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Core 4 again...i feel like i should know this.. hmm

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    show that [(x^2) / (x^2) -16] = 1 + [16 /(x^2)-16]

    It seems like something i should know..

    Urghh having a bad maths day
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    Notice that x^2 = (x^2-16)+16, so write the numerator of the fraction in this way. What do you notice?

    You could use partial fractions, writing \dfrac{x^2}{x^2-16} = A + \dfrac{Bx+C}{x^2-16}, but this will take a lot more time and effort than it's worth by spotting the above trick. (NB: most partial fractions with quadratics on the numerator, in a C4 exam, will use the above trick.)
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    \displaystyle \frac{x^2}{x^2-16} = \frac{x^2-16+16}{x^2-16}

    Does that help?

    Being confident with this kind of manipiulation is very useful at A Level.
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    It's the same as (x^2 - 16) + 16 / x^2 - 16
    You can then split it into: (x^2 - 16 / x^2 - 16) + (16 / x^2 - 16)
    Which should simplify to give your answer.
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    ooooh yes they cancel! sneaky... thank you! i would have never figured that out in an exam..

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    (Original post by sarah-madeline)
    ooooh yes they cancel! sneaky... thank you! i would have never figured that out in an exam..

    You should try to practice it.

    If the numerator looks like the denominator but with something missing, it's often the case that you can add and subtract something, leading to simplification like you've been shown above.
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    (Original post by sarah-madeline)
    ooooh yes they cancel! sneaky... thank you! i would have never figured that out in an exam..

    Actually before using partial fraction, ALWAYS consider this method, it is a lot faster.

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