How do you solve #(10y^2-13y-30)/(2y^5-50y^3)<0#?

1 Answer
Dec 18, 2016

Answer:

Solution is #-oo< y< -5# or #-6/5< y< 0# or #5/2< y< 5#.

Explanation:

#(10y^2-13y-30)/(2y^5-50y^3)#

= #(10y^2-25y+12y-30)/(2y^3(y^2-25)#

= #(5y(2y-5)+6(2y-5))/(2y^3(y+5)(y-5))#

= #((5y+6)(2y-5))/(2y^3(y+5)(y-5))#

Hence, #f(y)=(10y^2-13y-30)/(2y^5-50y^3)<0#

#hArrf(y)=((5y+6)(2y-5))/(2y^3(y+5)(y-5))<0#
#:.# zeros of numerator and denominators are

#{-5,-6/5,0,5/2,5}# and they divide real number line in #6# intervals.

In #-oo< y< -5#, all monomials are negative and hence #f(y)<0#

In #-5< y< -6/5#, while #(y+5)# is positive, all other monomials are negative and hence #f(y)>0#

In #-6/5< y< 0#, while #(y+5)# and #(5y+6)# are positive, all other monomials are negative and hence #f(y)<0#

In #0< y< 5/2#, while #(y+5)#, #(5y+6)# and #y# are positive, #(2y-5)# and #(y-5)# are negative and hence #f(y)>0#

In #5/2< y< 5#, while #(y+5)#, #(5y+6)#, #y# and #(2y-5)# are positive, #(y-5)# is negative and hence #f(y)<0#

and in #5< y< oo#, all monomials are positive and hence #f(y)>0#

Hence solution is #-oo< y< -5# or #-6/5< y< 0# or #5/2< y< 5#.
graph{(10x^2-13x-30)/(2x^5-50x^3) [-10, 10, -2, 2]}