Chapter 1: Problem 18
Solve the inequality. Express the solution as an interval or as the union of intervals. Mark the solution on a number line. $$\frac{x^{2}-4 x+4}{x^{2}-2 x-3} \leq 0$$
Short Answer
Expert verified
The solution to the equation \(\frac{(x-2)^{2}}{(x-3)(x+1)} \leq 0\) is \((-\infty, -1) \cup (2,3)\).
Step by step solution
01
Factoring the Polynomial in the Numerator and Denominator
We factor the numerator \(x^{2}-4x+4\) and the denominator \(x^{2}-2x-3\) of the rational expression, which gives \(\frac{(x-2)^{2}}{(x-3)(x+1)}\). This rewriting makes it easier to analyze and find the zeros and undefined points of the inequality.
02
Find the Zeros and Undefined Points
We set the numerator equal to 0 and solve to find the zeros of the expression, which gives \(x=2\). And we set the denominator equal to 0 and solve to find the undefined points of the expression, which gives \(x=-1\) and \(x=3\).
03
Create Test Points from the Relevant Intervals
The original inequality is not defined and also changes sign at \(x=-1\), \(x=2\), and \(x=3\). We thus create four intervals, each separated by those values. Choose a test point from each interval such as -2, 0, 2.5, and 4.
04
Test the Intervals
Plug each test point into the simplified inequality \(\frac{(x-2)^{2}}{(x-3)(x+1)} \leq 0\). This yields negative, positive, negative, and positive values respectively. So intervals where \(x<-1\) and \(1<x<3\) satisfy the inequality.
05
Final Solution
Combine the intervals and include the value where the expression equals to 0. The solution is thus \((-\infty, -1) \cup (2,3)\). To express this on a number line, draw a line marking the numbers -1, 2, and 3. Make an open circle at -1 and 3, and a closed circle at 2. Shade the area between -1 and 2, and also the area between 2 and 3.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factoring Polynomials
Factoring polynomials is a crucial step in solving inequalities, particularly when dealing with rational expressions. It involves breaking down a polynomial into a product of simpler polynomials. This process can make complex expressions much easier to handle. Let's take the expression from the exercise:
- The numerator is the quadratic polynomial \( x^{2} - 4x + 4 \).
- Notice that this quadratic can be expressed as a perfect square: \( (x-2)^{2} \).
- \( (x-3)(x+1) \).
Rational Expressions
When solving inequalities involving rational expressions, it is important to comprehend how they function. A rational expression is a fraction where both the numerator and the denominator are polynomials.
- The given inequality involves the rational expression \( \frac{(x-2)^{2}}{(x-3)(x+1)} \leq 0 \).
- Critical aspects are the zeros of the numerator (solutions making the expression zero) and undefined values from the denominator (solutions that make the denominator zero).
- This particular expression is zero at \( x=2 \), and is undefined at \( x=-1 \) and \( x=3 \).
Interval Notation
Interval notation is a shorthand method used to express solutions of inequalities or unions of intervals. It conveys a range of numbers and is important for simplifying complex sets that represent solutions.
- For instance, the solution from the exercise is written as \((-\infty, -1) \cup (2, 3)\).
- This means that \(x\) is in the solution set when it is less than \(-1\) or between \(2\) and \(3\).
- The use of parentheses denotes that these endpoints are not included in the solution set. However, a closed bracket, like \([\,2]\), would mean inclusion.