|x^2+4x|<12
To solve the inequality \(|x^2 + 4x| < 12\), we can break it down into two cases based on the property of absolute values.
Case 1: \(x^2 + 4x < 12\)
Rearrange the inequality: \[ x^2 + 4x - 12 < 0 \]
Factor the quadratic: \[ (x + 6)(x - 2) < 0 \]
Determine the critical points by setting the factors equal to zero: \[ x + 6 = 0 \quad \Rightarrow \quad x = -6 \] \[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
Test intervals created by the critical points (\(-\infty, -6\), \(-6, 2\), \(2, \infty\)):
- For \(x < -6\), choose \(x = -7\): \[ (-7 + 6)(-7 - 2) = (-1)(-9) = 9 > 0 \]
- For \(-6 < x < 2\), choose \(x = 0\): \[ (0 + 6)(0 - 2) = (6)(-2) = -12 < 0 \]
- For \(x > 2\), choose \(x = 3\): \[ (3 + 6)(3 - 2) = (9)(1) = 9 > 0 \]
The solution for this case is: \[ -6 < x < 2 \]
Case 2: \(x^2 + 4x > -12\)
Rearrange the inequality: \[ x^2 + 4x + 12 > 0 \]
Determine the discriminant of the quadratic: \[ D = b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot 12 = 16 - 48 = -32 \]
Since the discriminant is negative, the quadratic \(x^2 + 4x + 12\) is always positive (as it opens upwards). Thus, this inequality holds for all \(x\).
Combine Results
From Case 1, we found \(-6 < x < 2\), and from Case 2, the inequality holds for all \(x\). Therefore, the solution to the original inequality \(|x^2 + 4x| < 12\) is:
\[ \boxed{(-6, 2)} \]
