Problem 58 - Algebra and Trigonometry

Question

Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$ \frac{2 x+6}{x-2}<0 $$

Step-by-Step Solution

Answer

The solution is \((-3, 2)\).

Step 1: Identify Zero Points

To solve the inequality \( \frac{2x+6}{x-2}<0 \), first we identify the points where the numerator and the denominator are zero. The numerator \( 2x + 6 = 0 \) leads to \( x = -3 \). For the denominator, \( x - 2 = 0 \) gives \( x = 2 \). These are the critical points.

Step 2: Determine the Intervals

With the critical points \( x = -3 \) and \( x = 2 \), divide the number line into intervals: \( (-\infty, -3) \), \( (-3, 2) \), and \( (2, \infty) \). These intervals will help us determine where the expression is negative.

Step 3: Test Each Interval

Select test points from each interval to determine the sign of \( \frac{2x+6}{x-2} \) in that interval. For \( (-\infty, -3) \), choose \( x = -4 \); for \( (-3, 2) \), choose \( x = 0 \); for \( (2, \infty) \), choose \( x = 3 \). Compute the sign of the expression for each test point.

Step 4: Analyze the Sign of the Expression

Evaluate the expression at each test point:- For \( x = -4 \): \( \frac{2(-4)+6}{-4-2} = \frac{-8+6}{-6} = \frac{-2}{-6} > 0 \)- For \( x = 0 \): \( \frac{2(0)+6}{0-2} = \frac{6}{-2} = -3 < 0 \)- For \( x = 3 \): \( \frac{2(3)+6}{3-2} = \frac{6+6}{1} = 12 > 0 \)The inequality \( \frac{2x+6}{x-2} < 0 \) holds in \( (-3, 2) \).

Step 5: Express in Interval Notation and Graph

The solution in interval notation is \((-3, 2)\). To graph the solution, draw a number line. Place open circles at \(x = -3\) and \(x = 2\) because these points are not included in the solution set. Shade the region between \(-3\) and \(2\) on the number line, indicating that only this interval satisfies the inequality.

Key Concepts

Understanding Critical Points

Critical points in a nonlinear inequality are where the expression might change its sign. They are pivotal to simplifying complex inequalities. In the example of \[ \frac{2x+6}{x-2} < 0 \] we find critical points by identifying the values of \(x\) that make the numerator or denominator zero.

• The numerator \(2x + 6 = 0\) results in \(x = -3\).
• The denominator \(x - 2 = 0\) results in \(x = 2\).

These critical points, \(x = -3\) and \(x = 2\), split the number line into segments, marking potential transitions from positive to negative. Remember:

• Critical points do not automatically belong to the solution set, especially in strict inequalities (\(<\) or \(>\)).
• They help us test intervals to determine where the inequality holds true.

Using Interval Notation

Interval notation is a concise way to represent the solution set of inequalities. It expresses entire sections of the number line without itemizing each value. For the inequality \[ \frac{2x+6}{x-2} < 0 \], we found the solution was valid only within a single segment: \((-3, 2)\).

• The parentheses \(( \) and \() \) indicate that the endpoints \(-3\) and \(2\) are not included.
• This exclusion is due to the inequality being strict ("less than"), meaning the expression can't equal zero at these points.

Interval notation provides clarity and precision. It's critical in mathematics for simplifying the communication of complex solutions. Understanding:

• Parentheses (\((\), \()\)) for not including endpoints.
• Brackets (\([\), \()]\)) for including endpoints, though not applicable here.

The Role of the Number Line

A number line is an invaluable tool for visualizing inequalities. It helps us see where our expression holds true. By plotting critical points, we segment the number line into different intervals. For \[ \frac{2x+6}{x-2} < 0 \], these intervals are \((-\infty, -3)\), \((-3, 2)\), and \((2, \infty)\). We use the number line by:

• Placing open circles at critical points: \(-3\) and \(2\) in our case, because they are not included.
• Testing points from each interval (e.g., \(x = -4\), \(x = 0\), \(x = 3\)) to check where the inequality holds.
• Shading the solution interval, \((-3, 2)\), indicating that this is where the inequality is true.

This graphical representation provides a clear picture of solutions. It's a straightforward method to validate analytical findings with a visual tool.