Problem 69 - A Graphical Approach to College Algebra

Question

Solve each rational inequality by hand. Do not use a calculator. $$2-\frac{5}{x}+\frac{2}{x^{2}} \geq 0$$

Step-by-Step Solution

Answer

Solution: \([\frac{1}{2}, 1] \cup (1, \infty)\).

Step 1: Find Common Denominator

The inequality is \(2 - \frac{5}{x} + \frac{2}{x^2} \geq 0\). To combine these terms, identify a common denominator which is \(x^2\). Rewrite the expression as \(\frac{2x^2}{x^2} - \frac{5x}{x^2} + \frac{2}{x^2} \geq 0\).

Step 2: Combine Fractions

Now combine the fractions over the common denominator \(x^2\). This results in \(\frac{2x^2 - 5x + 2}{x^2} \geq 0\).

Step 3: Find Roots of the Numerator

To determine the intervals to test for inequality, find the roots of the numerator \(2x^2 - 5x + 2 = 0\). Use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 2\), \(b = -5\), and \(c = 2\). This gives roots \(x = 1\) and \(x = \frac{1}{2}\).

Step 4: Identify Critical Points

The critical points include the roots \(x = 1\), \(x = \frac{1}{2}\), and \(x = 0\) due to the denominator \(x^2\). These divide the number line into intervals: \(( -\infty, 0 )\), \((0, \frac{1}{2})\), \((\frac{1}{2}, 1)\), and \((1, \infty)\).

Step 5: Test the Intervals

Test each interval by selecting test points within each interval. For example, in \(( -\infty, 0 )\), test \(x = -1\). The computation gives a negative value. For \((0, \frac{1}{2})\), test \(x = \frac{1}{4}\). The computation gives a negative value. For \((\frac{1}{2}, 1)\), test \(x = \frac{3}{4}\). The computation gives a positive value. For \((1, \infty)\), test \(x = 2\). The computation gives a positive value.

Step 6: Include Boundary Points

Determine if boundary points are included by substituting them back into the inequality. \(x = 1\) gives the result \(0 \geq 0\), making it a valid boundary point. \(x = \frac{1}{2}\) also gives the result \(0 \geq 0\), so it's also a valid boundary point. \(x = 0\) results in division by zero; therefore, it's not included.

Step 7: Write the Solution Set

The solution set includes intervals where \( \frac{2x^2 - 5x + 2}{x^2} \geq 0 \) holds true: \(\left[ \frac{1}{2}, 1 \right] \cup (1, \infty)\).

Key Concepts

Quadratic Formula

When solving rational inequalities, such as our example, we often need to find roots of a quadratic equation. In this exercise, we encounter the quadratic expression in the numerator: \(2x^2 - 5x + 2\). Here, the quadratic formula becomes very useful:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In our specific case:

• \(a = 2\)
• \(b = -5\)
• \(c = 2\)

Substitute these values into the quadratic formula to find the roots. Solving this gives:

• \(x = 1\)
• \(x = \frac{1}{2}\)

These roots tell us the x-values where the numerator equals zero, which are important for identifying critical points along the number line.

Common Denominator

When dealing with rational expressions, especially inequalities, it's crucial to combine all terms into a single fraction. The first step involves identifying a common denominator. For the inequality \[2 - \frac{5}{x} + \frac{2}{x^2} \geq 0\]we notice the terms include \(x\) and \(x^2\) in the denominators. Here, the common denominator is \(x^2\).Once we establish that \(x^2\) is our common denominator, rewrite each term with it:

• \(2\) becomes \(\frac{2x^2}{x^2}\)
• \(\frac{5}{x}\) becomes \(\frac{5x}{x^2}\)
• \(\frac{2}{x^2}\) stays the same

This allows us to consolidate the inequality into a single expression: \[\frac{2x^2 - 5x + 2}{x^2} \geq 0\]

Critical Points

Critical points in a rational inequality are essential for understanding where the behavior of the inequality might change. These points arise from both the numerator and the denominator:1. **Numerator Roots:** - From the quadratic formula applied to \(2x^2 - 5x + 2 = 0\), we found the roots \(x = 1\) and \(x = \frac{1}{2}\).2. **Denominator Zero:** - The expression \(x^2\) implies that \(x = 0\) is a critical point, where the function is undefined due to division by zero.These critical points split the domain into distinct intervals for further testing. They represent locations where the expression might change its sign, which is why they are critical to the solving process.

Intervals Testing

Interval testing is a method used to determine the sign of the expression across intervals defined by critical points. After identifying critical points at \(x = 0\), \(x = \frac{1}{2}\), and \(x = 1\), we divide the number line into the following intervals:

• \((-\infty, 0)\)
• \((0, \frac{1}{2})\)
• \((\frac{1}{2}, 1)\)
• \((1, \infty)\)

For each interval, we choose a test point to evaluate the sign of the entire expression \(\frac{2x^2 - 5x + 2}{x^2}\):- **\((-\infty, 0)\):** Test \(x = -1\); results in a negative value.- **\((0, \frac{1}{2})\):** Test \(x = \frac{1}{4}\); results in a negative value.- **\((\frac{1}{2}, 1)\):** Test \(x = \frac{3}{4}\); results in a positive value.- **\((1, \infty)\):** Test \(x = 2\); results in a positive value.Knowing where the expression is positive or negative helps us determine the solution set. The true solution encompasses intervals and edges where the numerator is zero or makes the expression non-negative, excluding places where the denominator is zero. This gives us the solution set: \[\left[ \frac{1}{2}, 1 \right] \cup (1, \infty)\]