Right, maths, yeah? What the hell?

I've recently decided to embark upon a new career path involving mathematics, mechanics, engineering and so on. Ultimately I will build cool machines and stuff when I am all grown up but I've not studied any of these things beyond what they tried to teach us at GCSE level. So, I know how to calculate the hypotenuse of a triangle and stuff but things like quadatic equations (which I need to understand) are well out of my league.

The thing is, I'm not able to pay for a course. I need to teach myself but I can't seem to find a good (read: free) source to learn these fundamentals from.

So, all you mathemagicians out there, where can I teach myself how to maths? Failing that, can one of you explain what a quadratic equation is? I thought I understood them but my answers were wrong.

If one of you mods feel this is likely to generate too much on-topic chatter, I can move it to a thread of its own.

Moved from 100%OT to a separate thread - will be easier for us all then!

Can you post one of your quadratic questions and your answer - will probably be easier for us to help you.

I don't know how to write the equation in a text format, so there's an image of it attached.

My answers were -6.6 or 5.1 and I came to that conclusion by doing 1.5x1.5=2.25, 2.25+4=6.25, 6.25x22=137.5, the square root of 137.5 is 11.7, so either -1.5+11.7=10.2, 10.2/2=5.1, or -1.5-11.7=-13.2, -13.2/2=-6.6

However, having just used the calculator on this pc to do the equation, I come up with 4 or -5.5... so maybe it's my phone's calculator's fault. I'm confused. Help! :)

Order of operations! Multiplication comes before addition. so it's supposed to be evalutated as 1.5^2 + (4 * 1 * 22) whereas you're doing it as (1.5^2 + 4) * 1 * 22.

Phone calculators can't do order of operations and instead evaluate things in the order you enter them. Good scientific calculators can keep track of this stuff and you can enter it as written.

Using the basic form of the quadratic (I'm too lazy to open equation editor, so you'll have to do with a bit of text syntax :p):

t = (-b ± SQRT(b^2 - 4*a*c))/(2*a)

a = 1
b = 1.5
c = -22

t = -5.5
t = 4


I did it this way:

4*1*22 = 88
88+1.5^2 = 90.25
Answer 1: (-1.5 + SQRT(90.25))/2 = 4
Answer 2: (-1.5 - SQRT(90.25))/2 = -5.5

If your calculator can't do brackets, work out SQRT(90.25) first, then add/subtract -1.5, then divide that by 2.


Edit:

Hmmm typed all that out while SomePerson was typing :p Might be worth getting yourself a scientific calculator - makes things so much easier. They're quite cheap.

Good scientific calculators can keep track of this stuff and you can enter it as written.

Indeed, you'll find a scientific calculator very helpful, particularly as you progress. The fact it does the correct order of operations (by allowing you to type the whole thing in before calculating) is nice, but the extra operations it has become useful later on. You can usually pick one up for a fiver or mabye less.

Because you brought it up, I'm going to try to explain quadratic equations. I'm sure you know a lot of this (as you've gone on to solving), but I'll say it anyway.

A quadratic equaiton is an equation of the form: ax^2 + bx + c = 0 where a, b and c are constants.
http://upload.wikimedia.org/math/f/c/6/fc6366dd1fd484528b085dff42ba1027.png

These equations can be solved in a number of ways, depending on the constants. A sure way of solving them is to use the quadratic formula:
x=http://upload.wikimedia.org/math/d/c/d/dcd0cf8df040e453d640be0a92188ddd.png
This will give two solutions for x (one using the + sign, the other using the - sign).
Note that if, for any reason, the term under the square-root is negative, you cannot solve the equation - it has no real solutions. (You can technically solve it, but it involves complex numbers and you don't want to go there yet).


As for teaching yourself. Well the internet may help, but it could be tricky finding the right websites. The wikipedia is probably a bad place to look at first, because it can go a bit deep.
I'd recommend brushing up with GCSE stuff first, then mabye moving onto the AS Level stuff.
All I can say is, it will not be easy.

