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 equationx2+Bx+C=0is equivalent to a form(x+D)2=Efor some specific expressionsDandEin terms ofBandC. This is equivalent in the sense that the two equations have the same set of roots.
  2. The complete set of numbers that satisfyv2=Eis{E,E}, and no other numbers square toE.
  3. Therefore,(x+D)2=Eif and only ifx+Dis equal to one of±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 thatv3=8has at least one solution, because23=8. Many students initially think that there is only that one cube root. Only later do they find out that1±i3also have cubes equal to 8. So, the assumption that there are no other complex numbers that square toEother than±Eactually 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 toEare±E, it suffices to find all solutions tov2=E. This is equivalent to:v2E=0.This can be factored by finding numbers that sum to 0 and multiply toE. Selecting±Eworks. Therefore, this equation is equivalent to:(vE)(v+E)=0,and by the zero-product property,±Eare all the solutions.

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