Evil simultaneous equation!

3i is a number. Imaginary granted, but still a number. Squaring it gives:
(3i)^2 = (3^2).(i^2) = 9.(-1) = -9 which is negative.
Not all numbers squared are positive. Only if they are positive or negative. You may have been ignoring this for sake of clarity, but the 'no matter what it was' bit isn't true.
For those who look at this and think 'Wuh!?', the root of negative one (required in a lot of things from solving quadratic equations to engineering) is written as a lower case i.

shhh, you might be getting a bit ahead of them here. We're talking GCSEs, they haven't come on to complex numbers yet.

Reply 21

username582574

OP

Original post by mzfemalecupid

me tooo
but i like simultaneous equations :biggrin:

i just hate everything else :s-smilie:

phaha I thought I liked them until I came across this question! Normally they're fine though... I much prefer the substitution method to elimination though! Good luck with your exam!

Reply 22

mzfemalecupid

Original post by kpritchx

phaha I thought I liked them until I came across this question! Normally they're fine though... I much prefer the substitution method to elimination though! Good luck with your exam!

thnx

Reply 23

ginister

11

Original post by yaschaeffer

3i is a number. Imaginary granted, but still a number. Squaring it gives:
(3i)^2 = (3^2).(i^2) = 9.(-1) = -9 which is negative.
Not all numbers squared are positive. Only if they are positive or negative. You may have been ignoring this for sake of clarity, but the 'no matter what it was' bit isn't true.
For those who look at this and think 'Wuh!?', the square root of negative one (required in a lot of things from solving quadratic equations to engineering) is written as a lower case i.
If you take the n-th root of a number, there will be n answers. That's why there's the +/- root when you look at square (n=2) roots, ie 2 roots.
For example, there are four 4-th roots of one: 1, -1, i and -i.
If you haven't dealt with solutions of cubic functions, or quadratics where b^2-4ac is negative, then you probably haven't come across this stuff yet. You will.
Enjoy.

Don't forget we are only dealing with questions at GCSE level: For the sake of getting a point across for maximum marks tomorrow lets just keep the currently not required arithmetic out of the way, but thanks for the evidence. :smile:

Reply 24

yaschaeffer

1

Original post by The Mr Z

shhh, you might be getting a bit ahead of them here. We're talking GCSEs, they haven't come on to complex numbers yet.

My bad. Didn't know the GCSE exams were taking place. Good luck to those sitting them! In terms of the difficult concepts I raised, I remember a very very long time ago, my whole year being given questions from the 'Scottish maths Challenge'. I was about 14 (more than 1/2 my life away!). This was a series of maths puzzles produced by a consortium of universities to test early teen school pupil maths skills. These puzzle sheets (2 sides of A4) were produced twice a year, and had about 4 or 5 questions on them. And let me tell you, they were damned hard! One involved proving Fermat's little theorem(!!), and this was another one. Try it. Remember this was for normal 'run of the mill' 13-15yr olds to try to solve. It can't be that hard... can it? :colone:

'' In the attached pic you see a conical flask. This is not a scale diagram. The edges of the conical flask are straight. In the conical flask is a quantity of liquid. The diameter of the base of the flask is 12cm. The diameter of the narrowest part of the neck of the flask is 1cm. The diameter of the flask at the upper surface of the liquid is 9cm. The flask is turned upside down so that the liquid rushes into the (corked) neck of the flask. Calculate X, the diameter of the new upper surface of liquid. ''

And for those that think I've forgotten something, making it impossible to solve, I haven't: that's all the info there was. Looks easy, doesn't it? It isn't!

Hey! If they expected us to solve it at 14, then surely the current crop should be able to :biggrin:

If anyone wants to try this, I definitely recommend waiting till after you've sat your exam!!
I will just say this: I only know one method of solving this (there may be others), but it involves mathematics that is at least at A-level level!! So, I could post the solution if you wanted, unfortunately chances are, you wouldn't understand the proof!
Either way, enjoy your exams!

(edited 12 years ago)

Conical_Flask_Puzzle.jpg10.4KB

Reply 25

hrickards

2

Original post by yaschaeffer

My bad. Didn't know the GCSE exams were taking place. Good luck to those sitting them! In terms of the difficult concepts I raised, I remember a very very long time ago, my whole year being given questions from the 'Scottish maths Challenge'. I was about 14 (more than 1/2 my life away!). This was a series of maths puzzles produced by a consortium of universities to test early teen school pupil maths skills. These puzzle sheets (2 sides of A4) were produced twice a year, and had about 4 or 5 questions on them. And let me tell you, they were damned hard! One involved proving Fermat's little theorem(!!), and this was another one. Try it. Remember this was for normal 'run of the mill' 13-15yr olds to try to solve. It can't be that hard... can it? :colone:

'' In the attached pic you see a conical flask. This is not a scale diagram. The edges of the conical flask are straight. In the conical flask is a quantity of liquid. The diameter of the base of the flask is 12cm. The diameter of the narrowest part of the neck of the flask is 1cm. The diameter of the flask at the upper surface of the liquid is 9cm. The flask is turned upside down so that the liquid rushes into the (corked) neck of the flask. Calculate X, the diameter of the new upper surface of liquid. ''

And for those that think I've forgotten something, making it impossible to solve, I haven't: that's all the info there was. Looks easy, doesn't it? It isn't!

