c4countdown

My family have spent a lot of time doing their maths homework this week and and completed it bar the following equation which has baffled both my son who is good at maths + my wife who is also quite a good mathematician.

((x+1/2x-1) + (3x-2/2x+1)) -2 = 0

They have both tried various ways of solving it but to no avail.

Have any of you maths whizzes got any tips as to how to tackle it please?

Many thanks

Al

Assuming I've interpreted it correctly (your brackets are very dodgy), the question is You need to get rid of the fractions, so first multiply through by (2x-1) to get

(x+1) + (3x-2)(2x-1)/(2x+1) - 2(2x-1) = 0

and then multiply again by (2x+1) to get

(x+1)(2x+1) + (3x-2)(2x+1) - 2(2x+1)(2x-1) = 0

This is now a normal quadratic; multiply out the brackets and gather the x^2's, x's and numbers. The x^2's cancel and you get left with 5 - 4x = 0, which obviously has the solution x=1.25.

Incidentally how old is your son?

What Charlie said

Charlie Reams wrote:Assuming I've interpreted it correctly (your brackets are very dodgy), the question is

Code: Select all

x+1      3x-2
----  +  ----  - 2 = 0
2x-1     2x+1

Makes sense now, I didn't interpret the question like that.

Yeah, the bracketing implies that the question is

x + 1/(2x) - 1 + 3x - 2/(2x) + 1 = 0

but that's such a weird way to write the equation that I assumed it must be something else.

Funny, I leapt to completely the wrong interpretation, which we're all assuming is correct. It was only when you said about the bracketing being dodgy that I realised!

Jon Corby wrote:Funny, I leapt to completely the wrong interpretation, which we're all assuming is correct. It was only when you said about the bracketing being dodgy that I realised!

So are you saying you think the question means something else?

Charlie Reams wrote:

Jon Corby wrote:Funny, I leapt to completely the wrong interpretation, which we're all assuming is correct. It was only when you said about the bracketing being dodgy that I realised!

So are you saying you think the question means something else?

No! I agree with what you wrote. But that is, of course, not exactly what that specific question is asking, but what we assume the original question is asking, and the bracketing errors have arisen from Allan's attempt to type it here.

Clear?

Jon Corby wrote:

Charlie Reams wrote:

Jon Corby wrote:Funny, I leapt to completely the wrong interpretation, which we're all assuming is correct. It was only when you said about the bracketing being dodgy that I realised!

So are you saying you think the question means something else?

No! I agree with what you wrote. But that is, of course, not exactly what that specific question is asking, but what we assume the original question is asking, and the bracketing errors have arisen from Allan's attempt to type it here.

Clear?

I see I see.

Charlie Reams wrote:Assuming I've interpreted it correctly (your brackets are very dodgy), the question is

Code: Select all

x+1      3x-2
----  +  ----  - 2 = 0
2x-1     2x+1

You need to get rid of the fractions, so first multiply through by (2x-1) to get

(x+1) + (3x-2)(2x-1)/(2x+1) - 2(2x-1) = 0

and then multiply again by (2x+1) to get

(x+1)(2x+1) + (3x-2)(2x+1) - 2(2x+1)(2x-1) = 0

This is now a normal quadratic; multiply out the brackets and gather the x^2's, x's and numbers. The x^2's cancel and you get left with 5 - 4x = 0, which obviously has the solution x=1.25.

Incidentally how old is your son?

Thanks very much Charlie - This is very kind of you - Mike is 16 and on track for A* in his GCSE maths.

We now know where he went wrong when he was multiplying through and got positive x instead of -7x to add back the 3x and get -4x.

Thanks again

Al

ps sorry for setting it out so badly - I should have used separate lines like you did

Allan Harmer wrote:ps sorry for setting it out so badly - I should have used separate lines like you did

Or you could simply have bracketed it correctly:

(x+1)/(2x-1) + (3x-2)/(2x+1) - 2 = 0

A rectangular brick has a volume of 72 cubic inches and one edge twice the length of one of the others. What is its minimum surface area?

GCE O-Level 1964

David Williams wrote:A rectangular brick has a volume of 72 cubic inches and one edge twice the length of one of the others. What is its minimum surface area?

