Problem 47 - Algebra 2

Question

OPEN ENDED List three points you might test to find the solution of \((x+3)(x-5)<0\)

Step-by-Step Solution

Answer

The inequality \((x+3)(x-5)<0\) holds true for \(x\) in \((-3, 5)\).

Step 1: Identify Critical Points

We start by identifying the critical points of the inequality. Set each factor to zero: - For \((x+3) = 0\), solve for \(x\): \[ x = -3 \]- For \((x-5) = 0\), solve for \(x\): \[ x = 5 \]These critical points divide the number line into intervals, which we'll test in the next step.

Step 2: Determine Intervals

Using the critical points \(-3\) and \(5\), determine the intervals to test:1. \((-\infty, -3)\)2. \((-3, 5)\)3. \((5, \infty)\)

Step 3: Test an Interval Point

We'll pick a test point from each interval to determine where the inequality holds. - Choose \(x = -4\) from \((-\infty, -3)\): \[ (x+3)(x-5) = (-4+3)(-4-5) = (-1)(-9) = 9 \geq 0 \] This does not satisfy the inequality.- Choose \(x = 0\) from \((-3, 5)\): \[ (x+3)(x-5) = (0+3)(0-5) = (3)(-5) = -15 < 0 \] This satisfies the inequality.- Choose \(x = 6\) from \((5, \infty)\): \[ (x+3)(x-5) = (6+3)(6-5) = (9)(1) = 9 \geq 0 \] This does not satisfy the inequality.

Step 4: Conclusion

From the tests, we find that the inequality is satisfied in the interval:- \((-3, 5)\)This is where the inequality \((x+3)(x-5)<0\) holds true.

Key Concepts

Understanding Critical Points

When solving inequalities, a crucial first step is identifying the critical points. These are the values of \( x \) that make the expression zero. By determining where an inequality either touches or crosses the x-axis, we can gauge where the expression may switch from negative to positive, or vice versa.

For the inequality \((x+3)(x-5)<0\), we need to find the critical points by setting each factor to zero:

• For \(x + 3 = 0\), solve to get \(x = -3\).
• For \(x - 5 = 0\), solve to get \(x = 5\).

These solutions, \(-3\) and \(5\), are the critical points that divide the number line into intervals. These points are essentially the boundaries at which the product \((x+3)(x-5)\) may change sign.

Using Interval Notation

With the critical points identified, we next break the number line into intervals. Interval notation is a way to represent these segments that tell us where to test our inequality. After marking the critical points on the number line, we divide it into:

• \((-fty, -3)\)
• \((-3, 5)\)
• \((5, fty)\)

Interval notation uses parentheses \(( )\) for open intervals, indicating that endpoints are not included. This approach helps target regions to analyze whether the inequality will hold true.

By using interval notation, you get a tidy overview of sections where you may need to perform tests, offering structure and simplicity when tackling inequalities.

Choosing and Testing Points

After you have your intervals, the next step in solving an inequality is to test points from each of these regions. This helps check when the inequality \((x+3)(x-5)<0\) holds true.

To decide:

• In \((-fty, -3)\), pick \(x = -4\).
• In \((-3, 5)\), select \(x = 0\).
• In \((5, fty)\), choose \(x = 6\).

Plug these values back into the inequality to determine the sign of the product. For example:

• At \(x = -4\), \((x+3)(x-5) = 9\), which does not satisfy \(<0\).
• At \(x = 0\), \((x+3)(x-5) = -15\), which satisfies \(<0\).
• At \(x = 6\), \((x+3)(x-5) = 9\), which does not satisfy \(<0\).

This process helps you pinpoint exactly where the inequality is valid, which, in our example, happens in the interval \((-3, 5)\). Through testing points, you gain confidence about the true solution set for the inequality.