Problem 15 - Beginning and Intermediate Algebra

Question

Solve each quadratic inequality. Graph the solution set and write the solution in interval notation. $$7 p^{2}-4>12 p$$

Step-by-Step Solution

Answer

The solution to the quadratic inequality \(7p^2 - 12p - 4 > 0\) is: \((-∞, -\frac{2}{7}) \cup (2, ∞)\).

Step 1: Set up the inequality as a difference

Subtract \(12p\) from both sides of the inequality to have the inequality in the form of a difference: \[7p^2 - 12p - 4 > 0\]

Step 2: Factor the quadratic polynomial

Factor the quadratic expression on the left-hand side of the inequality: \[(7p + 2)(p - 2) > 0\]

Step 3: Find the critical points

The critical points are the roots of the quadratic polynomial. Set the factors equal to zero and solve for \(p\): \[7p + 2 = 0 \Rightarrow p = -\dfrac{2}{7}\] and \[p - 2 = 0 \Rightarrow p = 2\] So the critical points are \(-\frac{2}{7}\) and \(2\).

Step 4: Determine the intervals of the inequality

To find the intervals where the inequality holds true, we'll test some points from each interval created by the critical points. Choose a value less than \(-\frac{2}{7}\), a value between \(-\frac{2}{7}\) and \(2\), and a value greater than \(2\). Let's test \(p = -1\): \[(7(-1) + 2)((-1) - 2) > 0\] \[-5 \cdot (-3) > 0\] which is true. Let's test \(p = 0\): \[(7(0) + 2)((0) - 2) > 0\] \(2 \cdot (-2) > 0\) which is false. Let's test \(p = 3\): \[(7(3) + 2)((3) - 2) > 0\] \(23 \cdot 1 > 0\) which is true. Based on our tests, the inequality holds true for \(p < -\frac{2}{7}\) and \(p > 2\).

Step 5: Graph the solution set

To graph the solution set, plot the critical points \(-\frac{2}{7}\) and \(2\) on a number line, and indicate that the solution set exists outside these points (to the left of \(-\frac{2}{7}\) and to the right of \(2\)) using open circles and rays.

Step 6: Write the solution in interval notation

Finally, express the solution in interval notation: \[(-\infty, -\frac{2}{7}) \cup (2, \infty)\] This interval notation indicates that the solution to the quadratic inequality \(7p^2 - 12p - 4 > 0\) is valid for all values of \(p\) less than \(-\frac{2}{7}\) or greater than \(2\).

Key Concepts

Critical Points

In the context of quadratic inequalities, **critical points** play a crucial role. They are the values where the inequality "changes direction." These points are derived from setting the expression equal to zero. For the inequality \[7p^2 - 12p - 4 > 0,\] you first rearrange it into \[(7p + 2)(p - 2) = 0.\] By solving, you find the critical points:

• For \(7p + 2 = 0\), solve for \(p\) to get \(p = -\frac{2}{7}.\)
• For \(p - 2 = 0\), solve to find \(p = 2.\)

These critical points, \(-\frac{2}{7}\) and \(2\), are where the sign of the inequality expression changes. They divide the number line into intervals that need to be tested to understand where the inequality holds true.

Interval Notation

Once you've determined which intervals satisfy the inequality, using **interval notation** becomes a way to succinctly express the solution set. For the inequality \[7p^2 - 12p - 4 > 0,\] the testing process showed that it holds for \(p < -\frac{2}{7}\) and \(p > 2.\)**Interval notation** is a shorthand way of conveying these ranges:

• The interval \((-\infty, -\frac{2}{7})\) indicates all numbers less than \(-\frac{2}{7}.\)
• The interval \((2, \infty)\) indicates all numbers greater than \(2.\)

Thus, the solution in interval notation is written as:\[(-\infty, -\frac{2}{7}) \cup (2, \infty).\]The "\(\cup\)" symbol signifies a union of the intervals, meaning both are parts of the solution.

Graphing Solution Sets

**Graphing solution sets** provides a visual representation of where an inequality holds. For the quadratic inequality \[7p^2 - 12p - 4 > 0,\] graphing is a helpful way to see the solution intervals.**Steps for Graphing:**

• First, plot the critical points \(-\frac{2}{7}\) and \(2\) on a number line. As these are not part of the solution (since the expression must be greater than zero), use open circles to represent them.
• Then, use rays to highlight the interval solutions. A ray to the left of \(-\frac{2}{7}\) indicates \((-\infty, -\frac{2}{7})\), and a ray extending to the right of \(2\) shows \((2, \infty).\)

By doing this, students can now visually see the solution extends indefinitely in both directions outside the critical points, reinforcing the interval notation and the concept of the union of two distinct intervals.