Solving Absolute Value Equations A Step-by-Step Guide To |x^2 + 2x - 5| = 2

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In this article, we will delve into the process of solving the absolute value equation |x² + 2x - 5| = 2. Absolute value equations often present a unique challenge, as they require us to consider both positive and negative scenarios of the expression within the absolute value. This comprehensive guide will walk you through the steps necessary to find all possible solutions for x. We will explore the fundamental principles behind absolute value equations, break down the equation into manageable parts, and apply algebraic techniques to arrive at the final solutions. Whether you are a student grappling with algebra or simply seeking to enhance your problem-solving skills, this article aims to provide a clear and thorough understanding of how to tackle such equations. By the end of this guide, you will be equipped with the knowledge and confidence to solve similar absolute value problems independently. Our focus will be on ensuring each step is explained in detail, making it easy to follow along and grasp the underlying concepts. Let’s embark on this mathematical journey together and unravel the solutions to this intriguing equation.

Understanding Absolute Value Equations

Before we tackle the given equation, it is crucial to understand the fundamental concept of absolute value. The absolute value of a number is its distance from zero on the number line, regardless of direction. This means that the absolute value of a positive number is the number itself, and the absolute value of a negative number is its positive counterpart. For example, |3| = 3 and |-3| = 3. When solving absolute value equations, we need to consider both the positive and negative possibilities of the expression inside the absolute value symbols, as both cases will satisfy the equation. This is because if |A| = B, then either A = B or A = -B. Applying this principle to our equation, |x² + 2x - 5| = 2, we recognize that the expression x² + 2x - 5 could either be equal to 2 or -2. This bifurcation is the key to solving absolute value equations. We will set up two separate equations based on these possibilities and solve each one independently. This approach ensures we capture all possible values of x that satisfy the original equation. Understanding this initial step is vital, as it sets the stage for the subsequent algebraic manipulations and solutions. By grasping the dual nature of absolute values, we can systematically approach and solve a wide range of similar problems. The following sections will elaborate on how to handle each case effectively.

Setting Up the Equations

To solve the equation |x² + 2x - 5| = 2, we must consider two separate cases, as mentioned earlier. The first case is when the expression inside the absolute value is equal to 2, and the second case is when it is equal to -2. Let's break down these two cases:

  1. Case 1: x² + 2x - 5 = 2 This case considers the scenario where the expression x² + 2x - 5 is positive and equal to 2. This is a direct application of the definition of absolute value. We will rearrange this equation to set it equal to zero, forming a quadratic equation that can be solved using various methods, such as factoring, completing the square, or the quadratic formula. The resulting quadratic equation will provide us with potential solutions for x that satisfy this particular condition.
  2. Case 2: x² + 2x - 5 = -2 The second case addresses the scenario where the expression x² + 2x - 5 is negative, and its absolute value is equal to 2. This means the expression itself must be equal to -2. Similar to Case 1, we will rearrange this equation to obtain another quadratic equation. This quadratic equation will yield a different set of potential solutions for x. By considering both positive and negative scenarios, we ensure that we capture all possible solutions for the original absolute value equation. Each case will be solved independently to find the respective roots.

By setting up these two equations, we have transformed the original absolute value equation into two standard quadratic equations, which are easier to handle algebraically. The following sections will delve into the process of solving each of these quadratic equations to find the values of x.

Solving Case 1: x² + 2x - 5 = 2

Now, let's solve the first quadratic equation: x² + 2x - 5 = 2. The first step is to rearrange the equation by subtracting 2 from both sides to set it equal to zero. This gives us:

x² + 2x - 5 - 2 = 0 x² + 2x - 7 = 0

This is a quadratic equation in the standard form ax² + bx + c = 0, where a = 1, b = 2, and c = -7. Since this equation does not factor easily, we will use the quadratic formula to find the solutions for x. The quadratic formula is given by:

x = (-b ± √(b² - 4ac)) / (2a)

Plugging in the values, we get:

x = (-2 ± √(2² - 4(1)(-7))) / (2(1)) x = (-2 ± √(4 + 28)) / 2 x = (-2 ± √32) / 2 x = (-2 ± 4√2) / 2

Now, we can simplify by dividing both terms in the numerator by 2:

x = -1 ± 2√2

So, the two solutions for x in Case 1 are:

x₁ = -1 + 2√2 x₂ = -1 - 2√2

These are two real solutions that satisfy the first case of our absolute value equation. Next, we will move on to solving the second case and finding the remaining solutions.

