# The Physics Philes, lesson 18: Math Interlude!

*In which equations are factored, slope is determined, and quadratic equations are solved.*

The Physics Philes is a bit of a misnomer this week, because there is basically no physics in this post. You see, I want to do this right. While I remember enough math to muddle my way through some of the simple physics problems my textbook is posing, I’m not completely comfortable with it. So I’ve decided to take a couple of weeks off from physics to focus exclusively on reteaching myself everything I’ve forgotten about algebra and trigonometry. And maybe, if you’re lucky, I’ll get into a little calculus.

I wasn’t sure I was even going to write this post. As I was studying this week I kept thinking, Man, this is way too basic to include in the series after so many weeks. But dang it! When I started writing these weekly posts, the idea was to walk you through my (hopefully not in vain) attempt to lean physics, and this is very much part of that journey. So…here we go.

As one does who wants to eventually learn calculus, I cracked open The Complete Idiot’s Guide to Calculus. Evidently, linear equations are going to be really important in calculus so that’s where the book starts. And since the book starts there, so do I.

A linear equation has three components: two variables and one constant. In standard form, a linear equation looks like this:

Ax + By = C

Ax and By are the variable terms and C is the constant. And those A’s and B’s? Those are coefficients. The coefficients must be integers (a number without a fraction or decimal) and the A must be positive. Hey, don’t look at me. Them’s the rules.

There are a couple of ways to create the equation of a line, both of which I only vaguely remembered from high school. In the first method requires knowing the slope of the line and the y-intercept of the line. It’s appropriately called slope-intercept form.

Slope-intercept form looks like this:

y = mx + b

Where m represents the slope and b represents the y-intercept. Then just solve for y! (Pro-tip: Don’t forget that the y doesn’t represent the y-intercept. That doesn’t work out for anyone.) So if the slope of a line is -3 and the y-intercept is 5, the linear equation can be found using slope-intercept form:

m = -3

b = 5

y = -3x + 5

But what if you don’t have all the information you need for slope-intercept form? Never fear! Point-slope form has got your back. You can create a linear equation if you have the slope and any point (x, y) on the graph. The point-slope form equation looks like this:

The variables with the subscripts are the graph coordinates. We can use this equation to find the linear equation for line g if the line goes through point (-5, 2) and a slope of – 1/5.

We have the slope and we have the point which means we have have enough information to plug into the point-slope form equation.

y – 2 = – 1/5(x – (-5))

y – 2 = – 1/5(x + 5)

Ugh we have to get rid of that fraction if we want the linear equation to be in standard form. Let’s just get rid of it now by multiplying the whole equation by 5:

5(y – 2 = – 1/5(x + 5))

5y – 10 = – x – 5

Now all we have to do is combine the constants and get all the variables on the correct side and we’ve got ourselves a linear equation!

x + 5y = 5

I’ve been talking about slope like it’s a self-evident concept. And maybe it is. Slope is basically the slant of the line. The larger the value of the slope, the steeper the line. The sign of the slope value lets you know in what direction the line is going. The slope will be positive if the line rises from left to right; if it’s negative, the line falls from left to right. If the slope of a line is zero, then the line is horizontal. A vertical line has an undefined slope, aka no slope at all. To find the slope of a line, all you need are two points – (a, b) and (c, d) – on that line to plug into this formula:

The slope of a line equals the difference between y sub 2 and y sub 1 divided the difference between x sub 2 and x sub 1. It’s a little weird to see written out in words, but trust me, it’s easy. Let’s find the slope of a line going through points (3, 7) and (-1, 4).

4 – 7 / -1 – 3

-3 / -4

slope = 3/4

The slope of the line going through points (3, 7) and (-1, 4) is 3/4.

That’s much simpler than I remember it being. Hm. But no rest for the wicked! Time to move on to factoring polynomials!

Factoring is basically unmultiplying something. You break the number or expression down into parts that, if multiplied together, would give you the original expression. And easy number to factor is 6. Just break it down: 3 x 2 = 6.

The greatest common factor is the easiest way to factor, and it’s a method I actually remembered from high school algebra. All you have to do figure out what the numbers have in common. Let’s try one:

See? We can take out 7xy and multiply it to what’s left to get the original expression. It’s cake. But there are a few special factoring patterns any math wizard should know:

**Difference of perfect squares**

As you know, a perfect square is a number that can be created by multiplying another number by itself. For example, 16 is the square of 4. If you see an equation like the one above, you can automatically factor it using that pattern.

