Problem 62 - Algebra for College Students

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: $$ \begin{gathered} (x-2<0 \text { and } x+3<0) \text { or }(x-2>0 \text { and } x+3>0) \\ (x<2 \text { and } x<-3) \text { or }(x>2 \text { and } x>-3) \\ x<-3 \text { or } x>2 \end{gathered} $$ 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) \((-4, 8)\), (e) \((-\infty, -4) \cup (7, \infty)\), (f) \((-8, 5]\).

Step 1: Identify the Zero Points

For each expression, determine the points where the expression equals zero.(a) The zero points are at \(x = 2\) and \(x = -7\).(b) The zero points are at \(x = 3\) and \(x = -9\).(c) The zero points are at \(x = -1\) and \(x = 6\).(d) The zero points are at \(x = -4\) and \(x = 8\).(e) The zero point in the numerator is \(x = -4\), and in the denominator is \(x = 7\).(f) The zero point in the numerator is \(x = 5\), and in the denominator is \(x = -8\).

Step 2: Check Sign Changes Around Zero Points

Check the sign of the product or quotient in intervals around the zero points.(a) Evaluate the sign of \((x-2)(x+7)\) in the intervals: \((-\infty, -7)\), \((-7, 2)\), and \((2, \infty)\).(b) Evaluate the sign of \((x-3)(x+9)\) in the intervals: \((-\infty, -9)\), \((-9, 3)\), and \((3, \infty)\).(c) Evaluate the sign of \((x+1)(x-6)\) in the intervals: \((-\infty, -1)\), \((-1, 6)\), and \((6, \infty)\).(d) Evaluate the sign of \((x+4)(x-8)\) in the intervals: \((-\infty, -4)\), \((-4, 8)\), and \((8, \infty)\).(e) Evaluate the sign of \(\frac{x+4}{x-7}\) in the intervals: \((-\infty, -4)\), \((-4, 7)\), and \((7, \infty)\).(f) Evaluate the sign of \(\frac{x-5}{x+8}\) in the intervals: \((-\infty, -8)\), \((-8, 5)\), and \((5, \infty)\).

Step 3: Determine Valid Intervals Based on Solution Condition

Select the intervals where the product or quotient meets the given condition (positive, non-negative, negative, or non-positive).(a) The expression \((x-2)(x+7)>0\) is true for \(x < -7\) or \(x > 2\).(b) The expression \((x-3)(x+9) \geq 0\) is true for \(-9 \leq x \leq 3\).(c) The expression \((x+1)(x-6) \leq 0\) is true for \(-1 \leq x \leq 6\).(d) The expression \((x+4)(x-8)<0\) is true for \(-4 < x < 8\).(e) The expression \(\frac{x+4}{x-7}>0\) is true for \(x < -4\) or \(x > 7\).(f) The expression \(\frac{x-5}{x+8} \leq 0\) is true for \(-8 < x \leq 5\).

Step 4: Express Solution Sets

Write the solution intervals using union of intervals where applicable.(a) Solution set: \((-\infty, -7) \cup (2, \infty)\).(b) Solution set: \([-9, 3]\).(c) Solution set: \([-1, 6]\).(d) Solution set: \((-4, 8)\).(e) Solution set: \((-\infty, -4) \cup (7, \infty)\).(f) Solution set: \((-8, 5]\).

Key Concepts

Factoring Expressions

Factoring expressions involves breaking down a complex algebraic expression into simpler parts called factors. This is a crucial skill because it helps to simplify equations and solve algebraic problems more efficiently. For example, in the expression \((x-2)(x+3) > 0\), the factors are \((x-2)\) and \((x+3)\). These factors are obtained by setting each part of the quadratic to zero. It's essential to factor correctly because the solutions to the inequalities heavily rely on understanding the structure of these factors. When you factor expressions correctly, it makes solving and analyzing the inequality much easier because you can quickly determine when the expression is positive or negative.

Interval Notation

Interval notation is a method used to write the set of solutions or expressing ranges of values for inequalities. This notation uses brackets and parentheses to indicate inclusive or exclusive bounds, respectively. For example, if you have a solution set of a function that is defined for all values greater than 2 or less than -3, it can be expressed using interval notation as \((-fty,-3) \cup (2, \infty)\).

• \((-\infty, -3)\) uses parentheses because \(-3\) is not included in the solution.
• \((2, \infty)\) also uses parentheses because 2 is not included either.

Interval notation provides a concise way to show the set of all possible values (intervals) that satisfy an inequality.

Algebraic Solutions

Algebraic solutions involve manipulations of equations or inequalities to find unknown variables. The goal is to isolate the variable on one side of the inequality or equation. When dealing with quadratic inequalities like \((x-2)(x+7)>0\), the algebraic solution involves:

• Identifying zeros of each factor by setting each to zero; for instance, \(x-2=0\) yields \(x=2\) and \(x+7=0\) gives \(x=-7\).
• Creating test intervals based on these zeros to determine where the inequality holds.

The solution is found by selecting the intervals where the product is positive. This step is essential to find the solution set that satisfies the original inequality.

Inequalities in One Variable

Inequalities in one variable are expressions that compare a variable to a number or another expression using inequality symbols like \(>\), \(<\), \(\geq\), and \(\leq\). The main goal is to find all values of the variable that make the inequality true. Solving such inequalities involves several standard steps:

• Identifying critical points, which are the values that make each part of the expression equal to zero.
• Testing intervals between critical points to determine where the inequality is satisfied.
• Writing the solution using interval notation, which shows the range of values that satisfy the inequality.

Inequalities in one variable are fundamental because they form the basis for solving more complex algebraic problems. Understanding how these inequalities work lays the groundwork for handling more complicated expressions and equations.