Solving And Comparing Roots Of Quadratic Equations

This article delves into the process of solving quadratic equations and comparing their roots, focusing on two specific equations:

I). x² - 15√5 x + 180 = 0

II). y² + 3√3 y - 54 = 0

We will explore different methods for finding the roots of these equations and then compare them to determine the relationship between x and y. The solutions will be presented in a clear and concise manner, suitable for students and enthusiasts alike. Understanding the nuances of quadratic equations is fundamental in algebra, and this article aims to provide a comprehensive guide to tackling such problems. We will dissect each step, ensuring clarity and accuracy in our approach. This exploration not only helps in solving the given equations but also enhances the understanding of quadratic equations in general. Let's embark on this mathematical journey to unravel the intricacies of these equations and derive meaningful conclusions.

Understanding Quadratic Equations

Before diving into the solutions, it's crucial to understand the basics of quadratic equations. A quadratic equation is a polynomial equation of the second degree. The general form of a quadratic equation is ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0. The solutions to a quadratic equation are called roots, and they represent the values of the variable (in this case, x or y) that satisfy the equation. There are several methods to find the roots of a quadratic equation, including factoring, completing the square, and using the quadratic formula. Each method has its advantages and disadvantages, depending on the specific equation. Factoring is often the quickest method if the equation can be easily factored. Completing the square is a more general method that can be used for any quadratic equation. The quadratic formula is a direct formula that provides the roots regardless of the complexity of the equation. Understanding these methods is essential for solving various mathematical problems, and this article will demonstrate their application in detail. The nature of the roots can also be determined by the discriminant (b² - 4ac), which provides valuable insights into the solutions without actually solving the equation.

Solving Equation I: x² - 15√5 x + 180 = 0

To solve the first equation, x² - 15√5 x + 180 = 0, we can use the quadratic formula. The quadratic formula is given by: x = [-b ± √(b² - 4ac)] / 2a. In this case, a = 1, b = -15√5, and c = 180. Substituting these values into the formula, we get: x = [15√5 ± √((-15√5)² - 4 * 1 * 180)] / 2 * 1. Simplifying the expression under the square root: (-15√5)² = 225 * 5 = 1125, and 4 * 1 * 180 = 720. Thus, the discriminant is 1125 - 720 = 405. Now, x = [15√5 ± √405] / 2. We can simplify √405 as √(81 * 5) = 9√5. Therefore, x = [15√5 ± 9√5] / 2. The two possible values for x are: x₁ = (15√5 + 9√5) / 2 = 24√5 / 2 = 12√5 and x₂ = (15√5 - 9√5) / 2 = 6√5 / 2 = 3√5. These are the roots of the first quadratic equation. Understanding the steps involved in applying the quadratic formula is crucial for solving similar equations. The ability to simplify radicals and perform arithmetic operations with them is also essential for arriving at the correct solution. This detailed breakdown ensures a clear understanding of the process involved.

Solving Equation II: y² + 3√3 y - 54 = 0

Now, let's solve the second equation, y² + 3√3 y - 54 = 0. Again, we can use the quadratic formula: y = [-b ± √(b² - 4ac)] / 2a. In this equation, a = 1, b = 3√3, and c = -54. Substituting these values into the formula, we get: y = [-3√3 ± √((3√3)² - 4 * 1 * -54)] / 2 * 1. Simplifying the expression under the square root: (3√3)² = 9 * 3 = 27, and 4 * 1 * -54 = -216. Thus, the discriminant is 27 - (-216) = 27 + 216 = 243. Now, y = [-3√3 ± √243] / 2. We can simplify √243 as √(81 * 3) = 9√3. Therefore, y = [-3√3 ± 9√3] / 2. The two possible values for y are: y₁ = (-3√3 + 9√3) / 2 = 6√3 / 2 = 3√3 and y₂ = (-3√3 - 9√3) / 2 = -12√3 / 2 = -6√3. These are the roots of the second quadratic equation. Understanding how to handle negative values and simplify radicals is key to solving such equations accurately. This step-by-step approach provides a clear understanding of the solution process. The quadratic formula's versatility shines through as it handles both equations with ease, highlighting its importance in solving quadratic equations.

Comparing the Roots of the Equations

After finding the roots of both equations, we need to compare them to determine the relationship between x and y. The roots for equation I are x₁ = 12√5 and x₂ = 3√5. The roots for equation II are y₁ = 3√3 and y₂ = -6√3. To compare these values, it's helpful to approximate the square roots. √5 is approximately 2.236, and √3 is approximately 1.732. Thus, x₁ ≈ 12 * 2.236 ≈ 26.832, x₂ ≈ 3 * 2.236 ≈ 6.708, y₁ ≈ 3 * 1.732 ≈ 5.196, and y₂ ≈ -6 * 1.732 ≈ -10.392. Comparing x₁ and y₁: 26.832 > 5.196, so x₁ > y₁. Comparing x₁ and y₂: 26.832 > -10.392, so x₁ > y₂. Comparing x₂ and y₁: 6.708 > 5.196, so x₂ > y₁. Comparing x₂ and y₂: 6.708 > -10.392, so x₂ > y₂. From these comparisons, we can see that both roots of x are greater than both roots of y. Therefore, x > y. This careful comparison ensures a definitive conclusion about the relationship between the roots. The use of approximations helps in visualizing the magnitudes of the roots and making informed comparisons.

Conclusion: Determining the Relationship Between x and y

In conclusion, after solving the quadratic equations I). x² - 15√5 x + 180 = 0 and II). y² + 3√3 y - 54 = 0 and comparing their roots, we have determined that x > y. The roots of the first equation are x₁ = 12√5 and x₂ = 3√5, while the roots of the second equation are y₁ = 3√3 and y₂ = -6√3. By approximating the values of the square roots and comparing the corresponding roots, we definitively established that the values of x are greater than the values of y. This analysis demonstrates the importance of understanding quadratic equations and the methods for solving them, as well as the ability to compare roots effectively. The final answer, x > y, is a testament to the step-by-step approach and careful calculations performed throughout the solution process. This comprehensive exploration provides a solid understanding of how to solve quadratic equations and compare their roots, highlighting the significance of these mathematical skills in problem-solving.

Final Answer

A. x > y