Okay, for a story I'm working on I need two math equations. The first one easy. The second one complicated.

Okay, so I have the first one.

4+7=11 (four plus seven equals eleven)

Now... Does someone have a complicated one for me? Maybe something with lots of brackets and a square root or something. But if you give me a complicated one, I need the answer too.

Thanks,

ML

Does this help? <G>

From my current class, complete with the solution:

oops I'm too late!

How complicated? Square roots? Cube roots? Logs? Exponentials? Differentials? Matrixes? Partial derivatives? And should it concern some particular subject? Do you need a numerical answer, or are variables in the solution okay? Multiple variables?

Just how messy do you want the equation to look, and just how messy can the answer be?

Paul

Okay, Anne. Help me out here (it's been a long time since I've done any serious math).

If the square root of two x squared minus three x minus ten equals x, then x equals... what? (or is that even how you would read that equation out loud?)

Paul, let me explain what I'm using it for.

A guy wins a prize. He's practicing for his skill testing question. And here is a short portion of his practice.

"Two plus five equals eight. I mean nine. Or.. well, seven. Right. I meant to say seven. Okay, let me try another one. [And here is where I'm putting in a really complicated question which he then gets right]

So I need to be able to write it out and to give the answer.

How's this ...

Solve the square root of two x squared minus three x minus ten equal to x where x equals either positive five or negative two.

To solve you sqare both sides giving you the square root of two x squared minus three x minus ten squared equal to x squared.

Which is equal to two x squared minus three x minus ten equals x squared.

Subtract x squared from both sides leaving you with x squared minus three x minus ten equals zero.

You can factor that down to x minus 5 times x plus two. Finally, each factor can then be finshed off as x equals positive five and negative 2.

Yes. Although most people would probably flip that around. "If x equals the square root of..."

And that probably is a good one to use. I've been browsing around looking for more complex stuff (with solutions) because I'm kind of rusty and not that good at putting these things together off the top of my head.

I mean, it does get a lot more complex and tricky than the algebra in Annette's equation, but I'm not sure how much this guy can really do in his head. And some of the trickier ones look simple on paper.

I'll keep thinking. And maybe dig up some of my old math and engineering homework...

Or you could just use Annette's example, which does seem to fit the situation.

Paul

ML, you have it, basically. I would probably say "the square root of the quantity two x squared minus three x minus 10 is equal to x", just because that indicates that the entire trinomial is inside the square root rather than just the two x squared being under the radical.

Sheilah's got a good point. I was thinking that might be a problem, but forgot the proper way to specificy things so it would be correctly understood.

Meantime, I've stumbled on to a most interesting site... http://www.exampleproblems.com Just like the URL says, it has example problems in a bunch of different mathmatical subjects. I found it while looking for a good juicy partial differential equation, but I'm thinking now that something like that may be more trouble than it's worth (as, come to think, PDEs in general tend to be...).

What's interesting, though, is that there's a link off to the left (under "navigation") which takes you to a random page on the site. It's kind of funny. If you keep clicking it, you can get a simple and direct trig problem followed by some basic algebra and then find yourself staring at a jumble of multiple equations with differentials all over the place and Fourier transforms...

*click, click*

Ooo, here 's one of those things I was talking about that look so darned simple until you start trying to solve them... Take a look at what you're starting with, then scroll down to see the "solution."

*click, click*

Oh, this might work...

You'd read it as:

"If Y double prime plus four Y prime plus three Y equals zero, with initial conditions of Y equals one and Y prime equals zero, then Y as a function of t would be...

"Let's see... Roots of negative one and negative three... So the general solution would be of the form C One times e to the negative t plus C Two times E to the negative three t... Plug in the initial conditions... C One equals one minus C Two and... negative C One minus negative thee C Two equals zero... Sub in the first equation... Negative one plus C Two minus three C Two equals zero, so C Two equals negative one half, which means C One equals one plus a half, which is three halves. So... Y as a function of t would be...

"Three halves e to the negative t minus one half e to the negative three t. Simple."

You could, of course, drop the middle paragraph to avoid making your readers' eyes cross. You could also change "Y double prime" to "the second derivative of Y" and "Y prime" to "the first derivative of Y." You might also specify the initial conditions as "Y of zero" and "Y prime of zero" or "Y at time zero" or something similar.

You could also keep clicking and find a different equation. Or you could go back to using Annette's.

Oh, and the equation itself is a "second order ordinary differential equation" or "second order ODE" or "second order linear ODE" or even (if you want to get really snooty and specific) a "second order linear homogeneous ODE." Which, as ODEs go, is actually fairly simple and straighforward.

Paul

Man, a question about Math, and I am late! I agree with Paul about an ODE. They are commonly used to model many, many, many things, especially in the human body -- Here is the solution of a linear, homogeneous ordinary differential equation, solved by seperation of variables (this is pretty much the easiest differential equation there is).

dx/dt = x given x(0) = 0
S(0 --> x) dx/x = S(0 --> t) dt
ln(x)-ln(0) = t-0
(ln(x) - 1) = t
e^ln(x) = e^(t+1)
x = e^(t+1)


In words, d x by d t equals x with an initial condition of x = 0 at t - 0. You solve by seperation of variables. First, you split up the equation getting all terms with t on the rigth side and all terms with x on the left. Then, you take the integral of both sides, from 0 to x on the left and 0 to t on the right, using your initial condition. The integral on the left, with respect to x comes out as a natural log while the integral on the right with respect to t just comes out as t. Then you plug in your limits of integration, and you can solve by taking the exponential of both sides to get rid of the ln term.
QED

You guys are so smart!

Thanks everyone. I think I've got enough for the section now. Just one thing... After reading all your posts, my head is throbbing. Anyone got an asperin for my headache

Thanks again.

ML

Oh wow - I live for this stuff & it looks like I missed it!

Just a note on the example problem that was posted - in case you decided to use it. The "check for extraneous solutions" wasn't shown in the post - so it looks like both 5 & -2 are solutions to the original equation. (Checking would just involve "plugging in" each answer to see if it yields a true statement.) -2 can't be a solution, though, because the square root of a quantity can't be negative in a situation like this. The moral of the story is to always check for extraneous roots if you square both sides to solve an equation. Well - it should be done in other cases, too, but that's not exactly important now, is it?

Now if only I could get my students to check their answers!

shells