Problem 50 - College Algebra with Modeling and Visualization

Question

Solve the inequality. $$ 0.5 x^{2}-3.2 x>-0.9 $$

Step-by-Step Solution

Answer

The solution is \((-\infty, \frac{1}{5}) \cup (9, \infty)\).

Step 1: Move all terms to one side

To solve the inequality, we move all terms to one side of the inequality sign. Start by adding \(0.9\) to both sides of the inequality:\[0.5x^2 - 3.2x + 0.9 > 0\]

Step 2: Factor the quadratic inequality

Next, we need to factor the quadratic expression \(0.5x^2 - 3.2x + 0.9\). To make factoring easier, multiply through by 10 to eliminate the decimal and simplify further:\[5x^2 - 32x + 9 > 0\]Now, factor the expression: \[(5x - 1)(x - 9) > 0\]

Step 3: Find the critical points

The critical points, or roots, of the inequality occur where each factor equals zero. Solve for \(x\):\[5x - 1 = 0 \ x - 9 = 0\]The solutions are \(x = \frac{1}{5}\) and \(x = 9\). These are our critical points.

Step 4: Test intervals between critical points

Determine which intervals satisfy the inequality by selecting test points in each interval created: 1. Choose a point from the interval \((-\infty, \, \frac{1}{5})\), such as \(x = 0\): - Substitute into the inequality: \((5(0) - 1)(0 - 9) = -1 \,\cdot -9 = 9 > 0\). True.2. Choose a point from the interval \((\frac{1}{5}, \, 9)\), such as \(x = 5\): - Substitute into the inequality: \((5(5) - 1)(5 - 9) = 24\,\cdot -4 = -96 < 0\). False.3. Choose a point from the interval \((9, \, \infty)\), such as \(x = 10\): - Substitute into the inequality: \((5(10) - 1)(10 - 9) = 49\,\cdot 1 = 49 > 0\). True.

Step 5: Write the solution

The solution to the inequality, based on test intervals that satisfy the inequality, are the intervals where the test points yield a positive result. Therefore, the solution is: \[x \in (-\infty, \frac{1}{5}) \cup (9, \infty)\]

Key Concepts

Factoring Quadratics

Quadratic inequalities often involve quadratic equations, such as the one we have here: \( 0.5x^2 - 3.2x + 0.9 > 0 \). A powerful technique in solving these quadratic equations is "factoring." It simply involves expressing the quadratic equation in the form of two simpler expressions (binomials) multiplied together.
This process can break down what looks like a complex quadratic expression into parts that are easier to understand and solve.

• Start by ensuring all coefficients are integers, which may require multiplying through by a factor - in our case, multiplying the equation by 10 to obtain \( 5x^2 - 32x + 9 > 0 \).
• Next, identify pairs of factors for both the leading coefficient (5) and the constant term (9) to find combinations that add up to the middle term coefficient (-32).
• Through trial and error or using the quadratic formula, we determined that factoring results in \((5x - 1)(x - 9) > 0 \).

This allows us to think of the inequality in terms of its roots: where each of these factors equals zero.

Critical Points

The concept of "critical points" or "roots" is central to solving quadratic inequalities. Critical points are the solutions to the equation where each factor equals zero.
To identify these points, we solve the equations derived from each factor:

• \( 5x - 1 = 0 \) gives a solution \( x = \frac{1}{5} \).
• \( x - 9 = 0 \) gives a solution \( x = 9 \).

These solutions are important because they divide the x-axis into regions or intervals that we need to test in order to determine where the inequality holds true.
Think of critical points as boundaries in the solution space, indicating potential changes in sign or behavior of the inequality. Analyzing the intervals around these points helps us understand where the inequality is valid.

Interval Testing

Interval testing is a technique used in conjunction with critical points to determine where a quadratic inequality holds true.
Once we have our critical points \(x = \frac{1}{5} \) and \(x = 9 \), we must examine intervals divided by these points:

• \((-fty, \frac{1}{5})\)
• \((\frac{1}{5}, 9)\)
• \((9, \infty)\)

Selecting a test point within each interval allows us to substitute these values into the factored inequality \((5x - 1)(x - 9) > 0\).

• For an interval like \((-fty, \frac{1}{5})\), choose \(x = 0\). The calculation shows that this interval satisfies the inequality.
• For \((\frac{1}{5}, 9)\), \(x = 5\) results in a negative product, meaning this interval does not satisfy the inequality.
• For \((9, \infty)\), \(x = 10\) gives a positive result, satisfying the inequality.

Through these tests, we conclude the valid solution intervals are \(x \in (-\infty, \frac{1}{5}) \cup (9, \infty)\). This systematic testing ensures that only true parts of the inequality are included in the solution.