Introduction

Relevance of the Theme

The Basic Second Degree Equation is the backbone of elementary algebra and sets a key topic for the development of the ability to solve mathematical problems. It is a 'gateway' to more advanced mathematical concepts, as it is essential for solving infinite varieties of real-world problems. Understanding the elements that make up the equation and mastering solving strategies is fundamental for the progress of the study of mathematics.

Contextualization

In the 8th grade of Elementary School, the Basic Second Degree Equation is an extension of the study of first-degree equations, introducing the concept of the term of degree 2, or 'quadratic term,' which is crucial in the analysis of growth and decay phenomena in mathematics. This new perspective is a significant step in the evolution of students' understanding of the interactions between variables and the effects of manipulating these variables in an equation. In addition, the second-degree equation is also strongly connected to studies of geometry, especially when discussing the graphical representations of quadratic equations, the parabolas.

This topic is, therefore, a bridge that connects various mathematical concepts and paves the way for a deeper and more comprehensive understanding of mathematics. Mastering the Basic Second Degree Equation in the 8th grade also lays the groundwork for further studies in High School and in higher-level disciplines that use mathematical concepts, such as physics, engineering, economics, among others.

Now, let's dive into the world of second-degree equations!

Theoretical Development

Components

• Quadratic Term: In the second-degree equation, the quadratic term (x²) is the element that determines the non-linear growth or decay portion. This term is crucial because it differentiates the characteristics of parabolas, which are curves, from lines, which are first-degree pairs.

• Linear Term: The linear term (bx) represents the constant rate of change. When this term is zero, the equation reduces to an equation of the type x²=c, where the variable x is not multiplied by any coefficient, but is raised to the power of two. Understanding this component is essential to understand how the equation represents numerical relationships.

• Constant Term: The constant term (c) is the result of the equation. It is essential for solving mathematical problems, as it represents the final state or the initial situation, depending on the context.

• Equality ( = ): The equality sign is what unites all elements of the equation. Without it, the equation makes no sense, as it does not express an equality, but rather an inequality.

• Solve: The act of 'solving' a second-degree equation refers to finding the value (or values) of the variable x that satisfy the equality. Generally, the number of solutions is directly related to the discriminant term, a concept to be addressed later.

Key Terms

• Second Degree Equation: A second-degree (or quadratic) equation is an equality where the highest power of the unknown (variable) is 2. The general form of a second-degree equation is ax²+bx+c=0, where a, b, and c are constants, with a ≠ 0.

• Second Degree Inequality: A second-degree inequality is an equality where the highest power of the unknown (variable) is 2, but with the coefficients a, b, or c of signs exchanged or equal to zero, and with the inequality sign replacing the equality sign. It will not be the focus of our lesson, but it is important to differentiate it for a complete understanding of the topic.

• Roots of an Equation: The roots of a second-degree equation are the values of x that, when substituted into the equation, make it true. A second-degree equation has 0, 1, or 2 roots, depending on the value of the discriminant term.

Examples and Cases

• Example 1: Solving a basic second-degree equation: Suppose the equation x²-16=0. The solution to this equation is the square root of 16, which is 4. Therefore, x=±4 are the two possible solutions to this equation.

• Example 2: Understanding the roots of an equation: In the equation x²+2x+1=0, the coefficients are a=1, b=2, and c=1. To calculate the discriminant term (b²-4ac), we have: 2²-4(1)(1) = 4-4 = 0. Since the discriminant is zero, this equation has only one real root, which is -1.

• Case 1: Finding negative roots: Consider the equation 4x²-9=0. Again, we calculate the discriminant: b²-4ac = 0² - 4(4)(-9) = 144. Since the discriminant is positive, we have two real roots. We find the roots by calculating the square root of 144 and dividing by 2a: x=±√(144)/(2(4)) = ±3/2 = ±1.5. Therefore, the roots of the equation are 1.5 and -1.5, both negative.

• Case 2: Manipulating the equation to solve a problem: Suppose the perimeter of a square is 36 meters. If x represents the length of a side, we can express the perimeter as 4x. Now, using the information that the perimeter is 36, we have the second-degree equation 4x²-36=0. Solving this equation, we find that the roots are x=3 and x=-3. Although negative, x=-3 is a valid solution, as it represents the length of one side of the square.

Detailed Summary

Key Points

• Definition of a second-degree equation: It is an equality formed by a quadratic term, a linear term, and an independent term. The quadratic term is the one in which the unknown (variable) is squared, the linear term is the one in which the unknown appears only in a simple form (without power), and the independent term is the constant not associated with any variable.

• Discriminant Term: It is a mathematical formula used to determine the quantity and nature of the roots of a second-degree equation. It is calculated by subtracting the square of the linear term from the product of the quadratic term by the independent term.

• Nature of the Roots: Through the discriminant term, it can be determined whether the second-degree equation has two distinct real roots (if the discriminant is positive), one real root (if the discriminant is zero), or two complex roots (if the discriminant is negative).

• Solution of the second-degree equation: Through the application of the Bhaskara formula, it is possible to find the roots of a second-degree equation. The Bhaskara formula is x = (-b ± √(b² - 4ac)) / 2a, where a, b, and c are the coefficients of the equation.

Conclusions

• Relevance of the Quadratic Term: In the second-degree equation, the quadratic term is the basis that brings the curve characteristic to the equation. Its presence or absence completely alters the nature of the equation.

• Importance of Componentization of the Equation: Understanding the structure of the second-degree equation, that is, understanding how each term relates and contributes to the total expression, is fundamental for solving and interpreting problems.

• Connection with Real Context: The second-degree equation has applications in numerous real-life situations, from describing the motion of objects to analyzing costs and revenues in companies. Therefore, understanding this topic is essential to apply mathematics in everyday life.

Suggested Exercises

1. Exercise 1: Solve the equation x² - 5x + 6 = 0 and determine the nature of its roots.

2. Exercise 2: Write a second-degree equation that has roots -2 and 3.

3. Exercise 3: A parallelogram has a height of 5 cm. If the length of one base of the parallelogram is twice the other base, what is the length of each base? (Hint: Use the second-degree equation to solve this problem!)