How do you solve #3(2w^2-5)<w# using a sign chart?

1 Answer
Jan 31, 2017

Answer:

The answer is #w in]-3/2, 5/3 [#

Explanation:

Let's rewrite the inequality

#6w^2-w-6<0#

Let's factorise

#(3w-5)(2w+3)<0#

Let #f(w)=(3w-5)(2w+3)#

Now, we can build the sign chart

#color(white)(aaaa)##w##color(white)(aaaa)##-oo##color(white)(aaaa)##-3/2##color(white)(aaaa)##5/3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##2w+3##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##3w-5##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(w)##color(white)(aaaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(w)<0# when #w in]-3/2, 5/3 [#