Samples

Wouldn’t it be nice if you could get a feel for how I explain topics and see a glimpse of my personality before considering even a free consultation? If only there were a way to do that…

Well, what I can do is offer to let you peruse some of the answers I provided to math questions asked on Quora.com. The link is to my profile there if you’d like to read through a few in order to see how I explain topics and approach math in general. I will also provide a few of my favorites below.


problem 1

As with every answer, the first rule of being a mathematician is that we are lazy creatures. This is why we refer to breaking down a problem as “simplifying,” because there has to be an easier way to go about arriving at a solution.

In this case, we see the exponent and realize that this falls into the category of functions called quadratics. Now, for many mathematicians, quadratics do not mean the end of the world. After all, there are far worse equations to be dealing with, but that doesn’t mean that we can’t make it even simpler for ourselves.

Let’s begin by moving everything to the same side by subtracting 3x from both.

subtracting:

then,

Now we can see that everything on the left side has an x in common, which brings us to our old friend factoring. Particularly factoring out in this case, or undistributing if you would like to think of it that way. Remember, distributing is where you have something like:

You would distribute the 3 to both the x and 4 inside the parentheses by multiplying it to both terms, resulting in this:

In order to factor something that both terms have in common out, or undistribute and go backwards, we would have to recognize both 3x and 12 can be divided by 3, which would result in the 3(x+4) that we began with. Back to our resulting equation above.

Here, as said above, both terms have an x, so we would need to divide both x squared and 3x by x. Taking x from both terms would leave us with

Notice inside the parentheses that the exponent on the first term decreased by one since I took one x away, and in the second term, the x was removed because 3x divided by x canceled my x’s out.

It is also useful to note that if I multiplied the x through to both that x times x would be x squared and -3 times x would again be -3x. Thus we would be back to the equation we just came from, so we can be sure we correctly factored.

From here, we use something called the zero product property. Ew, more math speak. So let’s look at this again. We have to have factored for some reason. Well now we have only x’s, so no exponents. At least now we are looking at linear equations, which are the easiest to deal with. We are on track for our laziness so far.

Zero Product Property. Well, properties are things mathematicians notice over and over again about a broad range of specific types of equations. Zero means the number 0, and product means to multiply things together for an answer.

Ah! x and x-3 are being multiplied together and equal 0. Coincidence? I think not! The great thing about zero is that anything multiplied by zero is zero. So if we are looking at the x part, it doesn’t matter what x-3 is, if x is 0, the other part can be whatever it wants, but it will still multiply to equal zero. Look:

or

or

No matter what is left in the parentheses, if the x part outside is 0, the solution will be 0, so one solution is x=0. But before we started with a quadratic, which means we can have two solutions. Also remember, x might not be 0, so we need to check and see how to make x-3=0.

What number minus 3 = 0. We need to find the opposite of -3, which is 3.

Now we have the second part is zero, so the first part will not matter, which means our second solution is x=3.

Another way to look at it would be to set both terms equal to zero and solve.

and

The first is already given to us that x=0. The second we have to isolate x, or get x alone on the left. So after adding 3 to both sides we have:

Ta Da! Our two solutions are x=0 and x=3. How is that lazy, you ask? Well, we really only had four steps.

  1. Get everything on the same side.
  2. Factor x out of both terms.
  3. Find what makes the first term 0
  4. Find what makes the second term 0

Breaking the problem down step by step allows for a much easier time finding a solution. In this instance I broke the problem down very detailed which is the reason for the wall of text, but with practice, the process becomes the quickest way to arrive at an answer.

Other ways such as graphing on a graphing calculator or software and finding the intersections can occur as so:

You can see that the x values where the functions intersect are 0 and 3, hence x=0 and x=3.

Now, I could point out the old math teacher fallback here where they say “But what if you don’t have a calculator?” Realistically, that would never happen. It is true, however, that I can actually perform the before-mentioned factoring in my head faster than I can get a calculator and find the answer.

