### 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 can use a spreadsheet (i.e. NUMBERS) to help us.
Approach: we are going to substitute "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.

#### 67 comments:

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1. Wrong LOL

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4. Celine is correct as both equations are equal... as (a+b)^2=a^2+b^2

1. Make an attempt to expand (a+b)^2 to compare

2. (a+b)^2= a^2+b^2 hence they are equal

5. I think Benedict is correct, as when a is 2 and b is 3, (a+b)^2 is greater than a^2 + b^2. However when a is -2 and b is -3, (a+b)^2 is smaller than a^2 + b^2.

1. Try to evaluate the expressions with negative values of a and b to see if what you have concluded is correct?

6. I stand by Benidict's answer, that says (a+b)^ 2 < a^2 + b^2.

Since (a+b)^2 = a^2 + b^2 + 2ab,

if 2ab is a negative, the answer would be lesser than a^2 + b^2
if 2ab is a positive, than the answer would be more than a^2 + b^2

Sarah Chan

1. I think you are interpreting the question in the right track, however, what you've written does not explain correctly

2. if 2ab is a negative,

(a+b)^ 2= a^2 + b^2 -2ab

which means that (a+b)^ 2 >a^2 + b^2.

if 2ab is a positive,

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

which means (a+b)^ 2 >a^2 + b^2.

7. I stand by Benedict's answer that says (a+b)^2 is not always greater than a^2 + b^2 because if the letters are positive then it will be true BUT if the numbers are negative, then the statement is WRONG.

1. 1. What is the correct way of calling those "letters"?
2. Test out what you said for the second part when you say the numbers are negative.

8. (a+b)^2
= (a+b) (a+b)
= a^2 + ab + ba + b^2
= a^2 + b^2 + 2ab

(a+b)^2 will be greater than a^2 + b^2 if a and b are positive values ≥ 1 OR negative values ≤ -1.
They will be the same if a and b is 0.
(a+b)^2 will be lesser than a^2 + b^2 if one of the values, either a or b, are negative values ≤ -1.

1. Therefore, I agree with Benedict.

2. Very very good!
You have explained clearly by first showing the expanded version of the expression,
After which, considered all the possible combinations to explain each scenario.

To further improve the explanation so that those who could not understand through the algebraic interpretation,
You can include examples to illustrate each case.

3. For the First Example : Let a be 2 and b be 4. The difference will be 2ab, which is 2 * 2 * 4 = 16. Also, let a be -2 and b be -4. 2ab = 2 * -2 * -4 = 16. Therefore (a+b)^2 will be greater in this scenario.

For the Second Example : Let both a and b be 0. 2ab = 2 * 0 * 0 = 0. Therefore both will be the same.

For the Last Example : Let a be -2 and b be 4. 2ab = 2 * -2 * 4 = -16. Therefore (a+b)^2 will be lesser in this scenario.

9. (a+b)^2 is always greater than a^2 + b^2

Let a be 3 and b be 4.
(3+4)^2=7^2=49
3^2+4^2=9+16=25

Let a be -3 and b be 4.
(-3+4)^2=1^2=1
-3^2+4^2=-9+16=7

Therefore, I agree with Benedict.

1. You have provided examples to show that it is possible for (a+b)^2 to be either grater or lesser than a^2 +b^2.
The next step is to generalize.
On the other hand, have you explored all the possible scenarios already?

2. (a+b)^2 is always greater than a^2 + b^2

Let a be 3 and b be 4.
(3+4)^2=7^2=49
3^2+4^2=9+16=25

Let a be -3 and b be 4.
(-3+4)^2=1^2=1
-3^2+4^2=9+16=25

Let a be 3 and b be- 4.
[3+(-4)]^2=(-1)^2=1
3^2+(-4)^2=9+16=25

Let a be -3 and b be -4.
[-3+(-4)]^2=49
-3^2+(-4)^2=9+16=25

When both a and b are positive/negative, a will be greater than b.
When a and b are positive and negative, b will be greater than a.

10. I think Benedict is correct as if the a and b are positive numbers, (a+b)^2 is not greater than a^2 + b^2 , but when a and b are negative numbers, (a+b)^2 is greater than a^2 + b^2

1. Try using numbers to test out what you think before concluding.

2. If a=5 and b=10,
(a+b)^2
=(5+10)^2
=15^2
=225
a^2+b^2
=5^2+10^2
=25+100
=125

If a=-5 and b=-10
(a+b)^2
=((-5)+(-10))^2
=(-15)^2
=-225
a^2+b^2
=-5^2+-10^2
=25+100
=125

I meant that if a and b are positive, (a+b)^2 is greater than a^2 + b^2

11. I stand by Celine's post as if a was 5 and b was 10, it would be 225. And 225 does not equal to 5^2 + 10^2 which is 125
~Praveen~

1. HI PRAVEEN!!!

2. This is one possible scenario.
Try other combinations of numbers (remember the real number family?)

3. If the numbers are negative, the first part would be smaller than the second one, so I stand by Benedict's statement saying that it is only true sometimes.

12. I think benedict is right.
if a= 3 , b=3, (3+3)^2=36, 3^2+3^2=12 so (a+b)^2>a^2+b^2
but if a=6,b=-3, [6+(-3)]^2=9, 6^2+(-3^2)=45 so a^2+b^2>(a+b)^2
so sometimes (a+b)^2>a^2+b^2 and sometimes a^2+b^2>(a+b)^2

1. Good thinking and attempt to generalize.
However, the generalization is incomplete.
Also, are there other possibilities?

