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Math Help - completing the square

  1. #1
    Member
    Joined
    Aug 2007
    Posts
    148

    completing the square

    Hello everyone,

    Could someone please tell me if this is the correct way to complete the square? If not, could you please show me where I went wrong?

    11a^2+7a=6
    11a^2+7a-6=0
    11(a^2+7/11a-6/11)=0
    11(a+7/11/2)^2-7/11-6/11=0
    11(a+7/11/2)^2=13/11
    (a+49/121)^2=13/121
    a+49/121=square root(13/121)
    a=-.077

    Thank you very much
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  2. #2
    Newbie
    Joined
    Dec 2008
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    13
    11a^2+7a=6
    a^2+[7/11]a+[49/484]=[313/484]
    (a+[7/22])^2=[313/484]
    a+[7/22]=squrt[313/484]
    a=-[7/22]+/-squrt[313/484]

    dont know if thats any good to you, I'm waiting for a reply for something else so thought I'd give this ago
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  3. #3
    Member
    Joined
    Jan 2009
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    108
     11a^2 + 7a = 6
     a^2 + (7/11)a = (6/11)
     a^2 + (7/11)a + (7/22)^2 = (6/11) + (7/22)^2
     (a + (7/22))^2 = (264/484) + (49/484)
     (a + (7/22))^2 = 313/484
     a + (7/22) = (+/-) \sqrt{313}/\sqrt{484}
     a + (7/22) = (+/-) \sqrt{313}/22
     a = -7/22 (+/-) \sqrt{313}/22
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