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anonymous

  • 5 years ago

find two consecutive negative odd integers such that the square of the lesser integer is 40 more than the square of the greater integer

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  1. anonymous
    • 5 years ago
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    9 ,11. sounds weird :P

  2. anonymous
    • 5 years ago
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    nop

  3. anonymous
    • 5 years ago
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    i mean-9,-11

  4. anonymous
    • 5 years ago
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    how did you get that?

  5. anonymous
    • 5 years ago
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    it amounts to 81 and 121

  6. anonymous
    • 5 years ago
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    i think se said consecutive

  7. anonymous
    • 5 years ago
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    wait im sorry buit i dont understand like how you would do that prblem now..ciykd someone explain that? thanks

  8. anonymous
    • 5 years ago
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    ok for for a general formula try -x^2 =(-x+2)^2 + 40

  9. anonymous
    • 5 years ago
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    but then dont the x's cancel out?

  10. anonymous
    • 5 years ago
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    ok

  11. anonymous
    • 5 years ago
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    u start with -x^2=(-x+2)^2+40 multiply the square ull get -x^2=x^2-4x+4

  12. anonymous
    • 5 years ago
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    sorry -x^2=x^2-4x+4+40

  13. anonymous
    • 5 years ago
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    oh ok thank you guys!!

  14. anonymous
    • 5 years ago
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    wait where does the 4x come from?

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spraguer (Moderator)
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