Answer:
To factor the expression \( \sec^2x - \sec x - 2 \), let's denote \( u = \sec x \):
The expression becomes \( u^2 - u - 2 \).
Now, factor the quadratic expression:
\[ (u - 2)(u + 1) \]
Replace \( u \) with \( \sec x \):
\[ (\sec x - 2)(\sec x + 1) \]
So, \( \sec^2x - \sec x - 2 \) can be factored as \( (\sec x - 2)(\sec x + 1) \).To solve the inequality \(1.13x^2 - x + 6 < 2(x + 2)\), follow these steps:
1. Distribute the 2 on the right side of the inequality:
\(1.13x^2 - x + 6 < 2x + 4\)
2. Move all terms to one side of the inequality to set it to zero:
\(1.13x^2 - x - 2x + 6 - 4 < 0\)
Simplify the expression:
\(1.13x^2 - 3x + 2 < 0\)
3. Factor the quadratic expression:
\( (1.13x - 2)(x - 1) < 0 \)
4. Find the critical points by setting each factor equal to zero:
\(1.13x - 2 = 0 \) --> \( x = \frac{2}{1.13} \)
\(x - 1 = 0 \) --> \( x = 1 \)
5. Now, create intervals using these critical points and test each interval with a test point (e.g., \(x = 0\)) to determine the sign:
- For \( x < \frac{2}{1.13} \), use \( x = 0 \):
\( (1.13(0) - 2)(0 - 1) < 0 \) --> \( ( -2 )(-1) < 0 \), which is true.
- For \( \frac{2}{1.13} < x < 1 \), use \( x = \frac{3}{2} \):
\( (1.13(\frac{3}{2}) - 2)(\frac{3}{2} - 1) < 0 \) --> \( ( -\frac{1}{2})(\frac{1}{2}) < 0 \), which is true.
- For \( x > 1 \), use \( x = 2 \):
\( (1.13(2) - 2)(2 - 1) < 0 \) --> \( (0)(1) < 0 \), which is false.
6. Combine the intervals where the inequality is true:
\( x < \frac{2}{1.13} \) or \( \frac{2}{1.13} < x < 1 \)
So, the solution to \(1.13x^2 - x + 6 < 2(x + 2)\) is \( x < \frac{2}{1.13} \) or \( \frac{2}{1.13} < x < 1 \).
