How do you solve #(2y^2+3y-20)/(y^3-y^2)>0#?

1 Answer
Jan 11, 2017

Answer:

The answer is #y in [-4, 0[ uu] 0,1 [uu [5/2, +oo[#

Explanation:

Factorise the numerator and denominator

#2y^2+3y-20=(2y-5)(y+4)#

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

So,

#(2y^2+3y-20)/(y^3-y^2)=((2y-5)(y+4))/(y^2(y-1))#

Let #y=((2y-5)(y+4))/(y^2(y-1))#

The domain of #y# is #D_y=RR-{0,1}#

#y^2>0 AA y in D_y#

Let's do a sign chart

#color(white)(aaaa)##y##color(white)(aaaa)##-oo##color(white)(aaaa)##-4##color(white)(aaaaaaaa)##0##color(white)(aaaa)##1##color(white)(aaaaa)##5/2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##y+4##color(white)(aaaa)##-##color(white)(aaaaa)##+##color(white)(aaaa)##color(red)(∥)##color(white)()##+##color(red)(∥)##color(white)(a)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##y-1##color(white)(aaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##color(red)(∥)##color(white)()##-##color(red)(∥)##color(white)(a)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##2y-5##color(white)(aaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##color(red)(∥)##color(white)()##-##color(red)(∥)##color(white)(a)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(y)##color(white)(aaaaa)##-##color(white)(aaaaa)##+##color(white)(aaaa)##color(red)(∥)##color(white)()##+##color(red)(∥)##color(white)(a)##-##color(white)(aaaa)##+#

Therefore,

#f(y)>0#, when #y in [-4, 0[ uu] 0,1 [uu [5/2, +oo[#