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?
Best Answer
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!