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    EDIT : I managed to get there in the end. But I'll leave this question up anyway if others want to try.

    The quadratic equation x^2-5x+7=0 has roots \alpha and \beta. Find the quadratic equation with integer coefficents that has roots \alpha^3 and \beta^3.

    So I'm aware of how to use Vieta's formulas to solve this easily. I also know the substitution method for questions like this although I wouldn't consider it for this question because I know that the algebra isn't nice.

    I saw a markscheme today that mentioned the substitution method as an alternate approach (but didn't show the working) so I'm wondering if there's a simple way of doing this particular question using substitution that I'm missing? After substituting x=u^{\frac{1}{3}} it leads to

    u^{\frac{2}{3}} - 5u^{\frac{1}{3}}+7=0

    But I can't get from here to a quadratic by algebraic manipulation.
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    Sum of roots = + . =(a+b)(a^2-ab+b^2)=(a+b)[(a+b)^2 - 3ab] = 20
    Product of roots = 7^3 = 343
    --> x^2 - 20x + 343=0.

    Did you want it done using that specific u=x^1/3 method?
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    (Original post by thekidwhogames)
    Sum of roots = + . =(a+b)(a^2-ab+y^2)=(a+b)[(a+b)^2 - 3ab] = 20
    Product of roots = 7^3 = 343
    --> x^2 - 20x + 343=0.

    Did you want it done using that specific u=x^1/3 method?
    Yes as mentioned in my OP I was asking about the substitution method. I managed to get there in the end but it's still worth having a go
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    (Original post by Notnek)
    Yes as mentioned in my OP I was asking about the substitution method. I managed to get there in the end but it's still worth having a go
    Ah - that's a lot harder?

    How would that go about using u=x^1/3?
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    (Original post by Notnek)
    Yes as mentioned in my OP I was asking about the substitution method. I managed to get there in the end but it's still worth having a go
    Bit late so I can't be bothered to look at it further, but the quickest I found is to say:

    u^{1/3}(u^{1/3}-5)=-7 \Leftrightarrow u(u-15u^{2/3}+75u^{1/3}-125) = -343
    (cube both sides and just use the expansion (u+1)^3 = u^3+3u^2+3u+1)

    Then note that -15^{2/3} + 75 u^{1/3} = -15u^{1/3}(u^{1/3}-5) = -15(-7) = 105 so the expansion just simplifies down to u(u-20)=-343 hence u^2-20u+343=0
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    (Original post by Notnek)
    EDIT : I managed to get there in the end. But I'll leave this question up anyway if others want to try.

    The quadratic equation x^2-5x+7=0 has roots \alpha and \beta. Find the quadratic equation with integer coefficents that has roots \alpha^3 and \beta^3.

    So I'm aware of how to use Vieta's formulas to solve this easily. I also know the substitution method for questions like this although I wouldn't consider it for this question because I know that the algebra isn't nice.

    I saw a markscheme today that mentioned the substitution method as an alternate approach (but didn't show the working) so I'm wondering if there's a simple way of doing this particular question using substitution that I'm missing? After substituting u=x^{\frac{1}{3}} it leads to

    u^{\frac{2}{3}} - 5u^{\frac{1}{3}}+7=0

    But I can't get from here to a quadratic by algebraic manipulation.
    Spoiler:
    Show








    u^{\frac{2}{3}}-5u^{\frac{1}{3}}+7=0

    u^{\frac{1}{3}}(u^{\frac{1}{3}}-5)=-7

    u(u-15u^{\frac{2}{3}}+75u^{\frac{1}{  3}}-125)=-343

    u(u-15(u^{\frac{2}{3}}-5u^{\frac{1}{3}}+7)-20)=-343

    u^2-20u+343=0







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    (Original post by BuryMathsTutor)
    Spoiler:
    Show









    u^{\frac{2}{3}}-5u^{\frac{1}{3}}+7=0

    u^{\frac{1}{3}}(u^{\frac{1}{3}}-5)=-7

    u(u-15u^{\frac{2}{3}}+75u^{\frac{1}{  3}}-125)=-343

    u(u-15(u^{\frac{2}{3}}-5u^{\frac{1}{3}}+7)-20)=-343

    u^2-20u+343=0

    This is the way I did it. I made a slip in my working so gave up on this approach initially which is why I created the thread. But I saw the mistake straight after posting.

    I'd be interested to know how many students used this method for this exam question.
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    (Original post by Notnek)
    u=x^{\frac{1}{3}}
    Did you mean u=x^3 ?
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    (Original post by I hate maths)
    Did you mean u=x^3 ?
    Yes. I meant to post x=u^{\frac{1}{3}}.
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    (Original post by Notnek)
    This is the way I did it. I made a slip in my working so gave up on this approach initially which is why I created the thread. But I saw the mistake straight after posting.

    I'd be interested to know how many students used this method for this exam question.
    Where is the question from?
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    (Original post by BuryMathsTutor)
    Where is the question from?
    OCR FP1 Jan 2008 Q9

    Part i makes it even less likely that students would have used the substitution method. It's not mentioned in the examiners report.
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    (Original post by Notnek)
    OCR FP1 Jan 2008 Q9

    Part i makes it even less likely that students would have used the substitution method. It's not mentioned in the examiners report.
    Just a few then perhaps. We will never know.
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    (Original post by Notnek)
    OCR FP1 Jan 2008 Q9

    Part i makes it even less likely that students would have used the substitution method. It's not mentioned in the examiners report.
    Given part (i), I wouldn't be surprised if there wasn't a single correct solution using the substitution method. (Because you'd need to be either pretty clueless, or feeling particularly perverse, to do so).

    Giving 6 marks for (ii) seems pretty generous, also?
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    (Original post by DFranklin)
    Given part (i), I wouldn't be surprised if there wasn't a single correct solution using the substitution method. (Because you'd need to be either pretty clueless, or feeling particularly perverse, to do so).

    Giving 6 marks for (ii) seems pretty generous, also?
    I also wouldn't be surprised but I think there were probably some who attempted substitution. I've got a feeling there are some students who would think "Substitution always seems faster so I'll do it here". Hopefully these students switched to the Vieta's formula method once they realised the algebra was hard!

    6 marks does seem generous since i) has already been done.
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    (Original post by Notnek)
    I also wouldn't be surprised but I think there were probably some who attempted substitution.
    I'm sure there were some who *attempted* it, yes. But to think substitution is the way to go here, you'd need to be obvious to the fact that you've already done most of the work in (i). And if you were oblivious to that, I think it very unlikely you'd have the ability to work through the fairly gnarly substitution algebra.
 
 
 
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