### 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

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

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

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

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

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.

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

2. This comment has been removed by the author.

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

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

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

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

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

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

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

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

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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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

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

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

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

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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2. a = 3
b = 2

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

3. thus i agree

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

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

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

1. Justin....4+9=13 not 11!!

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

3. 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

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

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.

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