Problem 1 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-40: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ (3 x+1)(5-10 x)>0 $$

Step-by-Step Solution

Answer

Solution: \((-\frac{1}{3}, \frac{1}{2})\).

Step 1: Identify the Zero Product

First, recognize that the inequality is in the form \((A)(B) > 0\), where \( A = 3x+1 \) and \( B = 5-10x \). Our goal is to find when the product of these two expressions is positive. To do this, determine when each expression is zero: - Set \( A = 0 \): \( 3x + 1 = 0 \) gives \( x = -\frac{1}{3} \). - Set \( B = 0 \): \( 5 - 10x = 0 \) gives \( x = \frac{1}{2} \). Thus, the product's sign can change at these points.

Step 2: Analyze Critical Points

List the critical points where the expressions change sign: \( x = -\frac{1}{3} \) and \( x = \frac{1}{2} \). These points divide the number line into three intervals: 1. \( (-\infty, -\frac{1}{3}) \) 2. \( (-\frac{1}{3}, \frac{1}{2}) \) 3. \( (\frac{1}{2}, \infty) \). We will test each interval to determine where the product is positive.

Step 3: Test Intervals for Sign

Choose test points from each interval and substitute into \((3x+1)(5-10x)\) to check the sign: - For \((-\infty, -\frac{1}{3})\), choose \( x = -1 \): \((3(-1)+1)(5-10(-1)) = (-2)(15) = -30\). - For \((-\frac{1}{3}, \frac{1}{2})\), choose \( x = 0 \): \((3(0)+1)(5-10(0)) = (1)(5) = 5\). - For \((\frac{1}{2}, \infty)\), choose \( x = 1 \): \((3(1)+1)(5-10(1)) = (4)(-5) = -20\).

Step 4: Determine Solution Set

The product \((3x+1)(5-10x)\) is positive in the interval where testing provides a positive result. From our tests: - Negative in \((-\infty, -\frac{1}{3})\). - Positive in \((-\frac{1}{3}, \frac{1}{2})\). - Negative in \((\frac{1}{2}, \infty)\). Therefore, the solution to the inequality is \((-\frac{1}{3}, \frac{1}{2})\).

Key Concepts

Zero Product Property

The Zero Product Property is an essential concept when solving inequalities that involve products of expressions. This property states that if a product of two factors is zero, then at least one of the factors must be zero. This principle helps us break down complex expressions into simpler ones we can solve separately.
When dealing with inequalities in the format \(A \cdot B > 0\), like \( (3x+1)(5-10x) > 0\), the zero product property allows you to identify the points where each factor equals zero. By setting each factor to zero, we can solve for the critical points. In this case, \(A = 3x + 1 = 0\) results in \(x = -\frac{1}{3}\), and \(B = 5 - 10x = 0\) results in \(x = \frac{1}{2}\). These x-values are where the signs of the factors could change, which is crucial for further analysis.

Critical Points

Critical points are the values that make each factor of the product zero. In solving the inequality \( (3x+1)(5-10x) > 0\), we calculated the critical points to be \(x = -\frac{1}{3}\) and \(x = \frac{1}{2}\). These critical points divide the number line into distinct intervals.
Here's why critical points matter:

• They are potential points where the product's sign changes.
• Each interval between the critical points is worth exploring to determine the sign of the product.

By identifying critical points, you have a clear roadmap for determining where to test the intervals, which leads into our next concept: interval testing.

Interval Testing

Interval testing involves choosing test points from each interval that the critical points create, then determining the sign of the inequality within those intervals. This method is key to solving inequalities involving products.To test the intervals for \( (3x+1)(5-10x) > 0\):

• Choose a point from \((-\infty, -\frac{1}{3})\), like \(x = -1\), and calculate \( (3(-1)+1)(5-10(-1))\), which results in -30 (negative).
• Choose a point from \((-\frac{1}{3}, \frac{1}{2})\), like \(x = 0\), which yields \( (1)(5) = 5\) (positive).
• Choose a point from \(\left( \frac{1}{2}, \infty \right)\), like \(x = 1\), resulting in \( (4)(-5) = -20\) (negative).

After substituting these test points, we observe when the expression is positive, leading us directly to the solution set.

Solution Sets

Solution sets summarize where the product of an inequality holds true according to our interval testing results. After determining that \( (3x+1)(5-10x) > 0\) is only positive between \((-\frac{1}{3}, \frac{1}{2})\), we conclude this is the solution set.

• The intervals where the product is negative do not contribute to the solution set for this inequality.
• Hence, the solution set is the part of the number line where the product remains positive based on our test points.

This solution set can be expressed using interval notation, which concisely conveys all solutions of the inequality as the interval \( (-\frac{1}{3}, \frac{1}{2}) \). This format makes it easier to interpret and communicate the solutions.