Problem 36 - Calculus with Concepts in Calculus

Question

Solve the inequality. $$ \frac{t^{2}-2 t-3}{t^{2}-8 t+15}>0 $$

Step-by-Step Solution

Answer

The solution set is \((-\infty, -1) \cup (5, \infty)\).

Step 1: Factor the Numerator

The numerator of the inequality is a quadratic expression \( t^2 - 2t - 3 \). We need to factor this expression. \(t^2 - 2t - 3 = (t-3)(t+1)\)

Step 2: Factor the Denominator

The denominator of the inequality is another quadratic expression \( t^2 - 8t + 15 \). We will factor this as well.\(t^2 - 8t + 15 = (t-5)(t-3)\)

Step 3: Determine Critical Points

To solve the inequality \( \frac{(t-3)(t+1)}{(t-5)(t-3)} > 0 \), identify the critical points where the expression is either zero or undefined. These occur when any of the factors is zero.\((t-3) = 0 \Rightarrow t = 3 \quad \text{(but undefined because it's both numerator and denominator)}\)\((t+1) = 0 \Rightarrow t = -1\)\((t-5) = 0 \Rightarrow t = 5\)

Step 4: Create a Number Line

Plot the critical points on a number line: \( -1 \), \( 3 \) (not included in intervals), and \( 5 \). These points divide the number line into intervals \((-\infty, -1)\), \((-1, 3)\), \((3, 5)\), and \((5, \infty)\).

Step 5: Test Each Interval

Choose a test point in each interval to determine the sign of the expression. Remember \(t = 3\) is not a valid point.1. Interval \((-\infty, -1)\): Test with \(t = -2\) \( \frac{(-2-3)(-2+1)}{(-2-5)(-2-3)} = \frac{(-5)(-1)}{(-7)(-5)} = \frac{5}{35} > 0 \)2. Interval \((-1, 3)\): Test with \(t = 0\) \( \frac{(0-3)(0+1)}{(0-5)(0-3)} = \frac{(-3)(1)}{(-5)(-3)} = \frac{-3}{15} < 0 \)3. Interval \((3, 5)\): Test with \(t = 4\) \( \frac{(4-3)(4+1)}{(4-5)(4-3)} = \frac{(1)(5)}{(-1)(1)} = -5 < 0 \)4. Interval \((5, \infty)\): Test with \(t = 6\) \( \frac{(6-3)(6+1)}{(6-5)(6-3)} = \frac{(3)(7)}{(1)(3)} = 7 > 0 \)

Step 6: Write the Solution Set

From our interval tests, the expression is positive in the intervals \((-\infty, -1)\) and \((5, \infty)\). Since \(t = 3\) causes division by zero, it cannot be included.Therefore, the solution set is:\((-\infty, -1) \cup (5, \infty)\)

Key Concepts

Quadratic Expressions

A quadratic expression is a polynomial of degree two. It takes the general form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). In the provided problem, there are two quadratic expressions: the numerator \( t^2 - 2t - 3 \) and the denominator \( t^2 - 8t + 15 \). These expressions are made up of terms that involve the square of the variable \( t \), as well as linear terms and constant terms. Quadratic expressions are used in various mathematical problems and often involve finding roots or solving inequalities. Understanding how to manipulate these expressions is a crucial skill in solving algebraic problems effectively.

Factoring

Factoring is the process of rewriting a quadratic expression as a product of simpler expressions. This is a pivotal step in solving equations and inequalities because it reveals critical points of the expression.For instance, the quadratic expression in the numerator, \(t^2 - 2t - 3\), can be factored into \((t-3)(t+1)\). This simplifies handling quadratic inequalities by providing a clearer picture of when the expression changes sign. Similarly, the denominator \(t^2 - 8t + 15\) factors into \((t-5)(t-3)\).Factoring involves finding numbers that multiply to give the constant term while adding to give the middle coefficient. Mastery of factoring techniques is essential for success in algebra.

Critical Points

Critical points are values of \( t \) where the expression evaluates to zero or becomes undefined. These points demarcate the boundaries where the inequality may change sign.To find these critical points, we set each factor of our factored expression equal to zero. From \((t-3)(t+1)\) and \((t-5)(t-3)\), the critical points are determined by setting each factor to zero:

• \(t-3 = 0 \rightarrow t = 3\) (Note: this makes the expression undefined)
• \(t+1 = 0 \rightarrow t = -1\)
• \(t-5 = 0 \rightarrow t = 5\)

The critical point \( t = 3 \) is particularly significant as it results in a division by zero, making it an excluded point, even though it appears in both expressions.

Number Line Analysis

Number line analysis involves testing the intervals created by the critical points on a number line to determine where the inequality holds.For our problem, we analyze intervals determined by the critical points \( -1 \), \( 3 \), and \( 5 \):

• Interval \((-\infty, -1)\)
• Interval \((-1, 3)\)
• Interval \((3, 5)\)
• Interval \((5, \infty)\)

By selecting test points within these intervals and substituting them into the simplified inequality expression, we can identify where the inequality \( \frac{(t-3)(t+1)}{(t-5)(t-3)} > 0 \) is satisfied. This approach is effective in visually confirming regions of positivity or negativity and is a fundamental aspect of solving rational inequalities.