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

3 Answers
Feb 3, 2018

Answer:

#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

Answer:

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

Answer:

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]}