/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Let \(A=\left\\{x \in \mathbb{R}... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(A=\left\\{x \in \mathbb{R} \mid x(x+2)^{2}\left(x-\frac{3}{2}\right)=0\right\\} .\) Which of the following sets are equal to the set \(A\) and which are subsets of \(A ?\) (a) \\{-2,0,3\\} (b) \(\left\\{\frac{3}{2},-2,0\right\\}\) (c) \(\left\\{-2,-2,0, \frac{3}{2}\right\\}\) (d) \(\left\\{-2, \frac{3}{2}\right\\}\)

Short Answer

Expert verified
The sets (b) and (c) are equal to set A, while set (d) is a subset of set A.

Step by step solution

01

- Find the elements of set A

To find the elements of set A, we need to determine the roots of the given equation: \(x(x+2)^{2}\left(x-\frac{3}{2}\right)=0\) Since the equation is given in the form of a product of factors equal to 0, to find the roots, we just need to equate each factor of the equation to 0: 1. \(x=0\) 2. \(x+2=0\). Solving for x, we get \(x=-2\) 3. \(x-\frac{3}{2}=0\). Solving for x, we get \(x=\frac{3}{2}\) So, the set A is given by \(\left\\{0, -2, \frac{3}{2}\right\\}\).
02

- Compare elements

Now let's compare the elements of set A with the given sets: (a) \\{-2,0,3\\}: This set has two elements in common with set A (0 and -2) but also has an element (3) that is not in set A. Therefore it is neither equal nor a subset of A. (b) \(\left\\{\frac{3}{2},-2,0\right\\}\): This set has the exact same elements as set A, so it is equal to set A. (c) \(\left\\{-2,-2,0, \frac{3}{2}\right\\}\): This set contains all elements of set A and one duplicate element (-2). As a set cannot contain duplicate elements, this set is considered equal to set A. (d) \(\left\\{-2, \frac{3}{2}\right\\}\): This set has two elements of set A, but it is missing the element 0. Therefore, it is a subset of set A. In conclusion, the sets equal to A are (b) and (c), and the subset of A is (d).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Set Theory
The concept of set theory is foundational to modern mathematics. It’s a branch of mathematical logic that deals with sets, which are collections of objects. Think of it as a way of grouping things together based on common characteristics or properties. In the given exercise, we've been provided an equation and asked to find a specific set, denoted as set A.

Understanding how to translate the elements from an equation to form a concrete set is a fundamental skill in set theory. In this case, we identified that set A consists of numbers that make the equation equal to zero, essentially the roots of the equation. Once we find those roots, we can present them as the elements of our set, as seen in the step-by-step solution.

Another important concept is recognizing that sets, by definition, do not contain duplicate elements. This is why even if a set lists an element twice, like in choice (c), it is treated as having only one instance of that element. This concept is crucial when determining if sets are equal or if one is a subset of another.
Solving Equations
Equations are at the heart of algebra and are all about finding unknown values that make a statement true. In our example exercise, solving the equation involves finding values of x for which the product of three terms equals zero. This is based on the Zero Product Property, which states if the product of several factors is zero, at least one of the factors must be zero.

Concretely, we have the factors of an equation which are set to zero individually, as seen in the initial steps provided. It’s like unraveling a mystery, where every step is about isolating our 'suspect' variable, x, to find its true value. Once discovered, these values highlight the intersection points with the x-axis on a graph or, in the context of our set A, the elements that belong in our set.

It’s important to confirm each solution back into the original equation to ensure they're valid—this reinforces not only the practice of solving equations but also leads to a better understanding of the equation's behavior and its graphical representation.
Mathematical Proofs
Mathematical proofs are rigorously structured arguments that verify the truth of mathematical statements. They are the very backbone of mathematics, ensuring that our conclusions are not just guesses but are based on logical deduction. In the context of the exercise, showing how each option relates to set A constitutes the proof of each answer.

For example, in comparing set A with the other sets like (a) or (d), we’re essentially providing a proof for why they’re not equal to set A or why one is a subset. This process includes an examination of the elements to see if they match or are contained within set A. By doing so, we make logical deductions about their relationships.

Proofs aren't always presented in the form of written 'theorems' and their 'proofs'; they can also be implicit in exercises like these, where each step forms part of the argument leading to the final answer. The clear, logical progression from one step to the next is itself a form of proof.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use truth tables to prove each of the distributive laws from Theorem 2.8 . (a) \(P \vee(Q \wedge R) \equiv(P \vee Q) \wedge(P \vee R)\) (b) \(P \wedge(Q \vee R) \equiv(P \wedge Q) \vee(P \wedge R)\)

Use the roster method to specify the truth set for each of the following open sentences. The universal set for each open sentence is the set of integers \(\mathbb{Z}\). (a) \(n+7=4\). (b) \(n^{2}=64\). (c) \(\sqrt{n} \in \mathbb{N}\) and \(n\) is less than 50 . (d) \(n\) is an odd integer that is greater than 2 and less than 14 . (e) \(n\) is an even integer that is greater than 10 .

Suppose that Daisy says, "If it does not rain, then I will play golf." Later in the day you come to know that it did rain but Daisy still played golf. Was Daisy's statement true or false? Support your conclusion.

For each of the following sets, use English to describe the set and when appropriate, use the roster method to specify all of the elements of the set. (a) \(\\{x \in \mathbb{R} \mid-3 \leq x \leq 5\\}\) (b) \(\\{x \in \mathbb{Z} \mid-3 \leq x \leq 5\\}\) (c) \(\left\\{x \in \mathbb{R} \mid x^{2}=16\right\\}\) (d) \(\left\\{x \in \mathbb{R} \mid x^{2}+16=0\right\\}\) (e) \(\\{x \in \mathbb{Z} \mid x\) is odd \(\\}\) (f) \(\\{x \in \mathbb{R} \mid 3 x-4 \geq 17\\}\)

Suppose that \(P\) and \(Q\) are statements for which \(P \rightarrow Q\) is false. What conclusion (if any) can be made about the truth value of each of the following statements? (a) \(\neg P \rightarrow Q\) (b) \(Q \rightarrow P\) (c) \(P \vee Q\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.