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

May 26, 2015

Recall the definition of the absolute value:
$| C | = C$ for all $C \ge 0$ 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 ${x}^{2} - 2 x - 16 \ge 0$.
A2 is a set of all real numbers $x$, for which ${x}^{2} - 2 x - 16 < 0$.

Let's determine these areas. Graphically, ${x}^{2} - 2 x - 16$ 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 ${x}^{2} - 2 x - 16 = 0$ 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 ${x}^{2} - 2 x - 16 = 0$ are
${x}_{1 , 2} = \frac{2 \pm \sqrt{4 + 64}}{2} = 1 \pm \sqrt{17}$

So, area A1 where our quadratic polynomial is non-negative consists of two non-intersecting parts:
$x \le 1 - \sqrt{17}$ and
$x \ge 1 + \sqrt{17}$.

Area A2 where our quadratic polynomial is negative is characterized by a combined inequality
$1 - \sqrt{17} < x < 1 + \sqrt{17}$

Case 1:
$x \le 1 - \sqrt{17}$ OR $x \ge 1 + \sqrt{17}$
Since $\sqrt{17} \cong 4.123$, our area, approximately, consists of two parts:
$x \le - 3.123$ OR $x \ge 5.123$
Then our quadratic polynomial is non-negative and we can simply drop the absolute value sign obtaining an equation
${x}^{2} - 2 x - 16 = 8$ or, equivalently, ${x}^{2} - 2 x - 24 = 0$.
Its solutions are
${x}_{1 , 2} = \frac{2 \pm \sqrt{4 + 96}}{2} = \frac{2 \pm 10}{2}$, that is
${x}_{1} = 6$, ${x}_{2} = - 4$
Solution ${x}_{1} = 6$ is valid since $6 \ge 5.123$.
Solution ${x}_{2} = - 4$ is valid since $- 4 \le - 3.123$.

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
$- \left({x}^{2} - 2 x - 16\right) = 8$ or, equivalently, ${x}^{2} - 2 x - 8 = 0$.
Its solutions are
${x}_{3 , 4} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm 6}{2}$, 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.