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Chapter 5: Problem 48

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) \(\left(x^{2}+2 x\right)^{1 / 4}<3^{1 / 4}\)

Short Answer

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

Step by step solution

01

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: \[ x^2 + 2x - 3 = 0 \].
02

Factor the Quadratic Equation

Attempt to factor the quadratic equation \( x^2 + 2x - 3 = 0 \).Rewrite it as \( (x + 3)(x - 1) = 0 \).
03

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).
04

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 \( x \in (-\infty, -3) \cup (1, \infty) \).
05

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 \( x \in (-3, 1) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 \( x^2 + 2x > 3 \).

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