Also quite a few of us here love math and would be willing to help you through anything you don't understand if you chose to go through an online learning course.

I'm nthing a scientific calculator because although I'm not quite sure what you're going to be doing, it's one of the most solid investments you can make because it's an extraordinarily handy device. The TI-83+ is what I assume people are referring to : o)

As for teaching yourself. Well the internet may help, but it could be tricky finding the right websites. The wikipedia is probably a bad place to look at first, because it can go a bit deep.
I'd recommend brushing up with GCSE stuff first, then mabye moving onto the AS Level stuff.
All I can say is, it will not be easy.

So far I agree with all of the above. Wiki was way too much info for my level.
I'm going to try and get a GCSE past-paper and sit it one afternoon as a way of highlighting where I suck. Ultimately this is all down to me not paying attention at school (and rightly so, our maths teacher, I maintain to this day, was a psycho hose beast).

Also quite a few of us here love math and would be willing to help you

http://i229.photobucket.com/albums/ee292/surferjones/2yoshlh-1.png
Thank you Google Images for the above portrayal of zombie love

I suggest you go to your local university and check what basic maths books they are suggesting for e.g. physics freshmen.
Normally those books cover all the basics including differential and integral calculus.
A lot of those books also come with a working booklet with elaborated example exercises.

When I did two semesters of physics years ago, there was such a book written by two of my professors. Only cost ~20€.

You could of course check your local libraries for books.

I just looked up the TI-83+ and... heh... I don't need a calculator to work out I can't afford the £60+ it would cost.

This one (http://www.whsmith.co.uk/CatalogAndSearch/ProductDetails-Texet+Albert+Scientific+Calculator-00435741.html#) is much more likely to end up in my possession, although I would like a solar powered one.

I have a ti-89 and it's awesome. It can play games too. :cool: :p

That stuff was great for boring classes and for cheating on physics exams. :P

I just looked up the TI-83+ and... heh... I don't need a calculator to work out I can't afford the £60+ it would cost.

This one (http://www.whsmith.co.uk/CatalogAndSearch/ProductDetails-Texet+Albert+Scientific+Calculator-00435741.html#) is much more likely to end up in my possession, although I would like a solar powered one.
This is probably the best type of calculator to go for - really simple to use, doesn't have too many unnecessary features that you'll never use and is cheap!

http://www.whsmith.co.uk/CatalogAndSearch/ProductDetails-Casio+FX85ES+Scientific+Calculator-31291026.html

The Texas Instruments calculators in the TI-3x range also are sufficient enough and in the same price range.

Although, I prefer calculators where you can notate multiple operators and see the whole equation until you hit the equals button, instead of ones where everything is calculated immediately after each operator.

I managed a Maths AS Level with a Casio like thomasp linked, you really don't need a TI-anything.

Also, i found quadratics are easier if you can visualise what's going on. Generally speaking, if you draw a line on a graph based on a quadratic equation, you will get an upside-down parabola (ie, a "U" shape). The 2 results for x are the points where y=0, ie where the line passes through the x-axis of the graph.

AlienKing mentioned that if b² - 4ac is negative (or put differently, b² < 4ac), then there are no real solutions. This is because the line of that equation never goes below the x-axis, hence doesn't pass through it, so y is never 0. Also, if it turns out that b² = 4ac, there is just one answer for x, and this means the line just touches the x-axis at the very bottom of the parabola. Here's some MSPaint to demonstrate:

I'm with Liam on this one. That one Thomas linked to was the next one I looked at and it is my intention to purchase one as soon as this ******* from CityLink arrives to deliver my parcel.

I think I need to spend a fair bit of time with this. I get the fundamentals of quadratics, I think, but I don't understand them properly yet. That's just a case of doing enough with them to become familiar.

Still, up until Tuesday I'd not done any maths lessons or learning of any sort in 10+ years, so I think I'm doing OK for now. :)

Thanks, guys; I'm happy you all care so much!

