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M - (x -n)^2 = ?

  1. Sep 29, 2004 #1
    m - (x -n)^2 = ? HELP MEEEE

    Hi guys,

    Im new here and wasn't sure what section to post this in :)

    Doing as maths at the moment.


    m - (x - n)^2 =m - x^2 - N^2 I thought...

    But its

    m - (x - n)^2 = n^2 + 2nx - x^2

    Could someone explain why 2nx appears (It must be a basic rule that i have forgotten since doing gcse maths)

    If you could help that would be much apreciated :)
     
    Last edited: Sep 30, 2004
  2. jcsd
  3. Sep 29, 2004 #2
    the thing described above has really confused me!

    So say it is:


    = n^2 + 2nx - x^2

    Then the text book somehow gets it simplified to:

    = m - n^2






    When i tried to simplify it i got it to:

    m - 2nx - n^2

    So i thought n^2 would cancel 2n leaving 2x but cant do that...

    so im very confused :)

    Look forward to the answer

    Thanks,
     
  4. Sep 29, 2004 #3

    pig

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    (x - n)^2 = (x - n)(x - n) = (x*x - x*n - n*x + n*n) = x^2 - 2nx + n^2

    So,
    m - (x-n)^2 = m - (x^2 - 2nx + n^2) = m - x^2 + 2nx - n^2
     
  5. Sep 29, 2004 #4
    ok that makes sense cheers,

    but what about the simplyfying one?
     
  6. Sep 29, 2004 #5

    Chronos

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    Hmm, the algebraic solution pig gave is correct
    m - (x - n)^2 = m - x^2 + 2nx - n^2
    I don't follow how you can further simplify that expression without another equation. If the textbook says you can simplify this result to m - (x - n)^2 = m - n^2, then x = 0 and n can be any number [including 0].
     
  7. Sep 29, 2004 #6

    pig

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    Yes, and x can also be 2n.
     
  8. Sep 30, 2004 #7

    Chronos

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    Agreed. To further simplify m=n^2.
     
  9. Sep 30, 2004 #8

    Integral

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    PLease provide a complete statement of the problem. Then show us what you have done to arrive at a solution. With this information we can help you.
     
  10. Sep 30, 2004 #9
    Hey guys - Yeh sorry should of just shown you the full question (along with text book solution).

    Here is the question - Scroll down to see conclusion or dont look, do it yourself and if your arrive at same answer please explain as im confused :)

    Thanks.



    Question

    Find the numbers M and N such that:

    5 + 4x - x² = M - (x -n)²

    For all real values of x.




    Solution from textbook:

    m - (x - n)² = m - (x² - 2nx + n²)
    = m - n² + 2nx - x²

    Comparing coefficients, 5 = m - n² (... answer 1)
    and 4 = 2n (.... answer 2)
    From (answer 2), n = 2 ; putting this into (answer 1) gives 5 = m - 4
    Which means m = 9


    ---------------------------------------------------
    If you can't explain to well could you provide a link explaining these as my text book has confused me with this one.

    I got the correct answer to:

    x(1 + a + b) = 1 + 3 + 5 + 3 This was fairly straight forward (x = 1 then did x = 2) to find that a = 2 and b = 3.

    It then went on to solve them using a method called "coefficents" which i need to re read and find a better explanatioN!" Im sure once i learn this i will find these fine.

    Can i solve it using the "values" way to solve all Indentities or will i need to learn both ?


    Cheers :)
     
    Last edited: Sep 30, 2004
  11. Oct 1, 2004 #10
    BUMP For an answer
     
  12. Oct 1, 2004 #11

    Zurtex

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    I'm not sure about other people but I don't exactly know what problem you have anymore. Could you please briefly explain what you are don't know or are struggling with?
     
