The Maths Orgy Thread - The Bear and Badger

GooberTheHat

@workid, yep. See my edit.

WorKid

I think quadratics were pretty well explained earlier in this very Fred.

acemuzzy

The completing the squares thing was in fact elsewhere. But it ties into that, yup.

GooberTheHat

Yeah, I asked the completing the squares question to. It's this one problem that's throwing me though. I don't know if it is the -6x^2 that's throwing my but I'm drawing a blank when it comes to working out how to solve it.

acemuzzy

I shall attempt to clarify. It will all make sense. Just need the kids to go to sleep...

acemuzzy

So, I'll take a slightly circuitous route there, cos that's how I like to roll... We'll get to an magic formula at the end...

First things first, the stuff from the completing the square thread...

(x+b)^2 = (x+b)(x+b) = x.x + b.x + x.b + b.b = x^2 + 2xb + b^2

Now that's kinda handy, because if you have something like

x^2 + bx

you can cunningly go

(x+b/2)^2 = [from the formula above] x^2 + 2x.b/2 + (b/2)^2 = x^2 + bx + (b^2)/4

so, rearranging,

(x^2 + bx) = (x+b/2)^2 - (b^2)/4

That may seem irrelevant to your question. But does that vaguely make sense so far?

acemuzzy

right, so then what you do is to think about the case you have, with

ax^2 + bx + c = 0, and you want to work x out.

start by dividing by a:

[x^2 + b/a x] + c / a = 0

Then see my post above, and you have

[(x + b/2a)^2 - (b/a)^2/4] + c/a = 0

but everything apart from the left is like numbers, so

(x+b/2a)^2 = (b/a)^2/4 - c/a

but then that's easy, you can take a square root (remembering it could be the positive or negative root)!

x+b/2a = +/- sqrt((b/a)^2/4 - c/a)

and hence x = -b/2a +/- sqrt((b/a)^2/4 - c/a)

which you normally see as (from http://en.wikipedia.org/wiki/Quadratic_equation):

$bba46479d07af61b6e596219ce4e2cd5.png$

acemuzzy

So you had -6x^2 - x +7 = 0

So a = -6, b = -1, c = 7

hence, from the formula, x = (1+/-sqrt(1-4.-6.7)/2.-6

= -(1 +/- sqrt (1+24.7) / 12)
= -(1 +/- sqrt(1+168) / 12)
= -(1 +/- sqrt (169) ) / 12
= -(1+/-13) / 12
= -(1+13)/12 and -(1-13)/12
= -14/12 and 1

acemuzzy

Now the formula is a bit cheating. What I'd have actually done faced with

5x^2 -10x^2 = -7 +x +x^2

is tried to "find a root" - i.e. one solution for which is's true.

I'd have re-arranged to 6x^2+x-7=0 like you did (though I hate having a minus sign in front of the x^2), and I'd probably have noticed that putting x = 1 in works (6+1-7 = 0).

At that point, you can do some factorization...

given 1 is a solution, you can work out it must rearrange to (x-1)(something) = 0, as putting 1 into that obviously does make zero.

then it's a case of working out the something... start with the x^2 term... must be:

(x-1)(6x + something) to give 6x^2

but that multiplies out as 6x^2 -6x + something x - something

and you want that to be 6x^2+x-7, so something must be 7

hence you have

6x^2+x-7=0 ---->
(x-1)(6x+7) = 0

And for two things to give zero, either one or the other has to be zero. So, either
x = 1, or
6x+7 = 0 --> x = -7/6 (=-14/12)

acemuzzy

TL;DR.

Me + maths = LazyGunn + computer graphics

:-)

GooberTheHat

Cheers muzzy, I'd not come across this

$bba46479d07af61b6e596219ce4e2cd5.png$

Before, so that's a massive help. Thank you.

acemuzzy

it's a good'un. One of those things everyone in the generation above us had to rote learn. Not now we have calculators though

davyK

it only works if b^2 > 4ac though, otherwise you get into complex number world.

