Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    22
    ReputationRep:
    (Original post by throwawayayay)
    how do you do this?
    Expand the square and simplify: a^2 + 45 + a + (6a-b)\sqrt{5} = 51 \Rightarrow a^2 + a - 6 + (6a-b)\sqrt{5} = 0

    So you need a^2 + a - 6 = 0 and 6a-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.
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Expand the square and simplify: a^2 + 45 + a + (6a-b)\sqrt{5} = 51 \Rightarrow a^2 + a - 6 + (6a-b)\sqrt{5} = 0

    So you need a^2 + a - 6 = 0 and 6a-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:
    Offline

    22
    ReputationRep:
    (Original post by throwawayayay)
    sweet jesus :afraid:
    Was there anything in my explanation that wasn't clear enough?
    Offline

    2
    ReputationRep:
    (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
    Offline

    22
    ReputationRep:
    (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:
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    This 6 mark question came up in my GCSE exam and I quite enjoyed it.

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

    22
    ReputationRep:
    (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.
    Offline

    2
    ReputationRep:
    (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?
    Offline

    22
    ReputationRep:
    (Original post by theconfusedman)
    Could there be an irrational number that fits a and b?
    If you remove the condition that a and b are rational, I don't think the equation is solvable by conventional means or at all. It's like saying x+iy = a+ib and letting a, b, x, y \in \mathbb{C}.
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    If you remove the condition that a and b are rational, I don't think the equation is solvable by conventional means or at all. It's like saying x+iy = a+ib and letting a, b, x, y \in \mathbb{C}.
    lol sorry im not familiar with complex numbers
    Offline

    22
    ReputationRep:
    (Original post by theconfusedman)
    lol sorry im not familiar with complex numbers
    Haha, nevermind then. It's a useful analogy, that's all. :lol:
    Offline

    11
    ReputationRep:
    (Original post by Zacken)
    Spoiler:
    Show
    Call it y then \displaystyle y = \frac{1}{1 + \frac{2}{1 + y}} = \frac{1+y}{3+y} \Rightarrow y^2 + 2y - 1 = 0 \Rightarrow y = \sqrt{2} - 1.


    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]
    Offline

    22
    ReputationRep:
    (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.
    Offline

    13
    ReputationRep:
    (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.
    Offline

    3
    ReputationRep:
    Let me just insert the whole GCSE Maths syllabus...

    So glad I passed my maths last year!!
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    I think this thread may have gone a bit off-topic
    Offline

    11
    ReputationRep:
    (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.
    Offline

    11
    ReputationRep:
    (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 x_0 = 1, x_{n+1} = \frac{1}{1+x_n}
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (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.
    Offline

    11
    ReputationRep:
    (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.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What's your favourite Christmas sweets?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.