Problem 24 - Beginning and Intermediate Algebra

Question

Solve each quadratic inequality. Graph the solution set and write the solution in interval notation. $$1-d^{2}>0$$

Step-by-Step Solution

Answer

The solution set of the quadratic inequality \(1-d^2 > 0\) is \((-∞, -1) \cup (1, ∞)\). The graph highlights the intervals \((-∞, -1)\) and \((1, ∞)\) on a number line.

Step 1: Set up the inequality

The given inequality is \(1-d^2 > 0\). Our goal is to solve this inequality for the variable \(d\).

Step 2: Rewrite the inequality as an equation

To find the critical points, we rewrite the inequality as an equation and solve for \(d\): \(1 - d^2 = 0\).

Step 3: Factor the quadratic equation

Factoring the quadratic equation, we get \((1 - d)(1 + d) = 0\).

Step 4: Solve for critical points

Now, we solve for the critical points by setting each factor equal to zero: 1) \(1 - d = 0\) ⟹ \(d = 1\) 2) \(1 + d = 0\) ⟹ \(d = -1\) These two values, -1 and 1, will be used to divide the interval of our number line.

Step 5: Use interval test to find the solution

Using the interval test, we will test the three intervals that are formed by our critical points and the inequality: \((-∞, -1), (-1, 1),\) and \((1, ∞)\). 1) Test in the interval \((-∞, -1)\): Choose a test point, for example, \(d = -2\). Then, \((1 - (-2))(1 + (-2)) = (3)(-1) = -3 > 0\). Thus, this interval satisfies the inequality. 2) Test in the interval \((-1, 1)\): Choose a test point, for example, \(d = 0\). Then, \((1 - 0)(1 + 0) = (1)(1) = 1 > 0\). Thus, this interval does not satisfy the inequality. 3) Test in the interval \((1, ∞)\): Choose a test point, for example, \(d = 2\). Then, \((1 - 2)(1 + 2) = (-1)(3) = -3 > 0\). Thus, this interval satisfies the inequality. The solution set of the inequality is the union of \((-∞, -1)\) and \((1, ∞)\).

Step 6: Graph the solution and write in interval notation

We can represent the solution graphically by drawing a number line and highlighting \((-∞, -1)\) and \((1, ∞)\). The interval notation for this solution is \((-∞, -1) \cup (1, ∞)\).

Key Concepts

Interval Notation

Interval notation is a way of writing the set of all solutions to an inequality. It helps us describe the portion of the number line that satisfies the given condition. In our quadratic inequality, we found that the solution set lies in the intervals

• d \in (-\infty, -1)
• d \in (1, \infty)

We use parenthesis \( \) to indicate that the endpoints are not included in the solution set. Meaning, neither \(d = -1\) nor \(d = 1\) satisfy the inequality \(1 - d^2 > 0\). The use of the union symbol \(\cup\) between the two intervals signals that both intervals together form the complete solution. When writing your solution in interval notation, always be sure to list the smaller numbers first, moving towards larger numbers, from left to right.

Critical Points

Critical points are the values that make the equation \(1 - d^2 = 0\) true, or in simpler terms, the points where the graph of the inequality changes direction. To find these critical points, we first set the quadratic equation equal to zero:

• Factor the equation: \((1 - d)(1 + d) = 0\)
• Solve each factor:
  • \(1 - d = 0\) ⟹ \(d = 1\)
  • \(1 + d = 0\) ⟹ \(d = -1\)

These solutions, \(d = -1\) and \(d = 1\), divide the number line into intervals which help us determine where the inequality is true. It is crucial to test the intervals created by these critical points to find the solution set of the inequality.

Solution Set

The solution set is the collection of all values of \(d\) that satisfy the inequality. Once we found our critical points \(d = -1\) and \(d = 1\), we divided the number line into intervals. We then tested these intervals to see where the inequality holds true.

Interval \((-\infty, -1)\): Tested with \(d = -2\) and found satisfying.

Interval \((-1, 1)\): Tested with \(d = 0\) and found not satisfying.

Interval \((1, \infty)\): Tested with \(d = 2\) and found satisfying.

Combining these results, the solution set to the inequality \(1 - d^2 > 0\) is the union of the intervals \((-\infty, -1)\) and \((1, \infty)\). This immediately tells us where \(d\) values will satisfy the inequality, excluding the critical points themselves.

Graphing Inequalities

Graphing inequalities allows us to visually see where the solutions to an inequality lie on the number line. For the inequality \(1 - d^2 > 0\), we found that it is satisfied for intervals \((-\infty, -1)\) and \((1, \infty)\). To graph this on a number line:

• Mark the critical points \(d = -1\) and \(d = 1\) with open circles to indicate they are not part of the solution set.
• Shade the intervals \((-\infty, -1)\) and \((1, \infty)\) to show that any point within these intervals satisfies the inequality.

This shaded number line provides a quick visual reference for understanding where the inequality holds true, showing that the solution consists of all values less than \(-1\) and greater than \(1\).