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The Quadratic Equation Solution a Few Thousand Years in the Making

Everyone learns (and some readers maybe still remember) the quadratic formula. It’s a pillar of algebra and allows you to solve equations like Ax2+Bx+C=0. But just because you’ve used it doesn’t mean you know how to come up with the formula itself. It’s a bear to derive so the vast majority of us simply memorize the formula. A Carnegie Mellon mathematician named Po-Shen Loh didn’t expect to find a new way to derive the solution when he was reviewing math materials for middle school use to make them easier to understand. After all, people have been solving that equation for about 4,000 years. But that’s exactly what he did.

Before we look at the new solution, let’s talk about why you want to solve quadratic equations. They are used in many contexts. In ancient times you might use them to determine how much more crop to grow to cover pay tax payments without eating in to the crop you needed to subsist. In physics, it can describe motion. There’s seemingly no end to how many things you can describe with a quadratic equation.

Babylonians, in particular, would solve simultaneous equations to find the roots of a quadratic. Egyptians, Grecians, Indians, and Chinese peoples used graphical methods to solve the equations. The entire history is a bit much to get into, but still a great read. For this article, let’s dig into how the new derivation was discovered.

Setting Up the Equation

So what’s the method? You watch Loh explain it himself in the video, below. Suppose you have a standard quadratic equation for whatever reason that looks like this:

x2-6x+3=0

In this case, A=1, B=6, and C=3. Note that for Loh’s method to work, A should equal 1, but if it doesn’t you can always divide both sides by A to make that true. For example, consider this equation:

3x2-18x+9=0

It will have the same roots as the first one because if you divide both sides by 3, you get the first equation.

The Polynomial Factor

To factor a polynomial into two binomials we can use FOIL (First/Inner/Outer/Last). So we know that:

x2-6x+3=(x-S)(x-R)

Where we just made up S and R. They exist, but we don’t know what they are yet. However, for the answer to be zero, you can see that the equation will be zero when x=S or X=R so that means S and R are the roots of the equation.

So What’s S and R?

It sounds silly, but let’s multiply out the binomials into another polynomial. Remember FOIL:

x2-6x+3=x2-(S+R)x+SR

If you look at that for a second, you will probably realize that S times R must be 3. We also know they add up to 6. At this point, you can probably just guess the roots, but let’s make it formal.

The roots of a polynomial like this will be in the form of U±z. So if S=U+z and R=U-z, the only way to make (U+z)+(U-z) equal to 6 is if U is 6/2. Since B is -6 in this example, we can intuit that U must be -B/2 in the general case.

Home Stretch

So we now know the roots are -B/2+z and -B/2-z. We know that B is -6. We also know the answer is when we multiply those roots together must be C (3, for our example). So we can write:

(-(-6)/2+z)(-(-6)/2-z)=3

Or:

(3+z)(3-z)=3

Do the FOIL method again and get:

9+3z-3z-z2=3

The middle terms will always cancel out so you get:

9-z2=3

Subtract 3 from both sides and add z squared to both sides:

6=z2

Since the root must be -B/2±z we know that our two roots are 3+√6 and 3-√6.

The Key and the Catch

The point is, none of this was hard to remember or work out. Just remember that you rewrite the zero as (x-S)(x-R) and the rest follows very logically.

The only real problem is not many quadratic equations have A=1. For example, suppose you throw a ball straight up from 5 meters above the ground with a velocity of 15 m/s. We want to know when the ball will hit the ground.

Gravity is going to pull down on the ball at about 4.9 meters per second squared (assuming gravity accelerates at -9.8 m/s2). The physics formula is (at2)/2 so we get our first term of -4.8t2. The second term will represent the speed of the ball which is simply 15t. We also started 5 meters above the ground, so we will add 5 and wind up with:

-4.8t2+15t+5

This formula will tell you where the ball is assuming you throw at t=0. We want to know when it hits the ground so that will be one of the roots:

-4.8t2+15t+5=0

To use the new method, just divide it all through by -4.8 to get:

t2-3.125t-1.04=0

So now B=-3.125 and C=-1.04. We know the root will be -B/2±z or 1.5625 and that z2 will equal (1.5625)2+1.04.

That’s 3.48 (about) and the square root is about 1.866. The roots, then, are 1.5625±1.866. Working that out, we get -0.30 and 3.4285. The negative root is nonsense in this case but the ball will land almost 3.5 seconds after you throw it.

If you don’t trust our work, ask Wolfram Alpha or plug 3.4285 back into the original formula. Wolfram says 3.43 and the check your own work says 0.005 meters at that time, so given that I rounded a few times, that’s pretty close. Besides, I’m ignoring things like air resistance. This specific calculator says 3.36 seconds, which is still pretty close.

In Summary

If you prefer things in algorithmic steps, here you go:

  1. Find A, B, and C for the quadratic equation.
  2. If necessary, divide by A so that A now is 1.
  3. Write -(B/2)2-z2=C
  4. Solve for z (since you know B and C)
  5. Roots are -B/2±z

Everything Old is New Again

We get it. Many people may have figured this out before. But if they did, they apparently forgot to share it with anyone in any permanent form. If you read the actual paper, you’ll see how easy it is to symbolically derive the “old” standard formula. Compare that to the traditional method. If you read the proof, it seems simple enough, but go back in a few weeks and try to work it out yourself. Not so easy. This is also much easier to remember even if you don’t want to derive it every time.

Of course, making A=1 is part of the trick. It is well known, for example, that the product of the roots of a quadratic is C/A. Since we know A=1, it follows that the product of S and R are also C, using this method. Normalizing A to 1 is also an old trick and is sometimes called a reduced quadratic equation. However, it doesn’t look like anyone put all the pieces together until now.

If you don’t like Wolfram Alpha, there’s always Mathics (hint: use Solve[-4.8t^2+15t+5==0,t]).  There are also specific calculators. Or just bite the bullet and keep everything in a Jupyter Notebook.

source https://hackaday.com/2020/01/03/the-quadratic-equation-solution-a-few-thousand-years-in-the-making/

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