Solve the following inequation .
(viii) `log_10(x^2-2x-2)le0`

Answer

To solve the inequation \( \log_{10}(x^2 - 2x - 2) \leq 0 \), we will follow these steps:

### Step 1: Define the Domain
The expression inside the logarithm must be greater than zero:
\[ x^2 - 2x - 2 > 0 \]
To find the roots of the quadratic equation \( x^2 - 2x - 2 = 0 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -2, c = -2 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-2)^2 - 4(1)(-2) = 4 + 8 = 12 \]
Thus, the roots are:
\[ x = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3} \]
So, the roots are \( x_1 = 1 - \sqrt{3} \) and \( x_2 = 1 + \sqrt{3} \).

### Step 2: Analyze the Sign of the Quadratic
Next, we analyze the sign of \( x^2 - 2x - 2 \) using a number line:
- The quadratic opens upwards (as the coefficient of \( x^2 \) is positive).
- The intervals to test are \( (-\infty, 1 - \sqrt{3}) \), \( (1 - \sqrt{3}, 1 + \sqrt{3}) \), and \( (1 + \sqrt{3}, \infty) \).

Testing points in each interval:
1. For \( x < 1 - \sqrt{3} \): Choose \( x = -1 \) → \( (-1)^2 - 2(-1) - 2 = 1 + 2 - 2 = 1 > 0 \)
2. For \( 1 - \sqrt{3} < x < 1 + \sqrt{3} \): Choose \( x = 1 \) → \( 1^2 - 2(1) - 2 = 1 - 2 - 2 = -3 < 0 \)
3. For \( x > 1 + \sqrt{3} \): Choose \( x = 3 \) → \( 3^2 - 2(3) - 2 = 9 - 6 - 2 = 1 > 0 \)

Thus, the solution for the domain is:
\[ x \in (-\infty, 1 - \sqrt{3}) \cup (1 + \sqrt{3}, \infty) \]

### Step 3: Solve the Inequation
Now we solve the inequation:
\[ \log_{10}(x^2 - 2x - 2) \leq 0 \]
This implies:
\[ x^2 - 2x - 2 \leq 1 \]
Rearranging gives:
\[ x^2 - 2x - 3 \leq 0 \]
Factoring the quadratic:
\[ (x - 3)(x + 1) \leq 0 \]
Finding the roots:
\[ x = -1, \quad x = 3 \]
Testing intervals:
1. For \( x < -1 \): Choose \( x = -2 \) → \( (-2 - 3)(-2 + 1) = (-5)(-1) = 5 > 0 \)
2. For \( -1 < x < 3 \): Choose \( x = 0 \) → \( (0 - 3)(0 + 1) = (-3)(1) = -3 < 0 \)
3. For \( x > 3 \): Choose \( x = 4 \) → \( (4 - 3)(4 + 1) = (1)(5) = 5 > 0 \)

The solution for this part is:
\[ x \in [-1, 3] \]

### Step 4: Combine the Solutions
Now we combine the two solutions:
1. From the domain: \( (-\infty, 1 - \sqrt{3}) \cup (1 + \sqrt{3}, \infty) \)
2. From the inequation: \( [-1, 3] \)

Calculating \( 1 - \sqrt{3} \approx -0.732 \) and \( 1 + \sqrt{3} \approx 2.732 \):
- The intersection of \( [-1, 3] \) with \( (-\infty, 1 - \sqrt{3}) \) gives \( [-1, 1 - \sqrt{3}) \).
- The intersection of \( [-1, 3] \) with \( (1 + \sqrt{3}, \infty) \) gives \( (1 + \sqrt{3}, 3] \).

Thus, the final solution is:
\[ x \in [-1, 1 - \sqrt{3}) \cup (1 + \sqrt{3}, 3] \]