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TheSmartOne

  • one year ago

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 ;)

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  1. anonymous
    • one year ago
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    \(a^2+b^2=a^2-(-b^2)=a^2-(bi)^2=(a+bi)(a-bi)\)

  2. anonymous
    • one year ago
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    am I not supposed to post it? do I message or something?

  3. TheSmartOne
    • one year ago
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    well seems like it wasn't really hard at all, great job xD

  4. UsukiDoll
    • one year ago
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    \[ a^2 + b^2=?\] \(\sf a^2-b^2 = (a+b)(a-b)\) \(\sf a^2+2ab+b^2 = (a+b)^2\) (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

  5. anonymous
    • one year ago
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    this isn't rewriting but \(a^2+b^2=\frac12[(a-b)^2+(a+b)^2]\)

  6. anonymous
    • one year ago
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    err this isn't factoring but it is rewriting

  7. TheSmartOne
    • one year ago
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    ^

  8. UsukiDoll
    • one year ago
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    x( so it's like a proof... sort of.. T_T

  9. TheSmartOne
    • one year ago
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    Sort of... This equation blew my mind away when my Prof told us about it xD

  10. UsukiDoll
    • one year ago
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    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. . . .

  11. anonymous
    • one year ago
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    (a*a)+(b*b)? @TheSmartOne

  12. TheSmartOne
    • one year ago
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    It's already been answered :p

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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