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FP1 question is driving me mad Watch

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    RIGHT, the equation, x^3+4x+3 has roots alpha beta and gamma.(a) Use the subistution x= the square root of u to obtain a cubic equation in u.This supposedly equals u^3+8u^2+16u+9=0If you can solve this n show how you will be an absolute star even my teacher couldnt
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    (Original post by emilyp2206)
    RIGHT, the equation, x^3+4x+3 has roots alpha beta and gamma.(a) Use the subistution x= the square root of u to obtain a cubic equation in u.This supposedly equals u^3+8u^2+16u+9=0If you can solve this n show how you will be an absolute star even my teacher couldnt
    maybe it is late but I cannot make sense of this .... do you have a picture of the question?
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    Its not late i need all the help give us one second
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    (Original post by emilyp2206)
    RIGHT, the equation, x^3+4x+3 has roots alpha beta and gamma.(a) Use the subistution x= the square root of u to obtain a cubic equation in u.This supposedly equals u^3+8u^2+16u+9=0If you can solve this n show how you will be an absolute star even my teacher couldnt
    There's surely more to this. Can you post a picture of the question?

    Edit : too late
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    (Original post by emilyp2206)
    RIGHT, the equation, x^3+4x+3 has roots alpha beta and gamma.(a) Use the subistution x= the square root of u to obtain a cubic equation in u.This supposedly equals u^3+8u^2+16u+9=0If you can solve this n show how you will be an absolute star even my teacher couldnt
    Do you mean that you can't figure out how to get the cubic in u, or you can't go on from that?
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    (Original post by notnek)
    There's surely more to this. Can you post a picture of the question?

    Edit : too late
    posted, its not too late i just left my mock corrections v. late, sorry for the bad lighting
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    (Original post by C0pper)
    Do you mean that you can't figure out how to get the cubic in u, or you can't go on from that?
    i can get as far as substituting but yeah going from there to get the cubic, ive tried rearranging and squaring but i think that might be the wrong approach
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    if anyone has a worked solution i would appreciate it SO much if you could share i picture, all of my class mates and teacher are stumped strangely
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    on this right now
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    (Original post by emilyp2206)
    i can get as far as substituting but yeah going from there to get the cubic, ive tried rearranging and squaring but i think that might be the wrong approach
    firstly I got -9 as the last term of the cubic

    Secondly the cubic in u has solutions alpha2, beta2 and gamma2

    I will now be off... hope it all makes sense
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    (Original post by emilyp2206)
    i can get as far as substituting but yeah going from there to get the cubic, ive tried rearranging and squaring but i think that might be the wrong approach
    Is this the 2015 paper? Swear it's the one I sat last year :lol:

    I'm guessing you've got to u \sqrt u + 4 \sqrt u + 3 = 0 just from substituting in? Try taking the 3 to the other side and squaring, it comes out pretty easily from there. Took me way longer than I'd like to admit to spot that in the actual exam.
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    (Original post by emilyp2206)
    if anyone has a worked solution i would appreciate it SO much if you could share i picture, all of my class mates and teacher are stumped strangely
    Dropped you a PM with the solution.
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    (Original post by C0pper)
    Is this the 2015 paper? Swear it's the one I sat last year :lol:

    I'm guessing you've got to u \sqrt u + 4 \sqrt u + 3 = 0 just from substituting in? Try taking the 3 other to the other side and squaring, it comes out pretty easily from there. Took me way longer than I'd like to admit to spot that in the actual exam.
    yeah its the last question im still struggling
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    I subbed in sqrt(u) into x. Then to get a u^3 I squared the whole thing, but then I get u^3 + 8u^2 + 4u + 6u^3/2 + 24u^1/2 + 9 = 0. Also what exam board was this?
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    shurley if you sub u into the equation it equals
     u^{\frac{3}{2}}+4u^{\frac{1}{2}}  +3=0

    Then move the three over and square
     (u^{\frac{3}{2}}+4u^{\frac{1}{2}  })^2=-3^2
    then expand
     u^3+8u^2+16u=9

    sorry for the slow reply I'm really bad at LaTeX
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    (Original post by emilyp2206)
    yeah its the last question im still struggling
    u\sqrt{u}+4\sqrt{u}+3 =0 \Rightarrow \sqrt{u}(u+4)=-3 \Rightarrow ?

    How can you remove the root?
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    (Original post by NinjaOtter)
    I subbed in sqrt(u) into x. Then to get a u^3 I squared the whole thing, but then I get u^3 + 8u^2 + 4u + 6u^3/2 + 24u^1/2 + 9 = 0. Also what exam board was this?
    It's OCR. Subtract 3 from both sides before squaring.
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    (Original post by olegasr)
    Dropped you a PM with the solution.
    So sound of you, thank you v. much, you doing fp1 in summer?
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    (Original post by C0pper)
    It's OCR. Subtract 3 from both sides before squaring.
    Yeah that makes a lot more sense, I did Edexcel so this wasn't a section we covered. We only covered subbing u into an x at A2
 
 
 
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