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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*

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

ReplyDeleteAGREE*

Deleteif a =2 and B =4.

Delete(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

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

ReplyDeleteLet's agree to disagree as (a+b)^2 is equal to a^2 + b^2.

DeleteExample:

a= 10 and b= 6

(10+6)^2 = 16^2

=256

Whereas, 10^2+6^2 = 100+36

= 136

I disagree.

ReplyDelete(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.

I agree*

DeleteExample if a: 6

and b: 7

(6+7)^2 = 13^2

= 169

a^2 + b^2 = 6^2 + 7^2

= 36 + 49

= 85

This comment has been removed by the author.

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

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

Delete(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

DeleteIn 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

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

ReplyDelete(a+b) x (a+b) = (2+3) x (2+3)

Delete= 25

a x a + b x b = 2x2 + 3x3

= 4+9

= 13

i disagree. if "a/b" were to be a negative number, a^2 + b^2, will be greater than/same as(a+b)^2

ReplyDeleteeg. if a = (-5)

b = (0)

(a+b)^2 = 25

a^2 + b^2 = 25

if a = (-5)

Deleteb = (2)

(a+b)^2 = 9

a^2 + b^2 = 29

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

DeleteThis comment has been removed by the author.

ReplyDelete* I AGREE!

DeleteFor 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

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

ReplyDelete(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

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

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

ReplyDeleteThis comment has been removed by the author.

Deletea = 3

Deleteb = 2

( 3+2)^2 > 3^2 + 2^2

= 25 > ( 9+ 4 )

= 25 > 13

thus i agree

DeleteI 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.

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteI 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.

ReplyDeleteExample: 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.

This comment has been removed by the author.

ReplyDeleteI agree as (a+b)^2 = a^2 +2ab+ b^2 has 2ab more than a^2+b^2

ReplyDeleteexample : a=2 b=4

(a+b)^2 = (2+4) ^2 = 36

a^2+ b^2 = 2^2 + 4^2 = 4+ 16 = 20

I agree.

ReplyDelete(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

I agree with Antonio.

ReplyDeleteExample 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

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

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

DeleteI now disagree because if 'a' = 0

Delete'b' = 1

LHS = 0^2+2(0)(1)=1^2

= 1

RHS = 0+1^2

= 1

LHS+RHS

I agree.

ReplyDeletetake 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

This comment has been removed by the author.

ReplyDeleteI 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

ReplyDeleteE.g. A = 0

B=0

(a+b)^2 = 0

(a^2+b^2)=0

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).

ReplyDeleteSo 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.

This comment has been removed by the author.

ReplyDeleteI agree because:

ReplyDelete(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.

I AGREE

ReplyDeletebecause 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