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

Question

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?

Answer 1

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 $$\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-\frac{b}{2}-\sqrt{\frac{b^2-4ac}{4}}}{a}=\frac{-\frac{b}{2}-\sqrt{\left(\frac{b}{2}\right)^2-ac}}{a}=-\frac{b}{2}-\sqrt{\left(\frac{b}{2}\right)^2-c}.$$

Answer 2

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}

Answer 3

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

Answer 4

The teacher's conclusion at the end "guesswork has been replaced by a clever trick" implies that the main result of the proposed new method is the clever trick (change of unknowns $x_1$ and $x_2$) of solving the system of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

Answer 5

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?