Problem 62 - Intermediate Algebra

Question

The product \((x-2)(x+3)\) is positive if both factors are negative or if both factors are positive. Therefore, we can solve \((x-2)(x+3)>0\) as follows: \((x-2<0\) and \(x+3<0)\) or \((x-2>0\) and \(x+3>0)\) \((x<2\) and \(x<-3)\) or \((x>2\) and \(x>-3)\) $$ x<-3 \text { or } x>2 $$ The solution set is \((-\infty,-3) \cup(2, \infty)\). Use this type of analysis to solve each of the following. (a) \((x-2)(x+7)>0\) (b) \((x-3)(x+9) \geq 0\) (c) \((x+1)(x-6) \leq 0\) (d) \((x+4)(x-8)<0\) (e) \(\frac{x+4}{x-7}>0\) (f) \(\frac{x-5}{x+8} \leq 0\)

Step-by-Step Solution

Answer

(a) \((-\infty, -7) \cup (2, \infty)\); (b) \([-9, 3]\); (c) \([-1, 6]\); (d) \((-8, -4)\); (e) \((-\infty, -4) \cup (7, \infty)\); (f) \((-\infty, -8) \cup (5, \infty)\).

Step 1: Determine where each factor is zero

For each inequality, first determine where each factor is zero by solving for the values of \(x\) that make one or both of the factors zero. For example, for part (a), set each factor equal to zero: \(x - 2 = 0\) gives \(x = 2\), and \(x + 7 = 0\) gives \(x = -7\).

Step 2: Identify intervals between zeros

Using the zero points from Step 1, identify the intervals on the number line to test the sign of the product or quotient. For part (a), the intervals are \((-\infty, -7), (-7, 2), (2, \infty)\).

Step 3: Test each interval

In each interval, pick a test point to determine the sign of the product or quotient. For part (a), test \(x = -8\) for \((-\infty, -7)\), \(x = 0\) for \((-7, 2)\), and \(x = 3\) for \((2, \infty)\). Substitute these into \((x-2)(x+7)\) and determine if the result is positive or negative.

Step 4: Determine where the inequality holds

Compile the intervals where the inequality is satisfied based on your tests. For part (a), if the product is positive for \((x-8)(x+7) > 0\), choose the intervals where the test gives a positive result.

Step 5: Combine results for different cases

Repeat Steps 1 to 4 for each part, (b) through (f), taking note of whether the inequality is strict (\(>\) or \(<\)) or inclusive (\(\geq\) or \(\leq\)). Combine results as needed, remembering to include the boundary points if the inequality includes equals.

Step 6: Write final solutions

Write down the final solution for each part. For part (a), the solution is \((x \in (-\infty, -7) \cup (2, \infty))\). Repeat for all parts, ensuring to use union or intersection as necessary based on positive or zero results.

Key Concepts

Polynomial Inequalities

Polynomial inequalities involve expressions formed by summing terms of the form \(ax^n\) where \(n\) is a non-negative integer. These inequalities can express conditions like \((x-2)(x+3)>0\). Here, we seek values of \(x\) that make this expression true.
To solve polynomial inequalities, it's useful to factor the expression as shown in the example. Once factored, polynomial inequalities reveal the points where each factor equals zero. These points, called 'roots' or 'zero points', will help you identify which intervals to test.
Through factorization, we determine where the factors switch signs, providing insight into which intervals make the inequality true. Often roots divide the number line into distinct regions, which are useful for further analysis.

Rational Inequalities

Rational inequalities are expressions in the form \(\frac{P(x)}{Q(x)} \) where \(P(x)\) and \(Q(x)\) are polynomials, and conditions involve inequality signs such as \(>\), \(<\), \(\geq\), or \(\leq\). An example is \(\frac{x+4}{x-7}>0\).
When solving rational inequalities, the goal is to determine where the rational expression is positive or negative. First, identify the points where either the numerator or the denominator equals zero. These points are critical because they can change the sign of the expression or create undefined points (particularly roots of the denominator).
Identify the intervals formed by these critical points and then test the sign of the rational expression in each interval. Choose a test point from each interval, substitute it back into the expression, and determine whether it satisfies the inequality. This process will help establish the intervals where the inequality is true.

Interval Testing

Interval testing is a strategy used to solve inequalities by breaking the real number line into testable intervals. To do this, you'll first find the critical points for your inequality by solving for when each part of the expression is equal to zero.
Once you have the critical points, divide the number line into intervals using these points as boundaries. For example, if the critical points are \(-7\) and \(2\), the intervals would be \((-\infty, -7)\), \((-7, 2)\), and \((2, \infty)\).
Within each interval, select a "test point"—a simple number that lies inside the interval. Substitute this point into the original inequality and determine whether it makes the inequality true or false. This test helps identify which intervals satisfy the inequality condition. This method effectively uses the sign changes of the factors to determine where the entire inequality is strictly positive, negative, or zero.

Solution Sets

Solution sets are the range or collection of \(x\) values that satisfy a given inequality. After interval testing, you can conclude which intervals of \(x\) make the original inequality true.
Solution sets can be expressed using interval notation. For instance, a solution set like \((-\infty, -7) \cup (2, \infty)\) means the inequality is satisfied for all \(x\) values in these intervals but not including the points \(-7\) and \(2\). Include or exclude endpoints based on whether the inequality allows equality, as seen in \(\geq\) or \(\leq\).
Understanding solution sets is crucial as they provide the exact answer to the inequality question, mathematically defining the scope and limits of your solution.