# Po-Shen Loh's new way of solving quadratic equation

Quadratic equation, $$ax^2+bx+c=0$$ and its solution is quadratic equation, $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

Now setting $$a=1$$ then we have $$x^2+bx+c=0$$

$$x=\frac{-b\pm \sqrt{b^2-4c}}{2}$$ rewrite as

$$x=-\frac{b}{2}\pm \sqrt{\left(\frac{b}{2}\right)^2-c}$$

In this new video Dr. Loh claims to discover a new way of solving the quadratic equation! How? It is the same as the above formula, by using the quadratic formula, the only thing I see different, is he rewrite it in the above form!

Can someone please explain to me how this is a new way?

• Well, with all the respect, it is nothing new, for me at least.... – dmtri Dec 29 '19 at 8:00
• It's not a new way, just a 'new rebranding' of a classic way to solve quadratics. I still don't get why people find the quadratic formula hard, like show see the cubics and the quartics and then realize the true monsters. And the thing is that no matter, how many different methods you choose to pursue solving the quadratics, that will always lead you down the quad formula, directly or indirectly. – zoro Dec 29 '19 at 8:05
• This is how I learned it in high school in Germany in the 1990s. The $b$ was called $p$ and the $c$ was called $q$ and it was called "die p-q-Formel", and to be honest, in my head I still use it when solving quadratic equations. – Torsten Schoeneberg Dec 29 '19 at 8:21
• The result is equivalent to the Quadratic Formula (with $a=1$), because it has to be. What Mr. Loh is explaining is a different way to derive it. Typically, textbooks teach "completing the square" as the path. Loh is explaining that, in the special case where $a=1$, we can exploit the fact that "$b$" is the (negative of) the sum of the roots, and "$c$" is the product of the roots, to find an alternative path that may be less daunting to students. It's not a bad way to think of things; but, once you arrive at the ultimate Quadratic Formula, it doesn't really matter how you got there. – Blue Dec 29 '19 at 8:23
• Of course it is not new. Formula $x=\dfrac{-2c}{b\pm\sqrt{b^2-4ac}}$ also not new, but she work for $a=0$, when one root is $\infty$ and second root is solution of linear equation $bx+c=0$. – Dmitry Ezhov Dec 29 '19 at 9:30

For $$b^2-4ac\geq0$$ and $$a=1$$ they are the same: $$\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-\frac{b}{2}+\sqrt{\frac{b^2-4ac}{4}}}{a}=\frac{-\frac{b}{2}+\sqrt{\left(\frac{b}{2}\right)^2-ac}}{a}=-\frac{b}{2}+\sqrt{\left(\frac{b}{2}\right)^2-c}$$ and $$\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-\frac{b}{2}-\sqrt{\frac{b^2-4ac}{4}}}{a}=\frac{-\frac{b}{2}-\sqrt{\left(\frac{b}{2}\right)^2-ac}}{a}=-\frac{b}{2}-\sqrt{\left(\frac{b}{2}\right)^2-c}.$$

Let us illustrate by example. Consider the equation \begin{align} x^2-2019x-2020 = (x-r_1)(x-r_2) = x^2-(r_1+r_2)x+r_1r_2=0. \end{align} The key observation is that the roots $$r_1, r_2$$ adds up to $$2019$$, which means the average of $$r_1$$ and $$r_2$$ is $$\frac{2019}{2}$$. Hence the roots have the form $$r_\pm = \frac{2019}{2}\pm z$$ for some $$z$$.

Next, it follows \begin{align} \frac{(2019)^2}{4}-z^2 = r_1r_2 = -2020 \ \ \implies \ \ z = \sqrt{\frac{(2019)^2}{4}+2020}. \end{align} Thus, we have \begin{align} r = \frac{2019}{2}\pm \sqrt{\frac{(2019)^2}{4}+2020}. \end{align} The "new" aspect is that the derivation avoids completeing the square.

Here's another example \begin{align} x^2+212323x+24434 = 0 \end{align} Again, the roots should have the form \begin{align} r_\pm = -\frac{212323}{2}\pm z \ \ \implies \ \ \frac{(212323)^2}{4}- z^2= 24434 \end{align} which means \begin{align} z = \sqrt{\frac{(212323)^2}{4}-24434} \ \ \implies \ \ r_\pm = -\frac{212323}{2}\pm \sqrt{\frac{(212323)^2}{4}-24434}. \end{align}

Well, notice that the $$2$$ in the denominator has been absorbed into the radical. Thus instead of $$\sqrt{b^2-4c}$$, we have $$\sqrt{(b/2)^2-c}$$.

The teacher's conclusion at the end "guesswork has been replaced by a clever trick" implies that the main result of the proposed new method is the clever trick (change of unknowns $$x_1$$ and $$x_2$$) of solving the system of equations (which is the Vieta's theorem): $$\begin{cases}x_1+x_2=-b\\ x_1x_2=c\end{cases} \stackrel{x_1=\frac{-b}{2}-t\\x_2=\frac{-b}{2}+t}\Rightarrow x_1x_2=\frac{b^2}{4}-t^2=c \Rightarrow t=\pm \sqrt{\frac{b^2}{4}-c} \Rightarrow \\ x_1=\frac{-b}{2}-\sqrt{\frac{b^2}{4}-c}=\frac{-b-\sqrt{b^2-4c}}{2}\\ x_2=\frac{-b}{2}+\sqrt{\frac{b^2}{4}-c}=\frac{-b+\sqrt{b^2-4c}}{2}$$

Suppose that $$A$$ is the arithmetic mean of the roots of the quadratic $$P$$, and $$R$$ is the geometric mean [where $$P(x) = ax^2+bx+c = a(x-r_1)(x-r_2)$$], then we have this equation:

\begin{align} P(x) &= 0\\ ax^2+bx+c &= 0\\ x^2+\frac{b}{a}x+\frac{c}{a} &= 0\\ x^2+\beta x+\gamma &= 0\\ \end{align} But we can also look at it from another point of view. \begin{align} a(x-r_1)(x-r_2) &= 0\\ (x-r_1)(x-r_2) &= 0\\ x^2 -(r_1+r_2)x+r_1r_2 &= 0\\ x^2 -2\bigg(\frac{r_1+r_2}{2}\bigg)x+\sqrt{r_1r_2}^2 &= 0\\ x^2 -2Ax+R^2 &= 0\\ \end{align}

which can be solved via completing the square. \begin{align} x^2-2Ax+R^2 &= 0\\ (x^2-2Ax+A^2)+(R^2-A^2) &= 0\\ (x-A)^2+(R^2-A^2) &= 0\\ (x-A)^2-(A+R)(A-R) &= 0\\ \dots\\ \end{align}

But isn't that just Dr. Loh's method?