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Math Help - Finding the roots of cubic equations

  1. #1
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    Finding the roots of cubic equations

    Hi, I've tried the question below and got stuck halfway =) I'd really appreciate any help!! Thanks!!

    QUESTION: ax^3+bx^2+cx+d=0 has roots A, B and C. Find the new equation for roots (A-k), (B-k), (C-k).

    From eq given,
    SOR=A+B+C=-b/a
    Sum of product of roots taking 2 at a time=AB+BC+AC=c/a
    POR=ABC=-d/a

    Using roots (A-k), (B-k), (C-k),

    SOR=A+B+C-3k=-b/a -3k

    Sum of product of roots taking 2 at a time
    =(A-k)(B-k)+(B-k)(C-k)+(A-k)(C-k)
    =AB+BC+AC+(-A-B-B-C-A-C)k+3k^2
    =AB+BC+AC+(-2A-2B-2C)k+3k^2
    =c/a -2k(A+B+C)+3k^2
    =c/a -2k(-b/a)+3k^2
    =c/a +(2bk)/a + 3k^2

    POR
    =(A-k)(B-k)(C-k)
    =ABC+(-AB-AC-BC)k+(A+B+C)k^2-k^3
    =-d/a -k(c/a) + (b/a)k^2 - k^3

    After this, I got stuck. How do I get the cubic equation? Something just seems wrong. Help! =) Thanks everyone!
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  2. #2
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    Hi

    The equation is therefore

    x^3+(b/a + 3k)x^2+(c/a +(2bk)/a + 3k^2)x+(d/a + k(c/a) + (b/a)k^2 + k^3)=0

    I think that you have made a mistake in the last part
    POR
    =(A-k)(B-k)(C-k)
    =ABC+(-AB-AC-BC)k+(A+B+C)k^2-k^3
    =-d/a -k(c/a) - (b/a)k^2 - k^3
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  3. #3
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    Thanks running-gag for pointing out that mistake!

    The answer given in my book is
    a(x+k)^3 +b(x+k)^2 +c(x+k) +d=0

    how did they exactly get that answer?

    are we supposed to factorize or do something along that line?
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  4. #4
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    Suppose that A is a solution of ax^3+bx^2+cx+d=0
    then aA^3+bA^2+cA+d=0

    a[(A-k)+k]^3+b[(A-k)+k]^2+c[(A-k)+k]+d=0

    Therefore A-k is solution of a[x+k]^3+b[x+k]^2+c[x+k]+d=0
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  5. #5
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    Quote Originally Posted by running-gag View Post
    Suppose that A is a solution of ax^3+bx^2+cx+d=0
    then aA^3+bA^2+cA+d=0

    a[(A-k)+k]^3+b[(A-k)+k]^2+c[(A-k)+k]+d=0

    Therefore A-k is solution of a[x+k]^3+b[x+k]^2+c[x+k]+d=0
    So basically all that working out was redundant?
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