  13. Oct 1, 2004 #12
    i dont see how they got the above answer from the question - They seem to have got it down from:


    5 + 4x - x² = M - (x -n)²



    To :


    m - (x - n)² = m - (x² - 2nx + n²)
    = m - n² + 2nx - x²

    Comparing coefficients, 5 = m - n² (... answer 1)
    and 4 = 2n (.... answer 2)
    From (answer 2), n = 2 ; putting this into (answer 1) gives 5 = m - 4
    Which means m = 9
     
  14. Oct 1, 2004 #13

    pig

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    5 + 4x - x^2 = m - (x -n)^2
    5 + 4x - x^2 = m - x^2 + 2nx - n^2
    4x - x^2 + x^2 - 2nx = m - 5 - n^2
    4x - 2nx = m - 5 - n^2
    x(4 - 2n) = m - 5 - n^2

    In order for this to work for all values of x, it must not depend on x. And it won't depend on x if 4 - 2n = 0, because x * 0 will be 0 regardless of x, so:

    4 - 2n = 0
    n = 2

    And:

    x(4 - 2n) = m - 5 - n^2
    x(4 - 4) = m - 5 - 4
    0 = m - 9
    m = 9
     
  15. Oct 1, 2004 #14

    uart

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    Chris, the above is called "completing the square" and is a very usuful technique for dealing with quadratic equations. You should try to learn and understand it.

    I'm still not sure exactly what part of the given proof that you're struggling with, it seems very stright forward, expand and compare coefficients.

    Is it the expansion (x-n)^2 = x^2 - 2nx + n^2 that's confusing you or is it the subtraction of the bracketed term in the next line that you dont understand or is it the very concept of comparing coefficients that has you beat?
     
  16. Oct 2, 2004 #15
    Ok so how do you go from this:

    4x - x^2 + x^2 - 2nx = m - 5 - n^2

    to this:
    4x - 2nx = m - 5 - n^2

    You have somehow delete the
    " x^2 + x^2" From the equation.

    Then i see you have factorised this and got down to the final answer but how did you get rid of the thing shown above


    x(4 - 2n) = m - 5 - n^2



    --------------------

    The book took big steps aswell like somehow got:
    = m - n^2 + 2nx - x^2

    Then said "comparing coefficients, 5 = m-n^2 and it totally lost me!

    I hate it when books that are suppose to teach you things dont explain the stages clearly.

    I feel like a right stupid idiot
     
  17. Oct 2, 2004 #16
    4x - x^2 + x^2 - 2nx = m - 5 - n^2
    4x + (- x^2 + x^2) - 2nx = m - 5 - n^2

    since -x^2 + x^2 = 0

    then 4x + 0 - 2nx = m - 5 - n^2
    4x - 2nx = m - 5 - n^2
    x(4 - 2n) = m - 5 - n^2

    since there are no x on the right hand side of this equation,
    then 4 - 2n must be equal to zero for this equation to be valid

    4 - 2n = 0
    2n = 4
    n = 2

    we're left with:
    0 = m - 5 - n^2

    add both side with 5 + n^2

    m = 5 + n^2
    since n = 2, then
    m = 5 + 2^2
    m = 9

    Hope that helps! :smile:
     
  18. Oct 2, 2004 #17
    well about the book
    5 + 4x - x^2 = M - x^2 + 2nx - n^2
    and then add x^2 to both sides of equation,
    5 + 4x = M + 2nx - n^2

    by comparing coefficients, it means that in order to make this equation valid, you have to have the same coefficient in left hand side and right hand side for x^1, and also for x^0 (note: x^0 = 1 which means the constants have to be equal --> constants = terms that don't have x)
    therefore, the coefficient of x^1 on left side is 4 and on the right side is 2n, to have a valid equation, 4 must be equal to 2n...
    4 = 2n
    n = 2

    for x^0 (i.e. the constant) the term on left side is 5 and on the right side are M - n^2
    for this equation to be valid 5 must be equal to M - n^2
    5 = M - n^2
    we already calculated the value of n^2
    therefore,
    5 = M - 2^2
    5 = M - 4
    add both sides with 4.
    9 = M

    hope that helps! :smile:
     
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