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    How would u complete the square for this:
    2x^2 + 8x - 5

    i've got up to getting 2(x + 2)^2 + ????
    i can't get to the final bit, which is - 13. Can sum1 please explain how to do this.

    this is wat i'm doing:
    2(x + 2)^2 - 2^2 - 5
    2(x + 2)^2 - 9 (which is wrong).
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    Should be -2*2^2 instead.

    For future reference, just expand what you have and adjust accordingly.
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    Know that x2 + bx = (x + b/2)2 - b2/4

    => 2x2 + 8x - 5 = 2(x2 + 4x) - 5 = 2(x + 2x)2 - 5 - 2(42/4) = 2(x + 2x)2 - 13.
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    2x^2+8x-5
    = 2(x^2+4x) -5
    = 2(x-2)^2 -(2^2)2 -5
    =2(x-2)^2 -8-5
    =2(x-2)^2 -13

    ok so i basically did wat dvs did but i guess i just wanted to be overefficient or something!! :P
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    thanx a lot guyz, i completely forgot about the 2* part. lol...
 
 
 
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Updated: April 9, 2006
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