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    (Original post by ben-smith)
    Need there be?
    Not really but it never hurts.
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    (Original post by Edwin Okli)
    France or Belgium probably but I wouldn't mind studying a language ab initio. In fact, I would like to.
    Ooh cool. That sounds good
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    (Original post by ben-smith)
     x(x+2)(x+4)(x+6)+16=x^4+12x^3+44  x^2+48x+16=(x^2+6x+4)^2
    What method did you use to work out that x^4+12x^3+44x^2+48x+16=(x^2+6x+4  )^2? I didn't have a clue how to do that bit.

    (Think this is probably a really stupid question ).
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    hey does anyone know any step questions that don't need C3 trig or C4 knowledge (not done it yet)
    thanks
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    (Original post by anyone_can_fly)
    What method did you use to work out that x^4+12x^3+44x^2+48x+16=(x^2+6x+4  )^2? I didn't have a clue how to do that bit.

    (Think this is probably a really stupid question ).
    The question implies that it can be written in the form [quadratic]^2. It is also fairly obvious that the coefficients are going to be integers so right away we know that the first term is x^2 and the last is 4. We are left with one other coefficient to find, I was feeling lazy at the time so I just called it a, expanded the expression, equated coefficients to give me a=6. Some people can just factorise things like that in their heads.
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    (Original post by like_a_star)
    hey does anyone know any step questions that don't need C3 trig or C4 knowledge (not done it yet)
    thanks
    A small goat is tethered by a rope to a fixed point at ground level on a square barn which stands in a horizontal field of grass. The sides of the barn are of length 2a and the rope is of length 4a. Let A be the area of the grass that the goat can graze. Prove that A <(or equal to) 14π(a^2) (π=pi) and determine the minimum value of A.

    Here's a hint if you are struggling.

    Spoiler:
    Show
    Think using circles!


    The answer will require a diagram so I can't really type it up clearly, but if anyone wants my answer I will happily provide although it may not make much sense if you aren't thinking along the right lines.
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    (Original post by ben-smith)
    The question implies that it can be written in the form [quadratic]^2. It is also fairly obvious that the coefficients are going to be integers so right away we know that the first term is x^2 and the last is 4. We are left with one other coefficient to find, I was feeling lazy at the time so I just called it a, expanded the expression, equated coefficients to give me a=6. Some people can just factorise things like that in their heads.
    Oh of course, thanks
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    I'm defiantly applying to do maths finished my personal statement today . Currently thinking kings ucl royal holloway city and Kingston
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    (Original post by LoveLifeHate)
    I'm defiantly applying to do maths finished my personal statement today . Currently thinking kings ucl royal holloway city and Kingston
    Whom are you defying?
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    Definitely* damn iPhone spell corrections
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    (Original post by deejayy)
    A small goat is tethered by a rope to a fixed point at ground level on a square barn which stands in a horizontal field of grass. The sides of the barn are of length 2a and the rope is of length 4a. Let A be the area of the grass that the goat can graze. Prove that A <(or equal to) 14π(a^2) (π=pi) and determine the minimum value of A.

    Here's a hint if you are struggling.

    Spoiler:
    Show
    Think using circles!


    The answer will require a diagram so I can't really type it up clearly, but if anyone wants my answer I will happily provide although it may not make much sense if you aren't thinking along the right lines.
    I've done this one, it was my first every complete solution haha
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    Gcse's: 3A*s 3As 2Bs
    A levels: Maths A*(finished a year early), Further Maths, Physics, chemistry
    Applying to: Cambridge, Bath, Durham, Southampton, Lancaster
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    (Original post by ccfcadam36)
    Gcse's: 3A*s 3As 2Bs
    A levels: Maths A*(finished a year early), Further Maths, Physics, chemistry
    Applying to: Cambridge, Bath, Durham, Southampton, Lancaster
    Woo Lancaster!!!!!!
    Good luck!!
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    (Original post by ben-smith)
    The question implies that it can be written in the form [quadratic]^2. It is also fairly obvious that the coefficients are going to be integers so right away we know that the first term is x^2 and the last is 4. We are left with one other coefficient to find, I was feeling lazy at the time so I just called it a, expanded the expression, equated coefficients to give me a=6. Some people can just factorise things like that in their heads.
    That was obviously too easy

    This is one of my favourites: Prove
    \int^{\infty}_{-\infty} e^{-((x^2)/2)} \ dx = \sqrt{2\pi}
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    (Original post by TheMagicMan)
    That was obviously too easy

    This is one of my favourites: Prove
    \int^{\infty}_{-\infty} e^{-((x^2)/2)} \ dx = \sqrt{2\pi}
    Meh, bookwork. You turn it into a double integral and convert to polar coordinates. That result is too famous to ask as a brain teaser.
    Beautiful result though.
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    (Original post by TheMagicMan)
    That was obviously too easy

    This is one of my favourites: Prove
    \int^{\infty}_{-\infty} e^{-((x^2)/2)} \ dx = \sqrt{2\pi}
    Lol Gaussian Integral ... Brilliant question but I don't have the time now to write all that on LaTeX...
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    (Original post by LoveLifeHate)
    I'm defiantly applying to do maths finished my personal statement today . Currently thinking kings ucl royal holloway city and Kingston
    :woo:
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    Now that I am doing Maths in University, it's surprising how many idiots manage to get in there. It's like I'm in A-level again.
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    (Original post by ben-smith)
    Meh, bookwork. You turn it into a double integral and convert to polar coordinates. That result is too famous to ask as a brain teaser.
    Beautiful result though.
    Ok then. Prove it without any change of variable or transform.
    Spoiler:
    Show
    Consider \int^{\sqrt{n}}_{\sqrt-n} (1-{x^2}/n)^{n/2} \ dx as well as sine reduction formulae NB: I've just realised you need to change the variable to get to sines but it's a much nicer one than polar or inverses
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    (Original post by TheMagicMan)
    Ok then. Prove it without any change of variable or transform.
    I'd have to be Gauss to do that
 
 
 
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