2. when a and b are positive values ≥ 1 or negative values ≤ -1.
(a+b)^2 will be greater than a^2 + b^2 .They will be the same if a and b is 0,
but when one of a or b are negative values ≤ -1, (a+b)^2 will be lesser than a^2 + b^2

13. I stand by Antonio's answer. As I have used a range if negative and positive numbers and substituted them with the unknowns and I have found that it corresponds with Antonio's statement.

1. How about other combinations of real numbers? Test them out.

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15. I stand by Benedict answer. As the a and b can be any value, the statement [(a+b)^2 is greater than a^2+b^2] may not always be true. For example, if a=2 and b=1, (2+1)^2 equals 9 and 2^2 + 1^2 equals 5 so Antonio is correct. But, if the a=1 and b=0, (1+0)^2=1 and 1^2+0^2=1

1. This proves that they are both wrong

2. Good try.
There are more possible combination of numbers.
Test them out, then generalize.

3. So the conclusion gathered is that if a=any number expect 0 and b=any number except 0, Antonia is correct. But if not, Antonia and Celine is both wrong.

4. OR,
If a=0 or 1 and b=0 or 1 ,Celine and Antonio is both wrong.
But if a>1 and b>1, Anotio is wrong

16. I stand by Benedict's answer. This is because (a+b)^2=a^2+2ab+b^2, proving that (a+b)^2 is greater than a^2+b^2 by 2ab.This will always be the case even when negative numbers are involved, except when a is negative (-) and b is positive (+) and vice versa.

1. Good explanation. Can you include some illustrations?
On the other hand, have you tested all possible scenarios?

17. I stand by Antonio's answer as (a+b)^2 = a^2+2ab+b^2, which is greater than a^2+b^2.

1. Elaborate with illustrations.

18. I think benedict is correct because

Scenario 1) If 'a' is 3 and 'b' is 5. Then, (3+5)^2=64
3^2 + 5^2 =34
Then,what Antonio said will be correct.

Scenario 2) If 'a' is -3 and 'b' is -5.Then,{(-3)+(-5)}^2=64
(-3^2)+(-5^2)=-34

Then,what Antonio said will also be correct

Scenario 3)If 'a' is -3 and 'b' is 5.Then,(-3+5)^2=4
-3^2 + 5^2=16

Now,Antonio is wrong

Thus,benedict is correct

1. Good way of deducing the conclusion. Now, are there other scenarios?
Next, try to generalize what you've come up with.

2. If the numbers are both positive or negative.Then, antonio is correct
But the one number is positive and one number is negative.Then.benedict is correct

19. I support Benedict's claim, as seen here:

a= 5
b= 10

(5+10)^2 = 225

(5^2) + (10^2) = 125

However, if

a= -5
b= 10

(-5+10)^2 = 25

(-5^2) + (10^2) = 75

If 'a' is a negative number, then Benedict would be right, as (a+b)^2 is not always greater than a^2 + b^2.
If 'a' is a positive number, then Antonio is correct.
Celine is also correct.

1. Good way of using numbers to check. Now, try to generalize your answers.
Next, have you explored all the scenarios?

2. If a=0 and b=5

Then Antonio is wrong, as:

(0+5)^2 = 25
(0^2) + (5^2) = 25

This means that (a+b)^2 is equal to a^2 + b^2.

20. I stand by Benedict's answer. This is because, when 2 numbers are added then squared, it is definitely larger than when two numbers are squared before adding up. However, if a/b is 0, the answers will be the same.

1. I guess, when you type a/b, you mean a or b?
Next... do you think there are other possible scenario?

2. - a/b means a or b, not a÷b

- if both a and b are 0, the answers would also be the same.

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

21. I suport Benedtict's answer.
This is because that if one of the numbers, lets say a=-2, then the answer will definitely be less than a^2+b^2
But if the number is positive, then The answer will definitely be more than that.

1. Good attempt.
The next thing is to put your thoughts down systematically and try to surface the different scenarios.

2. So b=3 ,then if -2^2+3^2=4+9=13

22. I stand benedict's answer.
1) If both of the numbers are positive,
a = 3, b = 4
(3+4)^2 = 49
3^2 + 4^2 = 25

2) a = 5, b = 10
(5+10)^2 = 225
5^2 + 10^2= 125

1. Good illustrations.
Next step: Generalise what you can say from each example.
Also, are there other possible scenarios?

23. I stand by Benedict's answer as when a=5 and b=6,(5+6)^2 equals to 121 but when 5^2+6^2 equals to 61.But when it is 0 they get the same answer.

1. Try to generalize what you have come up with.
Next, try exploring other scenarios.

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3. When the number is negative (a+b)^2 will be smaller than a^2 + b^2 "

24. I stand at Benedict's answer as when a and b are positive numbers more than 0, (a+b)^2 will be greater than a^2 + b^2.
If they are 0, then the both are the same.
If they are negative numbers, then one would be lesser than the other.

25. You would like to elaborate (to explain clearly) what you mean by "If they are negative numbers, then one would be lesser than the other."
Probably examples would help?

1. I they are negative numbers like a is -2 and then b is -4 ,
(-2-4)^2 is 36
(-2^2)+(-4^2) is -20
36 is greater than -20.

26. I think that Benedict is Correct
If a= 3 , b=3, (3+3)^2=36, 3^2+3^2=12 so (a+b)^2>a^2+b^2
However if a=6,b=-3, [6+(-3)]^2=9, 6^2+(-3^2)=45 so a^2+b^2>(a+b)^2
so on sometimes (a+b)^2>a^2+b^2 and on sometimes a^2+b^2>(a+b)^2

27. I think Celine is right because if a and b are both positive/negative (a+b)^2 will become a^2+b^2