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Chapter 3: Problem 23

In \(15-26,\) find and graph the solution set of each inequality. $$ 2|x+2| > -3 $$

Short Answer

Expert verified
The solution is all real numbers, \(x \in \mathbb{R}\).

Step by step solution

01

Understanding the Inequality

The inequality given is \(2|x+2| > -3\). The absolute value of any expression is always non-negative (zero or positive). Thus, \(|x+2|\) is always greater than or equal to 0. Therefore, multiplying it by 2 will still yield a non-negative result. This expression \(2|x+2|\) will always be greater than \(-3\), regardless of the value of \(x\). This means the inequality holds true for all real numbers \(x\).
02

Graphing the Solution Set

Since we determined that the inequality is true for all real numbers, the graph will be a number line shaded for all \(x\). Draw a number line and shade it entirely to represent all real numbers. This indicates that every point on the number line is a solution to the inequality.

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Key Concepts

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

Absolute Value and Its Properties
The absolute value of a number is a measure of its distance from zero on a number line. This means it is always a non-negative number. For any real number \(x\), the absolute value, denoted by \(|x|\), is defined as follows:
  • \(|x| = x\) if \(x \geq 0\)
  • \(|x| = -x\) if \(x < 0\)
In the inequality \(2|x+2| > -3\), we see \(|x+2|\) inside the absolute value. The property of absolute values ensures \(|x+2|\) cannot be negative.
Indeed, any positive number multiplied with a non-negative number remains non-negative. Therefore, \(2|x+2|\) is always non-negative and certainly greater than \(-3\). This is an important property that makes it easy to solve inequalities involving absolute values.
Understanding Real Numbers in Context
Real numbers include every point along the number line, ranging from negative infinity to positive infinity. They consist of all rational and irrational numbers.
Knowing that all possible values of \(x\) in this exercise are real numbers, we mean every value of \(x\) without exception follows the inequality \(2|x+2| > -3\).
In practical terms, this implies that no matter what test value you choose for \(x\), the operation within \(2|x+2|\) will always satisfy the given inequality. This is why the solution includes all real numbers: the inequality is never violated.
Graphing Solutions on a Number Line
To graph the solution set of an inequality that holds for all real numbers, like \(2|x+2| > -3\), you can use a number line.
  • Draw a horizontal line to represent the number line.
  • Since the inequality is true for every real number, you shade the entire line to indicate all values are solutions.
Placing points or segmenting the number line is unnecessary because every possible value of \(x\) works. This shading communicates that there are no restrictions or exceptions 鈥� the solution is all-encompassing.

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Most popular questions from this chapter

In \(3-38,\) solve each equation for the variable, check, and write the solution set. $$ 2+\sqrt[3]{3 b-2}=6 $$

In \(3-41\) , express each product in simplest form. Variables in the radicand with an even index are non-negative. $$ \sqrt{2}(2+\sqrt{2}) $$

In \(43-46,\) solve each equation for the variable. $$ b^{2}=100 $$

In \(3-41\) , express each product in simplest form. Variables in the radicand with an even index are non-negative. $$ 8 \sqrt{6} \cdot \sqrt{\frac{5}{12}} $$

In \(3-38,\) solve each equation for the variable, check, and write the solution set. $$ 5+\sqrt[3]{a+2}=3 $$
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