Education & Reference Silly maths question

[Bugz]

I am having an argument with someone about the following equation- the 2's represent "squared"....

(a+b)2 = a2+b2

What are 'a' and 'b'.

 

[Kahuna]

Well it can mean squared or multiply by 2, its a little hard to tell with the bigfooty not letting you hypercase. However it looks like they mean squared.

The letters are just variables, so if 2 = 10 then b = ? .

 

[hawkaz1]

I don't know if I've interpreted this the right way, but

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

So the only way that (a + b)^2 could be equal to a^2 + b^2 would be if a = 0, b=0.

 

[spudmaster]

I am having an argument with someone about the following equation- the 2's represent "squared"....

(a+b)2 = a2+b2

What are 'a' and 'b'.

(a+b)^2 = a^2+2ab+b^2, not a^2+b^2


Working:
(a+b)*(a+b)
=a*a + a*b + b*a + b*b
(a*b and b*a are like terms so you have 2 lots of a*b)
(a*a =a^2 and b*b=b^2)

 

[spudmaster]

beat me to it, hawkaz.

I don't think he's trying to get the two to equal each other, I think he's been given an 'expand this' question for homework and has misinterpreted it.

 

[Bugz]

No, I am trying to get them to equal one another. Its my brothers question and it makes little sense -

"find numbers to take the place of the letters a and b in the two formulas below, so that both formula's give the same answer. (a must be the same number in both formulas and so must b)"

the formulas are:

(a+b)^2 & a^2+b^2


I am thinking one must be 0?!?

 

[Bugz]

I really should have paid more attention in High School....

 

[hawkaz1]

Actually, yeah. If b = 0 a could equal anything and it would be correct.

Eg. If a = 5.

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

5^2 = 5^2

25 = 25

So that would work.

 

[Simmonds2Hislop]

No, I am trying to get them to equal one another. Its my brothers question and it makes little sense -

"find numbers to take the place of the letters a and b in the two formulas below, so that both formula's give the same answer. (a must be the same number in both formulas and so must b)"

the formulas are:

(a+b)^2 & a^2+b^2


I am thinking one must be 0?!?

The way I'd do it would be:

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

therefore:

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

a^2 + 2ab + b^2 = a^2 + b^2

a^2 and b^2 are on both sides so cancel out.

therefore:

2ab = 0 (or a = 0/b)

So the solution is "at least one of a or b must equal 0. The other value can be any real number, including zero".

Edit: ^^^^^ I should've read the above post.

 

[Simple Jack]

They call that a quadratic if I recall correctly.

 

[mike_]

The way I'd do it would be:

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

therefore:

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

a^2 + 2ab + b^2 = a^2 + b^2

a^2 and b^2 are on both sides so cancel out.


therefore:

2ab = 0 (or a = 0/b)

So the solution is "at least one of a or b must equal 0. The other value can be any real number, including zero".

Edit: ^^^^^ I should've read the above post.

What if a = -1 and b = 1:
(a+b)^2 = a^2 + 2ab + b^2
(-1+1)^2 = -1^2 + 2(-1)(1) + 1^2
0 = 1 + (-2) + 1 = 0

Would also work if a = 1, b = -1.

EDIT: Disregard, didn't read OP': a^2 + b^2 = (a+b)^2

 

[obscura]

quadratics, psh

 

[DREAM TEAM BUFF]

As my maths teacher in year 9 told me there is never a silly maths question

 

[Andre]

As my maths teacher in year 9 told me there is never a silly maths question

And yet there are imaginary numbers. Maths may not be silly, but it can be strange.