Hey! If they expected us to solve it at 14, then surely the current crop should be able to :biggrin:

If anyone wants to try this, I definitely recommend waiting till after you've sat your exam!!
I will just say this: I only know one method of solving this (there may be others), but it involves mathematics that is at least at A-level level!! So, I could post the solution if you wanted, unfortunately chances are, you wouldn't understand the proof!
Either way, enjoy your exams!

Care to give any clues? I think it's solvable using a *really* horrendous (we're talking multiple lines with surds) simeltanous equation, but I'm sure your way is much cleaner.

Reply 26

OwenP277

Original post by ginister

Squared numbers always turn positive, no matter what they were before. So yes, its plus 1

a square root is an operator which finds all of the possible numbers that can be squared to give the subject of the square root as an outcome, in this context both negative one and positive one can be squared to give 1 so there are two answers

think about it, the question has two simultaneous equations involving a linear line and a circle, so they would meet at two individual points unless it is a tangent which in this case it is not:

Reply 27

yaschaeffer

1

Original post by hrickards

Care to give any clues? I think it's solvable using a *really* horrendous (we're talking multiple lines with surds) simeltanous equation, but I'm sure your way is much cleaner.

If only! I spent weeks picking away at this, scouring a whole pile of advanced maths textbooks from the local library, (and this was before the days of the internet!: best computer in the school ran at an unbelievably lightning fast 33 MHz - no typo!) looking for some way into this. This fairly easy problem to describe is a nightmare to solve.
Essentially solving this boils down (obviously!) to:
1) Calculate volume of liquid in normal flask... Hey! It's part of a cone!
2) Volume in upside down flask must be the same... Hey! it's also a bit of a cone!
3) What must x be to give same volume?

At a first glance, as the liquid is in the shape of a large cone with a smaller cone removed from above, and as I know the equation for finding the volume of a cone { V=(1/3).(pi).r^2.h }, I should be able to work out part 1 fairly quickly... or so I thought. The stumbling block is: What's the height!? You are deliberatly not told what the height is! And as its not a scale diagram, you cannot rely on the angles to calculate the height using trig!

After I'd come up with an answer, I showed it to my maths teacher (these questions were all to be done outside of class), and after looking over my proof, she said it was correct and I was to be congratulated. I then asked her ''Why on earth were we being given stuff that advanced!?'' (or words perhaps not as clean as that! :colondollar:

). She just shrugged and said: ''It's supposed to be hard.'' :eek:

Surds?: Some cube roots involved. That's all.
Simultaneous equations?: Just that V1 = V2 and then some rearranging
Horrendous looking equations?: Yep! Not too bad. It's knowing what the equations are!
Need to your head to 'wrap around' new types of mathematics you may cover in several years time?: Definitely!
And yes... that was a hint!

Unless someone can come up with a method that does not involve funny looking mathematical symbols?
And yes.. another hint.
Enjoy!
Mwahahaha! :colone:

Edit: Seriously though, don't spend time on this if you have exam revision to do! I really mean it!

(edited 12 years ago)

Reply 28

hrickards

2

Original post by yaschaeffer

Edit: Seriously though, don't spend time on this if you have exam revision to do! I really mean it!

Only History and Biology left (no Maths this year), and I needed a break anyway :biggrin:

I need to check my working, but because it's such a nice answer to such a horrible problem, I'm inclined to think the following might be right. Is this what you got?

Spoiler

Reply 29

The Mr Z

13

Original post by yaschaeffer

Challenging question

This isn't actually that difficult (that said, experience counts for a lot, no idea how I'd have handled it at that age), if you spot the trick. You don't need to know the angle at all, only that it's constant (which it must be)

Volume is proportional to the length-dimension cubed, the other factors are all just constant, and as the cones are all similar, the constant is the same.

hence x^3 - 1^3 = 12^3 - 9^3

x^3 = 12^3 - 9^3 + 1 = 1000, x=10

(the height and radius are, as you said, related, in that r/h = tan

\theta

where theta is the angle at the point of the cone. as theta is the same for all the cones, h is proportional to r, so volume is proportional to r^2.r=r^3)

These types of 'advanced' questions are always "spot the trick" questions.

Reply 30

yaschaeffer

1

Original post by The Mr Z

This isn't actually that difficult (that said, experience counts for a lot, no idea how I'd have handled it at that age),
These types of 'advanced' questions are always "spot the trick" questions.

Well, I certainly found it difficult at the time. Hmm.. I think I'll need to have a closer look at the first part of your method, to really get your reasoning. The first part certainly seems to have an elegance. Looking over it, the sums of powers you have is how my proof ends. The difference seems to be on how we get there.

Original post by hrickards

I'm inclined to think the following might be right. Is this what you got?

Yes. That is what I got. And I took weeks to work this out. And 2 people get in a few hours. :angry:

Grumble grumble! Although to be fair, most of those weeks was looking at learning the new tools I was using to help me.

For those that want to see what I cooked up all those years ago, here it is. Warning: pretty advanced to those only doing GCSE (and probably AS-level too). And bear in mind. I hadn't even started my GCSE equivalent when I came up with this! It took another 4 years before we covered this stuff in class!

Spoiler

See. Told you the proof I came up was 'a tad' involved!!

Good luck on those exams!

(edited 12 years ago)

Evil simultaneous equation!

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