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Double quadratic equation

  1. Jun 12, 2010 #1
    1. The problem statement, all variables and given/known data


    Find n+m

    2. Relevant equations

    3. The attempt at a solution

    I try to express the equation in the form of a(n+m)^2-b(n+m)+c=0 but i couldn't find a way to do so .
  2. jcsd
  3. Jun 12, 2010 #2


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    The expression can be written in the form of 4(n-a)^2+9(m-b)^2=0

  4. Jun 13, 2010 #3
    What calculations have you done so far? If you show what attempts you've made so far then can see how it's going. :smile:
  5. Jun 14, 2010 #4
    Thanks for replying here , i hv got it solved.

    Just express it in the form Ehild suggested and the values of m and n will be b and a respectively.
  6. Jun 14, 2010 #5
    OK great :smile:
  7. Jun 25, 2011 #6
    can someone pls show me the full step working?
  8. Jun 25, 2011 #7


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    Do you know how to complete the square?
  9. Jun 25, 2011 #8
    thanks for answering...
    not really...
    i'm just wondering how to get rid of the constant,18000,if we express it as 4(n-a)^2+9(m-b)^2=0
  10. Jun 25, 2011 #9


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    Sorry, but at this forum that is not how we give help.

    Try expanding that expression first. Then you can compare it to the original equation.
  11. Jun 25, 2011 #10


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    I wasn't answering anything, I was asking you a question. If you know how to complete the square then this problem isn't too difficult at all.

    I just wanted to know where to start helping you, but I don't think you really deserve any unless you clean up your attitude mate.
  12. Jun 25, 2011 #11
    i have solved it...
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