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anonymous

  • one year ago

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

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  1. xapproachesinfinity
    • one year ago
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    first do this \((x+y)^2=(x+y)(x+y)\)

  2. xapproachesinfinity
    • one year ago
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    now try to distribute

  3. anonymous
    • one year ago
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    |dw:1439924847191:dw|

  4. anonymous
    • one year ago
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    (x+y) (x+y) = (x^2 + xy + xy + y^2) @xapproachesinfinity @Crazyandbeautiful

  5. anonymous
    • one year ago
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    combine like terms, you have the right hand side

  6. anonymous
    • one year ago
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    so i just combine all thats on the left now?

  7. anonymous
    • one year ago
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    @Crazyandbeautiful x^2+xy+xy+y^2 = (x^2 + xy + xy + y^2)

  8. anonymous
    • one year ago
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    @Crazyandbeautiful @xapproachesinfinity is this right???

  9. anonymous
    • one year ago
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    \[xy+xy=2xy\] Just like how \[a+a=2a\]

  10. anonymous
    • one year ago
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    do you understand that @The_Yee_Dinosaur

  11. anonymous
    • one year ago
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    yea

  12. anonymous
    • one year ago
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    but it is not a true equation now such as 25=25 or 17=17 right? @Crazyandbeautiful

  13. anonymous
    • one year ago
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    @Nishant_Garg

  14. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Suppose x=3 and y=4 \[xy=3\times4=12\] \[12+12=24=2 \times 12\] \[xy+xy=2xy=2 \times xy\]

  17. phi
    • one year ago
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    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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