To solve the inequation \( \sqrt{x^2 - 3x - 10} > (8 - x) \), we will follow these steps:
### Step 1: Determine the domain of the square root
The expression inside the square root must be non-negative:
\[
x^2 - 3x - 10 \geq 0
\]
To factor this quadratic, we can find the roots using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1, b = -3, c = -10 \).
Calculating the discriminant:
\[
b^2 - 4ac = (-3)^2 - 4 \cdot 1 \cdot (-10) = 9 + 40 = 49
\]
Now applying the quadratic formula:
\[
x = \frac{3 \pm 7}{2}
\]
This gives us the roots:
\[
x_1 = \frac{10}{2} = 5, \quad x_2 = \frac{-4}{2} = -2
\]
Thus, the factored form is:
\[
(x - 5)(x + 2) \geq 0
\]
### Step 2: Analyze the intervals
The critical points are \( x = -2 \) and \( x = 5 \). We will test the intervals:
1. \( (-\infty, -2) \)
2. \( (-2, 5) \)
3. \( (5, \infty) \)
Testing:
- For \( x = -3 \) in \( (-\infty, -2) \):
\[
(-3 - 5)(-3 + 2) = (-8)(-1) > 0 \quad \text{(True)}
\]
- For \( x = 0 \) in \( (-2, 5) \):
\[
(0 - 5)(0 + 2) = (-5)(2) < 0 \quad \text{(False)}
\]
- For \( x = 6 \) in \( (5, \infty) \):
\[
(6 - 5)(6 + 2) = (1)(8) > 0 \quad \text{(True)}
\]
Thus, the solution for the domain is:
\[
x \in (-\infty, -2] \cup [5, \infty)
\]
### Step 3: Solve the inequation
Now we square both sides of the inequation:
\[
x^2 - 3x - 10 > (8 - x)^2
\]
Expanding the right side:
\[
(8 - x)^2 = 64 - 16x + x^2
\]
Now we have:
\[
x^2 - 3x - 10 > 64 - 16x + x^2
\]
Subtract \( x^2 \) from both sides:
\[
-3x - 10 > 64 - 16x
\]
Rearranging gives:
\[
16x - 3x > 64 + 10
\]
\[
13x > 74
\]
Dividing by 13:
\[
x > \frac{74}{13}
\]
### Step 4: Combine the results
Now we need to combine this with the domain found in Step 2. The value \( \frac{74}{13} \approx 5.69 \) is greater than 5, so we only need to consider the interval:
\[
x \in \left(\frac{74}{13}, \infty\right)
\]
### Final Solution
Thus, the solution to the inequation \( \sqrt{x^2 - 3x - 10} > (8 - x) \) is:
\[
x \in \left(\frac{74}{13}, \infty\right)
\]
