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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 $\frac{c}{a}$.

Square and rectangle.

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?

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    $\begingroup$ "... and it would make sense if it was geometrically possible. " Why? Algebra doesn't need a geometric interpretation to be valid. $\endgroup$
    – jjagmath
    Commented Oct 27, 2023 at 19:52
  • $\begingroup$ @jjagmath But isn't completing the square a geometric step? $\endgroup$
    – Jay Gatsby
    Commented Oct 27, 2023 at 19:58
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    $\begingroup$ Why would adding a number to both sides of an equation be a geometric step? $\endgroup$
    – jjagmath
    Commented Oct 27, 2023 at 20:03
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    $\begingroup$ 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. $\endgroup$
    – Paul
    Commented Oct 27, 2023 at 20:09
  • $\begingroup$ 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. $\endgroup$
    – Chris Duerschner
    Commented Oct 27, 2023 at 20:17

1 Answer 1

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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!

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