Problem 28 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-40: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ \frac{x+5}{x^{2}-7 x+12} \leq 0 $$

Step-by-Step Solution

Answer

The solution is \((-\infty, -5]\).

Step 1: Factor the Denominator

To solve the inequality \( \frac{x+5}{x^2 - 7x + 12} \leq 0 \), start by factoring the quadratic expression in the denominator. The expression \( x^2 - 7x + 12 \) can be factored into \( (x - 3)(x - 4) \).

Step 2: Identify Critical Points

The critical points come from the zeros of the numerator and the denominator. From the numerator \( x+5 = 0 \), we have \( x = -5 \). From the denominator \( (x - 3)(x - 4) = 0 \), we have \( x = 3 \) and \( x = 4 \). These points are \( x = -5, 3, 4 \).

Step 3: Determine Test Intervals

Using the critical points, the number line is divided into four intervals: \( (-\infty, -5) \), \( (-5, 3) \), \( (3, 4) \), and \( (4, \infty) \). We will test each interval to determine where the inequality \( \frac{x+5}{(x-3)(x-4)} \leq 0 \) holds true.

Step 4: Test Each Interval

- For \( (-\infty, -5) \), pick \( x = -6 \): \( \frac{-6+5}{(-6-3)(-6-4)} = \frac{-1}{90} < 0 \).- For \( (-5, 3) \), pick \( x = 0 \): \( \frac{0+5}{(0-3)(0-4)} = \frac{5}{12} > 0 \).- For \( (3, 4) \), pick \( x = 3.5 \): \( \frac{3.5+5}{(3.5-3)(3.5-4)} = \frac{8.5}{0.25} > 0 \).- For \( (4, \infty) \), pick \( x = 5 \): \( \frac{5+5}{(5-3)(5-4)} = \frac{10}{2} > 0 \).

Step 5: Consider Critical Points

Since we have a non-strict inequality (\( \leq \)), check where the inequality could be zero. The only point where the numerator is zero is at \( x = -5 \), which satisfies the inequality.

Step 6: Write Solution in Interval Notation

The inequality is satisfied when \( \frac{x+5}{(x-3)(x-4)} \leq 0 \) at \( x = -5 \) and the entire interval \( (-\infty, -5] \). Therefore, the solution is \( (-\infty, -5] \).

Key Concepts

Factoring Quadratics

In the given exercise, we are dealing with a rational inequality, where the denominator is a quadratic expression. Solving such inequalities often begins with factoring the quadratic. Factoring is crucial because it helps us understand the behavior of the function around specific values. Quadratics are commonly denoted as \( ax^2 + bx + c \). The goal is to rewrite it as a product: \( (x - p)(x - q) \), where \( p \) and \( q \) are the roots of the equation. Here, we factor \( x^2 - 7x + 12 \) into \( (x - 3)(x - 4) \).

• The sum of the roots \( p \) and \( q \) must equal the middle coefficient (\( b \)).
• The product of the roots should be the constant term (\( c \)).

By factoring correctly, it reveals when the denominator is zero, highlighting critical points that affect the solution of the inequality. A correctly factored quadratic is vital for simplifying the problem into manageable parts.

Critical Points

Identifying critical points is a pivotal step in solving rational inequalities. These are the values that make the numerator or denominator zero. For our inequality example \( \frac{x+5}{(x-3)(x-4)} \leq 0 \), the process involves:1. Setting the numerator equal to zero: \( x+5 = 0 \) gives \( x = -5 \).2. Setting the denominator equal to zero: \( (x-3)(x-4) = 0 \) results in \( x = 3 \) and \( x = 4 \).

• These values divide the number line into distinct segments.
• Critical points are not always part of the solution set, as you must test them against the inequality, depending on whether it is strict or non-strict.

Critical points help us determine intervals over which the inequality will be tested, thus guiding us into finding valid solutions.

Interval Notation

Interval notation is a way of expressing the subset of numbers (interval) that satisfies the inequality. Each interval is indicated by brackets:- Use \([a, b]\) for closed intervals where both ends are included.- Use \((a, b)\) for open intervals where neither end is included.- Mixed brackets like \((a, b]\) indicate inclusion of one endpoint.For our inequality \( \frac{x+5}{(x-3)(x-4)} \leq 0 \), after checking each divided segment of the number line, we find that:

• The solution lies in the interval \((-\infty, -5]\), including \(-5\) because it makes the numerator zero.

Using interval notation, we compactly represent the set of all \( x \) values fulfilling the inequality condition. This simplifies understanding and communication of the solution.

Non-Strict Inequality

Non-strict inequalities are types of inequalities represented by \( \leq \) or \( \geq \). These include the possibility of equality, meaning the function can equal zero. For our example \( \frac{x+5}{x^2 - 7x + 12} \leq 0 \), it suggests checking points where the expression becomes zero is essential.

• Such inequalities allow for solutions where the function equals zero; hence, points like \( x = -5 \) in our solution must be checked.
• This differs from strict inequalities (\( < \) or \( > \)), where equality is not allowed.

Understanding non-strict inequalities is crucial because they expand the solution set, potentially including endpoints which are often critical, especially when solving real-world problems.