How do you solve (3y^2-9y^3)/(12y^2+5y-2)>=0?

3 Answers
Feb 3, 2018

1/3>=y>1/4 or y"<" -2/3 or y=0

Explanation:

If we divide a negative number by another negative number, we get a positive number. If we divide a positive by a positive number, we get also a positive answer. Therefore:

3y^2-9y^3>0 and 12y^2+5y-2>0

or

3y^2-9y^3<0 and 12y^2+5y-2<0

3y^2-9y^3>0
y^2(3-9y)>0|:y^2 (y^2 will always be positive)
3-9y>0|+9y |:9

1/3>y if 3y^2-9y^3>0

Therefore

1/3"<"y if 3y^2-9y^3<0

12y^2+5y-2>0|:12
y^2+5/12y-1/6>0
(y+5/24)^2-25/24^2-1/6>0 |+25/24^2+1/6
(y+5/24)^2>121/576
Take the root
y+5/24>+-11/24|-5/24

y_1>1/4 or y_2< -2/3 if 12y^2+5y-2>0

Therefore:

1/4">"y_2">" -2/3 if 12y^2+5y-2<0

All in all:
1/3>y>1/4 or y"<" -2/3

Also y=0 and y=1/3 are answers because of
3y^2-9y^3=0

Feb 3, 2018

See below

Explanation:

(3y^2 - 9y^3)/(12y^2+5y-2)>=0

Multiplying by 12y^2+5y-2 on both sides,
3y^2-9y^3>=0

Dividing by 3y^2 on both sides,
1-3y>=0

That can be simplified as:
-3y>=-1

To eliminate the negative sign, we have to change the equality sign. Thus, we get it as:
y<=1/3 ; which is your answer.

Feb 3, 2018

The solutions are y in(-oo,-2/3)uu {0}uu(1/4,1/3]

Explanation:

Solve this inequality with a sign chart

The roots of the denominator

12y^2+5y-2=0 are

y_1=(-5+sqrt(5^2-4*12*(-2)))/(2*12)=(-5+11)/(24)=6/24=1/4

y_2=(-5-sqrt(5^2-4*12*(-2)))/(2*12)=(-5-11)/(24)=-16/24
=-2/3

The roots of the numerator

3y^2-9y^3=(3y^2)(1-3y)

y_3=0

and

y_4=1/3

Let f(y)=(3y^2-9y^3)/(12y^2+5y-2)

Construct the sign chart

color(white)(aaaa)ycolor(white)(aaaa)-oocolor(white)(aaaa)-2/3color(white)(aaaa)0color(white)(aaaa)1/4color(white)(aaaa)1/3color(white)(aaaa)+oo

color(white)(aaaa)y+2/3color(white)(aaaa)-color(white)(aaa)||color(white)(aa)+color(white)(aa)+color(white)(aaaa)+color(white)(aaaa)+

color(white)(aaaa)y^2color(white)(aaaaaaa)+color(white)(aaa)||color(white)(aa)+color(white)(a)0color(white)(a)+color(white)(aaaa)+color(white)(aaa)+

color(white)(aaaa)y-1/4color(white)(aaaa)-color(white)(aaa)||color(white)(aa)-color(white)(a)0color(white)(a)-color(white)(a)||color(white)(aa)+color(white)(aaa)+

color(white)(aaaa)1-3ycolor(white)(aaaa)+color(white)(aaa)||color(white)(aa)+color(white)(a)0color(white)(a)+color(white)(a)||color(white)(aa)+color(white)(a)0color(white)(a)-

color(white)(aaaa)f(y)color(white)(aaaaaa)+color(white)(aaa)||color(white)(aa)-color(white)(a)0color(white)(a)-color(white)(a)||color(white)(aa)+color(white)(a)0color(white)(a)-

Therefore,

f(y)>=0 when y in(-oo,-2/3)uu {0}uu(1/4,1/3]

graph{(3x^2-9x^3)/(12x^2+5x-2) [-8.02, 6.03, -2.8, 4.23]}