We all care so much about maths! ;)

We all care so much about maths! ;)
That's because we're a load of engineers and scientists with a few mathematicians thrown in for good measure and we all think that this makes an interesting topic of conversation :p

This looks nice: http://tutorial.math.lamar.edu/
Has PDF downloads of all lections, exercises and cheat sheets too.

Here's the first section for quadratic equations:
http://tutorial.math.lamar.edu/Classes/Alg/SolveQuadraticEqnsI.aspx

That would probably be a good equivalent of a basic maths book.

I'd also recommend this bit from Loughborough University's HELM (Helping Engineers Learn Mathematics) series of books (yes I know, shameless promotion of my uni...): http://www.mash.dept.shef.ac.uk/Resources/3_2_solving_quadratic_equatns.pdf

The Lboro site is only accessible by uni students, but Sheffield have theirs in the public domain - same book. That's pretty much A-level standard and are used to ensure all students on engineering degrees are at the same level in the first term of their degree.

This looks nice: http://tutorial.math.lamar.edu/
Has PDF downloads of all lections, exercises and cheat sheets too.

Here's the first section for quadratic equations:
http://tutorial.math.lamar.edu/Classes/Alg/SolveQuadraticEqnsI.aspx

That would probably be a good equivalent of a basic maths book.

Dude, that's awesome. Your Google skills far exceed my own!

Thanks :D

Your Google skills far exceed my own!
Huh?
http://www.google.at/search?hl=en&q=maths%20tutorials

I figured you would have stumbled across one of those.

We all care so much about maths! ;)
except for me, i hate math

OK, bonz, so your skills aren't that great but despite multiple Google searches I didn't find that particular page. Or, maybe it was in the search results but I missed it. Either way, you provided some damn good material right there and for that I thank you whole heartedly.

I realised just how difficult this is going to be last night. I thought I had a reasonable grasp on the basics of maths but it seems there's a whole load of stuff I had no idea about. Integer exponents, for example. I think I understand the principle but remembering all the little bits and bobs is the tricky bit. I'm getting there, though. I can't expect to learn these things overnight, can I? Or can I? :P

Don't worry, it took me millions of years to wrap my head around maths, and then choke it to death.

Unless you're talking about properties of integer exponents they're pretty straightforward, an integer is simply any whole number (including zero i'm 99% sure). You can't add or subtract terms with different integer exponents, though you can multiply/divide them (in which case you add or subtract them, respectively) and when you are raising a term to a power you multiply the two exponents. This all applies to non-integer exponents as well.

an integer is simply any whole number (including zero i'm 99% sure).

0 is an integer.
1 is an integer.
-1 is an integer.
0 is the integer between 1 and - 1. :p

Well acceleration kinda has me stumped at the minute. I've just fried my brain for a couple of hours over two questions in particular, one of which I just can't get my head around in the slightest:

A long jumper runs 30m reaching a speed of 10ms^-1 from a standing start. How long does it take him and what's his acceleration?

So, s=30, v=10, u=0 and I need to find t and then a.

It has me utterly stumped. I think I my brain might have shut down an hour ago, though.

Oh yeah, I have no idea whether these are standard letters to use or whatever, so just in case:
s=displacement (metres)
v=final velocity (metres per second)
u=initial velocity (metres per second)
t=time (seconds)
a=acceleration (metres per second squared)

Obvious, really, but I'm a total n00b at maths so I dunno, yeah.

Hmm, seems a bit odd.

v = u + at
at = v - u
t = (v - u)*a^-1
v = 10, u = 0
t = 10/a

s = ut + 0.5at^2
30 = 0.5 * a * (10/a)^2
30 = 0.5 * a * 100 * a^-2
30 = 50 / a
30a = 50
a = 5/3 metres per second^2

t = 10 / (5/3)
t = 30/5
t = 6 seconds.

Pretty sure that's how it's supposed to be done.


In general, if you have some formulas that don't seem to fit the situation, try looking for some substitutions to eliminate any extra unknown variables.

