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# Hard maths questions for higher maths GCSE watch

1. (Original post by throwawayayay)
how do you do this?
Expand the square and simplify:

So you need and 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.
2. (Original post by Zacken)
Expand the square and simplify:

So you need and 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
3. (Original post by throwawayayay)
sweet jesus
Was there anything in my explanation that wasn't clear enough?
4. (Original post by Zacken)
Was there anything in my explanation that wasn't clear enough?
No, it was fine, but seems like a horrible question for GCSE
5. (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.
6. (Original post by Zacken)
This 6 mark question came up in my GCSE exam and I quite enjoyed it.

Integers and are such that . Find the possible values of and the corresponding values of .
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
7. (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.
8. (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?
9. (Original post by theconfusedman)
Could there be an irrational number that fits a and b?
If you remove the condition that and are rational, I don't think the equation is solvable by conventional means or at all. It's like saying and letting .
10. (Original post by Zacken)
If you remove the condition that and are rational, I don't think the equation is solvable by conventional means or at all. It's like saying and letting .
lol sorry im not familiar with complex numbers
11. (Original post by theconfusedman)
lol sorry im not familiar with complex numbers
Haha, nevermind then. It's a useful analogy, that's all.
12. (Original post by Zacken)
Spoiler:
Show
Call it then .

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]
13. (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.
14. (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.
15. Let me just insert the whole GCSE Maths syllabus...

So glad I passed my maths last year!!
16. I think this thread may have gone a bit off-topic
17. (Original post by notnek)
I think this thread may have gone a bit off-topic
I think it has, but the OP hasn't been back to attempt any of the problems, anyway.
18. (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
19. (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.
20. (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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