**Sum of perfect cubs**

Perfect cubes are similar to perfect squares in that a perfect square is a number that is created by multiplying another number by itself three times. For example, 125 is the square of 5, because 5 x 5 x 5 = 125. This factoring pattern has a similar process to the next and last pattern…

**Difference of perfect cubes**

Well…you already know what a cube is, so…

Finally it’s time to look at quadratic equations. The highest exponent in a quadratic equation is 2. There are a few ways to solve a quadratic equation. The first one we know how to do already. It’s factoring. If the equation is factorable, this is the easiest way to go. Just set the equation equal to zero and factor away!

Now we need to set each of the sets of parentheses to zero:

Now for the second equation!

This is old hat for us by now. The next method for solving a quadratic equation is completing the square. This method has more steps than factoring. But it’ll get you to the answer. Let me demonstrate with an example:

First we need to get the constant on the opposite side of the equation from the variables.

To solve the equation, the coefficient of the x^2 variable needs to be one. In this case, that means we need to divide both sides of the equation by 2.

Here’s where it gets weird. Take half of the coefficient of the x term – in this case, 3 – square it, and add it to both sides of the equation. That means that we need to add 9 to both sides, as 9 is the square of 3.

The equation is now a perfect square! And it should be every time you use this method. Just factor it as you would a perfect square.

Now we take the square root of both sides. Because we square root both sides, we need to stick in that plus or minus sign.

The square root allows us to take away that pesky exponent.

Now we just need to solve for x!

Easy peasy lemon squeezy! Just lots of steps, but nothing too onerous. The final method of solving quadratic equations, however, is my favorite. All you have to do is plug in the values into a formula! Who wouldn’t like that? I’m of course talking about the quadratic formula. All you have to do set the equation equal to zero so it looks like this:

Literally all you need to do is plug the corresponding numbers into this equation:

I’m getting flashbacks from middle school and high school. I could have recited that equation in my sleep. And for good reason. Any quadratic equation can be solved using this method. Just plug in the values and solve for x.

This has been a detour on my journey into physics. I know it’s basic, but I can already feel my math confidence building. I’ll be continuing my math review next week, and possibly the week after. Thanks for being so patient. This will pay off, if only for me.

*Featured image credit: lednichenkoolga*

## 1 Comment

I have a… well, I’m not sure if you’d call it a correction. But nonetheless…

No, them’s not the rules. These “rules” are completely arbitrary.

Ax + By = C is a 100%

bona fidelinear equation as long as A, B and C are scalars. Apart from that, it matters not what they are. They can be integral, rational, irrational or even complex.If there wasn’t, there would be no way to describe a line like √2x – y = 5. This is a line (graph it and see), but there are no integers A, B and C which you could use to describe it.

It is sometimes manners to normalise the equation in some way. If A, B and C are rational numbers, for example, there is a way to describe the line using only integers, and that can have some advantages.

However, I’ve never heard the one that A is positive. If anything, C is the coefficient that should be positive, to preserve symmetry! Or perhaps C should be positive in the equation Ax+By+C=0.

Another constraint that you commonly see is A²+B²=1. This is a useful one, since it means that (A,B) is a unit vector perpendicular to the line, and C is the shortest (signed) distance from the line to the origin.

Incidentally, the equation y=5 is a valid line, too. (Graph it and see!) But if you try to express it in the form y=mx+c, you’ll find that it can’t be done, because y is not a function of x at all.

Describing lines mathematically turns out, as you can see, to be a trickier problem than you may have first thought. You may even be surprised to learn that this is

stillan active research area, thanks to the needs of computers. Computational geometry and its allied arts (e.g. computer modelling of physical systems, computer graphics, and video games) have a large set of requirements, and different line representations turn out to be useful in different application areas.Finally, here’s another way to solve the quadratic equation.

Take the general quadratic equation:

f(x) = ax² + bx + c

Its derivative is:

f'(x) = 2ax + b

Now we divide f(x) by f'(x), using whatever method you’ve been taught. (I was taught to do polynomial long division, but apparently something called “synthetic division” is popular these days.)

What we find is:

(ax² + bx + c)/(2ax + b) = (x/2 + b/4a) + (c – b²/4a)/(2ax+b)

The first term is the quotient, the second is the remainder. A little bit of manipulation later:

(ax² + bx + c)/(2ax + b) = 1/4a [ (2ax + b) + (4ac – b²)/(2ax+b) ]

⇔ ax² + bx + c = 1/4a [ (2ax + b)² + (4ac – b²) ]

The rest you should be able to work out. What I find interesting about this is that it shows that the discriminant is related to the remainder of f(x) divided by f'(x).