Algebra: Is Antonio right?

Antonio claims, " (a+b)^2 is always greater than a^2 + b^2 "

Do you agree?
  • State your stand: I "agree"/ "disagree"...
  • Give example(s) to illustrate your point
  • Clue: Consider the different possible scenarios
  • Try to generalize your point

    42 comments:

    1. i disagree because (a+b)^2 is equal to a^2 + b^2+ 2ab and a^2 + b^2+ 2ab is 2ab more than a^2 + b^2

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      1. if a =2 and B =4.
        (2+4)^2 = 2^2 +2(2+4)+4^2

        if a=3 and B=6
        (3+6)^2 = 3^2 +2(3+6) +6^2

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    2. I STRONGLY disagree!!! (a+b)^2 is EQUAL to a^2 + b^2.

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      1. Let's agree to disagree as (a+b)^2 is equal to a^2 + b^2.
        Example:
        a= 10 and b= 6
        (10+6)^2 = 16^2
        =256
        Whereas, 10^2+6^2 = 100+36
        = 136

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    3. I disagree.
      (a+b)^2 is a^2 + 2ab + b^2 while a^2 + b^2 is (a x a) + (b x b), which is much greater than (a+b)^2.

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      1. I agree*
        Example if a: 6
        and b: 7
        (6+7)^2 = 13^2
        = 169
        a^2 + b^2 = 6^2 + 7^2
        = 36 + 49
        = 85

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      2. This comment has been removed by the author.

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    4. I agree.If a=1 and b=2,(1+2)^2=9,1^2+2^2=5.

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      1. @Joshua,I think your explanation is wrong because a^2+b^2=-25(not 25) -5^2+0^2=-25

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      2. (a+b)^2>a^2+b^2 because if a=1 and b=2,(1+2)^2=9,1^2+2^2=5,the difference is 4
        In another example if a=2 and b=3,(2+3)^2=25,2^2+3^2,the difference is 12
        So,(a+b)^2>a^2+b^2

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    5. I agree as there is an additional 2ab in (a+b)^2 so (a+b)^2 should be bigger than a^2 + b^2

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      1. (a+b) x (a+b) = (2+3) x (2+3)
        = 25
        a x a + b x b = 2x2 + 3x3
        = 4+9
        = 13

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    6. i disagree. if "a/b" were to be a negative number, a^2 + b^2, will be greater than/same as(a+b)^2
      eg. if a = (-5)
      b = (0)
      (a+b)^2 = 25
      a^2 + b^2 = 25

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      1. if a = (-5)
        b = (2)

        (a+b)^2 = 9
        a^2 + b^2 = 29

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      2. but if all the numbers were to be positive, (a+b)^2 is always greater than a^2 + b^2, since (a+b)^2 a^2+b^2 + 2 ab. so there is an additional 2ab so (a+b)^2 is always greater than a^2 + b^2 when all nmbers is positive and not a 1 or 0

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    7. This comment has been removed by the author.

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      1. * I AGREE!

        For Example:
        a= 7
        b= 5

        So,
        (a+b)^2
        (7+5)^2= 13^2= 169

        a^2 + b^2
        7^2 + 5^2= 49 + 25= 74

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    8. I disagree with Antonio as (a+b)^2 is equals to a^2 + b^2 as (a+b)^2 is a^2+b^2 is same as a^2 + b^2 is

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    9. (a+b)^2 is always greater than a^2 + b^2- I sit on the fence as (a+b)^2 is 2a+2b but a^2 + b^2 is aa+bb

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      1. what if a=5 and b=6? ,(5+6)^2=121,5^2+6^2=55 so i agree

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    10. I agree because (a + b)^2 = 2ab + a^2+b^2 and it has 2ab more than a^2b^2

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      1. This comment has been removed by the author.

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      2. a = 3
        b = 2

        ( 3+2)^2 > 3^2 + 2^2
        = 25 > ( 9+ 4 )
        = 25 > 13

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    11. I disagree to Antonio's claim as a+b^2 is a^2 + 2ab + b^2 and a^2 + b^2 is a x a + b x b, which is greater than (a+b)^2.

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    13. I agree because the difference between (a+b)^2 and a^2 + b^2 is 2ab. Thus 2ab would make the difference that would make (a+b)^2 greater than a^2 + b^2.
      Example: Let a=3,b=2. (a+b)^2 = (3+2)(3+2) = 9+6+4+6=25
      a^2 + b^2=3x3+2x2=9+4=13.
      25 is greater than 13 thus I would agree.

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    15. I agree as (a+b)^2 = a^2 +2ab+ b^2 has 2ab more than a^2+b^2

      example : a=2 b=4
      (a+b)^2 = (2+4) ^2 = 36
      a^2+ b^2 = 2^2 + 4^2 = 4+ 16 = 20

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    16. I agree.
      (a+b)^2 = (a+b)(a+b) = a^2 + ab + ba +b^2 = a^2 + B^2 + ab^2

      a^2 + b^2

      if a = 5 and b = 10

      (a+b)^2 = 5^2 + 10^2 + 5x10^2

      a^2 + b^2 = 5^2 + 10^2

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    17. I agree with Antonio.
      Example if 'a' is (2) and 'b' is (3), (2+3)^2 equals to (2+3)(2+3) = 4+6+6+9=25
      a^2+b^2 = 2^2+3^2 =4+9=11

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      1. Sorry, plus u spelled my name wrongly. Please excuse the inconvenience.

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      2. I now disagree because if 'a' = 0
        'b' = 1
        LHS = 0^2+2(0)(1)=1^2
        = 1
        RHS = 0+1^2
        = 1
        LHS+RHS

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    18. I agree.
      take 3 as an example:
      (3+3)^2=9^2=81
      3^2+3^2=9+9=18
      take 8 as an example:
      (8+8)^2=16^2=256
      8^2+8^2=64+64=128

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    20. I disagree with Antonio as although (a+b)^2 is equal to a^2 + 2ab + b^2 which is greater than a^2 + b^2 but what if A=0 B=0

      E.g. A = 0

      B=0

      (a+b)^2 = 0
      (a^2+b^2)=0

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    21. I agree because (a+b)^2 equals to (a+b)(a+b) which equals to a(a+b)+b(a+b) and a^2+b^2 equals to (a*a)+(b*b).
      So a(a+b)+b(a+b) IS greater than (a*a)+(b*b)
      So if I replace the letters with numbers it will be something like:
      a=2 and b=3
      2(2+3)+3(2+3)= 25
      2*2+3*3= 4+9 = 13
      And obviously 25 is bigger so (a+b)^2 is greater than a^2+b^2.

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    23. I agree because:
      (a+b)^2 = (a+b)(a+b) = a^2 + 2ab +b^2

      For example, a=1 and b=2:
      (a+b)^2 = (1+2)(1+2) = 1 + 2 + 2 + 4 = 9

      a^2 + b^2 = (1x1) + (2x2) = 5

      Therefore, (a+b)^2 is greater than a^2 + b^2.

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    24. I AGREE
      because if a=2
      b=3

      (a+b)^2 = (2+3)(2+3)
      = 5X5 = 13
      = 25

      a^2 + b^2 = 4+9
      = 13
      im hoping you can understand the workings and stuff

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