Answer
All real numbers \( x \) such that \( x \leq 3 - \sqrt{21} \) or \( x \geq 3 + \sqrt{21} \).
Solution
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:
\[
(-\infty, \; 3 - \sqrt{21}] \cup [3 + \sqrt{21}, \; \infty)
\]
Answered by UpStudy AI and reviewed by a Professional Tutor
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