# Completing the square step for quadratic equation

I’m confused about the third step In the proof of the quadratic equation with the form $$ax^2 + bx + c = 0.$$ In the third step it is added $$\bigl(\frac{b}{2a}\bigr)^2$$ to both sides to complete the square, and it would make sense if it was geometrically possible.

Screenshot of derivation of quadratic formula.

I mean, if it was $$ax^2 + bx - c = 0$$, that is, $$ax^2 + bx = c$$, we can think of it as a square with both sides $$x$$, and a (maybe) rectangle with side $$\frac{b}{a}$$ and the other $$x$$, and they have an area of $$\frac{c}{a}$$.

so we can complete that square, and if we continue the rest of following algebra steps, we get a quite similar result that is, $$x = \frac{-b±\sqrt{b^2+4ac}}{2a}$$ So... Did they just apply the concept of completing the square to find a square with a negative area? is a negative area possible?

• "... and it would make sense if it was geometrically possible. " Why? Algebra doesn't need a geometric interpretation to be valid. Commented Oct 27, 2023 at 19:52
• @jjagmath But isn't completing the square a geometric step? Commented Oct 27, 2023 at 19:58
• Why would adding a number to both sides of an equation be a geometric step? Commented Oct 27, 2023 at 20:03
• Arabic mathematicians (Al-Kwarizmi?) discounted the solution which was not geometric, in the way you say. The method had its history in literally completing a square figure, but the algebra does not require it.
– Paul
Commented Oct 27, 2023 at 20:09
• Have you learned about imaginary numbers? If so, think about how a square with sides of length 'i' has an area of -1. Mathematicians would not have learned to use imaginary numbers if they had only considered problems with physical geometric interpretations. Commented Oct 27, 2023 at 20:17

A square in the context of algebra is an expression of the form $$u^2$$ for some $$u$$. To complete the square is a technique to add a (possibly negative, i.e. subtract) quantity to make the expression a square. Since we know that $$\biggl( x + \frac{b}{2a} \biggr)^2 = x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}$$ is a square by definition with $$u = x + \smash[b]{\dfrac{b}{2a}}$$, the natural step in this derivation is to add $$\smash[b]{\biggl( \dfrac{b}{2a} \biggr)^2}$$ to both sides of the equation to complete it!