In the end, it is up to you how best to approach a question. Build your confidence so that you believe you can handle the problem, then worry about alternative methods and speed.

Notice both let us arrive at the correct answer, so both ways are valid mathematically. Good luck, and happy math-ing!


problem 2

This is classified as an exponential equation. While working with students, I always enjoy giving a little vocabulary lesson as it truly does help with the understanding of concepts. Why? I’m so glad you asked.

My absolute favorite thing about being a mathematician is that the number one thing to keep in mind is that we are very lazy creatures. Ever wonder why we call solving simplifying? It has to do with the fact that we don’t enjoy doing work and would like to look at the easiest form of an equation or formula to save work.

Naming exemplifies this quite well. Let’s look at our question. Exponents are the superscripts up and to the right of a larger number, in this case the x. Others include:

In each case, the smaller values are called exponents. (The 2, 3, 1000, and 3 respectively.)

Letters, are referred to as variables…because they vary. That’s right, brilliant mathematicians call it that because it’s what the letters do. They substitute in for numbers that can be anything. When we have an equation (called so due to the equals sign) where a variable is in the exponent spot, we call it an exponential equation.

There’s a pattern here of lazy naming, which is great for students of mathematics as it quickly tells on itself. Here, we know that there are two ways for solving exponential equations. Yes, I am aware you may not need to know what it’s called, and I have heard that excuse numerous times when asking my pupils to write a definition. Here we can see, however, that the name helps identify the problem spot, the exponent.

So:

Method 1

We recognize that this literally translates to “7 raised to some power equals 49.” Another way to read this as a question would be to say “7 raised to what power equals 49?”

From here, the game begins as how many 7’s do I need to multiply before getting to 49? Let’s check.

That’s two 7’s, so 7 raised to the second power (or 7 squared for more math speak) will get us 49. In other words:

Thus, x must be 2.

Method 2

The second way is to utilize inverse functions. Inverse means opposite in math, so what function is the opposite of an exponential (or undoes an exponential)? Enter the logarithm.

Say what?!? Alright, if this is your first time looking at this animal, it can seem a bit intimidating, but all we need to remember is the circle rule. But let’s get some terminology out of the way first.

This is read as “The log base 7 of 49 equals 2.” Log is the type of function we are working with, hence the log part. The base is the subscript 7. But wait, we have heard that word before haven’t we. That’s right in our example of 7 raised to the x power, the 7 is called the base because we raise a base to an exponent power.

Clever mathematicians didn’t want to rename something, so they left it the same. The 49 inside is called the argument, or what we plug into a logarithmic function. And lastly, 2 is your answer.

Now the circle rule: The base raised to the answer equals the argument. Clears it up, right? No, didn’t think so. How about a visual:

Much better. We go from the 7 to the 2, then from the 2 to the 49. “7 raised to the 2 power equals 49.

But in our case, we did not know what the power was, did we? So we had:

Now we have two options yet again. In this case, we can follow the steps from the first method to ask ourselves how many times we need to multiply 7 by itself to reach 49. But that’s been done, so why bother with another setup? Because not all answers are so nice, which is why we have programmed calculators with functions like this in them.

Were this problem more difficult with a decimal answer or even irrational solution, we would have to rely on some computer power such as a calculator to arrive at the conclusion. Luckily, we can utilize a calculator for simple cases such as the one above as well.

On a graphing or scientific calculator you must first find the “log” button, or “logbase” area in the Texas Instruments graphing calculators. You would then type

logbase 7 ( 49 ) Enter

Which will spit out your answer of x=2!

Conclusion

I wrote this out assuming that any of the steps were new to a reader in order to give a complete walk-through. Hopefully anyone can read through this explanation and pick up the pieces they are missing to find the solution.

I also see in other answers that the question may have at one time appeared different. Possibly:

If this is the case, utilizing some exponent rules, we know

and thus

We understand that

so

Our new equation would then appear like this:

which would necessitate that our exponents are equivalent, eventually leading to the conclusions listed in the alternate form where x must be 0.

