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Math Help - Let's see if any of you have the same

  1. #1
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    Let's see if any of you have the same

    Find two consecutive even integers whose sum squared is 100.
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  2. #2
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    Find two consecutive even integers whose sum squared is 100.

    Now my two numbers will be x and x+2 and their sum -squared- will be 100.

    (x+x+2)^2=100

    (x+x+2)(x+x+2)=100

    x^2+x^2+2x+x^2+x^2+2x+2x+2x+4=100

    4x^2+4x+4-100=0

    4x^2+4x-96=0

    4x^2+4x-96=0 /4

    x^2+x-24=0

    Now I try to find the factor for middle term

    1(-24)=-24 -23
    2(-12)=-24 -10
    3(-8)=-24 -5
    4(-6)=-24 -2
    6(-4)=-24 +2

    Mmm...I Can't find two factors that add to +1...Grrr....
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  3. #3
    MHF Contributor
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    You made a mistake

    4x^2+8x+4-100=0

    4x^2+8x-96=0

    x^2+2x-24=0
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  4. #4
    Super Member
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    Quote Originally Posted by Alienis Back View Post
    Find two consecutive even integers whose sum squared is 100.

    Now my two numbers will be x and x+2 and their sum -squared- will be 100.

    (x+x+2)^2=100

    (x+x+2)(x+x+2)=100

    x^2+x^2+2x+x^2+x^2+2x+2x+2x+4=100

    4x^2+4x+4-100=0

    4x^2+4x-96=0

    4x^2+4x-96=0 /4

    x^2+x-24=0

    Now I try to find the factor for middle term

    1(-24)=-24 -23
    2(-12)=-24 -10
    3(-8)=-24 -5
    4(-6)=-24 -2
    6(-4)=-24 +2

    Mmm...I Can't find two factors that add to +1...Grrr....
    You could've avoided all this mess by simply taking the square root:

    (2x+2)^2=10^2 \implies 2x+2=\pm 10 \implies x = -1 \pm 5
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