How do you solve #x/(x-2)> -1/(x+3)# using a sign chart?

1 Answer
Dec 10, 2017

Answer:

The solution is #x in (-oo,-4.45] uu(-3, 0.45] uu (2, +oo)#

Explanation:

Simplify the inequality, we cannot do crossing over.

#x/(x-2) > -1/(x+3)#

#x/(x-2)+1/(x+3)>0#

#(x(x+3)+(x-2))/((x-2)(x+3))>0#

#(x^2+3x+x-2)/((x-2)(x+3)) >0#

#(x^2+4x-2)/((x-2)(x+3)) >0#

The roots of the numerator

#x^2+4x-2=0#

are

#x=(-4+-sqrt(16-4(1)(-2)))/(2)#

#=-2+-sqrt6#

#x_1=-2-sqrt6=-4.45#

#x_2=-2+sqrt6=0.45#

Let

#f(x)=((x-x_1)(x-x_2))/((x-2)(x+3))#

We can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##x_1##color(white)(aaaa)##-3##color(white)(aaaaa)##x_2##color(white)(aaaaa)##2##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##x-x_1##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)###

#color(white)(aaaa)##x-3##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aa)##||##color(white)(a)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-x_2##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aa)####color(white)(aa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-2##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aa)####color(white)(aa)##-##color(white)(aaaa)##-##color(white)(aa)##||##color(white)(a)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaaa)##-##color(white)(aa)##||##color(white)(a)##+##color(white)(aaaa)##-##color(white)(aa)##||##color(white)(a)##+#

Therefore,

#f(x) >0# when #x in (-oo,-4.45] uu(-3, 0.45] uu (2, +oo)#

graph{(x^2+4x-2)/((x-2)(x+3)) [-10, 10, -5, 5]}