How do you solve #x^2+3x-18>=0# by algebraically?

1 Answer
Jan 8, 2017

Answer:

#{x:x in [3" to "+oo)}#

#{x:x in [-6" to "-oo)}#

Explanation:

Solve for #x^2+3x-18=0# and the resulting values for #x# form the reference points for 2 sets of values (domains -> input)

Set # x^2+3x-18=0#

Note that #3xx6=18" and that "6-3=3# giving:

#(x+6)(x-3)=x^2+3x-18=0#

So for this condition #x=-6" and "x=+3#
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As the #x^2# term is positive the graph is of form #uu#. So we are looking for all those values of #x# that relate to the plot where it is cuts and is above the x-axis. In other words they will be 'traveling' away from the origin in both the positive and negative direction.

#x<=-6" and "x>=3#
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The set #x# such that it is all values from and including -6 to approaching #-oo#

The set #x# such that it is all values from and including +3 to approaching #+oo#

#{x:x in [3" to "+oo)}#

#{x:x in [-6" to "-oo)}#

Tony B