Solving Quadratic Equations With Real Number Coefficients And Rational/Irrational Solutions
This article delves into the intricate world of quadratic equations, offering a step-by-step guide to solving them. We will explore various techniques, including factoring and applying the quadratic formula, with detailed explanations and examples. Our primary focus will be on two specific quadratic equations: (1) 4x² - 4ax + (a² - b²) = 0, where a, b ∈ ℝ and (2) 5x² - 12x - 9 = 0, examining their solutions within the realms of irrational () and rational () numbers. Understanding these equations and their solutions is crucial for a strong foundation in algebra and its applications.
1. Solving 4x² - 4ax + (a² - b²) = 0
Let's embark on solving the first quadratic equation: 4x² - 4ax + (a² - b²) = 0, where a and b are real numbers. This equation might initially seem complex, but with the right approach, it can be solved elegantly. The hint provided suggests rewriting the middle term to facilitate factoring. Let's break down the process step by step.
1.1 Rewriting the Equation
The crucial first step involves rewriting the middle term, -4ax, using the hint. We aim to express it as a sum of two terms that will allow us to factor by grouping. Following the hint, we rewrite -4ax as -2(a + b)x - 2(a - b)x. This transformation is key to unlocking the factorization.
The equation now becomes:
4x² - 2(a + b)x - 2(a - b)x + (a² - b²) = 0
This seemingly simple manipulation sets the stage for the subsequent factoring steps. The goal is to identify common factors within the terms, allowing us to group them and simplify the equation.
1.2 Factoring by Grouping
Now, we proceed with factoring by grouping. This technique involves pairing terms strategically to extract common factors. We group the first two terms and the last two terms:
[4x² - 2(a + b)x] + [- 2(a - b)x + (a² - b²)] = 0
From the first group, we can factor out 2x, and from the second group, we can factor out -(a - b):
2x[2x - (a + b)] - (a - b)[2x - (a + b)] = 0
Notice that we now have a common factor of [2x - (a + b)] in both terms. This is the critical step in factoring by grouping.
1.3 Extracting the Common Factor
Next, we extract the common factor [2x - (a + b)]:
[2x - (a + b)][2x - (a - b)] = 0
We have successfully factored the quadratic equation into two linear factors. This is a significant achievement, as it allows us to easily find the solutions.
1.4 Finding the Solutions
To find the solutions, we set each factor equal to zero:
2x - (a + b) = 0 or 2x - (a - b) = 0
Solving for x in each case, we get:
2x = a + b => x = (a + b) / 2
2x = a - b => x = (a - b) / 2
Thus, the solutions to the quadratic equation 4x² - 4ax + (a² - b²) = 0 are x = (a + b) / 2 and x = (a - b) / 2. These solutions are expressed in terms of the real numbers a and b. The nature of these solutions (rational, irrational, or integer) depends on the specific values of a and b.
2. Solving 5x² - 12x - 9 = 0
Now, let's turn our attention to the second quadratic equation: 5x² - 12x - 9 = 0. We will explore its solutions within different number sets, specifically irrational numbers (\mathbb{I}) and rational numbers (\mathbb{Q}). This exploration will highlight how the solution set changes based on the domain we consider.
2.1 Factoring the Quadratic Equation
The first step is to attempt to factor the quadratic equation. We are looking for two numbers that multiply to (5)(-9) = -45 and add up to -12. These numbers are -15 and 3. Using these numbers, we can rewrite the middle term and factor by grouping:
5x² - 15x + 3x - 9 = 0
Now, we group the terms and factor out common factors:
5x(x - 3) + 3(x - 3) = 0
Extracting the common factor (x - 3), we get:
(5x + 3)(x - 3) = 0
2.2 Finding the General Solutions
To find the solutions, we set each factor equal to zero:
5x + 3 = 0 or x - 3 = 0
Solving for x in each case, we get:
5x = -3 => x = -3/5
x = 3
Thus, the general solutions to the quadratic equation 5x² - 12x - 9 = 0 are x = -3/5 and x = 3.
2.3 Solutions in Irrational Numbers (x ∈ \mathbb{I})
Now, let's consider the solutions within the set of irrational numbers (\mathbb{I}). An irrational number is a real number that cannot be expressed as a simple fraction (a/b, where a and b are integers and b ≠ 0). Examples of irrational numbers include √2, π, and e.
Looking at our solutions, x = -3/5 and x = 3, we see that both are rational numbers. -3/5 is a fraction, and 3 can be expressed as 3/1. Therefore, there are no solutions to the equation 5x² - 12x - 9 = 0 when x ∈ \mathbb{I}. The solution set in this case is empty.
2.4 Solutions in Rational Numbers (x ∈ \mathbb{Q})
Next, we consider the solutions within the set of rational numbers (\mathbb{Q}). A rational number is any number that can be expressed as a fraction (a/b, where a and b are integers and b ≠ 0). This includes integers, terminating decimals, and repeating decimals.
As we found earlier, the solutions to the equation are x = -3/5 and x = 3. Both of these numbers are rational, as -3/5 is a fraction and 3 can be expressed as 3/1. Therefore, the solutions to the equation 5x² - 12x - 9 = 0 when x ∈ \mathbb{Q} are x = -3/5 and x = 3. The solution set in this case is {-3/5, 3}.
Conclusion
In this article, we have explored the solutions to two distinct quadratic equations. For the equation 4x² - 4ax + (a² - b²) = 0, we successfully found the solutions x = (a + b) / 2 and x = (a - b) / 2 by factoring by grouping. For the equation 5x² - 12x - 9 = 0, we found the general solutions x = -3/5 and x = 3. We then analyzed these solutions within the context of irrational and rational numbers, demonstrating that the solution set depends on the domain considered. When x ∈ \mathbb{I}, there were no solutions, while when x ∈ \mathbb{Q}, the solutions were x = -3/5 and x = 3. This comprehensive analysis highlights the importance of understanding the properties of different number sets when solving mathematical problems. Mastering quadratic equations is a fundamental step in mathematics, paving the way for more advanced concepts and applications.