Formula for a square

  • #1
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[(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.
 
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Answers and Replies

  • #2


Originally posted by loop quantum gravity
[(x3-x2)+(x2-x1)]*[(x3+x2)+(x2-x1)]
the book expands it to:
(x3-x2)^2+2(x3-x2)*(x2-x1)+(x2-x1)^2
[(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
 
  • #3
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Originally posted by KL Kam
[(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
have you noticed the experssions on the right are the same?
 
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  • #4
Please read the expressions on the left hand sides carefully. I changed a "+" sign to a "-" sign in the third small bracket
 
  • #5
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yes you are right. i guess it was a type mistake )-:
 
  • #6
HallsofIvy
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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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