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Math Help - biquadratics

  1. #1
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    biquadratics

    Hi I cant seem to solve this equation. Wikipedia said to sub z=x^2 but I don't know how to then get the x results back. Once i have done this I get confused as to how to put it back into two quadratics which is what i've been aed to do.

    Q Express x^4-4x^2+16 in the form (x^2+Ax+B)(x^2+Cx+D)

    thnx.
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  2. #2
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    Quote Originally Posted by oxrigby View Post
    Hi I cant seem to solve this equation. Wikipedia said to sub z=x^2 but I don't know how to then get the x results back. Once i have done this I get confused as to how to put it back into two quadratics which is what i've been aed to do.

    Q Express x^4-4x^2+16 in the form (x^2+Ax+B)(x^2+Cx+D)

    thnx.
    (x^2+Ax+B)(x^2+Cx+D)
    =x^4 + Cx^3 + Dx^2 + Ax^3 + ACx^2 + ADx + Bx^2 + BCx + BD

    Comparing x^3 coefficient: 0 = C + A
    Comparing x^2 coefficient: -4 = D+AC+B
    Comparing x coefficient: 0 = AD + BC
    Comparing constant coefficient: 16 = BD

    Solve for A,B,C, D and substitute this value into: (x^2+Ax+B)(x^2+Cx+D)
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  3. #3
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    If u sub z into x^4-4x^2+16 u get z^2-4z+16 because z^2=x^4.
    z^2-4z+16 factorises into (z-2-i\sqrt12)(z-2+i\sqrt12)

    So
    factorises into (x^2-2-i\sqrt12)(x^2-2+i\sqrt12)

    So A=C=0
    B= -2-i\sqrt12
    and D= -2+i\sqrt12
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  4. #4
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    Question says,'where ABCD are real constants'''.??
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  5. #5
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    Quote Originally Posted by oxrigby View Post
    Question says,'where ABCD are real constants'''.??
    Then
    A= \sqrt12
    B=4
    C= -\sqrt12
    D=4

    (x^2-2-i\sqrt12)(x^2-2+i\sqrt12) = (x^2+4x-\sqrt12)(x^2+4x+\sqrt12) = x^4-4x^2+16
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  6. #6
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    Quote Originally Posted by oxrigby View Post
    Express x^4-4x^2+16 in the form (x^2+Ax+B)(x^2+Cx+D)
    Difference of two squares:

    \begin{aligned}x^4-4x^2+16 &= (x^2+4)^2 - 12x^2 \\ &= (x^2+4+\sqrt{12}x)(x^2+4-\sqrt{12}x) \\ &= (x^2+2\sqrt3x+4)(x^2-2\sqrt3x+4).\end{aligned}
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