GooberTheHat

I'm going sans calculator, so rote learning of formula is probably required.

acemuzzy

davyK wrote:

it only works if b^2 > 4ac though, otherwise you get into complex number world.

It still gives the right answer, though!

davyK

True

dynamiteReady

I'm not really big on maths, so don't rock me to hard, but...

Is there any real difference between a hierarchy and a nested set?

acemuzzy

Neither of those is really a "maths" term. I think I'd view them differently, but would say there's a trivial mapping of one to the other (i.e. set of children = set of contained sets). A hierarchy has an implied ordering, where as sets don't (you're effectively giving it via "contains").

dynamiteReady

acemuzzy wrote:

Neither of those is really a "maths" term. I think I'd view them differently, but would say there's a trivial mapping of one to the other (i.e. set of children = set of contained sets). A hierarchy has an implied ordering, where as sets don't (you're effectively giving it via "contains").

That's how I've been distinguishing between the two.
I guess I've just been explaining it in a really shit way. :[

Hodge360

Do you use this stuff regularly @acemuzzy ?
Had a few "oh yeah,I remember something about that" moments while reading through your breakdown there but I've no idea how I'd get back into the habit of seeing problems like that.

acemuzzy

Nah, I don't really get to do proper maths these days. Did a maths degree back in the day, and it's just kinda stuck since then. I'll occassionally go and read bits and bobs / attempt a puzzle of some sort to keep the grey cells ticking over, but most of my time is writing code and dickin about in project planning spreadsheets!

[Deleted User]

Basic for you folks, but maybe fun, I dunno...

Amazon Free App of the day (Android) is King of Maths. Grab it quick if you want FREE MATHS.

cockbeard

I have to recommend this blog here. Really good stuff, local teacher bloke, not updated mega often but well worth a look

http://matheminutes.blogspot.co.uk/

RamSteelwood

you may have seen/heard about people moaning about a 'hard' maths question in the maths GCSE...here it is. i was able to work it out and it's a long time since i did any maths, so assuming the question doesn't involve stuff completely not taught at GCSE level, it's a shame so many people seem to be moaning about it...the idea that not everyone will correctly answer every question in an exam would seem to be the point of them to me.

There are n sweets in a bag. Six of the sweets are orange. The rest of the sweets are yellow. Hannah takes a random sweet from the bag. She eats the sweet. Hannah then takes at random another sweet from the bag. She eats the sweet. The probability that Hannah eats two orange sweets is 1/3. Show that n²-n-90=0

voices

We were trying to work out what people were finding hard about it at work this morning. I think it's the bit where you write out the probabilities in algebra, but it doesn't seem like a hard question. I don't really know what level people are at for gcse maths though.

Maybe they should have said that n was the number of sweets in the bag

cockbeard

Wow, they did tell her what n was, and they also gave her the answer. All she had to do was choose a path from one to the other

Which means if you go wrong you know and can try again, rather than being overconfident in a wrong answer because you forgot to carry a zero or flip a negative when changing sides

I think they've held her hand there, amazed that they're complaining

acemuzzy

My guess is that people weren't expecting to have to algebra in a stats exam, or something along those lines. There's an element of rote learning that doesn't help cross-module thinking in maths, I suspect.

(My other suspicion was that "hang on, that doesn't have an integer solution". But it does.)

(I also initially ballsed the maths up by forgetting the first sweet had got eaten. Oops. That may be a pitfall some fell into - (6/n)^2 type stuff.)

voices

I think if you'd fallen into the (6/n)^2 = 1/3 pitfall, you'd hopefully have noticed that that 6 x 6 x 3 isn't quite 90, but 6 x 5 x 3 is 90. Giving the answer might have allowed them to set a more difficult question, since you can work at it from both angles. I don't know - I'm not a teacher.

WorKid

It's dead easy, fuck knows what they were on about.
Kids!

It's actually a very nice well written question. Plus I assume it was only for a couple of marks, probably next part was to solve the quadratic which obviously factors easily and conclude you can't have negative sweets in a bag.

voices

Yeah, back in my day we had to do proper sums on our graphics calculators.

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