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    (Original post by Nichrome)
    Yes, unless I'm missing something, when you multiply up and square both sides of the expression, you get a quartic, which if you plug it into a computer, gives the correct solution (out of four).
    When I did it yesterday it was a disguised quadratic, which means it wasn't too bad to solve by hand (although I left some nasty square root in the answer). Negative answers, or non-real ones can be disregarded of course.
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    (Original post by aKarma)
    If I were an eleven year old presented with this question i would have drawn a drawing. To test the accuracy of this I did this to a scale of 1cm to 1m (m are the units in the question). This easily found x to be 0.16m, 9m. Obviously these are approximate given on my diagram they were .16cm and 9 cm. Checking these finds them reasonably accurate. While this isn't a mathematical solution I think it is probably better than breaking down and crying, and would show potential for an 11 year old...
    yes... potential to FAIL!!!
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    (Original post by nota bene)
    When I did it yesterday it was a disguised quadratic, which means it wasn't too bad to solve by hand (although I left some nasty square root in the answer). Negative answers, or non-real ones can be disregarded of course.
    And how was a 1950's 13 year old supposed to evaluate those nasty square roots? Obviously there's something missing/dodgy given the context of the question. Could the OP please show us where he found it?
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    (Original post by ukdragon37)
    And how was a 1950's 13 year old supposed to evaluate those nasty square roots? Obviously there's something missing/dodgy given the context of the question. Could the OP please show us where he found it?
    Surds existed in the 1950s you know!

    As did tables of square roots and slide rules.
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    I was given this problem last year. The numbers were different but I ******* hated it. It took me an hour and a ******* half and I had to get 2 quartics and eliminate the x^3. I hated it so much I remember the solutions with my particular numbers were 30.1 and 7.0. I hated it, so much.
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    SOHCAHTOA baby
    got me through last year
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    (Original post by Hancock orbital)
    Saying that is just another way of saying you do not know how to do it - so it clearly is not straightforward.

    This might look like a straightforward integral:

    e^-x^2. I assure it is not. However, being like you, it would be all too easy to say - it looks straightforwawrd but have forgotten the knowledge of how to do it.

    You clearly have no idea how to do the question - how can you possibly say it is easy?
    No, I would compare it to evaluating the integral of 3cosx(e^2x). Then it would be perfectly reasonable to say "This question looks fairly straightforward, but I have forgotten how to do integration by parts exactly. Just look the formula up and do it"

    The original question seems similar to many other trig questions: you are given lots of lengths to work with. Your analogy is not similar to any other problems we know about. And comparing integration to simple trigonometry is a bit silly.

    (No doubt there will be times when something might look easily integrable and won't be, but the point is that we have a simple variation on a problem, and nothing conceptually harder.)
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    Note that if x is the distance between the box and one corner, then the distance between the box and the other corner is 1/x by similarity.

    Then (x + 1)^2 + (1/x + 1)^2 = 10^2

    So x^2 + 2x + 1 + 1/x^2 + 2/x + 1 = 10^2
    Or (x^2 + 2 + 1/x^2) + 2(x + 1/x) = 100
    (x + 1/x)^2 + 2(x+1/x) - 100 = 0

    Then solve this quadratic to find x+1/x, and solve the resulting quadratic to find x.

    Unless I'm missing something, you're not going to get a quartic which is just a quadratic in x^2 so I can't see an easier way of doing this.

    (Original post by Profesh)
    Not really. It strikes me as a relatively straightforward problem; I just don't recall the relevant trigonometry.
    I think it is actually fairly tricky.
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    (Original post by Mr M)
    Surds existed in the 1950s you know!

    As did tables of square roots and slide rules.
    Of course you would be able to solve it using uh, more primitive methods :p: but if I gave this question to some of the best maths people in my school they'll just go :eek: at it.

    I know standards have fallen but surely not that fallen.
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    (Original post by Hancock orbital)
    Saying that is just another way of saying you do not know how to do it - so it clearly is not straightforward.

    You clearly have no idea how to do the question - how can you possibly say it is easy?
    You are assuming Profesh is intelligent(probably reasonable) or, more precisely, good-at-trig.
    What he is saying is not silly. If a random person cant recall the method through temporary oblivion, does not mean the problem is hard.
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    I ended up with quartic equation too when I tried algebraic methods.

    I think I got the correct solution by making a geometric construction .
    Similar triangles anyone?
    I guess Euclidean geometry isn't taught anymore,but even for 11 years olds fifty years ago that's a shocker! :eek:

    edit:WRONG!
    Cut that...
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    (Original post by Chrrye)
    You are assuming Profesh is intelligent(probably reasonable) or, more precisely, good-at-trig.
    What he is saying is not silly. If a random person cant recall the method through temporary oblivion, does not mean the problem is hard.
    Yet that still does not imply the converse.
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    Have you missed something out of the problem Nichrome?
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    (Original post by Nichrome)
    Yes, unless I'm missing something, when you multiply up and square both sides of the expression, you get a quartic, which if you plug it into a computer, gives the correct solution (out of four).
    Yes if you wanted to obtain the correct answer to this question you would be required to solve the quartic, which I do agree is very demanding for an 11/13 year old!

    However, whilst many of you are stating that you think this question is too difficult for a student of that age I believe that the school merely wanted to see how the students thought and how they would approach the question even if they could not find the final answer.
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    (Original post by jboy184)
    Yes if you wanted to obtain the correct answer to this question you would be required to solve the quartic, which I do agree is very demanding for an 11/13 year old!

    However, whilst many of you are stating that you think this question is too difficult for a student of that age I believe that the school merely wanted to see how the students thought and how they would approach the question even if they could not find the final answer.
    I think the crucial thing people have failed to pick up on is that Nichrome reproduced the question from memory and admitted that he could not remember the actual numbers.

    The question can easily be manipulated so the quartic factorises and solves without difficulty. One way would be to ensure the hypotenuse is of the form \sqrt{k^2-1}.

    Might I suggest people try it again with a hypotenuse of 2\sqrt6 ? You will find it a much more satisfying experience!
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    Can I ask what quartic equation everyone ended up with, when they were solving the original?
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    (Original post by Robbie10538)
    Can I ask what quartic equation everyone ended up with, when they were solving the original?
    The original quartic was x^4+2x^3-98x^2+2x+1=0.

    It can easily be solved in an exact form as it is palindromic but the solutions are not particularly attractive!
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    Ahh nice. Never had to solve a quartic like that before, so wasn't sure of the method, so thanks for teaching me something new ^_^.

    I got the question from my maths supervisor, apparently he sometimes sets it on an entrance exam for admissions to natural sciences, and apparently hardly anyone solves it, but it's useful as the students try out lots of different methods, revealing how much mathematics they know in the process.
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    (Original post by Nichrome)
    Ahh nice. Never had to solve a quartic like that before, so wasn't sure of the method, so thanks for teaching me something new ^_^.

    I got the question from my maths supervisor, apparently he sometimes sets it on an entrance exam for admissions to natural sciences, and apparently hardly anyone solves it, but it's useful as the students try out lots of different methods, revealing how much mathematics they know in the process.
    Interesting. Yes it is a very rich question.
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    (Original post by Erdős)
    Attempt


     x > 0 .

    Hence:

     x = \frac{ \left( \sqrt{101} - 1 \right) \pm \sqrt{ 98 - 2 \sqrt{101}}}{2} .

    Can this be further simplified?

    .
    No that's the best you can do.
 
 
 
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