3. Worked Examples

📘 Example 1: Solve 𝑥² − 5𝑥 + 6 < 0 Step 1: Factor the expression 𝑥² − 5𝑥 + 6 = (𝑥 − 2)(𝑥 − 3) Step 2: Solve the equation 𝑥 − 2 = 0 ⇒ 𝑥 = 2 𝑥 − 3 = 0 ⇒ 𝑥 = 3 Step 3: Use sign analysis The roots divide the number line into intervals: • 𝑥 < 2 • 2 < 𝑥 < 3 • 𝑥 > 3 Choose test points: • 𝑥 = 1: (1 − 2)(1 − 3) = (−1)(−2) = +2 → not valid • 𝑥 = 2.5: (2.5 − 2)(2.5 − 3) = (+)(−) = − → valid • 𝑥 = 4: (+)(+) = + → not valid ✅ Final Answer: 2 < 𝑥 < 3

In this example, we solve the inequality x² - 5x + 6 < 0 step by step. We first need to factor the quadratic expression. Factoring gives us (x - 2)(x - 3). Next, we find the roots by setting each factor to zero, which tells us where the expression equals zero. From my calculations, we find that x = 2 and x = 3. These roots divide the number line into three intervals: values less than 2, between 2 and 3, and greater than 3. We choose a test point within each interval to determine whether the original inequality holds true. After testing the points, we find that the inequality is satisfied only in the interval 2 < x < 3.

You can think of the quadratic inequality as finding a range of temperatures for a specific chemical reaction to be effective. The reaction only works when the temperature is between 2 and 3 degrees (for example, Celsius), while at temperatures lower than 2 and higher than 3, the reaction doesn't occur. Hence, we can see that to achieve our desired result, we need to maintain our temperature within this range.

📘 Example 2: Solve 2𝑥² − 8𝑥 + 6 ≥ 0 Step 1: Factor or use quadratic formula −(−8)±√(−8)² − 4(2)(6)
8±√(64−48)
8±√16
𝑥 = 1, 3 Step 2: Test intervals: • 𝑥 < 1: Choose 𝑥 = 0: expression = 2(0)² − 8(0) + 6 = 6 → valid • 1 < 𝑥 < 3: Choose 𝑥 = 2: 8 − 16 + 6 = −2 → not valid • 𝑥 > 3: Choose 𝑥 = 4: 32 − 32 + 6 = 6 → valid ✅ Final Answer: 𝑥 ≤ 1 or 𝑥 ≥ 3

In this second example, we address the inequality 2x² - 8x + 6 ≥ 0. First, we can use the quadratic formula to solve for the roots, which gives us x = 1 and x = 3. These roots again split the number line into intervals for analysis. We then test each interval to see if the inequality holds. We find that for x < 1, the expression is valid, but between 1 and 3, it isn't. For values greater than 3, the inequality holds true again. Therefore, the solution to the inequality includes values of x less than or equal to 1 and greater than or equal to 3.

Imagine you're analyzing profit in a business scenario where the profits need to be above zero to ensure sustainability. The quadratic function represents the projected profits over time. In this case, x=1 can represent an initial threshold of sales, while x=3 indicates a substantial growth phase where sales can be sustained positively. Therefore, understanding the ranges satisfies a requirement for profit, just like finding valid intervals for our quadratic inequality.

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