensures that you state true things, and the latter equation is indeed true if the former is true. So to find the solutions of the original equation you could check through all the solutions of the latter equation.

Your "convention 1" last line is stating (B) but without proof. Again, you don't want just (B), because it only says "these are solutions" and does not guarantee "these are all the solutions".

Your "convention 2" does not make sense and is in fact not a convention in mathematics.

The above is far more important than wanting a formal expression but here it is:

$\forall x,a,b,c \in \mathbb{R}\ ( (x-a)(x+b)(x-c) = 0 \leftrightarrow x = a \lor x = -b \lor x = c )$.

After you learn logic, you will see crystal clearly how this sentence can be formally proven as a theorem of first-order logic from the field axioms for the reals, so learning logic should be your first priority!

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