Quadratic Method: Completing the Square

Some have expressed concerns that the method I shared is exactly the same as the traditional method of Completing the Square. The algebraic manipulations of expressions are very similar. However, there is a difference in the logical steps, which is evidenced by differing requirements on initial assumptions. This is a logical distillation of Completing the Square as commonly learned:

1. By adding and subtracting terms from both sides, and combining terms, the quadratic equation $x^2 + Bx + C = 0$ is equivalent to a form $(x+D)^2 = E$ for some specific expressions $D$ and $E$ in terms of $B$ and $C$. This is equivalent in the sense that the two equations have the same set of roots.
2. The complete set of numbers that satisfy $v^2 = E$ is $\{\sqrt{E}, -\sqrt{E}\}$, and no other numbers square to $E$.
3. Therefore, $(x+D)^2 = E$ if and only if $x+D$ is equal to one of $\pm \sqrt{E}$. We then get the complete set of solutions to the original quadratic.

When compared to the method I shared, Step 2 above assumes that the complete set of square roots of every number is known, whereas Step 4 in the method I shared just needs one working square root.

Depending on the context, existence assumptions can be much easier to settle than complete-set assumptions. As an analogous example: it is easy to observe that $v^3 = 8$ has at least one solution, because $2^3 = 8$. Many students initially think that there is only that one cube root. Only later do they find out that $-1 \pm i\sqrt{3}$ also have cubes equal to 8. So, the assumption that there are no other complex numbers that square to $E$ other than $\pm \sqrt{E}$ actually requires further justification.

If one wishes to make Completing the Square more complete (at least to match the method I shared), then some of the simplest ways return to the notions of factoring and the zero-product property. For example, to show that the only numbers that square to $E$ are $\pm \sqrt{E}$, it suffices to find all solutions to $v^2 = E$. This is equivalent to: \[ v^2 - E = 0. \] This can be factored by finding numbers that sum to 0 and multiply to $-E$. Selecting $\pm \sqrt{E}$ works. Therefore, this equation is equivalent to: \[ (v-\sqrt{E})(v+\sqrt{E}) = 0, \] and by the zero-product property, $\pm \sqrt{E}$ are all the solutions.