The solutiong set of `|x^(2)-10| le 6`, is

Options

A. `(2, 4)`
B. `(-4, -2)`
C. `(-4, -2) cup (2, 4)`
D. `[-4, -2] cup [2, 4]`

Answer

To solve the inequality \( |x^2 - 10| \leq 6 \), we can break it down into two separate inequalities. Here are the steps:

### Step 1: Remove the Absolute Value
The expression \( |x^2 - 10| \leq 6 \) implies:
\[ -6 \leq x^2 - 10 \leq 6 \]

### Step 2: Solve the Left Inequality
Starting with the left part:
\[ x^2 - 10 \geq -6 \]
Adding 10 to both sides gives:
\[ x^2 \geq 4 \]
This can be rewritten as:
\[ x^2 \geq 2^2 \]
Taking the square root of both sides, we find:
\[ x \leq -2 \quad \text{or} \quad x \geq 2 \]

### Step 3: Solve the Right Inequality
Now, for the right part:
\[ x^2 - 10 \leq 6 \]
Adding 10 to both sides gives:
\[ x^2 \leq 16 \]
This can be rewritten as:
\[ x^2 \leq 4^2 \]
Taking the square root of both sides, we find:
\[ -4 \leq x \leq 4 \]

### Step 4: Combine the Results
Now we have two sets of inequalities:
1. From \( x^2 \geq 4 \): \( x \leq -2 \) or \( x \geq 2 \)
2. From \( x^2 \leq 16 \): \( -4 \leq x \leq 4 \)

Now we need to find the intersection of these two sets.

- The first inequality gives us two ranges: \( (-\infty, -2] \) and \( [2, \infty) \).
- The second inequality gives us the range \( [-4, 4] \).

### Step 5: Find the Intersection
The intersection of these ranges is:
- From \( (-\infty, -2] \) and \( [-4, 4] \), we get \( [-4, -2] \).
- From \( [2, \infty) \) and \( [-4, 4] \), we get \( [2, 4] \).

### Final Solution
Thus, the solution set of the inequality \( |x^2 - 10| \leq 6 \) is:
\[ [-4, -2] \cup [2, 4] \]