x2 + 4х - 12 < 0
To solve the inequality \( x^2 + 4x - 12 < 0 \), we first need to determine the roots of the corresponding equation \( x^2 + 4x - 12 = 0 \).
We can use the quadratic formula to find the roots:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 4 \), and \( c = -12 \).
Calculating the discriminant:
\[ b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot (-12) = 16 + 48 = 64 \]
Now substituting back into the quadratic formula:
\[ x = \frac{-4 \pm \sqrt{64}}{2 \cdot 1} = \frac{-4 \pm 8}{2} \]
Calculating the two solutions:
- \( x = \frac{-4 + 8}{2} = \frac{4}{2} = 2 \)
- \( x = \frac{-4 - 8}{2} = \frac{-12}{2} = -6 \)
We have the roots \( x = 2 \) and \( x = -6 \).
Next, we need to determine where the quadratic \( x^2 + 4x - 12 \) is less than 0. We analyze the intervals defined by the roots:
- \( (-\infty, -6) \)
- \( (-6, 2) \)
- \( (2, \infty) \)
To do this, we can test points from each interval:
For \( x = -7 \) in \( (-\infty, -6) \): \[ (-7)^2 + 4(-7) - 12 = 49 - 28 - 12 = 9 \quad (\text{greater than } 0) \]
For \( x = 0 \) in \( (-6, 2) \): \[ 0^2 + 4(0) - 12 = -12 \quad (\text{less than } 0) \]
For \( x = 3 \) in \( (2, \infty) \): \[ 3^2 + 4(3) - 12 = 9 + 12 - 12 = 9 \quad (\text{greater than } 0) \]
From this analysis, the quadratic is negative (i.e., \( x^2 + 4x - 12 < 0 \)) in the interval \( (-6, 2) \).
Therefore, the solution to the inequality \( x^2 + 4x - 12 < 0 \) is:
\[ \boxed{(-6, 2)} \]
