Problem 13 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-40: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ x^{2}<16 $$

Step-by-Step Solution

Answer

The solution is \((-4, 4)\).

Step 1: Understand the Inequality

The inequality given is \( x^2 < 16 \). This implies we want the values of \( x \) for which its square is less than 16.

Step 2: Set Up the Boundary Equation

To find the critical points, we solve the equation \( x^2 = 16 \), which gives us the boundary cases where equality holds. Solving this gives \( x = 4 \) and \( x = -4 \). These are the points where the inequality changes direction.

Step 3: Test Intervals

The critical points divide the number line into three intervals: \((-\infty, -4)\), \((-4, 4)\), and \((4, \infty)\). We need to determine in which intervals the inequality \( x^2 < 16 \) holds true. To do this, we select test points from each interval:- For \((-\infty, -4)\), choose \( x = -5 \). Here, \((-5)^2 = 25 \), so it does not satisfy the inequality.- For \((-4, 4)\), choose \( x = 0 \). Here, \(0^2 = 0 \), which satisfies the inequality.- For \((4, \infty)\), choose \( x = 5 \). Here, \(5^2 = 25 \), so it does not satisfy the inequality.

Step 4: Write the Solution in Interval Notation

Based on the testing of intervals, the solution to the inequality \( x^2 < 16 \) is within the open interval \((-4, 4)\), since it is the only interval where the inequality holds true.

Key Concepts

Quadratic Inequality

Quadratic inequalities involve expressions where a squared term forms part of the inequality, such as the classic form \( ax^2 + bx + c < 0 \), \( ax^2 + bx + c > 0 \), and their inclusive counterparts. In the example \( x^2 < 16 \), we are dealing with a simple case of a quadratic inequality where only the squared term \( x^2 \) is compared to a constant.The solution process begins by comparing \( x^2 \) to a perfect square, 16 in this case. To understand the inequality \( x^2 < 16 \), we first determine the points where \( x^2 = 16 \), called the boundary points. Solving \( x^2 = 16 \) gives us \( x = 4 \) and \( x = -4 \). These solutions are pivotal because they are the points where the inequality might switch from true to false or vice versa.Once we have these boundaries, the next vital step is to determine the sets of values within the intervals formed by these points that satisfy the inequality. Quadratic inequalities when graphed can look like U-shaped curves (parabolas opening upwards), making it intuitive to grasp that solutions are often found between the x-values for which the parabola hits a specific height (in this case, the line \( y=16 \)).

Interval Notation

Interval notation is a mathematical syntax used to express a range of values. It succinctly represents the solution set of inequalities, making it an efficient method to communicate results. When solving the inequality \( x^2 < 16 \), we identified that this inequality holds for all values of \( x \) strictly between the boundary points -4 and 4.The notation uses parentheses \(( )\) to denote that endpoints are not included (open interval) and square brackets \([ ]\) to show that endpoints are included (closed interval). For our problem, since both -4 and 4 are where the inequality \( x^2 < 16 \) does not hold (because \( x^2 = 16 \) is not less than 16), they are excluded from the solution set.Thus, the solution for the inequality \( x^2 < 16 \) can be neatly expressed in interval notation as \((-4, 4)\). This notation clarifies that any \( x \) within this range, but not exactly -4 or 4, satisfies the inequality.

Test Intervals

To determine where a quadratic inequality holds true, we need to check the intervals created by boundary points, using them as section dividers on a number line. By introducing 'test intervals', we can systematically find which intervals make the inequality valid.Each critical point splits the number line into several segments or intervals. For \( x^2 < 16 \), the critical points found were -4 and 4, leading to three distinct intervals: \((-\infty, -4)\), \((-4, 4)\), and \((4, \infty)\). We then select a random test point from within each interval and substitute it into the inequality to check if it holds.For instance:

• In the interval \((-\infty, -4)\), choosing \( x = -5 \) results in \( (-5)^2 = 25 \), which does not satisfy \( x^2 < 16 \).
• Choosing \( x = 0 \) from \((-4, 4)\) yields \( 0^2 = 0 \), satisfying the inequality.
• In the interval \((4, \infty)\), testing \( x = 5 \) gives \( 5^2 = 25 \), again not satisfying the inequality.

Thus, only the interval \((-4, 4)\) satisfies \( x^2 < 16 \), indicating that within this region, the inequality is true.