How do you solve #x(x-1)(x+2)>0#?

2 Answers
Dec 7, 2016

Answer:

The answer is #x in ] -2,0 [ uu ] 1,+oo [ #

Explanation:

Let #f(x)=x(x-1)(x+2)#

We do a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaa)##0##color(white)(aaaa)##1##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+2##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x##color(white)(aaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-1##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

#f(x)>0# when #x in ] -2,0 [ uu ] 1,+oo [ #

graph{x(x-1)(x+2) [-8.89, 8.89, -4.444, 4.445]}

Dec 7, 2016

Answer:

The solution for the inequality is #-2 < x < 0# or #x > 1#.

Explanation:

The inequality given is #x(x-1)(x+2)>0# i.e. product of all the terms is positive. It is apparent that sign of terms #(x+2)#, #x# and #(x-1)# will change around the values #-2#, #0# and #1# respectively. In sign chart we divide the real number line using these values, i.e. below #-2#, between #-2# and #0#, between #0# and #1# and above #1# and see how the sign of #x(x-1)(x+2)# changes.

Sign Chart

#color(white)(XXXXXXXXXXX)-2color(white)(XXXXX)0color(white)(XXXXX)1#

#(x+2)color(white)(XXXX)-ive color(white)(XXXX)+ive color(white)(XX)+ive color(white)(XXX)+ive#

#xcolor(white)(XXXXXXX)-ive color(white)(XXXX)-ive color(white)(XX)+ive color(white)(XXX)+ive#

#(x-1)color(white)(XXXX)-ive color(white)(XXXX)-ive color(white)(XX)-ive color(white)(XXX)+ive#

#x(x-1)(x+2)#
#color(white)(XXXXXXXX)-ive color(white)(XXXX)+ive color(white)(XX)-ive color(white)(XXX)+ive#

It is observed that #x(x-1)(x+2)> 0# when either #-2 < x < 0# or #x > 1#, which is the solution for the inequality.