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?
$\endgroup$ 76 Answers
$\begingroup$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 $$\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-\frac{b}{2}-\sqrt{\frac{b^2-4ac}{4}}}{a}=\frac{-\frac{b}{2}-\sqrt{\left(\frac{b}{2}\right)^2-ac}}{a}=-\frac{b}{2}-\sqrt{\left(\frac{b}{2}\right)^2-c}.$$
$\endgroup$ $\begingroup$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}
$\endgroup$ $\begingroup$Well, notice that the $2$ in the denominator has been absorbed into the radical. Thus instead of $\sqrt{b^2-4c}$, we have $\sqrt{(b/2)^2-c}$.
$\endgroup$ $\begingroup$Nothing about the standard quadratic formula is really intuitive. Sure, you can derive it by completing the square, but that gets complicated, and isn't really an accessible proof on the level of those learning to solve quadratics for the first time. However, Loh's method builds on an understanding of both factoring and graphing.
For example, $x^2 – 10x + 21$ factors as $(x-3)(x-7)$ and therefore has the solutions $3$ and $7$. Notice that $B=-10=-(3+7)$ and $C=21=(3)(7)$. Therefore $-B$ is the sum of the solutions, and $C$ is the product of the solutions. Both of these facts will be needed.
Now consider the graph of $y=x^2 – 10x + 21$ shown below. To use Loh's method, we'll need two other variables: $m$ and $d$. Where $(m, 0)$ is the midpoint of the zeros, $m$ is the average of the solutions. Then $d$ is the distance each zero is from the midpoint. Therefore, we can represent the solutions as $m-d$ and $m+d$ or as just $m \pm d$. If we could calculate $m$ and $d$ simply from $B$ and $C$, we'd have an easy way to solve a quadratic. And we can!
Let's get to Loh's method. We'll begin by assuming we have a quadratic of the form $Ax^2+Bx+C=0$ where $A=1$. We've already established that $-B$ is the sum of our solutions. Since the mean of the solutions is their sum divided by 2, $m=\frac{-B}{2}$. Also recall that $C$ is the product of the solutions. Therefore, $C=(m-d)(m+d)=m^2-d^2$. If we rearrange this as $d^2=m^2-C$, we have an easy way to find $d$ from $m$ and $C$. We can then write our solutions as $m \pm d$.
Here's how it works out with $y=x^2 – 10x + 21$.
$m=\frac{-B}{2}=\frac{10}{2}=5$
$d^2=m^2-C=(5)^2-21=4$
Therefore, $d=\pm \sqrt 4=\pm 2$
Since $m\pm d=5\pm 2$, the solutions are 3 and 7.
That's Loh's method! Again, it's far more accessible to students just learning how to solve quadratics.
I will admit, no one talks much about the case where $A\neq 1$. Sure, you can divide through by $A$ and not affect the roots, but it means fractions, accompanying fraction arithmetic, and the possible need to rationalize denominators--all of which is not necessary if just using the commonly memorized quadratic formula. Consider just trying to solve $3x^2 + 3x + 1 = 0$ and you'll see what I mean. You end up with fractions all the way through with denominators of 2, 3, 4, 6, and 12 at some point in the process. Plus, the connection made from here to the actual quadratic formula isn't nearly as intuitive and accessible as the rest of Loh's method. Before I posted this answer, I posted a related question and answer here that I think is better for when $A\neq 1$.
$\endgroup$ $\begingroup$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):$$\begin{cases}x_1+x_2=-b\\ x_1x_2=c\end{cases} \stackrel{x_1=\frac{-b}{2}-t\\x_2=\frac{-b}{2}+t}\Rightarrow x_1x_2=\frac{b^2}{4}-t^2=c \Rightarrow t=\pm \sqrt{\frac{b^2}{4}-c} \Rightarrow \\ x_1=\frac{-b}{2}-\sqrt{\frac{b^2}{4}-c}=\frac{-b-\sqrt{b^2-4c}}{2}\\ x_2=\frac{-b}{2}+\sqrt{\frac{b^2}{4}-c}=\frac{-b+\sqrt{b^2-4c}}{2}$$
$\endgroup$ $\begingroup$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?
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