Solving Case 2: x² + 2x - 5 = -2

Now, let's tackle the second quadratic equation: x² + 2x - 5 = -2. Similar to Case 1, we begin by rearranging the equation to set it equal to zero. We add 2 to both sides of the equation, which gives us:

x² + 2x - 5 + 2 = 0 x² + 2x - 3 = 0

We now have a quadratic equation in the standard form ax² + bx + c = 0, where a = 1, b = 2, and c = -3. This equation is factorable, which will simplify the solution process. We need to find two numbers that multiply to -3 and add up to 2. These numbers are 3 and -1. Therefore, we can factor the quadratic equation as:

(x + 3)(x - 1) = 0

To find the solutions for x, we set each factor equal to zero:

x + 3 = 0 or x - 1 = 0

Solving for x in each equation gives us:

x₃ = -3 x₄ = 1

So, the two solutions for x in Case 2 are x₃ = -3 and x₄ = 1. We now have a total of four potential solutions from both cases. In the next section, we will summarize these solutions and verify their correctness.

Summarizing and Verifying the Solutions

After solving both cases of the absolute value equation |x² + 2x - 5| = 2, we have obtained four potential solutions for x. Let's summarize them:

  • From Case 1: x₁ = -1 + 2√2 and x₂ = -1 - 2√2
  • From Case 2: x₃ = -3 and x₄ = 1

To ensure that these solutions are correct, we need to verify them by plugging each value back into the original absolute value equation. This is a crucial step because sometimes, extraneous solutions may arise during the solving process, especially when dealing with absolute value or radical equations. Let's verify each solution:

  1. Verify x₁ = -1 + 2√2: |(-1 + 2√2)² + 2(-1 + 2√2) - 5| =|1 - 4√2 + 8 - 2 + 4√2 - 5| =|2| = 2 (Correct)
  2. Verify x₂ = -1 - 2√2: |(-1 - 2√2)² + 2(-1 - 2√2) - 5| =|1 + 4√2 + 8 - 2 - 4√2 - 5| =|2| = 2 (Correct)
  3. Verify x₃ = -3: |(-3)² + 2(-3) - 5| =|9 - 6 - 5| =|-2| = 2 (Correct)
  4. Verify x₄ = 1: |(1)² + 2(1) - 5| =|1 + 2 - 5| =|-2| = 2 (Correct)

Since all four values satisfy the original equation, we can confidently state that they are the correct solutions. Therefore, the solutions to the equation |x² + 2x - 5| = 2 are x = -1 + 2√2, x = -1 - 2√2, x = -3, and x = 1.

Conclusion

In conclusion, we have successfully solved the absolute value equation |x² + 2x - 5| = 2 by systematically breaking it down into two cases. The key to solving absolute value equations lies in recognizing that the expression inside the absolute value can be either positive or negative, leading to two separate equations. We first set up the two cases: x² + 2x - 5 = 2 and x² + 2x - 5 = -2. Each of these cases resulted in a quadratic equation. We solved the first quadratic equation using the quadratic formula, obtaining solutions x = -1 + 2√2 and x = -1 - 2√2. The second quadratic equation was solved by factoring, yielding solutions x = -3 and x = 1. Finally, we verified each of these solutions by substituting them back into the original absolute value equation. This verification step confirmed that all four solutions were correct and that no extraneous solutions were present.

This comprehensive approach demonstrates the importance of understanding the properties of absolute values and the methods for solving quadratic equations. By applying these techniques, you can confidently solve a wide range of absolute value equations. Remember to always consider both positive and negative cases and to verify your solutions to ensure accuracy. This methodical process not only helps in finding the correct answers but also enhances your problem-solving skills in algebra and beyond. The ability to tackle such equations is a valuable asset in various mathematical and real-world contexts. We hope this guide has provided you with a clear understanding and the necessary tools to solve similar problems effectively.