TheSmartOne
  • TheSmartOne
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 ;)
Mathematics
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\(a^2+b^2=a^2-(-b^2)=a^2-(bi)^2=(a+bi)(a-bi)\)
anonymous
  • anonymous
am I not supposed to post it? do I message or something?
TheSmartOne
  • TheSmartOne
well seems like it wasn't really hard at all, great job xD

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UsukiDoll
  • UsukiDoll
\[ a^2 + b^2=?\] \(\sf a^2-b^2 = (a+b)(a-b)\) \(\sf a^2+2ab+b^2 = (a+b)^2\) (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
  • anonymous
this isn't rewriting but \(a^2+b^2=\frac12[(a-b)^2+(a+b)^2]\)
anonymous
  • anonymous
err this isn't factoring but it is rewriting
TheSmartOne
  • TheSmartOne
^
UsukiDoll
  • UsukiDoll
x( so it's like a proof... sort of.. T_T
TheSmartOne
  • TheSmartOne
Sort of... This equation blew my mind away when my Prof told us about it xD
UsukiDoll
  • UsukiDoll
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
  • anonymous
(a*a)+(b*b)? @TheSmartOne
TheSmartOne
  • TheSmartOne
It's already been answered :p

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