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  1. #1
    Junior Member alwaysalillost's Avatar
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    Solve

    Solve. Leave in simplest radical form.

    1. x^2 + 8x = -16

    2. 3x^2 = 2x + 5

    Please help me solve. Thanks!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by alwaysalillost View Post
    Solve. Leave in simplest radical form.

    1. x^2 + 8x = -16
    I presume you aren't allowed to use the quadratic formula. So I'll solve this using the "complete the square" method.

    x^2 + 8x = -16

    The form of a perfect square is
    x^2 + (2a)x + a^2 = (x + a)^2

    Comparing the LHS of both expressions (ie. comparing the coefficients of the x terms) I get that
    2a = 8

    a = 4

    So we want to add a^2 = 4^2 = 16 to both sides of the equation. Thus
    x^2 + 8x + 16 = -16 + 16

    (x + 4)^2 = 0

    x + 4 = 0

    x = -4

    -Dan
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  3. #3
    Junior Member alwaysalillost's Avatar
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    Thanks Topsquark!
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by alwaysalillost View Post
    2. 3x^2 = 2x + 5
    This one's a bit tougher:

    3x^2 - 2x = 5

    Divide through by the 3:
    x^2 - (2/3)x = (5/3)

    Now complete the square:
    2a = -(2/3) ==> a = -(1/3)

    So we need to add a^2 = 1/9 to both sides:
    x^2 - (2/3)x + (1/9) = (5/3) + (1/9)

    (x - 1/3)^2 = 16/9

    x - 1/3 = (+/-)sqrt(16/9) = (+/-)(4/3)

    x = (1/3) (+/-) (4/3)

    So
    x = 1/3 + 4/3 = 5/3
    or
    x = 1/3 - 4/3 = -1

    -Dan
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  5. #5
    Junior Member alwaysalillost's Avatar
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    Thanks topsquark!
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