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Thread: biquadratics

  1. #1
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    biquadratics

    Hi I cant seem to solve this equation. Wikipedia said to sub $\displaystyle z=x^2$ 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 $\displaystyle x^4-4x^2+16$ in the form $\displaystyle (x^2+Ax+B)(x^2+Cx+D)$

    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 $\displaystyle z=x^2$ 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 $\displaystyle x^4-4x^2+16$ in the form $\displaystyle (x^2+Ax+B)(x^2+Cx+D)$

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

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

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

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

    So A=C=0
    B=$\displaystyle -2-i\sqrt12$
    and D=$\displaystyle -2+i\sqrt12$
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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=$\displaystyle \sqrt12$
    B=4
    C=$\displaystyle -\sqrt12$
    D=4

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

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