Completing the Square

The quadratic equation above, that can be solved by using comparison of areas, is not a general quadratic equation. In order to interpret the numbers as lengths, they must not be negative. It is possible to make a similar geometric construction to solve \(x^2-bx=a\), yielding \((x-b/2)^2=a+(b/2)^2\).

As long as we use a geometrical interpretation of the problem, \(a\) must not be negative and the geometrical solution will not yield negative \(x\)-values. In order to get all roots in the general case, it is better to use an

Algebraic approach

\[ax^2+bx+c=0 \Leftrightarrow ax^2+bx=-c \Leftrightarrow x^2+\frac{b}{a}=-\frac{c}{a}\]

We can assume that \(a\neq 0\), otherwise it would be a linear equation.

In order to make the left-hand-side of the equation a square, add \(\left(\frac{b}{2a}\right)^2\) to both sides of the equation

\[x^2+\frac{b}{a}+\left(\frac{b}{2a}\right)^2=\left(\frac{b}{2a}\right)^2-\frac{c}{a} \Leftrightarrow \left(x+\frac{b}{2a}\right)^2=\left(\frac{b}{2a}\right)^2-\frac{c}{a}\]

Rearranging the right-hand-side, we get

\[\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\]

If the right-hand-side is non-negative, we can solve the equation by using the square root. If the right-hand-side is negative, we can use complex numbers to find the roots.

The first person to find a method for solving quadratic functions was

Brahmagupta

who also made one of the greatest contributions to mankind ever!