How do you solve the system of inequalities: #-x ^ { 2} + 4< 0 and x ^ { 2} - 2x - 3\geq 0#?

1 Answer
Feb 8, 2018

#3<=x<-2#

Another way of writing this: #(-oo,-2)uu[3,+oo)#

Where the brackets ( or ) means 'excluding' and the brackets [ or ] means including. The #uu# means 'in union with' ( all together ).

Explanation:

You have two quadratics. The solutions are:

#color(brown)("Condition 1")#

For the cut off point set #x^2-2x-3=0 -> (x+1)(x-3)=0#
#=>x=-1 and x=+3# for #x^2-2x-3=0#

But we need #x^2-2x-3>=0 =>x>=-1 or x>=+3#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(brown)("Condition 2")#

For the cut of point set #-x^2+4=0 ->(-x+2)(x+2)=0 #
#=>x=+2 and x=-2# for #(-x+2)(x+2)=0 #

But we need #-x^2+4<0 => x<2 or x<-2#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Deciding which to use")#

Depending on the context of this type of problem you either need the 'boundaries' that are closest going inward or the furthest going outwards to construct the domains. The key point is that they do not cross over within the context of this question

Tony B

Tony B

#3<=x<-2#