Challenge Question:

Factor $\sf a^2 + b^2=?$

Note: This isn't the Pythagorean Theorem

The other similar formulas to that one, if you need it, are:

$\sf a^2-b^2 = (a+b)(a-b)$
$\sf a^2+2ab+b^2 = (a+b)^2$
$\sf a^2 - 2ab+b^2 = (a-b)^2$

Hint: Think outside the box ;)

Question

Challenge Question:

Factor $\sf a^2 + b^2=?$

Note: This isn't the Pythagorean Theorem

The other similar formulas to that one, if you need it, are:

$\sf a^2-b^2 = (a+b)(a-b)$
$\sf a^2+2ab+b^2 = (a+b)^2$
$\sf a^2 - 2ab+b^2 = (a-b)^2$

Hint: Think outside the box ;)

anonymous · Answer

$a^2+b^2=a^2-(-b^2)=a^2-(bi)^2=(a+bi)(a-bi)$

anonymous · Answer

am I not supposed to post it? do I message or something?

thesmartone · Answer

well seems like it wasn't really hard at all, great job xD

usukidoll · Answer

$$ a^2 + b^2=?$$

$\sf a^2-b^2 = (a+b)(a-b)$
$\sf a^2+2ab+b^2 = (a+b)^2$ (I'm taking this one) 
$\sf a^2 - 2ab+b^2 = (a-b)^2$

Thick outside the box? LIke this?

$\sf a^2+2ab+b^2 = (a+b)^2$
$\sf (a+b)(a+b) = (a+b)^2$ [ using factoring]
$\sf (a+b)^2 = (a+b)^2$  [same terms, so rewrite ]
I tried T_T

anonymous · Answer

this isn't rewriting but $a^2+b^2=\frac12[(a-b)^2+(a+b)^2]$

anonymous · Answer

err this isn't factoring but it is rewriting

thesmartone · Answer

^

usukidoll · Answer

x( 

so it's like a proof... sort of.. T_T

thesmartone · Answer

Sort of...

This equation blew my mind away when my Prof told us about it xD

usukidoll · Answer

it's like.. what I'm trying to apply is that for those problems... start at the left to achieve the right or vice versa. . . .

anonymous · Answer

(a*a)+(b*b)? @TheSmartOne

thesmartone · Answer

It's already been answered :p