How do you graph and solve |4x+8|>=20?

Jun 30, 2016

The solution is a union of two intervals:
$x \le - 7$ or $x \in \left(- \infty , - 7\right]$
UNION with
$x \ge 3$ or $x \in \left[3 , + \infty\right)$

Explanation:

The value of $| 4 x + 8 |$ is defined as follows:
if $4 x + 8 \ge 0$ then $| 4 x + 8 | = 4 x + 8$
otherwise (that is, id $4 x + 8 < 0$), $| 4 x + 8 | = - \left(4 x + 8\right)$.

Let's consider a point $x = - 2$, where term $4 x + 8$ changes from negative (to the left of $x = - 2$) to positive (to the right of $x = - 2$).

From definition of absolute value mentioned above,
if $x < - 2$, 4x+8 < 0 and, therefore, $| 4 x + 8 | = - \left(4 x + 8\right)$ and our initial inequality looks like this:
$- \left(4 x + 8\right) \ge 20$ or
$- 4 x - 8 \ge 20$ or
$- 8 - 20 \ge 4 x$ or
$4 x \le - 28$ or
$x \le - 7$
Since $x \le - 7$ lies inside the interval $x < - 2$ that we consider, all $x \le - 7$ are solutions.

From definition of absolute value mentioned above,
if $x \ge - 2$, 4x+8 >= 0 and, therefore, $| 4 x + 8 | = 4 x + 8$ and our initial inequality looks like this:
$4 x + 8 \ge 20$ or
$4 x \ge 12$ or
$x \ge 3$
Since $x \ge 3$ lies inside the interval $x \ge - 2$ that we consider, all $x \ge 3$ are solutions.

So, we have two separate intervals that represent the solutions to this inequality:
$x \le - 7$ and $x \ge 3$.

Graphically, it looks like this:
graph{|4x+8| [-24, 24, -23.12, 23.12]}
It can be observed that this graph is above the line $y = 20$ when $x \le - 7$ or $x \ge 3$.