Problem 47 - College Algebra with Modeling and Visualization

Question

Solve the polynomial inequality. $$ 7 x^{4}>14 x^{2} $$

Step-by-Step Solution

Answer

The solution is \((-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)\).

Step 1: Simplify the Inequality

Start by simplifying the given polynomial inequality: \[ 7x^4 > 14x^2 \]Divide both sides by 7 to simplify:\[ x^4 > 2x^2 \]

Step 2: Rearrange and Factor the Inequality

Move all terms to one side to facilitate factoring:\[ x^4 - 2x^2 > 0 \]Now, factor the left side:\[ x^2(x^2 - 2) > 0 \]

Step 3: Find Critical Points

Set each factor equal to zero to find critical points:1. \( x^2 = 0 \) gives \( x = 0 \)2. \( x^2 - 2 = 0 \) gives \( x = \pm \sqrt{2} \)The critical points are \( x = 0 \), \( x = \sqrt{2} \), and \( x = -\sqrt{2} \).

Step 4: Test Intervals Around Critical Points

Test intervals determined by critical points: - Interval \(( -\infty, -\sqrt{2})\)- Interval \(( -\sqrt{2}, 0)\)- Interval \(( 0, \sqrt{2})\)- Interval \(( \sqrt{2}, \infty)\)Choose a test point from each interval and substitute into the inequality \( x^2(x^2 - 2) > 0 \):- For \( x = -2 \) (Interval: \( -\infty, -\sqrt{2} \)), \((-2)^2(((-2)^2 - 2) = 4 > 0\)- For \( x = -1 \) (Interval: \( -\sqrt{2}, 0 \)), \((-1)^2(((-1)^2 - 2) = -1 < 0\)- For \( x = 1 \) (Interval: \( 0, \sqrt{2} \)), \((1)^2((1)^2 - 2) = -1 < 0\)- For \( x = 2 \) (Interval: \( \sqrt{2}, \infty \)), \((2)^2((2)^2 - 2) = 4 > 0\)The inequality is satisfied in the intervals \(( -\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)\).

Step 5: Write the Final Solution

The solution to the inequality \( 7x^4 > 14x^2 \) is:\[(-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)\]where \(x\) does not include \(0\) or \(\pm\sqrt{2}\) since the inequality is strict (> not \geq).

Key Concepts

Factoring

When solving polynomial inequalities, factoring plays a pivotal role. It's the process of breaking down a complex polynomial expression into the product of simpler factors. This makes it easier to solve inequalities like the one given: \(x^4 - 2x^2 > 0\). After rearranging, it can be factored further. The expression becomes \(x^2(x^2 - 2) > 0\), which reveals the underlying simpler expressions that can be solved individually.
Factoring transforms an equation into multiple zero-determined forms which is easier for finding solutions:

• It helps to find critical points by setting each factor to zero.
• It simplifies the testing process by breaking down into manageable parts.
• It aids in visualizing the polynomial's behavior as each factor reaches zero.

Mastering factoring is crucial for efficiently solving polynomial inequalities and is a key step for finding critical points and testing intervals.

Critical Points

Critical points are the values of \(x\) where the polynomial expression equals zero. These points are significant because they indicate potential changes in the inequality's sign, which helps define solution intervals. In our example, we set each factor from the factored form \(x^2(x^2 - 2)\) to zero:

• For \(x^2 = 0\), the solution is \(x = 0\).
• For \(x^2 - 2 = 0\), it gives \(x = \pm \sqrt{2}\).

Identifying critical points is essential as they are boundary markers between regions where the polynomial may change from positive to negative or vice versa. These points are where we "pivot" in our analysis of the polynomial's sign, critical for the next step: testing intervals.

Interval Testing

After identifying critical points, we use interval testing to determine where the polynomial inequality holds true. This method involves selecting test points from intervals divided by these critical points to check the sign of the inequality. For \(x^2(x^2 - 2) > 0\), our intervals are:

• \((-\infty, -\sqrt{2})\)
• \((-\sqrt{2}, 0)\)
• \((0, \sqrt{2})\)
• \((\sqrt{2}, \infty)\)

By choosing points like \(-2, -1, 1\), and \(2\) respectively, we test the inequality's sign:

• In \((-\infty, -\sqrt{2})\) and \((\sqrt{2}, \infty)\), the expression is positive, satisfying the inequality.
• In \((-\sqrt{2}, 0)\) and \((0, \sqrt{2})\), the expression is negative, not satisfying the inequality.

Through interval testing, we clearly understand where the inequality is valid, enabling us to express the final solution confidently: \((-\infty, -\sqrt{2}) \cup ( \sqrt{2}, \infty)\). This step completes the critical analysis and guides us to the correct answer.