Worked Examples
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Solving the First Inequality
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Let’s begin by solving the inequality x² - 5x + 6 < 0. Who can tell me how we should start?
We could factor the expression!
Exactly! When we factor, we rewrite it as (x - 2)(x - 3). Now, what should we do next?
We solve the equation to find the roots, right?
Yes! The roots are x = 2 and x = 3. Now we can identify the intervals. What are those?
They are x < 2, 2 < x < 3, and x > 3.
Correct! Next, choose test points in each interval to see where the inequality holds. What can we use for x < 2?
We can use x = 1.
Great choice! Now, when we plug that into the expression, what do we get?
It gives a positive result, so that part doesn't count for our solution.
Exactly! What about the interval 2 < x < 3?
We could try x = 2.5.
And what do we find?
That one gives a negative result, so it satisfies the inequality!
Right! Finally, for x > 3, what test point should we use?
We could use x = 4, and it gives a positive result.
Perfect! So what is our final answer?
2 < x < 3!
Let's summarize. We factored, found roots, analyzed intervals, and chose test points. Great work, everyone!
Solving the Second Inequality
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Moving on to our second example, we have 2x² - 8x + 6 ≥ 0. What are our initial steps?
Do we need to factor it or use the quadratic formula?
Yes! Let's use the quadratic formula. What is our quadratic formula?
x = [−b ± √(b² - 4ac)] / 2a.
Exactly! Here, a is 2, b is -8, and c is 6. Can someone calculate the roots?
We find x = 1 and x = 3!
Right! Now, we have our roots. Next, let’s find our intervals. What do we have?
We have x < 1, 1 < x < 3, and x > 3.
Great! Now, who would like to test the interval x < 1?
I can! We can use x = 0, and that gives a valid solution.
Excellent! And between 1 and 3, what will we use?
Let’s use x = 2. I think that will give a negative result.
Correct! And for x > 3?
Using x = 4 returns a valid result as well!
Exactly! So what is our final solution set?
x ≤ 1 or x ≥ 3!
Fantastic! Remember, we used the quadratic formula, tested intervals, and found our solution set. Well done, everyone!
Introduction & Overview
Read summaries of the section's main ideas at different levels of detail.
Quick Overview
Standard
The section presents two comprehensive examples of solving quadratic inequalities. Each example is broken down step by step, demonstrating how to factor expressions, find roots, analyze sign changes, and represent solutions correctly. The goal is to solidify the understanding of solving such inequalities through practical application.
Detailed
Worked Examples in Quadratic Inequalities
In this section, we explore worked examples that illustrate the process of solving quadratic inequalities, fundamental in mastering algebraic concepts involving quadratic expressions.
Example 1: Solve x² - 5x + 6 < 0
- Factor the expression: We seek a factorization resulting in zero, thus we rewrite the expression as
(x - 2)(x - 3).
- Solve the equation: Setting each factor to zero gives roots x = 2 and x = 3, dividing the number line into intervals:
- x < 2
- 2 < x < 3
- x > 3
- Use sign analysis:
- For values less than 2 (choose x = 1): the expression evaluates to positive.
- For values between 2 and 3 (choose x = 2.5): the expression evaluates to negative.
- For values greater than 3 (choose x = 4): the expression evaluates to positive.
In conclusion, the solution set is 2 < x < 3.
Example 2: Solve 2x² - 8x + 6 ≥ 0
- Factor or use the quadratic formula: The roots are calculated using
x = (8 ± √(64 - 48)) / (4) = 1 or 3.
- Test intervals:
- x < 1: Choose 0; this results in a valid expression.
- 1 < x < 3: Choose 2; this results in a negative expression.
- x > 3: Choose 4; this gives a valid response.
Thus, the solution set is x ≤ 1 or x ≥ 3.
Audio Book
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Example 1: Solve x² - 5x + 6 < 0
Chapter 1 of 2
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Chapter Content
📘 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
Detailed Explanation
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.
Examples & Analogies
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
Chapter 2 of 2
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Chapter Content
📘 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
Detailed Explanation
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.
Examples & Analogies
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.
Key Concepts
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Quadratic Inequalities: Algebraic expressions involving inequalities with a squared variable.
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Interval Notation: Method to express a range of values.
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Root Finding: Determining the values of x that make the equation true.
Examples & Applications
Example 1: Solve the inequality x² - 5x + 6 < 0 by factoring, finding the roots x = 2 and x = 3, and testing the intervals.
Example 2: Use the quadratic formula to solve 2x² - 8x + 6 ≥ 0, finding the roots x = 1 and x = 3 and testing the valid intervals.
Memory Aids
Interactive tools to help you remember key concepts
Rhymes
Factor, find the roots, test the sign, to solve inequalities in record time!
Stories
Imagine two friends, 2 and 3, meeting at a party under the light of 'x'. They hold hands, making x stay between them and the party starts when they’re apart!
Memory Tools
F-R-M: Factor, Root, Measure intervals - steps to solve quadratic inequalities!
Acronyms
S-I-R
Solve
Identify roots
Read intervals.
Flash Cards
Glossary
- Quadratic Inequality
An inequality that involves a quadratic expression and can take the forms of <, ≤, >, or ≥.
- Roots
The values of x that satisfy the equation ax² + bx + c = 0.
- Interval Notation
A mathematical notation used to represent a range of values, typically defined by two endpoints.
Reference links
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