Those are pretty standard letters. If it's constant accelleration then the average speed is ½(u+v)=5m/s. That gives you t, and then you can work out a because at=v-u (because change in speed over time is acceleration).

Edit: I'm opposed to doing this by memorising equations. Understanding concepts and deriving equations is a much better idea.

Hey, I managed to get a 2:1 in a Physics Degree, and those SUVAT equations really came in handy for the two General Physics paper I had to take in years 2 and 3. Pretty useless anywhere else though as they only work for constant acceleration, which is fairly unlikely in any advanced area of Physics. Still, they are handy for A-Level and lower level (i.e. GCSE) mechanics...

The acceleration is constant, yes.

Andrew, if I've understood them, by your calculations t=5 meaning a=2:

2 * 5 = 10 - 0

But in the words of Isaac Washington: **** man, that ain't right!

However, Alien King got it right. t=6s, a=1.6666r ms^-1.

It might be too early in the morning for me to understand it yet but nevertheless I'm a bit lost with your workings, AK. Here's what I do understand of them:

s = ut + 0.5at^2 --- equation to work out the area of the velocity-time graph.
30 = 0.5 * a * (10/a)^2 --- drop the ut (0*x=0)
30 = 0.5 * a * 100 * a^-2 --- square the 10
30 = 50 / a --- 0.5 * 100 = 50
30a = 50 --- 50/30=a
a = 5/3 metres per second^2 --- 50/30ms^-2, or 5/3, or 1 and 2 thirds as my book says.

Obviously there's a lot going on I don't understand. Hnnnngggg...

Oh, and cyph3r, how the hell did Babel not break your spirit? You know it's basically my time there that has made me want to get out of the games industry. Still, building machines sounds way more fun than what I was doing anyway :)

lol, well my stint at Babel made me never want to be a tester, but programming is so much more satisfying and creative, I don't just find problems, I have to fix them, which can be frustrating sometimes, but is also rewarding when it works.. To be honest though, I far prefer the beginning of a project, when you have to come up with things from scratch rather than just fixing bugs.

Heh, sounds fair. I wanted to learn programming but I keep failing at it. I also wanted to be an animator but whilst I love the idea of it and I've enjoyed animating in the past I just get ****ed off with it these days. I'm happier spending 2 hours over building a LEGO linear actuator than I am drawing, these days. It makes me sad on the inside but eventually my army of robots will soothe my emotions.

I like the problem solving aspect of it. I think that's my major interest; if there's a problem and it needs solving creatively I'm all over it like a rash. Maths is a little too well structured for my liking but I have to learn it anyway. ;)

Andrew, if I've understood them, by your calculations t=5 meaning a=2:

2 * 5 = 10 - 0

But in the words of Isaac Washington: **** man, that ain't right!

No, I mean that once you have the average speed (which is 5) and the distance (which is 30), the time is 30÷5=6. Once you have the time (6) and the change in speed (10) you know the acceleration is 10÷6=1⅔.

That seems like the 'natural' order to solve it to me, although armed with the standard equations it's not the easiest way to do it. But that's because I like to understand what's happening and work out simple equations like SUVAT on the fly as and when they're needed.

lol, well my stint at Babel made me never want to be a tester, but programming is so much more satisfying and creative, I don't just find problems, I have to fix them, which can be frustrating sometimes, but is also rewarding when it works.. To be honest though, I far prefer the beginning of a project, when you have to come up with things from scratch rather than just fixing bugs.

I'm the same way. Although I don't mind debugging, because sometimes it's interesting to see where bugs creep in. But it's also monumentally frustrating when I can't replicate the problem and I have to send off code halfway around the world and wait a day every time I attempt a fix. Doing it on my own machine with all the fancy debugging tools in .NET is so easy it's practically a joy.

No, I mean that once you have the average speed (which is 5) and the distance (which is 30), the time is 30÷5=6. Once you have the time (6) and the change in speed (10) you know the acceleration is 10÷6=1⅔.

Oh, okay, that makes sense.

