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Hard maths questions for higher maths GCSE

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Reply 60
Original post by throwawayayay
how do you do this?


Expand the square and simplify: a2+45+a+(6ab)5=51a2+a6+(6ab)5=0a^2 + 45 + a + (6a-b)\sqrt{5} = 51 \Rightarrow a^2 + a - 6 + (6a-b)\sqrt{5} = 0

So you need a2+a6=0a^2 + a - 6 = 0 and 6ab=06a-b = 0 by comparing the rational and irrational parts of both sides of the equation. Tripped up plenty of people in the exam, took me some time to work it out and the only reason I did was because I'd looked up complex numbers for fun and was used to dealing with imaginary and real parts, this was analogous.
Original post by Zacken
Expand the square and simplify: a2+45+a+(6ab)5=51a2+a6+(6ab)5=0a^2 + 45 + a + (6a-b)\sqrt{5} = 51 \Rightarrow a^2 + a - 6 + (6a-b)\sqrt{5} = 0

So you need a2+a6=0a^2 + a - 6 = 0 and 6ab=06a-b = 0 by comparing the rational and irrational parts of both sides of the equation. Tripped up plenty of people in the exam, took me some time to work it out and the only reason I did was because I'd looked up complex numbers for fun and was used to dealing with imaginary and real parts, this was analogous.


sweet jesus :afraid:
Reply 62
Original post by throwawayayay
sweet jesus :afraid:


Was there anything in my explanation that wasn't clear enough? :redface:
Original post by Zacken
Was there anything in my explanation that wasn't clear enough? :redface:


No, it was fine, but seems like a horrible question for GCSE
Reply 64
Original post by throwawayayay
No, it was fine, but seems like a horrible question for GCSE


Ah, yeah. It was my IGCSE which is notoriously tough. It was something like 156/160 raw for an A* that year as well. :lol:
Original post by Zacken
This 6 mark question came up in my GCSE exam and I quite enjoyed it.

Integers aa and bb are such that (a+35)2+ab5=51(a + 3\sqrt{5})^2 + a - b\sqrt{5} = 51. Find the possible values of aa and the corresponding values of bb.


Nice question, it can only be done if you know that its an integer, thats pretty clever lol. Question like that came in my mocks
Reply 66
Original post by theconfusedman
Nice question, it can only be done if you know that its an integer, thats pretty clever lol. Question like that came in my mocks


Being integers is far too strong a condition. You need only that they be rational. Same cardinalities, but oh well.
Original post by Zacken
Being integers is far too strong a condition. You need only that they be rational. Same cardinalities, but oh well.

Could there be an irrational number that fits a and b?
Reply 68
Original post by theconfusedman
Could there be an irrational number that fits a and b?


If you remove the condition that aa and bb are rational, I don't think the equation is solvable by conventional means or at all. It's like saying x+iy=a+ibx+iy = a+ib and letting a,b,x,yCa, b, x, y \in \mathbb{C}.
Original post by Zacken
If you remove the condition that aa and bb are rational, I don't think the equation is solvable by conventional means or at all. It's like saying x+iy=a+ibx+iy = a+ib and letting a,b,x,yCa, b, x, y \in \mathbb{C}.


lol sorry im not familiar with complex numbers
Reply 70
Original post by theconfusedman
lol sorry im not familiar with complex numbers


Haha, nevermind then. It's a useful analogy, that's all. :lol:
Original post by Zacken

Spoiler



D'you have a better way to justify taking the positive root?


I'm not very familiar with the theory of continued fractions, so maybe it's done with a more general argument there, but if I had to do it here, I would replace the continued fraction with an equivalent recursive series, then show that the series is decreasing and bounded below by some positive number. That proves than the sequence converges to a +ve limit.

That would then be enough to justify taking the +ve root, and the trick that you've used above (which implicitly assumes convergence)

Maybe someone like Gregorius knows how this is done more generally?

[edit: Have I tagged him properly? If not, something is effed up on this site]
Reply 72
Original post by atsruser
I'm not very familiar with the theory of continued fractions, so maybe it's done with a more general argument there, but if I had to do it here, I would replace the continued fraction with an equivalent recursive series, then show that the series is decreasing and bounded below by some positive number. That proves than the sequence converges to a +ve limit.


Ah, thanks. I'll go have a read up on that in a sec because I'm not quite sure what you mean by an equivalent recursive series. How would you go about constructing that recursive series?

@Gregorius - tagged him properly for you.
Original post by atsruser
I'm not very familiar with the theory of continued fractions, so maybe it's done with a more general argument there, but if I had to do it here, I would replace the continued fraction with an equivalent recursive series, then show that the series is decreasing and bounded below by some positive number. That proves than the sequence converges to a +ve limit.

That would then be enough to justify taking the +ve root, and the trick that you've used above (which implicitly assumes convergence)

Maybe someone like Gregorius knows how this is done more generally?

[edit: Have I tagged him properly? If not, something is effed up on this site]


Only here for a couple of minutes, so haven't read the question properly! But what I think you're looking for is the idea of a convergent to a continued fraction.
Let me just insert the whole GCSE Maths syllabus...

So glad I passed my maths last year!!
Reply 75
I think this thread may have gone a bit off-topic :smile:
Original post by notnek
I think this thread may have gone a bit off-topic :smile:


I think it has, but the OP hasn't been back to attempt any of the problems, anyway.
Original post by Zacken
Ah, thanks. I'll go have a read up on that in a sec because I'm not quite sure what you mean by an equivalent recursive series. How would you go about constructing that recursive series?


I meant something like x0=1,xn+1=11+xnx_0 = 1, x_{n+1} = \frac{1}{1+x_n}
Reply 78
Original post by atsruser
I think it has, but the OP hasn't been back to attempt any of the problems, anyway.

There were other posters who were looking for GCSE level questions. It would have been nice if this thread was kept at GCSE level.

There are plenty of other threads for posting hard A Level questions.
Original post by notnek
There were other posters who were looking for GCSE level questions. It would have been nice if this thread was kept at GCSE level.

There are plenty of other threads for posting hard A Level questions.

All of the questions that I've posted can be done with GCSE knowledge + insight. It's the insight that makes them hard, of course.

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