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Math Help - Completing the square in quadratics of the form x^2 + bx + c

  1. #1
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    Completing the square in quadratics of the form x^2 + bx + c

    I am trying to understand where I am getting these types of questions wrong, it seems algebra is not my best subject at the moment.

    x^2 - 8x - 5
    (x - 4)^2 - 16 + 5 = (x - 4)^2 - 11

    Check it;

    (x - 4)^2 - 11 = x^2 - 8x + 5 + 11 = x^2 - 8x + 16

    I think the problem is in the (x - 4) area as this does not work back out to the original question. I have also tried (x - 2)^2 but this also results in x^2 + 4x + 4, which is not the same.

    My course book talks a lot about halfing the coefficients of x, but I am missing something somewhere?
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    Re: Completing the square in quadratics of the form x^2 + bx + c

    (x-4)^2=x^2-8x+16.

    So, (x-4)^2-11=x^2-8x+16-11\dots
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    Re: Completing the square in quadratics of the form x^2 + bx + c

    Why did you change the original -5 to 5 in the second line?
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    Re: Completing the square in quadratics of the form x^2 + bx + c

    Quote Originally Posted by SammyS View Post
    (x-4)^2=x^2-8x+16.

    So, (x-4)^2-11=x^2-8x+16-11\dots
    So if I am reading your example correctly, you are saying that it is correct to carryout a subtraction from the LHS and Add - 11 to the RHS.

    (x - 4)^2 - 11 - 11 = x^2 - 8x + 16 - 11 = x^2 - 8x - 5

    I tried another method of algebra;

    x2 8x 5
    (x 4)2 16 + 5
    (x 4)2 = 11
    Check;

    (x 4)2 11 = x2 8x - 5 11
    = x2 8x - 16
    (x 4)2 11 = x2 8x - 16 + 11
    = x2 8x 5
    Last edited by David Green; August 8th 2011 at 11:43 AM. Reason: I tried a different method of algebra
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    MHF Contributor Siron's Avatar
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    Re: Completing the square in quadratics of the form x^2 + bx + c

    16-11=5, not -5.
    I would write:
    x^2-8x-5=(x-4)^2-21
    Because:
    (x-4)^2-21=x^2-8x+16-21=x^2-8x-5

    Or does it has to be x^2-8x+5? In that case: (x-4)^2-11 is correct.
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    Re: Completing the square in quadratics of the form x^2 + bx + c

    Quote Originally Posted by Siron View Post
    16-11=5, not -5.
    I would write:
    x^2-8x-5=(x-4)^2-21
    Because:
    (x-4)^2-21=x^2-8x+16-21=x^2-8x-5

    If (x - 4)^2 gives - 16 and I add 5 then this = - 11.

    Or does it has to be x^2-8x+5? In that case: (x-4)^2-11 is correct.
    The original equation was as you say; x^2 - 8x - 5. I understand the half term (x - 4)^2 but -4^2 = - 16 which is why I changed the sign convention from + 5 to - 5.

    So in your example above because you are saying x^2 - 8x + 5, you are saying that (x - 4)^2 - 11 is correct, but I changed the - 5 to a + 5 from - 16 to get - 11.

    when I add -11 to both sides I get x^2 - 8x - 5, which is the original equation, so are you saying that I am wrong?

    Thanks

    David
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    MHF Contributor Siron's Avatar
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    Re: Completing the square in quadratics of the form x^2 + bx + c

    Notice: (x-4)^2=x^2-8x+16, you're right about -4^2=-16 but in this case it's (-4)^2 which is 16.
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    Re: Completing the square in quadratics of the form x^2 + bx + c

    Quote Originally Posted by David Green View Post
    The original equation was as you say; x^2 - 8x - 5. I understand the half term (x - 4)^2 but -4^2 = - 16 which is why I changed the sign convention from + 5 to - 5.
    -16 is the right thing to put, but it looks like you got it for the wrong reason. When you expand (x-4)^2, you should do (-4)^2, which equals 16. But that has nothing to do with the 5.

    What completing the square does is to represent algebraically what it would mean to literally fill in an incomplete square. You're on the right track with (x-4)^2, but this gives x^2-8x+16. But you can't just go adding 16 to one side of an equation, so you have to add it to the other side or subtract it again from the same side, which is why you're right by having -16 there.

    This has nothing to do with the -5 though, and I'm not sure why you tried to change it.
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    Re: Completing the square in quadratics of the form x^2 + bx + c

    Quote Originally Posted by Siron View Post
    Notice: (x-4)^2=x^2-8x+16, you're right about -4^2=-16 but in this case it's (-4)^2 which is 16.
    So would I be right along these lines;

    x^2 - 8x - 5

    (x - 4)^2 16 - 5

    (x - 4)^2 = 11

    (x - 4)^2 16 = x^2 - 8x + 11

    (x - 4)^2 16 - 16 = x^2 - 8x + 11 - 16

    = x^2 - 8x - 5

    Thanks

    David
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    Re: Completing the square in quadratics of the form x^2 + bx + c

    Quote Originally Posted by David Green View Post
    So would I be right along these lines;



    (x - 4)^2 16 - 5
    What does (x-4)^2 16 means?
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    Re: Completing the square in quadratics of the form x^2 + bx + c

    Quote Originally Posted by Siron View Post
    What does (x-4)^2 16 means?
    (x - 4)^2 = 16 as you said previously, it's just the way I have written it out.
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  12. #12
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    Re: Completing the square in quadratics of the form x^2 + bx + c

    You've to be careful with notations like that, it can be very confusing.
    Start with:
    (x-4)^2=x^2-8x+16
    Well, you want to have x^2-8x-5 so you have to substract a term -21 from each side, because 16-21=-5 and this what you want, so:
    (x-4)^2-21=x^2-8x+16-21 \Leftrightarrow (x-4)^2-21=x^2-8x-5
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    Re: Completing the square in quadratics of the form x^2 + bx + c

    Quote Originally Posted by Siron View Post
    You've to be careful with notations like that, it can be very confusing.
    Start with:
    (x-4)^2=x^2-8x+16
    Well, you want to have x^2-8x-5 so you have to substract a term -21 from each side, because 16-21=-5 and this what you want, so:
    (x-4)^2-21=x^2-8x+16-21 \Leftrightarrow (x-4)^2-21=x^2-8x-5
    Very much appreciated for all your help.

    So writing the quadratic equation x^2 - 8x - 5 in completed square form is;

    (x - 4)^2 = 16
    Last edited by David Green; August 8th 2011 at 02:11 PM.
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    Re: Completing the square in quadratics of the form x^2 + bx + c

    Quote Originally Posted by Siron View Post
    Notice: (x-4)^2=x^2-8x+16, you're right about -4^2=-16 but in this case it's (-4)^2 which is 16.
    Hi Siron, I have been on a learning curve with this, but now I understand the method used to find the square. We had different ideas regarding the value 16?

    This is what I have learned;

    x^2 - 8x - 5

    (x - 4)^2 - 16 - 5
    (x - 4)^2 - 21
    (x - 4)^2 - 21 = x^2 - 8x + 16 - 21

    = x^2 - 8x - 5

    However all though all the above has been a learning curve, I did not find the solution in completed square form?

    (x - 4)^2 - 16


    David
    Last edited by David Green; August 9th 2011 at 10:44 AM. Reason: put wrong square root value in
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  15. #15
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    Re: Completing the square in quadratics of the form x^2 + bx + c

    Can you be more clear about your question? We've found the complete square of x^2-8x-5 which is (x-4)^2-21.
    Where are you stuck? What do you not understand? ...
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