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Formula for a square

  1. Jul 25, 2003 #1
    [(x3-x2)+(x2-x1)]*[(x3+x2)+(x2-x1)]
    the book expands it to:
    (x3-x2)^2+2(x3-x2)*(x2-x1)+(x2-x1)^2

    i didnt get it so can someone please help me in this, i think there is a mistake in the book.
     
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. Jul 25, 2003 #2
    Re: question

    [(x3-x2)+(x2-x1)]*[(x3+x2)+(x2-x1)] does not equal to(x3-x2)^2+2(x3-x2)*(x2-x1)+(x2-x1)^2

    However,
    [(x3-x2)+(x2-x1)]*[(x3-x2)+(x2-x1)] = (x3-x2)^2+2(x3-x2)*(x2-x1)+(x2-x1)^2
     
  4. Jul 25, 2003 #3
    Re: Re: question

    have you noticed the experssions on the right are the same?
     
    Last edited: Jul 25, 2003
  5. Jul 25, 2003 #4
    Please read the expressions on the left hand sides carefully. I changed a "+" sign to a "-" sign in the third small bracket
     
  6. Jul 25, 2003 #5
    yes you are right. i guess it was a type mistake )-:
     
  7. Jul 25, 2003 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    With the negative, it is simply the formula for a square:

    (a+b)*(a+b)= a2+ 2ab+ b2

    with a= x3-x2 and b= x2- x1
     
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