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anonymous

  • 5 years ago

18m^2-1 factor plz!

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  1. anonymous
    • 5 years ago
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    \[81m^2-1\]

  2. anonymous
    • 5 years ago
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    wait which one, or both

  3. anonymous
    • 5 years ago
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    both plz!

  4. anonymous
    • 5 years ago
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    if you set them to zero just have 18m^2 -1=0 18m^2 =1 m^2 = 1/18 \[m = \sqrt{1/18}\]

  5. anonymous
    • 5 years ago
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    so the same with the other you will have \[(m - \sqrt{1/18})^{2}\] and \[(m - \sqrt{1/81})^{2}\]

  6. anonymous
    • 5 years ago
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    THANKZ YOUZ!

  7. anonymous
    • 5 years ago
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    your welcome

  8. asnaseer
    • 5 years ago
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    @Davidjohn - I /think/ your last steps where you factorized the expressions has a mistake in it. I believe you can use the rule \(a^2-b^2=(a+b)(a-b)\) to get:\[81m^2-1=9^2m^2-1^2=(9m)^2-1^2=(9m+1)(9m-1)\]

  9. anonymous
    • 5 years ago
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    @asnaseer that makes a whole lot more sense!

  10. anonymous
    • 5 years ago
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    oh yeah, i generalized that one too much, youre right

  11. anonymous
    • 5 years ago
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    but the answer would be right if expanded because it would simplify down and you could multiply them to get the same answer

  12. asnaseer
    • 5 years ago
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    glad I could help - and no worries @Davidjohn - we all make mistakes - that's what makes us human :-)

  13. anonymous
    • 5 years ago
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    ah except it would be plus or minus then, not just minus minus

  14. asnaseer
    • 5 years ago
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    \[(m - \sqrt{1/81})^{2}=m^2-2\sqrt{1/81}+1/81\]

  15. anonymous
    • 5 years ago
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    right my bad :) thanks for correcting

  16. anonymous
    • 5 years ago
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    Woah... the first thing you showed me was a lot easier to understand...

  17. asnaseer
    • 5 years ago
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    :)

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