Solving Predicates: Finding Solution Sets For Inequalities

Let's dive into the fascinating world of predicates and how to determine their solution sets. If you're scratching your head wondering what that even means, don't worry! We're going to break it down step-by-step with some real examples. We'll be tackling inequalities involving absolute values, which might sound intimidating, but trust me, it's like pie, guys, once you get the hang of it.

Understanding Predicates and Solution Sets

First, let’s get the basics down. A predicate is basically a statement that can be either true or false depending on the value of a variable. Think of it as a question that needs an answer. The solution set of a predicate is the set of all values that make the predicate true. So, our mission is to find those values for each of the given predicates. This involves a mix of algebraic manipulation, understanding absolute values, and a bit of logical reasoning. We'll work through each example methodically, explaining every step so you can follow along easily. The key is to remember the properties of absolute values and inequalities, and how they interact with each other. We'll also highlight common pitfalls to avoid, ensuring you not only get the correct answers but also understand the underlying principles. So, grab a pen and paper, and let's get started on this mathematical journey together!

a) p1(x): |4x^2 + 1(3 - |2x + 5|)| ≥ 0, x ∈ R

Let's kick things off with the first predicate. The predicate p1(x) is defined as |4x^2 + 1(3 - |2x + 5|)| ≥ 0, where x is a real number (x ∈ R). The core idea here is to understand absolute values. Remember, the absolute value of any expression is always non-negative. This is because the absolute value represents the distance from zero, and distance can't be negative. So, no matter what's inside those absolute value bars, the result will always be zero or a positive number. This is a crucial point to grasp, as it simplifies our problem significantly. Now, let's look at the expression inside the absolute value: 4x^2 + 1(3 - |2x + 5|). We have a quadratic term (4x^2), a constant (1), and another absolute value expression (|2x + 5|). The quadratic term 4x^2 is also always non-negative since squaring any real number results in a non-negative value. So, we have a non-negative term plus something else. Let’s focus on the 3 - |2x + 5| part. Since |2x + 5| is always non-negative, subtracting it from 3 will give us a value that could be positive, negative, or zero. However, the entire expression is inside an absolute value, which means the result will always be non-negative. Therefore, the predicate |4x^2 + 1(3 - |2x + 5|)| ≥ 0 is true for all real numbers. This might seem surprising at first, but it highlights a fundamental property of absolute values. No matter what value of x we plug in, the expression will always be greater than or equal to zero. Thus, the solution set for p1(x) is the set of all real numbers, often denoted as R. Understanding this principle allows us to quickly solve similar problems involving absolute values. It’s not just about crunching numbers; it’s about recognizing patterns and properties that make the math easier. With this knowledge, we're well-equipped to tackle the next predicate.

b) p2(x): |x + 6|(|3x - 7| - 8) ≤ 0, x ∈ Z

Alright, let's move on to the second predicate, p2(x), defined as |x + 6|(|3x - 7| - 8) ≤ 0, where x is an integer (x ∈ Z). This one's a bit more involved but still totally manageable. The first part we see is |x + 6|. As we learned in the previous example, absolute values are always non-negative. So, |x + 6| is either zero or positive. This is an important piece of information. Now, let's consider the entire predicate. We have a product of |x + 6| and (|3x - 7| - 8), and this product must be less than or equal to zero. Since |x + 6| is non-negative, the only way the entire expression can be less than or equal to zero is if (|3x - 7| - 8) is less than or equal to zero, or if |x + 6| is zero. Let's tackle the easier case first: |x + 6| = 0. This happens when x + 6 = 0, which means x = -6. So, -6 is one solution. Now, for the more interesting part: (|3x - 7| - 8) ≤ 0. Let's rewrite this as |3x - 7| ≤ 8. This inequality tells us that the distance between 3x and 7 is at most 8. To solve this, we can break it into two separate inequalities: -8 ≤ 3x - 7 ≤ 8. We need to solve for x in both parts. First, let's add 7 to all parts of the inequality: -1 ≤ 3x ≤ 15. Now, divide by 3: -1/3 ≤ x ≤ 5. Since we're looking for integer solutions (x ∈ Z), the possible values for x are 0, 1, 2, 3, 4, and 5. Don't forget the solution we found earlier, x = -6. Combining all these integer solutions, the solution set for p2(x) is {-6, 0, 1, 2, 3, 4, 5}. This example demonstrates how breaking down a complex inequality into smaller, manageable parts can lead to the solution. We used the properties of absolute values and inequalities, and the fact that x must be an integer, to narrow down the possibilities. It’s a good exercise in logical thinking and algebraic manipulation.

