Problem 24
Question
In \(15-26,\) find and graph the solution set of each inequality. $$ \left|\frac{5}{2} x+2\right| \leq 0 $$
Step-by-Step Solution
Verified Answer
The solution set is \( x = -\frac{4}{5} \), graphed as a single point at \( x = -\frac{4}{5} \).
1Step 1: Understand the Structure of Absolute Value Inequality
The absolute value inequality given is \( \left|\frac{5}{2} x+2\right| \leq 0\). The key understanding with absolute values is that they are non-negative. Therefore, \( |A| \leq 0 \) means \( A = 0 \). This is because absolute values can't be negative.
2Step 2: Set the Expression Inside the Absolute Value to Zero
Since \( \left|\frac{5}{2} x + 2\right| \leq 0 \) implies \( \frac{5}{2} x + 2 = 0 \), we solve for \( x \) by setting the expression inside the absolute value to zero.
3Step 3: Solve the Equation
Solve the equation \( \frac{5}{2} x + 2 = 0 \).Subtract 2 from both sides:\[ \frac{5}{2} x = -2 \]Multiply both sides by \( \frac{2}{5} \) to solve for \( x \):\[ x = -\frac{2 \cdot 2}{5} x = -\frac{4}{5} \]
4Step 4: Check the Solution
Plug \( x = -\frac{4}{5} \) back into the expression to verify:\( \frac{5}{2}(-\frac{4}{5}) + 2 = -2 + 2 = 0 \)The left side equals zero, confirming the solution is correct.
5Step 5: Graph the Solution Set
Since the inequality holds true only for \( x = -\frac{4}{5} \), the graph is a single point at \( x = -\frac{4}{5} \) on the number line. This means the solution set is \( x = -\frac{4}{5} \), represented by a closed dot at this point.
Key Concepts
Inequality SolutionsAbsolute Value EquationsGraphing Inequalities
Inequality Solutions
Inequality solutions involve finding all possible values of a variable that make an inequality true. Unlike equations, which specify exact values, inequalities can cover a range or set of values. However, in some special cases, like the inequality \( |A| \leq 0 \), they can pinpoint a specific value. For an inequality involving absolute values, the solution often requires setting the expression inside the absolute value equal to zero.
Here is why: The absolute value expression \(|A|\) is always zero or positive. So, when asked to solve \(|A| \leq 0\), the only possibility is that \(A\) itself must be zero. Solving for \(x\) then becomes the focus of the exercise.
Here is why: The absolute value expression \(|A|\) is always zero or positive. So, when asked to solve \(|A| \leq 0\), the only possibility is that \(A\) itself must be zero. Solving for \(x\) then becomes the focus of the exercise.
- If the expression within the absolute value results in zero, then the inequality holds only for that specific value of the variable.
- For more common inequalities, such as \(|A| < b\) or \(|A| > b\), the solution set provides a range of possible values.
Absolute Value Equations
Absolute value equations are concerned with expressions within absolute value bars. An absolute value denotes the distance of a number from zero on the number line, always resulting in a non-negative value.
When dealing with equations like \( \left|\frac{5}{2} x + 2\right| = 0 \), it's important to realize that the absolute value being zero occurs only when its contents are zero. Thus, the expression \(\frac{5}{2} x + 2\) equals zero.
When dealing with equations like \( \left|\frac{5}{2} x + 2\right| = 0 \), it's important to realize that the absolute value being zero occurs only when its contents are zero. Thus, the expression \(\frac{5}{2} x + 2\) equals zero.
- To solve, set the equation inside the absolute value equal to zero and solve for \(x\).
- Subtract, add, or multiply to isolate \(x\).
Graphing Inequalities
Graphing inequalities involves visually representing the solution set on a number line. For the equation \( \left|\frac{5}{2} x + 2\right| \leq 0 \), we determined that \(x = -\frac{4}{5}\) is the only solution.
In this case, the graph on the number line is just a single point. Here's how graphing would proceed:
In this case, the graph on the number line is just a single point. Here's how graphing would proceed:
- First, identify the solution from solving the inequality or equation.
- Place a closed dot on the number line at \(-\frac{4}{5}\) to denote the solution's inclusion.
- A closed dot signifies that the endpoint itself is part of the solution set.
- No shading is required here since the solution is not a range.
Other exercises in this chapter
Problem 24
In \(3-29\) write each quotient in simplest form. Variables in the radicand with an even index are non-negative. Variables occurring in the denominator of a fra
View solution Problem 24
In \(3-38\) , write each radical in simplest radical form. Variables in the radicand of an even index are non-negative. Variables occurring in the denominator o
View solution Problem 25
In \(3-41\) , express each product in simplest form. Variables in the radicand with an even index are non-negative. $$ \sqrt{2}(2+\sqrt{2}) $$
View solution Problem 25
In \(3-38,\) solve each equation for the variable, check, and write the solution set. $$ x=1+\sqrt{x+11} $$
View solution