subtract 6y ... 9y^2 + 12y - 12 = 0 ... divide by 3 ... 3y^2 + 4y - 4 = 0
factor ... (3y - 2) (y + 2) = 0
to find y , set each factor equal to zero
9y^2 + 18y - 12 = 6y
Can someone help me solve this. I am struggling. 😢
2 answers
9y^2 + 18y - 12 = 6y
9y^2 + 18y - 6y -12 = 0
9y^2 + 12y - 12 = 0
3y^2 + 4y - 4 = 0
A*C = 3(-4) = -12 = -2*6. sum = -2+6 = 4 = B.
3y^2 + (6y-2y) - 4 = 0
3y(y+2) -2(y+2) = 0
(y+2)(3y-2) = 0.
y+2 = 0, Y = -2.
3y-2 = 0, Y = 2/3.
9y^2 + 18y - 6y -12 = 0
9y^2 + 12y - 12 = 0
3y^2 + 4y - 4 = 0
A*C = 3(-4) = -12 = -2*6. sum = -2+6 = 4 = B.
3y^2 + (6y-2y) - 4 = 0
3y(y+2) -2(y+2) = 0
(y+2)(3y-2) = 0.
y+2 = 0, Y = -2.
3y-2 = 0, Y = 2/3.
1 answer