Algebra: Who is right: Antonio, Benedict or Celine?

Antonio claims, " (a+b)^2 is always greater than a^2 + b^2 "
Benedict thinks, " Sometimes "
Celine says, " No! "

Now who is right?

You have been assigned the role of Antonio, Benedict and Celine.
You are going to tell people why you make the claims or think in such a manner:
• If you are Antonio, give examples to tell people that U are correct about your claim.
• Benedict sits on the fence. If you are Benedict, tell people why you think so?
• Celine claims that (a+b)^2 cannot be greater than a^2 + b^2. If you are Celine, then you have to tell people why you think in this manner.
Clue:
We will use a spreadsheet (i.e. NUMBERS) to help us.

Approach, we are going to substituting "a" and "b" with a range of numbers to "prove" our point.
Remember, one set of number is not enough to prove that what you say is always true. You will need to test it out with several sets of numbers.
• Set the headings for: a, b, a^2, b^2, a^2+b^2, a+b, 2ab
• Apart from a and b, use "formula" feature in the spreadsheet to help you compute the numbers.

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2. I stand by Antonio because when we expand we can get a^2+ 2ab + b^2 , which is greater than a^2+b^2

1. I was wrong . Now I stand by Benedict as he is right because if i substitute 'a' and 'b' with 1 the answer would be smaller than a^2+b^2 . If 'a' and 'b' is lager than 1 the answer is greater than a^2+b^2

3. Benedict is right. In numbers, (a+b)^2 and a^2+b^2 is sometimes more or less than each other.

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2. SO if A=0 and B=1, then (A+B)^2=1 but A^2+B^2=1
But in another case, if both nos are equal to 1, (A+B)^2=4 and A^2+B^2 is equal to 2

3. But is either a or b is negative, (a+b)^2 is less than a^2+b^2.

4. I Stand By Benedict' s answer as It depends whether what is a and what is b. If a and b is 1 , it is the same. if they are above 1, (a+b)^2 is more than a^2+ b^2.

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6. I stand by Benedict's answer because it is mostly correct as any number plus any number squared will mostly be bigger except for when one number is 0 as 0^2 is always 0 and adding them won't make a difference.
(0+5)^2=25
0^2+5^2=25
(1+1)^2=4
1^2+1^2=1

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8. I stand by Celine's answer because (a+b)^2 after factorisation, (a+b)^2 will equal to (a+b) times (a+b) and that will equal to 2a+2b, so Antonio & Benedict are wrong and Celine is correct.

1. I mean only Celine's reply is correct and Benedict's reply is wrong.

9. I support Antonio as no matter what the numbers are, when the numbers added together and then squared, are bigger than the numbers are individually.

10. I stand by Benedict's answer because if a and b are positive, (a+b)^2 will be greater than a^2+b^2. If one of the numbers is negative, a^2+b^2 will be greater than (a+b)^2.

11. I stand by Antoniot's answer because if a=5 and b=6,then (a+b)^2 is 121. And a^2 + b^2 is 61.

12. I stand by Benedict's answer because sometimes (a+b)^2 is greater than a^2 + b^2, but sometimes it is vice versa or equal. Example : Replace a by 2. Replace b by 3. 25 is greater than 13 in this case. Replace a by 0. Replace b by 2. In this case, (a+b)^2= 4. a^2 + b^2 = 4. Hence, both are equal.

13. I support Benedict.
If a=2 and b=5, then (2+5)^2=49
2^2+5^2=29
If a=-2 and b=-5, then (-2+-5)^2=49
-2^2+-5^2=-29
If a=2 and b=-5, then (2+-5)^2=9
2^2+-5^2=-21
If a=-2 and b=5, then (-2+5)^2=9
-2^2+5^2=21

14. I stand by Antonio's answer because it depends on the value on a and b to prove that (a+b)^2 is should always be greater than a^2 + b^2.
For example: If a=2, b=3,
(a+b)^2
= (2+3)^2
= 25
a^2 + b^2
= 2^2 + 3^2
= 13

15. Celine is correct. It is because (A+B)^2= (A+B)(A+B)= A^2+B^2.

16. I support Benedict, as I believe he is right.

If a=2 and b=3

then (a+3)^2 would equal to (2+3)^2 which equals to 25

and a^2 + b^2 would equal to 2^2 + 3^2 which equal to 13

So (a+b)^2 is greater than a^2 + b^2

BUT THEN

If a= -2 and b= 3

the (a+b)^2 would equal to (-2+3)^2 which equals to 1

and a^2 + b^2 would be equal to -2^2 + 3^2 which equals to 5

So now a^2 + b^2 is greater

17. I stand by Benedict answer because sometimes (a+b)^2 is bigger than a^2+b^2 for example if a=3 and b=5 (5+3)^2=64 and 5^2+3^2 is 34 however they can be the same if a=1 and b=0 both will equal 0

18. I stand by Antonio because
Take 'a' to be 1 and 'b' to be 2.
So (1+2)^2=(3)^2=9 but 1^2+2^2=1+4=5

19. I stand by Benedict's answer because a and b may be negative which causes the two equations to be different.

20. I stand by benedict's answer because a/b could be negative or positive or anything, which will change the answer, but according to some law ( i forgot the name) the answer is a^2+ ab +b^2

21. I stand by benedict's answer because f the values of a and b are the same, (a+b)^2 is is not greater than a^2 + b^2. But if a and b are different values, (a+b)^2 is always greater than a^2 + b^2. Examples: a=9 b=8 so (9+8)^2 is 289 and 9^2+8^2=145.

22. i stand by benedict's answer's because if the number is small, (a+b)^2 is always greater than a^2 + b^2 ,example if a=-3 while b=-5 , a^2 + b^2 =34 and (a+b)^2 =64

23. I stand by Benedict's answer because if a=-1 and b=-2, (-1+-2)^2=9, and (-1)^2+(-2)^2=5; but if a=0 and b=2, (0+2)^2=4, and 0^2+2^2=4, so the formula depends on what numbers a and b are. :)

24. A=Antonio
B=Benedict
C=Celine Forever Alone XD
D=Demon Dun know
E=EEEEEEEVVIIIIIIILLL XD