Problem 73 - College Algebra with Modeling and Visualization

Question

Solve the rational inequality. $$ 2-\frac{5}{x}+\frac{2}{x^{2}} \geq 0 $$

Step-by-Step Solution

Answer

The solution is \((-\infty, 0) \cup [1, \infty)\).

Step 1: Move all terms to one side

Start by moving all terms to one side of the inequality to compare with zero. Our initial inequality is \( 2 - \frac{5}{x} + \frac{2}{x^2} \geq 0 \). Rewrite this as \( \frac{2}{x^2} - \frac{5}{x} + 2 \geq 0 \).

Step 2: Combine into a single fraction

To have a single expression, find a common denominator. The common denominator is \( x^2 \). Rewrite each term with this denominator: \( \frac{2}{x^2} = \frac{2}{x^2} \), \( -\frac{5}{x} = \frac{-5x}{x^2} \), and \( 2 = \frac{2x^2}{x^2} \). Combine them into \( \frac{2 - 5x + 2x^2}{x^2} \geq 0 \).

Step 3: Find the critical points

To find where the expression changes sign, set the numerator and denominator equal to zero separately. The numerator is \( 2x^2 - 5x + 2 = 0 \). Solve using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2, b = -5, c = 2 \). This gives \( x = 1 \) and \( x = \frac{2}{2} = \frac{1}{2} \). The denominator \( x^2 = 0 \) gives \( x = 0 \). Thus, critical points are \( x = 0, \frac{1}{2}, 1 \).

Step 4: Test intervals around critical points

Break the number line into intervals around critical points: \(( -\infty, 0 )\), \(( 0, \frac{1}{2} )\), \(( \frac{1}{2}, 1 )\), \(( 1, \infty )\). Test a point from each interval in the inequality \( \frac{2x^2 - 5x + 2}{x^2} \geq 0 \).- Test \( x = -1 \) from \( (-\infty, 0) \); expression is positive.- Test \( x = \frac{1}{4} \) from \( (0, \frac{1}{2}) \); expression is negative.- Test \( x = \frac{3}{4} \) from \( (\frac{1}{2}, 1) \); expression is negative.- Test \( x = 2 \) from \((1, \infty) \); expression is positive.

Step 5: Determine the solution intervals

From the previous analysis, the inequality \( \frac{2x^2 - 5x + 2}{x^2} \geq 0 \) holds in the intervals where the expression is non-negative, i.e., \( (-\infty, 0) \cup [1, \infty) \). However, note that \( x eq 0 \) due to the denominator restriction. Therefore, the solution set is \(( -\infty, 0 ) \cup [1, \infty )\).

Key Concepts

Critical Points

When solving rational inequalities, the critical points are essential in determining where the expression changes its sign. Critical points are those where the numerator or the denominator of the combined expression becomes zero. These points help to divide the number line into intervals that we need to test for defining the inequality's solution.
To find these critical points, start by setting the numerator equal to zero. Solving this provides potential values of the variable where the expression might change from positive to negative or vice versa. In our example, the numerator is a quadratic equation, given by \(2x^2 - 5x + 2 = 0\). Using this, we identify two critical points: \(x = 1\) and \(x = \frac{1}{2}\).
Moreover, the denominator \(x^2 = 0\) also gives us a critical point at \(x = 0\). However, it's crucial to remember this value cannot be a part of the solution set because it would make the original expression undefined. Critical points alert us to crucial intervals on the number line that need examination to determine where the inequality holds true.

Quadratic Formula

The Quadratic Formula is a powerful tool used to solve quadratic equations of the form \(ax^2 + bx + c = 0\). This formula is particularly helpful when factoring is not straightforward or possible. The formula is:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In this framework, \(a\), \(b\), and \(c\) are coefficients from the quadratic equation. In the context of the exercise, the expression \(2x^2 - 5x + 2 = 0\) was solved using this formula with \(a = 2\), \(b = -5\), and \(c = 2\).
Plugging these values into the quadratic formula:

• Calculate the discriminant \( \Delta = b^2 - 4ac \): \((-5)^2 - 4 \cdot 2 \cdot 2 = 1\).
• Substitute to find the roots: \(x = \frac{-(-5) \pm \sqrt{1}}{4} = \frac{5 \pm 1}{4}\).
• Determine the solutions: \(x = 1\) and \(x = \frac{1}{2}\).

Using the quadratic formula allows us to obtain the critical points, enhancing our ability to solve rational inequalities. Understanding this process ensures you are well-equipped to tackle any similar problems.

Solution Intervals

After identifying the critical points, the next crucial step is determining the solution intervals. This involves examining intervals derived from the critical points to evaluate where the original inequality holds.
Divide your number line based on critical points into different segments. For example, with critical points at \(x = 0\), \(x = \frac{1}{2}\), and \(x = 1\), we obtain the following intervals: \((-\infty, 0)\), \((0, \frac{1}{2})\), \((\frac{1}{2}, 1)\), and \((1, \infty)\).
Testing a point from within each interval in the rearranged inequality expression \(\frac{2x^2 - 5x + 2}{x^2} \geq 0\) helps to check whether the inequality holds:

• In \((-\infty, 0)\), try \(x = -1\); the expression is positive.
• In \((0, \frac{1}{2})\), try \(x = \frac{1}{4}\); the expression is negative.
• In \((\frac{1}{2}, 1)\), try \(x = \frac{3}{4}\); the expression is negative.
• In \((1, \infty)\), try \(x = 2\); the expression is positive.

From these tests, we determine that the solution intervals where the inequality is satisfied are \((-\infty, 0)\) and \([1, \infty)\). Notably, \(x = 0\) is excluded because it causes division by zero in the original expression.