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    (Original post by Zacken)
    You'll want to spell maths properly.
    LOL.
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    (Original post by The gains kinggg)
    You'll want to check the math before you try to 'help' people
    We all make mistakes don't we?
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    Actually looking at the quartic equation, if we set  e^x=w and divide equation by  w^2 and then making a substitution of  z=w+w^{-1} .It might work but I haven't tried it yet.

    Oh, I didn't realise someone had posted this before me.
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    (Original post by The gains kinggg)
    You'll want to check the math before you try to 'help' people
    Don't be mean to Zacken please.
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    (Original post by Farhan.Hanif93)
    By the latter option, I presume they didn't expect you to solve the quartic via direct factorisation. Instead, I suspect you're supposed to note:

    \left(e^x-e^{-x}\right)^2 = e^{2x}+e^{-2x}-2

    And let w=e^x - e^{-x}, so that the equation reduces to the quadratic:

    w^2 -7w -8=0

    Which you can solve for w and then solve further quadratics from the definition of w for e^x.

    However, it's worth pointing out that this is 'secretly' rebuilding the hyperbolic identity for \cosh 2x in terms of \sinh x so it pays to know those instead.
    Hmm, the MS just glosses over the factorisation, But what youre saying makes a lot of sense. Thanks for the help!

    (Original post by TeeEm)
    Use identities which is very easy

    or using exponentials you get what is known as a symmetric polynomial
    I see, but I was just wondering how people solved it if they used the exponential definitions.

    (Original post by 13 1 20 8 42)
    Usually if some polynomial of degree more than 2 comes up and its factorisation isn't immediately obvious you probably made a wrong turn..
    I see, but I was just wondering how people solved it if they used the exponential definitions.
    Zacken
    Thanks for all your help people!
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    (Original post by P____P)
    Hmm, the MS just glosses over the factorisation, But what youre saying makes a lot of sense. Thanks for the help!
    I find it amusing how the mark scheme awards marks for the correct quartic, but fails to give the correct quartic itself...
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    (Original post by Farhan.Hanif93)
    I find it amusing how the mark scheme awards marks for the correct quartic, but fails to give the correct quartic itself...

    What do you mean?

    The second line?

    Posted from TSR Mobile
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    (Original post by Farhan.Hanif93)
    By the latter option, I presume they didn't expect you to solve the quartic via direct factorisation. Instead, I suspect you're supposed to note:

    \left(e^x-e^{-x}\right)^2 = e^{2x}+e^{-2x}-2

    And let w=e^x - e^{-x}, so that the equation reduces to the quadratic:

    w^2 -7w -8=0

    Which you can solve for w and then solve further quadratics from the definition of w for e^x.

    However, it's worth pointing out that this is 'secretly' rebuilding the hyperbolic identity for \cosh 2x in terms of \sinh x so it pays to know those instead.
     -7w=-7e^x+7e^{-x} . So you would have to be  w=e^x+e^{-x} .
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    (Original post by P____P)
    What do you mean?

    The second line?

    Posted from TSR Mobile
    Yes, the e^x term should have a positive coefficient.

    Anyhow, to spot the factorisation, you could superficially let u=e^x and v=e^{2x}. Then, putting these into the quartic in a particular way:

    v^2 - (7u+2)v-(8u^2-7u-1)=0

    (Where I've chosen to write the middle term -10e^{2x} = -(2v+8u^2) because then the term that is (implicitly) independent of v is 1. a quadratic; and 2. factorable as (8u+1)(u-1), allowing us to be able to factorise the quadratic in v more neatly.)

    Then note that we can factorise the quadratic in v as:

    (v+[u-1])(v-[8u+1])=0

    Which unwraps as (e^{2x}+e^x-1)(e^{2x}-8e^x-1)=0, satisfying the mark scheme's factorisation.

    [Note: I don't think such a factorisation is reasonable in exam conditions.]

    (Original post by B_9710)
     -7w=-7e^x+7e^{-x} . So you would have to be  w=e^x+e^{-x} .
    Careful. -7e^x+7e^{-x} = -7(e^x-e^{-x}), so w must be chosen as written in my post.
 
 
 
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