Posted Dec 15, 2019

# 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$\left(x+D{\right)}^{2}=E$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$\left\{\sqrt{E},-\sqrt{E}\right\}$, and no other numbers square to$E$.
3. Therefore,$\left(x+D{\right)}^{2}=E$if and only if$x+D$is equal to one of$±\sqrt{E}$. 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$-1±i\sqrt{3}$also have cubes equal to 8. So, the assumption that there are no other complex numbers that square to$E$other than$±\sqrt{E}$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$±\sqrt{E}$, 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$±\sqrt{E}$works. Therefore, this equation is equivalent to:$\left(v-\sqrt{E}\right)\left(v+\sqrt{E}\right)=0,$and by the zero-product property,$±\sqrt{E}$are all the solutions.