GCE O-Level 1964

I make it 106 square inches.

Kai Laddiman wrote:

David Williams wrote:A rectangular brick has a volume of 72 cubic inches and one edge twice the length of one of the others. What is its minimum surface area?

GCE O-Level 1964

I make it 106 square inches.

Sure you don't mean 108? Or have you found a way of slicing two off?

David Williams wrote:A rectangular brick has a volume of 72 cubic inches and one edge twice the length of one of the others. What is its minimum surface area?

GCE O-Level 1964

I make it 108 square inches. I did it the good old fashioned way (no calculator) so chances are that it's wrong.

Charlie Reams wrote:

David Williams wrote:A rectangular brick has a volume of 72 cubic inches and one edge twice the length of one of the others. What is its minimum surface area?

GCE O-Level 1964

I make it 108 square inches. I did it the good old fashioned way (no calculator) so chances are that it's wrong.

It would make me T_T if anyone thought they needed a calculator for this...

Michael Wallace wrote:It would make me T_T if anyone thought they needed a calculator for this...

Well not so much a calculator, as a simple thing on tinternet where I put height, width and breadth and it pops out the answer And I have an A level in maths... oh dear!

I wonder how you were supposed to do this at the time - I just differentiated my expression for the surface area once I'd used the other information to simplify it, but it occurs to me that maybe you're meant to presume that the dimensions are all integers and so you're limited to only 3 options, and then just check both to see which yields the smaller area.

Is there a nice way to do it without calculus without presuming the dimensions are integers?

Ah yes! I forgot about the other possible dimensions... assumed there was only one possibility.

Yeah I just used calculus (which they probably had to do back in those days).

I got the lengths as 3, 4 and 6. This meant a surface area of 36+24+48=108, as already answered.

Elementary calculus was GCSE material at the time so it's possible that that's what was expected.

For anyone interested, the solution goes like this. Call the lengths of the sides a, b and c then:

One side is double the length of another, so a = 2b (it doesn't matter which pair we pick, but let's say we do it this way.)
We know the volume of the cube, so abc = 72.
The surface area S is 2(ab+bc+ac). Rearranging the above two equations allows you to eliminate first a and then c to get S = 4b^2 + 216/b.
Then differentiate S with respect to b to get dS/db = 8b - 216/b^2.
To find the minimum, set dS/db equal to 0 and solve, to give b = 3.
Then plug this value of b into the formula for S to get 108.

Would be interested to see if Harmer Jnr can do this question.

Kirk Bevins wrote:Yeah I just used calculus (which they probably had to do back in those days).

I got the lengths as 3, 4 and 6. This meant a surface area of 36+24+48=108, as already answered.

What I meant.

I just remember this as a neat little problem that illustrates the power of calculus.

Log tables are provided. (What's a calculator?)

Charlie Reams wrote:Elementary calculus was GCSE material at the time so it's possible that that's what was expected.

Would be interested to see if Harmer Jnr can do this question.

I'll get him working on it and report back.

I did it by finding the prime factors of 72 (2, 2, 2, 3, 3), finding the sets that fit the condition a=2b (2, 4, 9 and 3, 4, 6) and seeing which 2(ab + ac + bc) gives the lowest value. Nothing complicated or anything.

Kai Laddiman wrote:I did it by finding the prime factors of 72 (2, 2, 2, 3, 3), finding the sets that fit the condition a=2b (2, 4, 9 and 3, 4, 6) and seeing which 2(ab + ac + bc) gives the lowest value. Nothing complicated or anything.

You've assumed the dimensions are integers, though, which is why I was wondering if there was a neater way than calculus that didn't make this assumption (oh, and you missed 1, 6 and 12).

Michael Wallace wrote:

Kai Laddiman wrote:I did it by finding the prime factors of 72 (2, 2, 2, 3, 3), finding the sets that fit the condition a=2b (2, 4, 9 and 3, 4, 6) and seeing which 2(ab + ac + bc) gives the lowest value. Nothing complicated or anything.

You've assumed the dimensions are integers, though, which is why I was wondering if there was a neater way than calculus that didn't make this assumption (oh, and you missed 1, 6 and 12).

But it worked, OK?

Allan Harmer wrote:

Charlie Reams wrote:Elementary calculus was GCSE material at the time so it's possible that that's what was expected.

