Problem 48 - A Graphical Approach to Precalculus with Limits

Question

Use an analytic method to solve each equation in part (a). Support the solution with a graph. Then use the graph to solve the inequalities in parts (b) and (c). (a) \(\left(x^{2}+2 x\right)^{1 / 4}=3^{1 / 4}\) (b) \(\left(x^{2}+2 x\right)^{1 / 4}>3^{1 / 4}\) (c) \(\left(x^{2}+2 x\right)^{1 / 4}<3^{1 / 4}\)

Step-by-Step Solution

Answer

(a) Solutions: \( x = -3 \) and \( x = 1 \). (b) \( x \in (-\infty, -3) \cup (1, \infty) \). (c) \( x \in (-3, 1) \).

Step 1: Solve the Equation

First, isolate the expression under the fourth root: \[ (x^2 + 2x)^{1/4} = 3^{1/4} \]Raise both sides to the power of 4 to eliminate the roots: \[ x^2 + 2x = 3 \]This reduces to a quadratic equation: \[ x^2 + 2x - 3 = 0 \].

Step 2: Factor the Quadratic Equation

Attempt to factor the quadratic equation \( x^2 + 2x - 3 = 0 \).Rewrite it as \( (x + 3)(x - 1) = 0 \).

Step 3: Find the Roots

Set each factor equal to zero: \( x + 3 = 0 \) or \( x - 1 = 0 \).Solve to find \( x = -3 \) and \( x = 1 \). These are the solutions of the equation in part (a).

Step 4: Solve the Inequality Part (b)

For the inequality \( (x^2 + 2x)^{1/4} > 3^{1/4} \), it simplifies to \( x^2 + 2x > 3 \).Rewrite as \( x^2 + 2x - 3 > 0 \).Using the solutions from Step 3, identify the intervals: \(( -\infty, -3), (-3, 1), (1, \infty)\).Select a test point from each interval and determine the sign:- For \( x \in (-\infty, -3) \), test point \( x = -4 \) gives \( (-4)^2 + 2(-4) - 3 = 5 > 0 \).- For \( x \in (-3, 1) \), test point \( x = 0 \) gives \( 0^2 + 2(0) - 3 = -3 < 0 \).- For \( x \in (1, \infty) \), test point \( x = 2 \) gives \( 2^2 + 2(2) - 3 = 5 > 0 \).So the solution for part (b) is \( x \in (-\infty, -3) \cup (1, \infty) \).

Step 5: Solve the Inequality Part (c)

For the inequality \( (x^2 + 2x)^{1/4} < 3^{1/4} \), it simplifies to \( x^2 + 2x < 3 \).This is the opposite condition of part (b): \( x^2 + 2x - 3 < 0 \). Using the test points from Step 4, note:- The interval \( (-\infty, -3) \) is false.- The interval \( (-3, 1) \) is true.- The interval \( (1, \infty) \) is false.Thus, the solution for part (c) is \( x \in (-3, 1) \).

Key Concepts

Quadratic Equations

Quadratic equations are one of the fundamental topics in algebra. They take the general form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Solving quadratic equations typically involves finding the values of \( x \) that make the equation true. These solutions are referred to as roots of the equation. There are several methods to solve quadratic equations:

• **Factoring**: This involves expressing the quadratic as a product of two binomials if possible. For example, \( x^2 + 5x + 6 \) can be factored to \( (x + 2)(x + 3) \).
• **Quadratic Formula**: When factoring is challenging, the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) can be used to find the roots.
• **Completing the Square**: This method involves rewriting the quadratic in the form \( (x + d)^2 = e \), then solving for \( x \).

In our exercise, we used factoring to solve the equation \( x^2 + 2x - 3 = 0 \) and found the roots \( x = -3 \) and \( x = 1 \). These roots will assist in solving the inequalities that follow.

Inequalities

When working with inequalities, we are seeking a range of values that satisfy an inequality rather than just specific points as in equations. Inequalities such as \( x^2 + 2x > 3 \) require careful consideration to find these ranges.

• **Testing Intervals**: Once you have the roots from the equivalent equation, you can break down the number line into intervals using these roots.
• **Choosing Test Points**: Within each interval, choose a test point to determine if it satisfies the inequality.
• **Determining Solutions**: If the test makes the inequality true, then that interval is part of the solution.

In our example, the quadratic \( x^2 + 2x > 3 \) reduced to \( x^2 + 2x - 3 > 0 \). The roots from the equation \( x = -3 \) and \( x = 1 \) were used to create the intervals \(( -\infty, -3), (-3, 1), (1, \infty)\). By testing points from each interval, we determined the solution for the inequality: \( x \in (-\infty, -3) \cup (1, \infty) \).

Graphical Representation

Graphs can vividly demonstrate the relationship between algebraic expressions and their solutions. When solving quadratic equations or inequalities, graphing the quadratic function \( y = x^2 + 2x - 3 \) provides a clear visual representation of where the function meets certain conditions.

• **Intersection Points**: The points where the graph intersects the x-axis are the roots of the equation. Here, these are \( x = -3 \) and \( x = 1 \).
• **Above or Below the Axis**: For inequalities, observe whether parts of the graph are above or below the x-axis. For instance, if the graph is above the axis, the inequality \( y > 0 \) holds true.
• **Visualizing Intervals**: Graphs make it easy to see which intervals are solutions as they show where the graph lies above or below the x-axis within identified intervals.

Using a graph, students can confirm the solutions obtained analytically and develop a deeper understanding of how quadratic equations behave. In our scenario, plotting the graph allowed us to visualize and verify that the function was positive in the intervals \(( -\infty, -3)\) and \((1, \infty)\), reinforcing the analytical solution for the inequality \( x^2 + 2x > 3 \).