# 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. Po-Shen Loh is an associate professor of mathematics at Carnegie Mellon University and currently the United States International Math Olympiad team’s national coach.

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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