Factorising a² + b² using real numbers. (the outcome would be (a + b - √2ab)(a + b + √2ab))

[yuy]

Hello, I've been wondering for several months, we all know that a² + b² = (a + b)² - 2ab

Now I've always wanted to factorise it because (a + b)² - 2ab is not a factorised expression.

Therefore, I came up with a² + b² = (a + b)² - 2ab = (a + b - √2ab)(a + b + √2ab)

How did I do this? Well, I took into consideration that (a + b)² - 2ab is somewhat the same as x² - y² which is equal to (x - y)(x + y). Therefore, I assigned (a + b)² to 'x' and 2ab to 'y'. So the outcome would be (a + b - √2ab)(a + b + √2ab)


What'd you guys think about this?

 

[Dr.Peterson]

Hello, I've been wondering for several months, we all know that a² + b² = (a + b)² - 2ab

Now I've always wanted to factorise it because (a + b)² - 2ab is not a factorised expression.

Therefore, I came up with a² + b² = (a + b)² - 2ab = (a + b - √2ab)(a + b + √2ab)

How did I do this? Well, I took into consideration that (a + b)² - 2ab is somewhat the same as x² - y² which is equal to (x - y)(x + y). Therefore, I assigned (a + b)² to 'x' and 2ab to 'y'. So the outcome would be (a + b - √2ab)(a + b + √2ab)


What'd you guys think about this?

It isn't wrong; but I don't think it's very significant. Factorizing over the reals is trivial; if all you want is to factorize it, you could just as well write a² + b² = 2 * (a² + b²)/2. The significant thing is, what sort of factors would you want, and why?

Note, though, that using typed expressions, you should be writing (a + b - √(2ab))(a + b + √(2ab)).

To see what your result looks like in reality, consider 3² + 4², which we know is 5². Your method gives

3² + 4² = (3 + 4 - √(2*3*4))(3 + 4 + √(2*3*4)) = (7 - √24)(7 + √24) ≈ 11.899 * 2.101 = 25​

Is there anything significant about that particular factorization? Is it more meaningful than, say, (6 - √11)(6 + √11) ≈ 2.683 * 9.317 = 25?

Similarly, if you want to factorize x² + 1, your method gives (x + 1 + √x)(x + 1 - √x); how would you use that result?

 

[blamocur]

Therefore, I came up with a² + b² = (a + b)² - 2ab = (a + b - √2ab)(a + b + √2ab)

Typesetting in LaTeX makes it is less ambiguous:

[math]a^2 + b^2 = \left(a+b-\sqrt{2ab}\right)\left(a+b+\sqrt{2ab}\right)[/math]

 

[Steven G]

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

 

[yuy]

It isn't wrong; but I don't think it's very significant. Factorizing over the reals is trivial; if all you want is to factorize it, you could just as well write a² + b² = 2 * (a² + b²)/2. The significant thing is, what sort of factors would you want, and why?

Note, though, that using typed expressions, you should be writing (a + b - √(2ab))(a + b + √(2ab)).

To see what your result looks like in reality, consider 3² + 4², which we know is 5². Your method gives

3² + 4² = (3 + 4 - √(2*3*4))(3 + 4 + √(2*3*4)) = (7 - √24)(7 + √24) ≈ 11.899 * 2.101 = 25​

Is there anything significant about that particular factorization? Is it more meaningful than, say, (6 - √11)(6 + √11) ≈ 2.683 * 9.317 = 25?

Similarly, if you want to factorize x² + 1, your method gives (x + 1 + √x)(x + 1 - √x); how would you use that result?

I wouldn't actually use the result but was just wondering if it was correct. Thanks though!

 

[yuy]

Typesetting in LaTeX makes it is less ambiguous:

[math]a^2 + b^2 = \left(a+b-\sqrt{2ab}\right)\left(a+b+\sqrt{2ab}\right)[/math]

How can I type like that?

 

[stapel]

How can I type like that?

You can get started here and here.

 

[khansaheb]

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

Where did that come from?

 

[Steven G]

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

Where did that come from?

You got me! That c should be a b.