Problem 48 - College Algebra

Question

Solve each rational inequality in Exercises \(29-48,\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{x^{2}-3 x+2}{x^{2}-2 x-3}>0 $$

Step-by-Step Solution

Answer

The solution to the inequality \(\frac{x^{2}-3 x+2}{x^{2}-2 x-3} > 0 \) in interval notation is \((-1,1)\cup(2,3)\cup(3, \infty)\).

Step 1: Factor the Polynomials

The numerator \(x^{2} - 3x + 2\) can be factored into \((x-2)(x-1)\), and the denominator \(x^{2} - 2x - 3\) can be factored into \((x-3)(x+1)\). The inequality becomes \((x-2)(x-1)/((x-3)(x+1)) > 0\).

Step 2: Find the Intervals

Find the zeroes of both the numerator and denominator by setting them equal to zero. The zeroes are \(x = 2,1,3,-1\). This creates the intervals \((-\infty, -1),(-1,1),(1,2),(2,3), (3, \infty)\).

Step 3: Test the Intervals

Pick a test number from each interval and substitute it into the inequality, like \(x=-2,0,1.5,2.5,4\). If the result is positive, the interval satisfies the inequality, if negative, it does not.

Step 4: Solution in Interval Notation

The solution from the intervals that satisfy the inequality is \((-1,1)\cup(2,3)\cup(3, \infty)\).

Key Concepts

Factoring Polynomials

To solve rational inequalities like \(\frac{x^{2}-3x+2}{x^{2}-2x-3}>0\), we first tackle factoring polynomials. This involves breaking down a polynomial into simpler binomial expressions.
The given polynomial in the numerator is \(x^{2} - 3x + 2\). We need to find two numbers that multiply to 2 and add up to -3. These numbers are -1 and -2, giving us the factors \((x-2)(x-1)\).
Similarly, the denominator \(x^{2} - 2x - 3\) can be factored by finding two numbers that multiply to -3 and add to -2, which are -3 and 1. Hence, this factors into \((x-3)(x+1)\).
These factored forms are crucial because they illuminate the values of \(x\) that cause the numerator or denominator to equal zero. Such values define the boundaries between intervals.

Interval Notation

Interval notation is a method of writing down the set of all solutions to an inequality. It uses parentheses and brackets to describe intervals on the number line.
After factoring our expression, the critical points are \(x = 1, 2, 3, -1\). These points divide the number line into distinct segments:

• \((-\infty, -1)\)
• \((-1, 1)\)
• \((1, 2)\)
• \((2, 3)\)
• \((3, \infty)\)

Each interval and endpoint in interval notation denotes whether the boundary is included or excluded based on whether it satisfies the inequality (shown through brackets \([ \ ]\) for inclusion and parentheses \(( \ )\) for exclusion). The final solution set can be depicted as \((-1,1)\cup(2,3)\cup(3, \infty)\), with each segment where the inequality holds true.

Number Line Graphing

After identifying the critical points from the factorization, we can utilize these to delineate the intervals on a number line. Number line graphing is a visual strategy to determine where a polynomial or rational expression changes sign.
By plotting the critical points \(-1, 1, 2, 3\) on a number line, we mark these points, usually with either open or closed circles depending on inclusion (open for excluded, closed for included). Now, the number line is segmented into parts reflecting our identified intervals:

• From negative infinity to -1,
• -1 to 1,
• 1 to 2,
• 2 to 3,
• 3 to infinity.

This graph provides a clear picture of the intervals available for testing and those that satisfy the inequality.

Testing Intervals

Testing intervals helps determine which segments of the number line satisfy the given rational inequality. Once intervals are specified, choose a test point from each interval and substitute it back into the factored inequality.
Consider the intervals:

• For \((-\infty, -1)\), try \(x = -2\).
• For \((-1, 1)\), use \(x = 0\).
• For \((1, 2)\), pick \(x = 1.5\).
• For \((2, 3)\), select \(x = 2.5\).
• For \((3, \infty)\), try \(x = 4\).

Plug these values into the inequality \((x-2)(x-1)/((x-3)(x+1)) > 0\). If the result is positive, then the whole interval satisfies the inequality. For this case, the solution includes the intervals \((-1,1)\cup(2,3)\cup(3, \infty)\). The testing ensures only valid sections are included in the solution.