Solving Quadratic Equations with Piecewise Functions: A Detailed Approach

Quadratic equations can appear quite daunting at first glance, but they can be tackled systematically with a detailed approach. In this article, we will explore a specific quadratic equation through the lens of piecewise functions. We will break down each step in the process, identify potential solutions, and verify our findings using a series of cases. This approach emphasizes the importance of clear, methodical problem-solving in mathematics, which is crucial for both academic and practical applications.

Introduction to the Problem

We will begin by examining the equation:

Original Equation: x2-22-x/32x-4-5

This equation gives us an opportunity to apply our understanding of algebraic manipulation and piecewise functions. We will go through the steps of solving this equation step by step, ensuring that we do not overlook any possible solutions and that we double-check our work for any possible errors.

Step-by-Step Solution

First, we will simplify and rewrite the equation in a more manageable form:

Simplified Equation: 3x2-4x-212x-2-30

Case 1: x

When we consider x , we need to evaluate each term in the equation:

x - 2 implies x - 2 -x - 2 2 - x implies 2 - x -x - 2 2x - 4 implies 2x - 4 -x - 4

Therefore, the equation transforms to:

3x2 -16x - 2 - 30

Further simplification gives:

3x2 -16x - 32

Solving for x yields:

3x2 16x 32 0

Using the quadratic formula:

x [-16 ± √(162 - 4 * 3 * 32)] / (2 * 3)

This simplifies to:

x [-16 ± √(256 - 384)] / 6

This results in:

x [-16 ± √(-128)] / 6

The discriminant being negative means there are no real solutions for this case.

Case 2: -2 ≤ x ≤ 2

When we consider -2 ≤ x ≤ 2, we have:

x - 2 ≤ 0 implies x - 2 -x - 2 2 - x ≥ 0 implies 2 - x x 2 2x - 4 implies 2x - 4 -x - 4

Therefore, the equation transforms to:

3x2 16x - 2 - 30

Further simplification gives:

3x2 16x - 32

Solving for x yields:

3x2 - 16x 32 0

Using the quadratic formula:

x [16 ± √(162 - 4 * 3 * 32)] / (2 * 3)

This simplifies to:

x [16 ± √(256 - 384)] / 6

The discriminant being negative means there are no real solutions for this case as well.

Case 3: x > 2

When we consider x > 2, we have:

x - 2 > 0 implies x - 2 x - 2 2 - x implies 2 - x -x - 2 2x - 4 > 0 implies 2x - 4 x 4

Therefore, the equation transforms to:

3x2 16x - 2 - 30

Further simplification gives:

3x2 16x - 32

Solving for x yields:

3x2 - 16x 32 0

Using the quadratic formula:

x [16 ± √(162 - 4 * 3 * 32)] / (2 * 3)

This simplifies to:

x [16 ± √(256 - 384)] / 6

The discriminant being negative means there are no real solutions for this case as well.

Conclusion

After systematically evaluating each case, we find that there are no real solutions to the given equation. This highlights the importance of careful examination and verification in solving mathematical problems.

Related Keywords

quadratic equations piecewise functions algebraic solutions

Additional Resources

For further exploration of solving quadratic equations and piecewise functions, you can refer to the following:

Math Is Fun: Quadratic Equations Khan Academy: Algebra II Coursera: College Algebra and Problem Solving