Anyway, as the question appears now, Methods 1 and 2 would be the correct ways to tackle the process. Hope this helps!


problem 3

Anyone who’s seen my answers before knows that I always emphasize the number one rule or requirement for being a mathematician: we are lazy creatures. There’s a lot going on here, so first we should try to make it look prettier.

There are a lot of a’s and b’s floating around, which does not make me happy. How about you? No? Good, then let’s get rid of some by eliminating the b’s. We are given the following:

and

I say we add them together because the coefficients in front of our b’s are opposites. (-1 in front of b on the first equation and +1 in front of b in the second). Thus

Notice that we can add straight down, combining like terms. a’s with a’s, b’s with b’s, and numbers with numbers. So we end up with

or

From here, to solve for a we have a two-step equation. First we have to get rid of the 2 by using inverse operations (fancy math speak for using the opposite of what we see). Since we are multiplying 2 and a square, we have to divide both sides by 2 like this

which gives us

because the 2’s “cancelled (or divided to equal 1 if you are interested in more fancy math talk). But we aren’t finished yet. What is the opposite of squaring a number? Taking the square root. So…..

On the left, we have a by itself finally, and something neat happens on the right. The square root asks us “what multiplied by itself equals 25?” Our answer is of course that 5×5=25, but we also know two negatives multiply to make a positive. So we’ve also got -5x-5=25.

After all those shenanigans we found that a=5 or a=-5. So we’re done…..

But wait, what about b? Dang. I guess we probably better find that, too.

Well, if a=5 or a=-5, let’s do some plugging in, shall we? I’m going to use the second equation because it has less minus signs (laziness, remember?).

to

Here I would like to point out the importance of the parentheses when evaluating an answer. For this instance, it would not alter our result, but later on in the negative, it could. Why? Well, if you had -2 squared, that would be -2x-2 which is 4 after the two negatives make a positive. If you had -(2 squared), that would be saying 2×2=4 then applying the negative for an answer of -4. Completely different signs.

This is especially important for those following along using their calculators. You know who you are. That’s right, I can see you through the screen clicking away. No shame here, just make sure to utilize your tools correctly so as not to make an error. Where were we? Ah yes.

Then subtracting 25 from both sides since it’s the opposite of adding:

to

Again, we have a square we need to get rid of, so back to the square root.

Now b is isolated on the left all by itself, and we have to ask ourselves what two numbers multiplied by themselves will give us 4. Yep, b=2 and b=-2. Just like before, we have a positive and negative answer here (hmm wonder if that is a pattern rule that always works).

Soooooo we have that a=5 or a=-5 and b=2 or b=-2. Finally…

Wait…Didn’t my teacher always say there was something to do at the end of a problem. Checking my answer or some nonsense. Bah! The question wants ab? That means a times b! But I have four numbers, which means four combinations! (Sobbing noises heard from behind the screen.)

Sorry about that, let’s finish this.

or

or

or

Phew. Guess it wasn’t that bad comparatively. And look at that, only two answers. We have that ab=10 or ab=-10. Huzzah!

Now, for those of you out there with calculators who are always looking for shortcuts and an even lazier way, all you really need to do is to graph your two equations like so

For those curious, this would be the Desmos calculator online. Completely free, so use that bad boy for cool stuff like this. It likes to use x and y for variables when graphing, so I substituted x for a and y for b to get the same thing.

For even more curious little students, the first equation has a subtraction in the middle, so it is a hyperbola that opens horizontally because y comes second. The second equation is a circle as it has an addition in the middle. (Yay conic sections!)

You’ll see the points where the two intersect give us the same a and b values of 5 and -5 and 2 and -2. From there you would perform the same multiplication as in the last step from the first method (algebraically).

Tada! Happy math-ing folks. I’ll be here all week. Or on my website blogging about the same stuff.

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