Question

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

Answer 1

`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`

Answer 2

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.

Answer 3

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)` `y` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-2/3` `color(white)(aaaa)` `0` `color(white)(aaaa)` `1/4` `color(white)(aaaa)` `1/3` `color(white)(aaaa)` `+oo`

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

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

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

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

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