I have been working on this problem for an hour can someone please help.
2006-07-05 19:00:54 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(z^2)/-1 = (z)/3 + (2)/-3
left and right times -3
3z2 = -z + 2
3z2 +z -2 = 0
3z2 + 3z -2z - 2 = 0 see below
3z(z + 1) - 2(z + 1) = 0
(3z - 2)(z + 1) = 0
z = 2/3 or z=-1 Ready
3z2 +z -2 = 0
3 x -2 = -6
-6 = -1 x 6 = 1 x -6 = 2 x -3 = -2 x 3 bingo
the last one is good because -2 + 3 = +1, the co factor of the middle term +z
3z2 + 3z -2z - 2 = 0
2006-07-05 19:11:35 · answer #1 · answered by Thermo 6 · 1⤊ 0⤋
z = 2/3 or -1
Take original equation times -3 and move everything to one side.
3z^2 = -z + 2
3z^2 +z - 2 = 0
(3z - 2)(z + 1) = 0
z = 2/3 or -1
2006-07-06 02:08:23 · answer #2 · answered by Jason 3 · 0⤊ 0⤋
z^2/-1 is -(z^2)
now -z^2=z/3-2/3
this becomes -3z^2=z-2
take -3z^2 to the right side
3z^2+z-2=0
factorising this equation the factors are
(z+1)(3z-2)=0
therefore z=-1 or z=2/3
2006-07-06 02:08:14 · answer #3 · answered by priya 2 · 0⤊ 0⤋
-z^2=z/3-2/3
-3z^2=z-2
3z^2 +z -2=0
(3z-2)(z+1)=0
z=2/3 or -1
2006-07-06 02:05:01 · answer #4 · answered by Rohit C 3 · 0⤊ 0⤋
Really wish I could.
Sorry
2006-07-06 02:04:03 · answer #5 · answered by Rick A 5 · 0⤊ 0⤋
the backstreet boyz
2006-07-06 02:02:26 · answer #6 · answered by Chimichangaz4life 1 · 0⤊ 0⤋
zero
2006-07-06 02:06:56 · answer #7 · answered by slammerx1 1 · 0⤊ 0⤋
damn........
2006-07-06 02:04:30 · answer #8 · answered by Cha 3 · 0⤊ 0⤋