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Quadratics using Pascal's triangle

  1. May 18, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-5-18_10-28-50.png
    The problem equation is contained in the picture.

    2. Relevant equations
    Pascal's Triangle is useful is this one.

    3. The attempt at a solution

    The difficulty I'm having is in going between lines 2 and 3 which I've marked with a little red dot.

    upload_2017-5-18_10-31-50.png

    The closest I get to simplifying it is = a4 + B4 + 4aB(a2+B2) + 6a2B2 . From there I can't figure out the way in which to reduce it further.

    P.S I also used the quadratic formula to solve this one (the one with the b2 - 4ac) and my answer came to 748.52 but not quite 752. Why does the quadratic formula not work here?

    Thank you for your help,
    Ben
     

    Attached Files:

  2. jcsd
  3. May 18, 2017 #2
    ##\alpha^4 + \beta^4 + 4\alpha^3\beta + 6\alpha^2\beta^2 + 4\alpha \beta^3 = \alpha^4 + \beta^4 + 4\alpha^3\beta + \color{red}{8}\alpha^2\beta^2 + 4\alpha \beta^3 - \color{blue}{2\alpha^2 \beta^2} = \alpha^4 + \beta^4 + 4\alpha\beta(\alpha^2 + 2\alpha\beta + \beta^2) - {2\alpha^2 \beta^2}##

    Now use ##a^2 + 2ab + b^2 = (a+b)^2##
     
  4. May 18, 2017 #3
    Amazing! Thank you so much Buffu :)
     
  5. May 18, 2017 #4
  6. May 18, 2017 #5

    Mark44

    Staff: Mentor

    I have a minor gripe with the author of this problem.
    What is shown is not an equation, since the symbol = is not present.

    Again, my gripe is with the author, not the person who started this thread.
     
  7. May 18, 2017 #6
    Number of people confusing between expression and equation is surprisingly high.
     
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