Loh’s Method: A Different Way to Solve Quadratic Equations

I will talk about a new method of solving quadratic equations, which a maths professor recently announced at Carnegie Mellon University called Po-Shen Loh.

You might disagree that this is precisely a new method because it’s more like a new way of looking at the old way of solving quadratic equations. There’s nothing new to be discovered as far as quadratics go.

Anyway, it’s very interesting, and I want to describe it to you. So let’s go and discover the maths.

The quadratic equation has the form y=ax² +bx +c where a, b, and c are real numbers. The simplest case is when a=1 and b and c are both 0, which leaves us with just y=x². When we plot out any quadratic equation as a graph, it produces a curve known as a parabola with the turning point at the bottom if a is positive and at the top if a is negative.

To solve an equation means looking for values of x that make y=0. These correspond to the points where the curve meets or crosses the x-axis. There are three possibilities the curve may never reach the x-axis, in which case there are no real solutions.

It may just touch the x-axis, in which case there’s a single real solution, or if you want to think of it that way as two real solutions that coincide. Or it may cross the x-axis in two places, in which case there are two distinct real solutions.

We learn in school a couple of different ways of solving quadratics. The first is factorization, which only works if the equation is relatively simple, and the solutions are whole numbers.

For example, if the equation to be solved is,y = x² — x — 6, we put x² — x — 6 = 0.Then we factorize the left-hand side to give,(x-3)(x+2)= 0. So that the solutions are x=3 or x=-2.

If the quadratic doesn’t factorize easily, we can use the general formula for solving quadratic,

x = (-b ± √-4ac)/2a and just plug in the values for a, b, and c.

You may wonder where this general formula comes from. It comes from the process known as completing the square. Here’s how.

We start from ax² +bx +c = 0.Then we move the c to the other side. So we have,ax²+bx = -c.Now we divide through by a. So we have1/a [ax²+ bx) = -c/a.That simplifies down to x²+ b/a x= -c/a.Then we take the coefficient of x, which is the b/a, divide it by two, and square it. So we have[(b/a)/2]² = b²/4a².And now, we add this to both sides of the equation to complete the square. So what we get isx² +b/ax +(b/2a)² = -c/a + b²/4a².Now let’s simplify the right-hand side. We havex² +b/ax +(b/2a)² =( -4ac /4a²) +(b²/4a²).And the right-hand side now simplifies further, putting it all over the common denominator (b² -4ac)/4a².Now let’s factorize the left-hand side, which we’ve made a squared expression to give(x+b/2a)² = (b² — 4ac)/4a².Take the square root of both sides, and we havex + b/2a = (±√(b²-4ac))/2a.Finally moving that b/2a to the right-hand side, we havex =[ -b ±√(b² — 4ac)]/2a.There’s our general formula for solving a quadratic.

Now let’s look at Dr. Loh’s new method.

We’ll start with a quadratic in the form x² + bx +c = 0. If the coefficient of x² is a number other than 1, we can simply divide through by this to start with.Now suppose this factorizes as follows(x-m)•(x-n) = 0.So that m and n are the solutions that we’re looking for. Multiplying out, we getx²-(m+n)•x + m•n = 0.So we’re looking for values of m and n that multiply to give c, and that add to give -b.

Well, there’s nothing new so far. Dr. Loh’s insight was to realize that we don’t need to guess what m and n are and go through a process of trial and error.

Starting from m+n=-b, it’s clear that the average of m and n is(m+n)/2 = -b/2.Let’s say the difference between m and -b/2, which is the average is z. In other wordsm,n = (-b/2) ± z.Knowing their product is c, we can write(-b/2 + z) (-b/2 -z) =c.So that(b²/4) — z² = c.So,z² = (b²/4 ) — c.Therefore,z = ±√[(b²/4)-c].So the solutions arem,n = (-b/2) ± √(b²/4 -c).

Loh’s method’s big advantage is that you don’t have to remember this formula or the more familiar one [-b±√(b²-4ac)]/2a. Instead, all you need to remember is the method or the algorithm. And this is in three steps.

1- The solutions must add to give -b so their average is-b/2, and they can be written as -b/2 ±z.2- Write down the product. (-b/2 + z) (-b/2 -z) = b²/4 -z² =c. And solve this for z.3- The solutions are -b/2 + z and -b/2 — z.

Whether you think this is easier than trial factorization or using the well-known formula for finding a general solution is a matter of personal preference.

Loh’s method isn’t fundamentally new because there aren’t any discoveries to be made about quadratic equations. It may be a helpful way to help students who haven’t done much algebra before, so perhaps it will find its way into classrooms. Most of us old-timers, I suspect, will carry on using trial and error factorization or the traditional general formula.

Anyway, there you are, a new or at least a new-looking way of solving quadratics.

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