2007-10-29 09:04:56 · 8 answers · asked by Gotti 1 in Science & Mathematics ➔ Mathematics
y3 + 6y2 = 27y
=> y3 + 6y2 - 27y =0
=> y( y^2 + 6y - 27) =0
=> y(y+9)(y-3)=0
=> y = 0 , y = -9 , y = 3
QED
2007-10-29 09:08:52 · answer #1 · answered by harry m 6 · 2⤊ 0⤋
y^3+ 6y^2 - 27y = 0
y( y^2 +6y -27) =0
y^2 +6y -27 =0
Discriminant = 6^2 +4(27) =144 = 12^2
y1 = (-6 -12)/2 =-9
y2 = (-6 +12)/2 = 3
solutions:
y =0
y =3
y =-9
y3 + 6y2 -27y = y(y-3)(y+9)
2007-10-29 09:10:06 · answer #2 · answered by Anonymous · 1⤊ 0⤋
y3 + 6y2 = 27y
y^3 + 6y^2 - 27y = 0
y(y^2 + 6y - 27) = 0
y(y + 9)(y - 3) = 0
y = 0, -9, 3
2007-10-29 09:10:06 · answer #3 · answered by Anonymous · 1⤊ 0⤋
Divide through by y.
y2 + 6y = 27
Subtract 27 from both sides
y2 + 6y - 27 = 0
(y + 9)(y - 3) = 0
y + 9 = 0
y = -9
y - 3 = 0
y = 3
solution: 0, -9, 3
2007-10-29 09:12:01 · answer #4 · answered by papastolte 6 · 0⤊ 0⤋
y = 0, y = -9, y = 3
Steps:
Set equal to zero. ( y^3 + 6y^2 - 27y = 0 )
Factor out a y. ( y * ( y^2 + 6y - 27 ) = 0 )
Factor the remaining polynomial. ( y * (y + 9) * (y - 3) = 0 )
Solve. ( y = 0, y = -9, y = 3 )
2007-10-29 09:11:14 · answer #5 · answered by oc_xc 2 · 1⤊ 0⤋
Y^3 + 6Y^2 - 27y = 0
Y(Y^2 + 6Y - 27) = 0
Y[(Y +9)(Y - 3)] = 0
Hence 'y' = 3, 0 & -9.
2007-10-29 09:14:20 · answer #6 · answered by lenpol7 7 · 0⤊ 0⤋
y^3+6y^2-27y=0
y(y^2+6y-27)=0
y(y+9)(y-3)=0
y=0, y=-9, y=3
2007-10-29 09:09:28 · answer #7 · answered by aba 2 · 1⤊ 0⤋
divide the whole equation by y. Then apply the Quadratic formula.
2007-10-29 09:08:26 · answer #8 · answered by dongskie mcmelenccx 3 · 1⤊ 1⤋