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Math Help - need some quick help

  1. #1
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    need some quick help

    Solve for p:

    (p^2+3p-28)/(p-9)=0
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  2. #2
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    Hello, mathlete!

    It's pretty straight forward . . . exactly where is your difficulty?


    Solve for p\!:\;\;\frac{p^2+3p-28}{p-9}\:=\:0

    Providing p \neq 9, we can multiply by (p-9)

    . . and we have: . p^2 + 3p - 28 \:=\:0

    . . which factors: . (p-4)(p+7)\:=\:0

    . . and has roots: . p \;=\;4,\:-7

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