The Greek mathematician Heron (100? B.C.) is credited by some as the first (historically known) person who studied quadratic equations. He considered questions like: “There is a square. The sum of its area and circumference is 896. What is its length (or width)?” But some simple quadratic equations were already known to Mesopotamians 4,000 years ago. Along with linear equations, quadratic equations are among the oldest subjects of mathematics that have been studied. So, it is amazing and even somewhat amusing that there is still something new to find out about the solution of quadratic equations. I recently came across an article on the arXiv titled “A Simple Proof of the Quadratic Formula” by Po-Shen Loh at Carnegie Mellon. His proof also provides an alternative way to solve quadratic equations without dealing with factoring quadratic polynomials or having to remember the quadratic formula. I believe students will find it much easier than the conventional methods (factoring, completing the square, or the quadratic formula). First, here is the proof. Any quadratic equation can be written as a monic equation $$x^2+bc+c=0$$ Suppose that its solutions are $\alpha$ and $\beta$. Then $$x^2+bx+c=(x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha\beta=0$$ By comparing the coefficients, we have $\alpha+\beta=-b$ and $\alpha\beta=c$. Thus, $\alpha$ and $\beta$ can be written as $u-\frac{b}{2}$ and $-u-\frac{b}{2}$, respectively, where $u$ is an unknown quantity that is to be determined. Since $\alpha\beta=c$, we have $u^2-\frac{b^2}{4}=-c$. Solving this equation for $u$, we obtain $u=\pm\frac{\sqrt{b^2-4ac}}{2}$ and hence $\alpha$ and $\beta$ are given by $-\frac{b}{2}\pm\frac{\sqrt{b^2-4ac}}{2}$.
Example. The above question by Heron is equivalent to solving the quadratic equation $x^2+4x-896=0$. $\frac{b}{2}=2$ so we have $u^2-4=896$ i.e. $u^2=900$. Therefore, $u=\pm 30$ and the two solutions are $x=-2\pm 30$. Since $x>0$, the answer is $x=28$.
The method still applies to quadratic equations that have complex solutions. For example,
Example. Consider $x^2+x+1=0$. $\frac{b}{2}=\frac{1}{2}$, so we have the equation $u^2-\frac{1}{4}=-1$ i.e. $u^2=-\frac{3}{4}$. Therefore, $u=\pm\frac{\sqrt{3}i}{2}$ and the solutions are $x=-\frac{1}{2}\pm\frac{\sqrt{3}i}{2}$.