Would be interested to see if Harmer Jnr can do this question.

I'll get him working on it and report back.

He got it right by using trial and error - He hasn't done Calculus yet, but is doing differentiation and integration, which I guess is leading up to Calaculus, which is on his syllabus for the next 1/2 term.

Kai Laddiman wrote:But it worked, OK?

Clearly you are not a mathematician

Michael Wallace wrote:

Kai Laddiman wrote:I did it by finding the prime factors of 72 (2, 2, 2, 3, 3), finding the sets that fit the condition a=2b (2, 4, 9 and 3, 4, 6) and seeing which 2(ab + ac + bc) gives the lowest value. Nothing complicated or anything.

You've assumed the dimensions are integers, though, which is why I was wondering if there was a neater way than calculus that didn't make this assumption (oh, and you missed 1, 6 and 12).

And 36. My first trial and error stab was to assume the brick was 1 inch, 2 inches wide and 36 inches long...

Phil Reynolds wrote:

Michael Wallace wrote:

Kai Laddiman wrote:I did it by finding the prime factors of 72 (2, 2, 2, 3, 3), finding the sets that fit the condition a=2b (2, 4, 9 and 3, 4, 6) and seeing which 2(ab + ac + bc) gives the lowest value. Nothing complicated or anything.

You've assumed the dimensions are integers, though, which is why I was wondering if there was a neater way than calculus that didn't make this assumption (oh, and you missed 1, 6 and 12).

And 36. My first trial and error stab was to assume the brick was 1 inch, 2 inches wide and 36 inches long...

Ha, good point. All this just goes to show how sexy calculus is. (I tried to use this to demonstrate to CF how sexy calculus is, I'm not sure if he was convinced...)

Michael Wallace wrote:I tried to use this to demonstrate to CF how sexy calculus is, I'm not sure if he was convinced...

CF

Allan Harmer wrote:Mike is 16 and on track for A* in his GCSE maths.

based on his classwork or the homework that you do for him?

Roxanne wrote:

Allan Harmer wrote:Mike is 16 and on track for A* in his GCSE maths.

based on his classwork or the homework that you do for him?

ooooh, Roxanne

Roxanne wrote:

Allan Harmer wrote:Mike is 16 and on track for A* in his GCSE maths.

based on his classwork or the homework that you do for him?

Based on the results of the modules he has already passed at A* Cheeky twat! Sorry Kai.

Allan Harmer wrote:Based on the results of the modules he has already passed at A* Cheeky twat!

GCSEs are modular these days?

Allan Harmer wrote:

Roxanne wrote:

Allan Harmer wrote:Mike is 16 and on track for A* in his GCSE maths.

based on his classwork or the homework that you do for him?

Based on the results of the modules he has already passed at A* Cheeky twat!

Haha, the idea of my dad calling somebody named Roxanne a twat on an online forum.

Michael Wallace wrote:

Allan Harmer wrote:Based on the results of the modules he has already passed at A* Cheeky twat!

GCSEs are modular these days?

I know Michael - not like in our day when you had to cram for the exam at the end of the course!

They have 3 modules to sit - He has passed the two taken so far at A* and has the third to take in April/May + the Additional Maths he has been put in for.

Allan Harmer wrote:He hasn't done Calculus yet, but is doing differentiation and integration, which I guess is leading up to Calaculus, which is on his syllabus for the next 1/2 term.

I'm not sure what distinction you're drawing here, since calculus basically is differentiation and integration (at least until you get into the heavy stuff at degree level) but in any case differentiation is all that's needed here.

Incidentally I did Additional Maths back in the day, I'm surprised it's still going given the existence of AS levels, and the fact that no one recognised it as a qualification.

Jimmy Gough wrote:

Allan Harmer wrote:

Roxanne wrote:based on his classwork or the homework that you do for him?

Based on the results of the modules he has already passed at A* Cheeky twat!

Haha, the idea of my dad calling somebody named Roxanne a twat on an online forum.

Yeah, come on, there's no need to use that sort of language when talking to an 18 year old girl. For shame.

Jon Corby
Yeah, come on, there's no need to use that sort of language when talking to an 18 year old girl. For shame.