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

1 Answer
Jan 8, 2017

{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
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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 BTony B