x²+x-6/x²+x-2≥0 in rational inequalities with graph

To solve the inequality \(\frac{x^2 + x - 6}{x^2 + x - 2} \geq 0\), we can follow these steps:

### Step 1: Factor the Numerator and Denominator
First, let's factor the numerator and the denominator:

- **Numerator**: \(x^2 + x - 6\) factors as:
  \[
  x^2 + x - 6 = (x - 2)(x + 3)
  \]
- **Denominator**: \(x^2 + x - 2\) factors as:
  \[
  x^2 + x - 2 = (x - 1)(x + 2)
  \]

So the inequality becomes:
\[
\frac{(x - 2)(x + 3)}{(x - 1)(x + 2)} \geq 0
\]

### Step 2: Determine the Critical Points
The critical points are where the numerator or the denominator is equal to zero. These points divide the number line into intervals that we need to test.

- \(x - 2 = 0 \Rightarrow x = 2\)
- \(x + 3 = 0 \Rightarrow x = -3\)
- \(x - 1 = 0 \Rightarrow x = 1\)
- \(x + 2 = 0 \Rightarrow x = -2\)

The critical points are \(x = -3\), \(x = -2\), \(x = 1\), and \(x = 2\).

### Step 3: Test Intervals
We will test the sign of the expression in the intervals formed by these critical points:

1. **Interval**: \((-\infty, -3)\)
   - Test with \(x = -4\)
   - \(\frac{(x - 2)(x + 3)}{(x - 1)(x + 2)} = \frac{(-4 - 2)(-4 + 3)}{(-4 - 1)(-4 + 2)} = \frac{(-6)(-1)}{(-5)(-2)} = \frac{6}{10} > 0\)
   
2. **Interval**: \((-3, -2)\)
   - Test with \(x = -2.5\)
   - \(\frac{(x - 2)(x + 3)}{(x - 1)(x + 2)} = \frac{(-2.5 - 2)(-2.5 + 3)}{(-2.5 - 1)(-2.5 + 2)}\)
   - The numerator is negative, and the denominator is positive, so \(\frac{\text{negative}}{\text{positive}} < 0\).

3. **Interval**: \((-2, 1)\)
   - Test with \(x = 0\)
   - \(\frac{(0 - 2)(0 + 3)}{(0 - 1)(0 + 2)} = \frac{(-2)(3)}{(-1)(2)} = \frac{-6}{-2} = 3 > 0\)

4. **Interval**: \((1, 2)\)
   - Test with \(x = 1.5\)
   - \(\frac{(x - 2)(x + 3)}{(x - 1)(x + 2)} = \frac{(1.5 - 2)(1.5 + 3)}{(1.5 - 1)(1.5 + 2)}\)
   - The numerator is negative, and the denominator is positive, so \(\frac{\text{negative}}{\text{positive}} < 0\).

5. **Interval**: \((2, \infty)\)
   - Test with \(x = 3\)
   - \(\frac{(3 - 2)(3 + 3)}{(3 - 1)(3 + 2)} = \frac{(1)(6)}{(2)(5)} = \frac{6}{10} > 0\)

### Step 4: Consider the Sign Changes and Critical Points
From the intervals, the expression is non-negative (positive or zero) in the intervals:
\[
(-\infty, -3) \cup (-2, 1) \cup (2, \infty)
\]
However, we need to exclude the points where the denominator is zero because the expression is undefined at those points.

So the solution is:
\[
(-\infty, -3) \cup (-2, 1) \cup (2, \infty)
\]

### Step 5: Graph the Solution
In the graph, we should plot the regions where the inequality is satisfied. The points \(x = -2\) and \(x = 1\) should be excluded (open circles), while \(x = -3\) and \(x = 2\) can be included (solid circles) if the inequality is non-strict.

Let me know if you'd like to see the graph or need more details!

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**5 Related Questions:**
1. How would the solution change if the inequality were strict \(\frac{x^2 + x - 6}{x^2 + x - 2} > 0\)?
2. What is the impact on the solution if the denominator had an additional factor, such as \((x-3)\)?
3. How can you use sign charts to solve rational inequalities more efficiently?
4. What are the steps to determine the vertical asymptotes in the graph of a rational function?
5. How would you solve this inequality if the numerator or denominator had complex roots?

**Tip:** Always verify the critical points by substituting them back into the original inequality, as it helps to confirm if they belong to the solution set.

Learn how to solve the rational inequality (x² + x - 6) / (x² + x - 2) ≥ 0 with a detailed step-by-step guide. Understand the process of factoring polynomials, testing intervals, and determining critical points. Suitable for high school students.

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