Problem 66 - College Algebra

Question

Which one of the following is true? a. The solution set of \(x^{2}>25\) is \((5, \infty)\) b. The inequality \(\frac{x-2}{x+3}<2\) can be solved by multiplying both sides by \(x+3\), resulting in the equivalent inequality \(x-2<2(x+3)\) c. \((x+3)(x-1) \geq 0\) and \(\frac{x+3}{x-1} \geq 0\) have the same solution set. d. None of these statements is true.

Step-by-Step Solution

Answer

The correct option is D. None of the provided statements are true.

Step 1: Evaluate Statement A

This statement suggests the solution set of \(x^{2}>25\) is (5, \(\infty)\). Solve the inequality. We get: \(x<-5\) or \(x>5\), which leads to the intervals \((- \infty, -5)\cup(5, \infty)\). Hence, statement A is false.

Step 2: Evaluate Statement B

In this statement, the inequality \(\frac{x-2}{x+3}<2\) was manipulated by multiplying both sides by \((x+3)\), supposedly getting \(x-2<2(x+3)\). This method would only hold true if \((x+3)\) is systematically positive (as multiplying an inequality by a negative number changes the direction of the inequality), which is not always guaranteed. So, Statement B is false.

Step 3: Evaluate Statement C

Here, it is suggested that \((x+3)(x-1) \geq 0\) and \(\frac{x+3}{x-1} \geq 0\) have the same solution set. Solving each inequality, we get: For \((x+3)(x-1) \geq 0\) the solution set is \((-3, 1)\cup(1, \infty)\), and for \(\frac{x+3}{x-1} \geq 0\) the solution set is \((- \infty, -3)\cup(1, \infty)\). So, statement C is false.

Step 4: Final conclusion

All provided statements (A, B, C) were evaluated and found to be false. Therefore, the only remaining option is that none of the statements are true. Hence, option D is correct

Key Concepts

Quadratic Inequality

A quadratic inequality occurs when you have a quadratic expression compared with a value using an inequality sign, like greater than (\(>\)), less than (\(<\)), greater than or equal to (\(\geq\)), or less than or equal to (\(\leq\)). A classic form is \(ax^2 + bx + c > 0\). They can be solved by first finding the roots of the corresponding equation, and then determining the intervals where the inequality holds true. For example, consider the equation \(x^{2} > 25\). To solve it, rewrite it as \(x^{2} - 25 > 0\), which factors into \((x-5)(x+5) > 0\). The roots are \(x = 5\) and \(x = -5\), serving as boundaries for our intervals. With these roots, test the intervals:

• \((-\infty, -5)\)
• \((-5, 5)\)
• \((5, \infty)\)

Substitute a test value from each interval into the inequality to find where it holds. Here, \((-\infty, -5)\) and \((5, \infty)\) satisfy \(x^{2} > 25\). This shows the solution set is \((-\infty, -5) \cup (5, \infty)\).

Rational Inequality

A rational inequality involves a fraction with a variable in the numerator, the denominator, or both. Solving rational inequalities can be tricky due to the possibility of undefined terms. Consider solving \(\frac{x-2}{x+3} < 2\). First, rewrite the inequality as \(\frac{x-2}{x+3} - 2 < 0\). This common form involves combining terms into a single fraction: \(\frac{x-2 - 2(x+3)}{x+3}\), which simplifies to \(\frac{-x-8}{x+3} < 0\).
To resolve the inequality, find values where the expression equals zero or is undefined: the numerator \(-x-8 = 0\) at \(x = -8\) and the denominator \(x+3 = 0\), so \(x = -3\). These points divide the number line, creating intervals to test whether each interval satisfies the inequality.

• \((-\infty, -8)\)
• \((-8, -3)\)
• \((-3, \infty)\)

Choose a test point from each interval, substitute it back, and check the sign of the resulting value. For \(\frac{-x-8}{x+3}\), it is less than zero only within \((-8, -3)\).

Solution Set

The solution set of an inequality consists of all the values of the variable that make the inequality true. It is often expressed using interval notation to simplify representing continuous ranges of solutions. For instance, in the context of \(x^2 > 25\), after solving the quadratic inequality, we found that \((-\infty, -5)\) and \((5, \infty)\) are the pieces of the solution set, since any \(x\) in those intervals satisfies the condition. The complete solution set is not stated simply as numbers, as they represent a range of infinite values. In contrast, for the rational inequality \(\frac{x-2}{x+3} < 2\), the solution set after testing is \((-8, -3)\), which includes all real numbers between \(-8\) and \(-3\), not including these endpoints, as using them would make the inequality undefined.

Interval Notation

Interval notation offers an efficient way to describe a set of numbers between two endpoints. It's particularly useful in expressing solution sets of inequalities. Interval notation uses parentheses \(()\) for exclusive boundaries and brackets \([]\) for inclusive ones. For example, \((a, b)\) represents all numbers between \(a\) and \(b\), but not including \(a\) and \(b\) themselves. In some instances, you might encounter open or closed intervals represented as either half-open \((a, b]\) or \([a, b)\). These intervals illustrate continuous subsets of real numbers while indicating whether the endpoints are part of the solution at hand. Exploring \((-\infty, -5) \cup (5, \infty)\), the solution set of the inequality \(x^2 > 25\), shows two separate intervals. The use of parenthesis indicates that neither \(-5\) nor \(5\) is part of the set, resonating the nature of quadratic solutions where boundary values themselves do not meet the inequality's condition. Similarly, \((-8, -3)\) is captured without including \(-8\) and \(-3\) for the rational inequality's set, as these points either zero out the numerator or make the denominator zero, rendering the expression undefined.