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27 March, 22:15

Let a = x^2 + 4. Rewrite the following equation in terms of a and set it equal to zero. (x^2+4) ^2+32=12x^2+48 what are the solutions for x

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  1. 28 March, 00:57
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    Just replace x^2+4 with a

    factor out on left side

    (x^2+4) ^2+32=12 (x^2+4)

    replace x^2+4 with a

    a^2+32=12a

    minus 12a both sides

    a^2-12a+32=0

    solve

    factor

    what 2 numbers mulitply to get 32 and add to get - 12

    -4 and - 8

    (a-4) (a-8) = 0

    set each to zero

    a-4=0

    a=4

    a-8=0

    a=8

    remember

    a=x^2+4

    so

    4=x^2+4

    0=x^2

    0=x

    8=x^2+4

    4=x^2

    +/-2=x

    x=-2, 0, 2
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