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Math Help - Further maths. Cubics. What?

  1. #1
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    Further maths. Cubics. What?

    Ok this 1 confused my maths teacher! You might need a pen and paper for this....

    We have the equation 3x^3 + 6x^2 - 4x +7 = 0 with roots a,b,c
    we gotta find the equation with root b+c, c+a, a+b
    We did quite a lot of it this is how far we got...

    written in the form x^3 - (sum)x^2 + (pairs)x - (product) = 0

    sum = -6/3 = -2
    pairs = -4/3
    product = -7/3

    New sum = (b+c) + (c+a) + (a+b)
    = 2a + 2b + 2c
    = 2(a+b+c)
    =2*-2 = -4 We know this from the SUM worked out earlier

    New Pairs = (b+c)(c+a) + (c+a)(a+b) + (b+c)(a+b)
    = bc + ab + y^2 + ac + ac +bc + a^2 + ba + b^2 + ba + bc + ac
    = a^2 + b^2 + c^2 + 3(cb + ca + ab)
    = (a+b+c)^2 - 2(ab + ac + bc) + 3(-4/3) we know this from the PAIRS worked out earlier
    =(-2)^2 - 2(-4/3) + (-4)
    = 4 + 8/3 - 4
    =8/3

    Here comes the challenge...

    New product = (b+c)(c+a)(a+b)
    =(bc + ba + c^2 + ac)(a+b)
    = bca + yb^2 + ba^2 + ab^2 + ac^2 + bc^2 + ca^2 + abc
    = 2abc + a^2(b+c) + c^2(a+b) + b^2(c+a)
    = 2(-7/3) .............

    Please Help!!
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  2. #2
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    Quote Originally Posted by djmccabie View Post
    Ok this 1 confused my maths teacher! You might need a pen and paper for this....

    We have the equation 3x^3 + 6x^2 - 4x +7 = 0 with roots a,b,c
    we gotta find the equation with root b+c, c+a, a+b
    We did quite a lot of it this is how far we got...
    It would be [x-(b+c)][x - (a+c)] [x -  (a+b)] = 0.

    Note, (b+c)+(a+c)+(a+b) = 2(a+b+c) = -2\cdot \tfrac{6}{3} = -4

    And, (b+c)(a+c)+(b+c)(a+b)+(a+c)(a+b) = (a^2+b^2+c^2) + 3(ab+ac+bc)
    This becomes (a^2+b^2+c^2+2ab+2ac+2bc)+ab+ac+bc = (a+b+c)^2 + (ab+ac+bc) = (-\tfrac{6}{3})^2 - \tfrac{4}{3}

    Finally, (a+b)(a+c)(b+c)=(a+b+c)(ab+bc+ac) - abc = (-\tfrac{6}{3})(-\tfrac{4}{3}) - (-\tfrac{7}{3})

    Use that to set up the cubic.
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  3. #3
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    Thanks a lot!! you made a bit more work for yourself by repeating most of my working out though but thanks for the last part!!
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  4. #4
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    Hello, djmccabie!

    You did some awesomely excellent work!



    Here's one more fact we can dig up . . .


    We know: . \begin{array}{cccc}a + b + c &=& \text{-}2 & {\color{blue}[1]}\\ \\[-4mm] ab+bc + ac &=& \text{-}\frac{4}{3} & {\color{blue}[2]}\\ \\[-4mm] abc &=& \frac{7}{3} & {\color{blue}[3]} \end{array}


    Square [1]: . (a + b + c)^2 \;=\;(\text{-}2)^2 \quad\Rightarrow\quad a^2 + 2ab + 2ac + b^2 + 2bc + c^2\:=\:4


    \text{We have: }\;a^2+b^2+c^2 + 2\underbrace{(ab + bc + ac)}_{\text{This is -}\frac{4}{3}} \:=\:4 \quad\Rightarrow\quad a^2 + b^2 + c^2 + 2\left(\text{-}\frac{4}{3}\right) \:=\:4


    Therefore: . a^2+b^2+c^2\;=\;\frac{20}{3}

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  5. #5
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    just a quick check now. Do you get the answer 3x^3 + 12x^2 + 8x - 15 = 0 ?

    (note coefficients must be integers)
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