`f(x)=x^(2)+6, omega (x)=5x`. Solve the equation `f(x)=|omega (x)|`.

Answer

To solve the equation \( f(x) = | \omega(x) | \) where \( f(x) = x^2 + 6 \) and \( \omega(x) = 5x \), we will consider two cases based on the definition of the absolute value.

### Step-by-step Solution:

1. **Set up the equation**:
We need to solve \( f(x) = | \omega(x) | \).
This gives us two cases to consider:
- Case 1: \( \omega(x) \geq 0 \)
- Case 2: \( \omega(x) < 0 \)

2. **Case 1: \( \omega(x) \geq 0 \)**:
- Here, \( | \omega(x) | = \omega(x) \).
- Thus, we have:
\[ f(x) = \omega(x) \implies x^2 + 6 = 5x \]
- Rearranging gives:
\[ x^2 - 5x + 6 = 0 \]
- We can factor this quadratic equation:
\[ (x - 2)(x - 3) = 0 \]
- Therefore, the solutions are:
\[ x = 2 \quad \text{and} \quad x = 3 \]
- Since \( \omega(x) = 5x \) is non-negative for \( x \geq 0 \), both solutions \( x = 2 \) and \( x = 3 \) are valid.

3. **Case 2: \( \omega(x) < 0 \)**:
- Here, \( | \omega(x) | = -\omega(x) \).
- Thus, we have:
\[ f(x) = -\omega(x) \implies x^2 + 6 = -5x \]
- Rearranging gives:
\[ x^2 + 5x + 6 = 0 \]
- We can use the quadratic formula to solve this:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} \]
\[ = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm 1}{2} \]
- This gives us two solutions:
\[ x = \frac{-4}{2} = -2 \quad \text{and} \quad x = \frac{-6}{2} = -3 \]
- Both solutions \( x = -2 \) and \( x = -3 \) are valid since \( \omega(x) < 0 \) for \( x < 0 \).

4. **Final Solutions**:
Combining both cases, the solutions to the equation \( f(x) = | \omega(x) | \) are:
\[ x = 2, \quad x = 3, \quad x = -2, \quad x = -3 \]