Question

How do you solve `abs(x^2 - 2x - 16) = 8`?

Answer 1

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.