If `|x^2- 2x- 8| + |x^2+ x -2|= 3|x +2|`, then the set of all real values of x is

Options

A. `[1,4] uu {-2}`
B. `[1,4]`
C. `[-2,1] uu [4,oo)`
D. `(-oo,-2] uu [1,4]`

Explanation

To solve the equation \( |x^2 - 2x - 8| + |x^2 + x - 2| = 3|x + 2| \), we will break it down step by step.

### Step 1: Identify the critical points
First, we need to find the points where each expression inside the absolute values equals zero.

1. **For \( |x^2 - 2x - 8| \)**:
\[ x^2 - 2x - 8 = 0 \implies (x - 4)(x + 2) = 0 \implies x = 4 \text{ or } x = -2 \]

2. **For \( |x^2 + x - 2| \)**:
\[ x^2 + x - 2 = 0 \implies (x - 1)(x + 2) = 0 \implies x = 1 \text{ or } x = -2 \]

So, the critical points are \( x = -2, 1, 4 \).

### Step 2: Test intervals defined by critical points
We will test the intervals defined by these critical points: \( (-\infty, -2) \), \( (-2, 1) \), \( (1, 4) \), and \( (4, \infty) \).

#### Interval 1: \( (-\infty, -2) \)
Choose \( x = -3 \):
\[ |(-3)^2 - 2(-3) - 8| + |(-3)^2 + (-3) - 2| = |9 + 6 - 8| + |9 - 3 - 2| = |7| + |4| = 7 + 4 = 11 \]
\[ 3|-3 + 2| = 3|-1| = 3 \]
Not equal, so no solutions in this interval.

#### Interval 2: \( (-2, 1) \)
Choose \( x = 0 \):
\[ |0^2 - 2(0) - 8| + |0^2 + 0 - 2| = |-8| + |-2| = 8 + 2 = 10 \]
\[ 3|0 + 2| = 3|2| = 6 \]
Not equal, so no solutions in this interval.

#### Interval 3: \( (1, 4) \)
Choose \( x = 2 \):
\[ |2^2 - 2(2) - 8| + |2^2 + 2 - 2| = |4 - 4 - 8| + |4 + 2 - 2| = |-8| + |4| = 8 + 4 = 12 \]
\[ 3|2 + 2| = 3|4| = 12 \]
Equal, so \( x = 2 \) is a solution.

#### Interval 4: \( (4, \infty) \)
Choose \( x = 5 \):
\[ |5^2 - 2(5) - 8| + |5^2 + 5 - 2| = |25 - 10 - 8| + |25 + 5 - 2| = |7| + |28| = 7 + 28 = 35 \]
\[ 3|5 + 2| = 3|7| = 21 \]
Not equal, so no solutions in this interval.

### Step 3: Check critical points
1. **At \( x = -2 \)**:
\[ |(-2)^2 - 2(-2) - 8| + |(-2)^2 + (-2) - 2| = |4 + 4 - 8| + |4 - 2 - 2| = |0| + |0| = 0 \]
\[ 3|-2 + 2| = 3|0| = 0 \]
Equal, so \( x = -2 \) is a solution.

2. **At \( x = 1 \)**:
\[ |1^2 - 2(1) - 8| + |1^2 + 1 - 2| = |1 - 2 - 8| + |1 + 1 - 2| = |-9| + |0| = 9 + 0 = 9 \]
\[ 3|1 + 2| = 3|3| = 9 \]
Equal, so \( x = 1 \) is a solution.

3. **At \( x = 4 \)**:
\[ |4^2 - 2(4) - 8| + |4^2 + 4 - 2| = |16 - 8 - 8| + |16 + 4 - 2| = |0| + |18| = 0 + 18 = 18 \]
\[ 3|4 + 2| = 3|6| = 18 \]
Equal, so \( x = 4 \) is a solution.

### Final Solution
The set of all real values of \( x \) that satisfy the equation is:
\[ \{ -2, 1, 2, 4 \} \]