Recall the definition of the absolute value:
#|C|=C# for all #C>=0# and
#|C|=-C# for all #C<0#.
Let's divide the real numbers into two separate areas:
A1 is a set of all real numbers #x#, for which #x^2-2x-16 >= 0#.
A2 is a set of all real numbers #x#, for which #x^2-2x-16 < 0#.
Let's determine these areas. Graphically, #x^2-2x-16# is represented by a parabola with its two "horns" directed upwards. Therefore, it is greater than #0# to the left of the left (smaller) solution of an equation #x^2-2x-16=0# and to the right of the right (bigger) solution of this equation. In between the solutions this quadratic polynomial is less than #0#.
graph{x^2-2x-16 [-10, 10, -25, 25]}
The two solutions to the equation #x^2-2x-16=0# are
#x_(1,2)=(2+-sqrt(4+64))/2=1+-sqrt(17)#
So, area A1 where our quadratic polynomial is non-negative consists of two non-intersecting parts:
#x <= 1-sqrt(17)# and
#x >= 1+sqrt(17)#.
Area A2 where our quadratic polynomial is negative is characterized by a combined inequality
#1-sqrt(17) < x < 1+sqrt(17)#
Case 1:
#x <= 1-sqrt(17)# OR #x >= 1+sqrt(17)#
Since #sqrt(17)~=4.123#, our area, approximately, consists of two parts:
#x <= -3.123# OR #x >= 5.123#
Then our quadratic polynomial is non-negative and we can simply drop the absolute value sign obtaining an equation
#x^2-2x-16=8# or, equivalently, #x^2-2x-24=0#.
Its solutions are
#x_(1,2)=(2+-sqrt(4+96))/2=(2+-10)/2#, that is
#x_1=6#, #x_2=-4#
Solution #x_1=6# is valid since #6 >= 5.123#.
Solution #x_2=-4# is valid since #-4 <= -3.123#.
Case 2:
#1-sqrt(17) < x < 1+sqrt(17)#
Approximately, it means
#-3.123 < x < 5.123#
Then our quadratic polynomial is negative and, if we want to get rid of absolute value sign, we have to change the sign of this polynomial getting the equation
#-(x^2-2x-16)=8# or, equivalently, #x^2-2x-8=0#.
Its solutions are
#x_(3,4)=(2+-sqrt(4+32))/2=(2+-6)/2#, that is
#x_3=4#, #x_4=-2#
Solution #x_3=4# is valid since #-3.123 < 4 < 5.123#.
Solution #x_4=-2# is valid since #-3.123 < -2 < 5.123#.
Solution:
#x_1 = 6#
#x_2 = -4#
#x_3= 4#
#x_4 = -2#
Graphically, the absolute value of a given polynomial looks like
graph{|x^2-2x-16| [-10, 10, -25, 25]}
If you draw a line #y=8# on this graph, it will intersect the graph at four points listed above as solutions.
