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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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9 ,11. sounds weird :P
nop
i mean-9,-11
how did you get that?
it amounts to 81 and 121
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i think se said consecutive
wait im sorry buit i dont understand like how you would do that prblem now..ciykd someone explain that? thanks
ok for for a general formula try -x^2 =(-x+2)^2 + 40
but then dont the x's cancel out?
ok
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u start with -x^2=(-x+2)^2+40 multiply the square ull get -x^2=x^2-4x+4
sorry -x^2=x^2-4x+4+40
oh ok thank you guys!!
wait where does the 4x come from?
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