here is the problem
9p^2=-21p-6
heres what i did. i made it -9p^2-21p-6=0
i'm doing the "if it isnt a "1" before the variable technique and getting nowhere close to what the answer in the the back of the book says.
can anyone explain how to solve this? thanks.
2007-08-31 09:51:20 · 6 answers · asked by brobrahbro 1 in Science & Mathematics ➔ Mathematics
how did you get the 3p+1 at the end to get to a -1/3??
2007-08-31 10:12:49 · update #1
9 p ² + 21 p + 6 = 0
3 p ² + 7 p + 2 = 0
(3 p + 1) (p + 2) = 0
p = (- 1 / 3) , p = ( - 2 )
2007-09-04 05:51:48 · answer #1 · answered by Como 7 · 0⤊ 0⤋
9p^2 + 21p + 6 = 0 [ all 3 have a common 3 factor it out ]
3p^2 + 7p + 2 = 0
( 3p + 1) ( p + 2) = 0
when p = -1/3 or when p = -2 this equation is solved
2007-08-31 17:02:18 · answer #2 · answered by Will 4 · 0⤊ 0⤋
taken to the other side, 9p^2+21p+6=0
dividing out 3 gives 3p^2+7p+2=0
which factors into (3p+1)(p+2)
leaving p= -2 or -1/3
2007-08-31 16:58:41 · answer #3 · answered by Kenneth H 3 · 0⤊ 0⤋
multiply with -1 on both sides,
therefore, 9p^2 + 21p+6=0
therefore9p^2+18p+3p+6=0
which implies that 9p(p+2) + 3(p+2) = 0
which implies that (9p+3)(p=2) = 0.
Q.I.D.
2007-08-31 16:58:25 · answer #4 · answered by artp1991 1 · 0⤊ 0⤋
9p^2 + 21p + 6 = 0
3(3p^2 + 7p + 2) = 0
3p^2 + 7p + 2 = 0
3p^2 + 6p + p + 2 = 0
3p(p + 2) + 1(p +2) = 0
(3p + 1)(p + 2) = 0
p = -1/3, p = -2
2007-08-31 16:57:55 · answer #5 · answered by Anonymous · 0⤊ 0⤋
cant you just use the easy way...-b+-(sq root of b^2-4ac)all over 2a?
2007-08-31 16:59:41 · answer #6 · answered by Soran 4 · 0⤊ 0⤋