# Formula for a square

Gold Member
[(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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KLscilevothma

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

Gold Member

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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KLscilevothma
Please read the expressions on the left hand sides carefully. I changed a "+" sign to a "-" sign in the third small bracket

Gold Member
yes you are right. i guess it was a type mistake )-:

Homework Helper
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