quadatic inequalites \( x^{2}-6 x-12 \geq 0 \)

To solve the quadratic inequality: \[ x^{2} - 6x - 12 \geq 0 \] we'll follow a systematic approach involving the following steps: 1. **Find the Roots of the Quadratic Equation** 2. **Determine the Sign of the Quadratic Expression in Each Interval** 3. **Express the Solution in Interval Notation** Let's go through each step in detail. --- ### 1. Find the Roots of the Quadratic Equation First, solve the corresponding quadratic equation to find its roots: \[ x^{2} - 6x - 12 = 0 \] We can use the **quadratic formula**: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] where \( a = 1 \), \( b = -6 \), and \( c = -12 \). **Calculate the Discriminant (\( \Delta \))**: \[ \Delta = b^{2} - 4ac = (-6)^{2} - 4 \cdot 1 \cdot (-12) = 36 + 48 = 84 \] Since \( \Delta > 0 \), there are two distinct real roots. **Compute the Roots**: \[ x = \frac{6 \pm \sqrt{84}}{2} = \frac{6 \pm 2\sqrt{21}}{2} = 3 \pm \sqrt{21} \] So, the roots are: \[ x = 3 + \sqrt{21} \quad \text{and} \quad x = 3 - \sqrt{21} \] **Approximate Values** (for better understanding): \[ \sqrt{21} \approx 4.5837 \] \[ x \approx 3 + 4.5837 = 7.5837 \quad \text{and} \quad x \approx 3 - 4.5837 = -1.5837 \] --- ### 2. Determine the Sign of the Quadratic in Each Interval The roots divide the real number line into three intervals: 1. **Interval I**: \( x < 3 - \sqrt{21} \) (approximately \( x < -1.5837 \)) 2. **Interval II**: \( 3 - \sqrt{21} < x < 3 + \sqrt{21} \) (approximately \( -1.5837 < x < 7.5837 \)) 3. **Interval III**: \( x > 3 + \sqrt{21} \) (approximately \( x > 7.5837 \)) Since the coefficient of \( x^{2} \) is positive (\( a = 1 \)), the parabola opens **upwards**. This means: - **Outside the Roots (Intervals I and III)**: The quadratic expression is **positive**. - **Between the Roots (Interval II)**: The quadratic expression is **negative**. --- ### 3. Express the Solution in Interval Notation We are looking for where the quadratic expression is **greater than or equal to zero**: \[ x^{2} - 6x - 12 \geq 0 \] From our analysis: - The expression is **positive** in **Intervals I and III**. - It equals **zero** exactly at the roots. Therefore, the solution includes: - \( x \leq 3 - \sqrt{21} \) - \( x \geq 3 + \sqrt{21} \) **In Interval Notation**: \[ (-\infty, \; 3 - \sqrt{21}] \cup [3 + \sqrt{21}, \; \infty) \] --- ### **Final Answer** All real numbers x with x at most 3 – √21 or at least 3 + √21. In symbols, x ≤ 3 − √21 or x ≥ 3 + √21 Alternatively, using interval notation: \[ (-\infty, \; 3 - \sqrt{21}] \cup [3 + \sqrt{21}, \; \infty) \]