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.