|x^2+2x-4|>4

To solve the inequality \(|x^2 + 2x - 4| > 4\), we need to break it down into two cases based on the definition of absolute value.

1. Case 1: \(x^2 + 2x - 4 > 4\)

   This simplifies to: \[ x^2 + 2x - 8 > 0 \] Now we find the roots of the equation \(x^2 + 2x - 8 = 0\) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 1\), \(b = 2\), and \(c = -8\): \[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2} \] This gives: \[ x_1 = \frac{4}{2} = 2, \quad x_2 = \frac{-8}{2} = -4 \] Now we can test the intervals determined by the roots, which are \((- \infty, -4)\), \((-4, 2)\), and \((2, +\infty)\).

   • Test \(x = -5\) (in \((- \infty, -4)\)): \[ (-5)^2 + 2(-5) - 8 = 25 - 10 - 8 = 7 > 0 \quad \text{(True)} \]
   • Test \(x = 0\) (in \((-4, 2)\)): \[ 0^2 + 2(0) - 8 = -8 < 0 \quad \text{(False)} \]
   • Test \(x = 3\) (in \((2, +\infty)\)): \[ 3^2 + 2(3) - 8 = 9 + 6 - 8 = 7 > 0 \quad \text{(True)} \]

   Therefore, for the first case, the solution is \(x \in (-\infty, -4) \cup (2, +\infty)\).

2. Case 2: \(x^2 + 2x - 4 < -4\)

   This simplifies to: \[ x^2 + 2x > 0 \] The roots of \(x^2 + 2x = 0\) are: \[ x(x + 2) = 0 \Rightarrow x = 0, -2 \] We now test the intervals \((- \infty, -2)\), \((-2, 0)\), and \((0, +\infty)\).

   • Test \(x = -3\) (in \((- \infty, -2)\)): \[ (-3)(-3 + 2) = (-3)(-1) = 3 > 0 \quad \text{(True)} \]
   • Test \(x = -1\) (in \((-2, 0)\)): \[ (-1)(-1 + 2) = (-1)(1) = -1 < 0 \quad \text{(False)} \]
   • Test \(x = 1\) (in \((0, +\infty)\)): \[ (1)(1 + 2) = (1)(3) = 3 > 0 \quad \text{(True)} \]

   Therefore, for the second case, the solution is \(x \in (-\infty, -2) \cup (0, +\infty)\).

3. Combine the Solutions: From both cases, we combine the results:

   • Case 1: \(x \in (-\infty, -4) \cup (2, +\infty)\)
   • Case 2: \(x \in (-\infty, -2) \cup (0, +\infty)\)

   The combined solution is: \[ x \in (-\infty, -4) \cup (2, +\infty) \]

The final solution to the inequality \(|x^2 + 2x - 4| > 4\) is: \[ \boxed{(-\infty, -4) \cup (2, +\infty)} \]