Apply the distributive formula to (x+y)^2 = x^2 + 2xy + y^2

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Apply the distributive formula to (x+y)^2 = x^2 + 2xy + y^2

Mathematics
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first do this \((x+y)^2=(x+y)(x+y)\)
now try to distribute
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(x+y) (x+y) = (x^2 + xy + xy + y^2) @xapproachesinfinity @Crazyandbeautiful
combine like terms, you have the right hand side
so i just combine all thats on the left now?
@Crazyandbeautiful x^2+xy+xy+y^2 = (x^2 + xy + xy + y^2)
@Crazyandbeautiful @xapproachesinfinity is this right???
\[xy+xy=2xy\] Just like how \[a+a=2a\]
do you understand that @The_Yee_Dinosaur
yea
but it is not a true equation now such as 25=25 or 17=17 right? @Crazyandbeautiful
It is true, an equation is what an equation is, the left side is equal to the right side try it for your self take an example \[(3+4)^2\] Let's solve it directly and using the identity Directly first \[(3+4)^2=7^2=49\] Now we apply the identity \[(3+4)^2=3^2+4^2+2(3)(4)=9+16+24=49\]
Think of it like this, x and y are numbers so if you multiply x and y, that is calculate \[xy\](the product of x and y) you will get another number) now suppose you add that same final number you got that itself what do you get? It's 2 times the number itself Math will work whether you take letters or numbers
Suppose x=3 and y=4 \[xy=3\times4=12\] \[12+12=24=2 \times 12\] \[xy+xy=2xy=2 \times xy\]
  • phi
they may want you do use the distributive property twice \[ (x+y)(x+y) \] if we call the first (x+y) "A" (to make it clearer) we would do this (distribute) \[ A (x+y) = Ax + Ay \] and A is really (x+y), so this is \[ (x+y)(x+y)=(x+y)x + (x+y)y \] now distribute again. in other words the first (x+y)x becomes x^2 +xy and the second term becomes xy+y^2 \[ (x+y)x + (x+y)y = x^2 +xy+xy+y^2 \] finally , add xy+xy (1 xy plus another xy is 2xy) \[ (x+y)(x+y)= x62+2xy+y^2 \]

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