## anonymous one year ago Apply the distributive formula to (x+y)^2 = x^2 + 2xy + y^2

1. xapproachesinfinity

first do this $$(x+y)^2=(x+y)(x+y)$$

2. xapproachesinfinity

now try to distribute

3. anonymous

|dw:1439924847191:dw|

4. anonymous

(x+y) (x+y) = (x^2 + xy + xy + y^2) @xapproachesinfinity @Crazyandbeautiful

5. anonymous

combine like terms, you have the right hand side

6. anonymous

so i just combine all thats on the left now?

7. anonymous

@Crazyandbeautiful x^2+xy+xy+y^2 = (x^2 + xy + xy + y^2)

8. anonymous

@Crazyandbeautiful @xapproachesinfinity is this right???

9. anonymous

$xy+xy=2xy$ Just like how $a+a=2a$

10. anonymous

do you understand that @The_Yee_Dinosaur

11. anonymous

yea

12. anonymous

but it is not a true equation now such as 25=25 or 17=17 right? @Crazyandbeautiful

13. anonymous

@Nishant_Garg

14. anonymous

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 $(3+4)^2=3^2+4^2+2(3)(4)=9+16+24=49$

15. anonymous

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

16. anonymous

Suppose x=3 and y=4 $xy=3\times4=12$ $12+12=24=2 \times 12$ $xy+xy=2xy=2 \times xy$

17. 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) $(x+y)(x+y)= x62+2xy+y^2$