Solving inequalities. How to solve #(x+5)/(3-x^2)≥ 0#?

2 Answers
Apr 20, 2018

See details below

Explanation:

A fraction is positive or zero if and only if numerator and denominator have the same sign

Case 1.- Both positives
#x+5>=0# then #x>=-5# and
#3-x^2>0# (imposible to be zero) then #3>x^2# that is

#-sqrt3 < x < sqrt3#

The intersection of both sets of values is #[-5,oo)nn(-sqrt3,sqrt3)=(-sqrt3,sqrt3)#

Case 2.- Both negatives

Similarly the solutions are #(-oo,-5]nn((-oo,-sqrt3)uu(sqrt3,+oo))=#

#=[-5,-sqrt3)uu(sqrt3,+oo)#

Now, the union of both cases will be the final result

#[-5,-sqrt3)uu(-sqrt3,sqrt3)uu(sqrt3,+oo)#

Apr 20, 2018

The solution is #x in (-oo,-5]uu(-sqrt3, sqrt3)#

Explanation:

The inequality is

#(x+5)/(3-x^2)>=0#

#(x+5)/((sqrt3-x)(sqrt3+x))>=0#

Let #f(x)=(x+5)/((sqrt3-x)(sqrt3+x))#

Let's build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-5##color(white)(aaaa)##-sqrt3##color(white)(aaaa)##+sqrt3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+5##color(white)(aaaa)##-##color(white)(aaa)##0##color(white)(aaa)##+##color(white)(aaaaa)##+##color(white)(aaaaa)##+#

#color(white)(aaaa)##sqrt3+x##color(white)(aaa)##-##color(white)(aaa)####color(white)(aaa)##-##color(white)(aaa)##||##color(white)(aa)##+##color(white)(aaaaa)##+#

#color(white)(aaaa)##sqrt3-x##color(white)(aaa)##+##color(white)(aaa)####color(white)(aaa)##+##color(white)(aaa)####color(white)(aaa)##+##color(white)(aa)##||##color(white)(aa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaa)##0##color(white)(aa)##-##color(white)(aaa)##||##color(white)(aa)##+##color(white)(aa)##||##color(white)(aa)##-#

Therefore,

#f(x)>=0# when #x in (-oo,-5]uu(-sqrt3, sqrt3)#

graph{(x+5)/(3-x^2) [-12.66, 12.66, -6.33, 6.33]}