[...]armed with the standard equations it's not the easiest way to do it. But that's because I like to understand what's happening and work out simple equations like SUVAT on the fly as and when they're needed.

I'm trying to do the same. I have no problem doing it other than a lack of practice and the frequent minor errors (resulting in catastrophic failures) which come with it. I really don't want to memorise equations and pass an exam that way. I want to understand the problem and how to solve it. The good thing about learning the equations now is that they help me understand what's happening.

Heh, all of this actually makes me feel really stupid, like a really dense idiot. :)

Edit: I'm opposed to doing this by memorising equations. Understanding concepts and deriving equations is a much better idea.

Agreed. Although when I'm demonstrating something somebody is struggling with, I'll most likely refer to a general equation that they do know.

Yeah, and thanks for that.

Could you explain your fourth step, though, please?

Where did the a^-2 go? Do I fail at some basic principle here? :/

30 = 0.5 * a * 100 * a^-2
to
30 = 50 / a ?

Sorry, I'm a little inconsistent with my labelling here.
a^-2 = 1 / (a^2)

Here, I've done something a bit neater:

http://img53.imageshack.us/img53/4569/forvader.png

Don't be disheartend by being unable to follow someone else's calculations initially. It's common.

I always get annoyed when they only give you three or four of the SUVAT equations. There's five, dammit!

Poor s = vt - ½at²... everyone forgets it...

I always get annoyed when they only give you three or four of the SUVAT equations. There's five, dammit!

Poor s = vt - ½at²... everyone forgets it...

Picky. :p
I don't recall ever using that though, or finding a situation where it would be useful.

Picky. :p
I don't recall ever using that though, or finding a situation where it would be useful.

Well it's just that there's five variables, and the idea of each equation is that you can use it even if you don't have a clue what one of the variables is.

So there should be five equations, dammit. What if I don't know u, huh? WHAT IF I DON'T KNOW U?

EDIT: Okay, specific example. I'm travelling along, but I don't know my speed (say I'm in a go-cart or something), and I need to decelerate to a stop. My acceleration is -2m/s², and I take 3 seconds to stop. How far do I go in this time?

s = vt - ½at²

= 0*t - ½(-2)3² = 9m.

Well it's just that there's five variables, and the idea of each equation is that you can use it even if you don't have a clue what one of the variables is.

So there should be five equations, dammit. What if I don't know u, huh? WHAT IF I DON'T KNOW U?

EDIT: Okay, specific example. I'm travelling along, but I don't know my speed (say I'm in a go-cart or something), and I need to decelerate to a stop. My acceleration is -2m/s², and I take 3 seconds to stop. How far do I go in this time?

s = vt - ½at²

= 0*t - ½(-2)3² = 9m.

By the time you've done the calculations you've already run over the old lady.

Ah, but by how much?

By 27 Werthers.

I always get annoyed when they only give you three or four of the SUVAT equations. There's five, dammit!

Poor s = vt - ½at²... everyone forgets it...

You'll be pleased to know that that equation was the cornerstone of my physics education, then :p

Well, it turns out I fail at GCSE maths, by my reckoning. It's a bit disheartening and makes me very really dim but I'm sure I can fast track it or something. :P

Basically I don't undertsnad quadratic equations. Well, I don't understand how they apply to the questions I'm being asked. In fact, I can only do it with a bunch of examples I do understand in front of me.

Self-teaching sucks. Unless you're a good teacher. Or a good student. I've never been either.

A couple of years ago, I barely got them either (it kinda scares me how you could go through GCSE maths understanding barely anything) then I did 2 maths A-levels. The sheer quantity of stuff and the speed it was thrown at you at A-level meant I got the hang of quadratics, algebraic manipulation (all the stuff I'd successfully bluffed at GCSE), etc, very very quickly. So I suppose finding a textbook and spending a while trudging through dozens of questions might well help you too.

Yeah, and that's what I intend to do.

Right now I'm a bit preoccupied with trying to find a job/apprenticeship.