c) p3(x): (|5x - 4| - 6) / |3x + 8| ≥ 0, x ∈ R+

Now, let's tackle p3(x): (|5x - 4| - 6) / |3x + 8| ≥ 0, where x is a positive real number (x ∈ R+). This predicate involves a fraction, so we need to be a bit careful about the denominator. Remember, we can't divide by zero! So, the first thing we need to consider is the denominator, |3x + 8|. For the fraction to be defined, |3x + 8| must not be equal to zero. This means 3x + 8 ≠ 0, so x ≠ -8/3. Since we're only considering positive real numbers (x ∈ R+), this condition is automatically satisfied because -8/3 is negative. Now, let's focus on the inequality itself. For a fraction to be greater than or equal to zero, either both the numerator and denominator are positive, or both are negative, or the numerator is zero. In our case, the denominator |3x + 8| is always positive (except when x = -8/3, which we've already excluded). Therefore, we only need to consider when the numerator is greater than or equal to zero. So, we need to solve |5x - 4| - 6 ≥ 0. Let's rewrite this as |5x - 4| ≥ 6. This inequality tells us that the distance between 5x and 4 is at least 6. Just like before, we can break this into two separate inequalities: 5x - 4 ≥ 6 or 5x - 4 ≤ -6. Let's solve the first one: 5x - 4 ≥ 6. Add 4 to both sides: 5x ≥ 10. Divide by 5: x ≥ 2. Now, let's solve the second one: 5x - 4 ≤ -6. Add 4 to both sides: 5x ≤ -2. Divide by 5: x ≤ -2/5. However, remember that we're only considering positive real numbers (x ∈ R+). So, the solution x ≤ -2/5 is not relevant in this case. Therefore, the solution set for p3(x) is x ≥ 2, where x is a positive real number. In interval notation, this is [2, ∞). This example highlights the importance of considering the domain of the variable. We started with a fraction, so we had to make sure the denominator wasn't zero. We then used the properties of absolute values and inequalities to find the solution, but we always kept in mind that x must be positive. It’s a great demonstration of how different mathematical concepts work together to solve a problem.

d) p4(x): (x^2 - 25) / |6x - 7| - 11 ≤ 0, x ∈ Z

Finally, let's tackle the last predicate, p4(x): (x^2 - 25) / |6x - 7| - 11 ≤ 0, where x is an integer (x ∈ Z). This one looks a bit intimidating, but we'll break it down piece by piece, just like before. First, we have a fraction, so we need to make sure the denominator isn't zero. We have |6x - 7| in the denominator, so we need to ensure that |6x - 7| ≠ 0. This means 6x - 7 ≠ 0, so x ≠ 7/6. Since 7/6 is not an integer, we don't need to worry about this condition affecting our integer solutions. Now, let's focus on the inequality: (x^2 - 25) / |6x - 7| - 11 ≤ 0. To make things easier, let's add 11 to both sides: (x^2 - 25) / |6x - 7| ≤ 11. This still looks a bit messy, but we're getting there. The next step is to realize that |6x - 7| is always positive (or zero, but we've already excluded the case where it's zero). So, we can multiply both sides of the inequality by |6x - 7| without changing the direction of the inequality: x^2 - 25 ≤ 11|6x - 7|. Now, we have an inequality involving an absolute value. To solve this, we'll consider two cases: 6x - 7 ≥ 0 and 6x - 7 < 0. Case 1: 6x - 7 ≥ 0, which means x ≥ 7/6. In this case, |6x - 7| = 6x - 7, so our inequality becomes: x^2 - 25 ≤ 11(6x - 7). Simplify: x^2 - 25 ≤ 66x - 77. Rearrange: x^2 - 66x + 52 ≤ 0. This is a quadratic inequality. To solve it, we first find the roots of the quadratic equation x^2 - 66x + 52 = 0. Using the quadratic formula, we find the roots to be approximately x ≈ 0.79 and x ≈ 65.21. Since we're looking for x such that the quadratic is less than or equal to zero, the solution in this case is 0.79 ≤ x ≤ 65.21. However, we also have the condition x ≥ 7/6, and we're only considering integers. So, the integer solutions in this case are x = 2, 3, 4, ..., 65. Case 2: 6x - 7 < 0, which means x < 7/6. In this case, |6x - 7| = -(6x - 7) = 7 - 6x, so our inequality becomes: x^2 - 25 ≤ 11(7 - 6x). Simplify: x^2 - 25 ≤ 77 - 66x. Rearrange: x^2 + 66x - 102 ≤ 0. Again, we have a quadratic inequality. Find the roots of x^2 + 66x - 102 = 0. Using the quadratic formula, we find the roots to be approximately x ≈ 1.49 and x ≈ -67.49. So, the solution in this case is -67.49 ≤ x ≤ 1.49. We also have the condition x < 7/6, and we're only considering integers. So, the integer solutions in this case are x = -67, -66, ..., 1. Combining the solutions from both cases, the integer solutions for p4(x) are x = -67, -66, ..., 1, 2, 3, ..., 65. That's a lot of integers! This example was definitely the most challenging, but we tackled it systematically. We handled the fraction, the absolute value, and the quadratic inequalities step by step. It’s a testament to the power of breaking down complex problems into smaller, more manageable parts. And hey, you made it through! Give yourself a pat on the back, guys!

Conclusion

So, there you have it! We've journeyed through the world of predicates and their solution sets, tackling inequalities with absolute values and fractions along the way. Remember, the key to solving these types of problems is to break them down into smaller, more manageable steps. Understand the properties of absolute values, inequalities, and fractions, and don't forget to consider the domain of the variable. With practice, you'll become a predicate-solving pro in no time! Keep practicing, and you'll find that even the most complex problems become less daunting. Math is like a puzzle, and each piece you solve brings you closer to the bigger picture. So, keep puzzling, keep learning, and keep enjoying the challenge!