Solve for p:

(p^2+3p-28)/(p-9)=0

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- Jan 18th 2008, 01:24 PMmathleteneed some quick help
Solve for p:

(p^2+3p-28)/(p-9)=0 - Jan 18th 2008, 02:56 PMSoroban
Hello, mathlete!

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

Quote:

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

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

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

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

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