Solving Quadratic Equations Step-by-Step Guide
In the realm of mathematics, quadratic equations hold a prominent position, appearing in various fields ranging from physics and engineering to economics and computer science. Understanding how to solve these equations is a fundamental skill for anyone pursuing studies or a career in these areas. This comprehensive guide delves into the intricacies of solving quadratic equations, offering step-by-step solutions and explanations to enhance your understanding. We will explore different methods for finding the roots of quadratic equations, including factoring, completing the square, and the quadratic formula. This guide aims to equip you with the knowledge and confidence to tackle any quadratic equation that comes your way. Mastering the art of solving quadratic equations not only strengthens your mathematical foundation but also enhances your problem-solving abilities, which are crucial in various aspects of life. Whether you are a student preparing for an exam or a professional seeking to refresh your knowledge, this guide provides a valuable resource for understanding and solving quadratic equations.
The significance of quadratic equations extends beyond the classroom, influencing real-world applications such as trajectory calculations, optimization problems, and curve fitting. For instance, engineers use quadratic equations to design bridges and structures, while economists apply them to model supply and demand curves. The ability to manipulate and solve quadratic equations is therefore a highly sought-after skill in many industries. By understanding the underlying principles and techniques discussed in this guide, you will be well-prepared to apply this knowledge to practical scenarios. Let's embark on this journey to unravel the mysteries of quadratic equations and equip ourselves with the tools to solve them effectively. This guide will not only provide you with step-by-step solutions but also offer insights into the logic behind each method, ensuring a deeper understanding of the subject matter. With practice and dedication, you can master the art of solving quadratic equations and unlock their potential in various mathematical and real-world contexts. So, let's begin our exploration of these fascinating equations and discover the methods to find their solutions.
When faced with a quadratic equation like 4√3x² + 5x - 2√3 = 0, the first step is to identify the coefficients a, b, and c. In this case, a = 4√3, b = 5, and c = -2√3. The standard form of a quadratic equation is ax² + bx + c = 0. Recognizing these coefficients is crucial for applying various methods to solve the equation, such as factoring, completing the square, or using the quadratic formula. Each method has its advantages, and the choice depends on the specific characteristics of the equation. Factoring is often the quickest method when the equation can be easily factored, while the quadratic formula provides a universal solution applicable to all quadratic equations. Understanding these methods and when to apply them is key to efficiently solving quadratic equations. This problem provides an excellent opportunity to demonstrate the application of these methods and highlight their strengths. By carefully examining the coefficients, we can determine the most efficient approach to find the solutions for x. Let's delve into the process of solving this equation, breaking down each step to ensure clarity and understanding. Remember, the goal is not just to find the answers but to grasp the underlying principles that make the solution possible.
Method 1: Factoring
To solve the quadratic equation 4√3x² + 5x - 2√3 = 0 by factoring, we look for two numbers that multiply to (4√3)(-2√3) = -24 and add up to 5. These numbers are 8 and -3. Rewriting the middle term, we get 4√3x² + 8x - 3x - 2√3 = 0. Next, we factor by grouping: 4x(√3x + 2) - √3(√3x + 2) = 0. This simplifies to (4x - √3)(√3x + 2) = 0. Setting each factor equal to zero gives us two possible solutions: 4x - √3 = 0 and √3x + 2 = 0. Solving for x in each case, we find x = √3/4 and x = -2/√3. These are the roots of the quadratic equation, representing the values of x that make the equation true. Factoring is a powerful technique for solving quadratic equations, but it requires the ability to identify the correct factors. This method is particularly efficient when the coefficients are integers and the factors are readily apparent. However, for more complex equations with irrational or large coefficients, other methods like the quadratic formula might be more suitable. The key to mastering factoring is practice and familiarity with different patterns and techniques.
Method 2: Quadratic Formula
The quadratic formula is a versatile tool for solving any quadratic equation of the form ax² + bx + c = 0. The formula is given by x = (-b ± √(b² - 4ac)) / (2a). In our case, a = 4√3, b = 5, and c = -2√3. Substituting these values into the formula, we get x = (-5 ± √(5² - 4(4√3)(-2√3))) / (2(4√3)). Simplifying the expression under the square root, we have 5² - 4(4√3)(-2√3) = 25 + 96 = 121. Thus, x = (-5 ± √121) / (8√3) = (-5 ± 11) / (8√3). This gives us two possible solutions: x = (6 / (8√3)) = (3 / (4√3)) = √3/4 and x = (-16 / (8√3)) = (-2 / √3). These solutions match the ones we found by factoring, confirming the accuracy of both methods. The quadratic formula is a reliable method that always provides the solutions, even when factoring is difficult or impossible. It is particularly useful for equations with complex or irrational roots. Understanding and applying the quadratic formula is an essential skill in algebra, allowing you to solve a wide range of quadratic equations with confidence. The formula might seem daunting at first, but with practice, it becomes a straightforward and powerful tool in your mathematical arsenal.
Solutions
Therefore, the solutions to the quadratic equation 4√3x² + 5x - 2√3 = 0 are x = √3/4 and x = -2/√3. We obtained these solutions using both the factoring method and the quadratic formula, demonstrating the consistency and reliability of these techniques. It is always a good practice to verify the solutions by substituting them back into the original equation to ensure they satisfy the equation. In this case, plugging in x = √3/4 and x = -2/√3 into 4√3x² + 5x - 2√3 = 0 will confirm that both values are indeed the roots of the equation. Understanding how to solve quadratic equations is a fundamental skill in algebra, and mastering different methods provides flexibility and efficiency in problem-solving. The ability to choose the most appropriate method for a given equation, whether it's factoring, using the quadratic formula, or completing the square, is a sign of strong mathematical proficiency. This problem has illustrated the application of two common methods, showcasing their strengths and how they lead to the same solutions. Remember, practice is key to mastering these techniques and building confidence in your ability to solve quadratic equations.
In this problem, we are tasked with finding the value of p for which one root of the quadratic equation px² - 14x + 8 = 0 is 6 times the other. Let's denote the roots as α and 6α. Understanding the relationship between the roots and the coefficients of a quadratic equation is crucial for solving this type of problem. According to Vieta's formulas, the sum of the roots (α + 6α) is equal to -(-14)/p = 14/p, and the product of the roots (α * 6α) is equal to 8/p. These relationships provide us with two equations involving α and p, which we can solve simultaneously to find the value of p. Vieta's formulas are a powerful tool for relating the roots and coefficients of polynomials, and they are particularly useful in problems involving the roots of quadratic equations. By applying these formulas, we can avoid the need to explicitly solve for the roots, which can be more complex and time-consuming. This problem highlights the elegance and efficiency of Vieta's formulas in solving problems related to the roots of quadratic equations. Let's proceed with solving the equations to find the value of p that satisfies the given condition. Remember, the key to success in mathematics is to understand the underlying principles and apply them strategically to solve problems.
Applying Vieta's Formulas
Using Vieta's formulas, we know that the sum of the roots α + 6α = 7α is equal to -(-14)/p = 14/p, and the product of the roots α * 6α = 6α² is equal to 8/p. This gives us two equations: 7α = 14/p and 6α² = 8/p. From the first equation, we can express α in terms of p: α = (14/p) / 7 = 2/p. Substituting this expression for α into the second equation, we get 6(2/p)² = 8/p. Simplifying this equation, we have 6(4/p²) = 8/p, which leads to 24/p² = 8/p. Multiplying both sides by p² and dividing by 8, we get 3 = p. Therefore, the value of p that satisfies the condition that one root is 6 times the other is p = 3. Vieta's formulas provide a powerful and elegant way to relate the roots and coefficients of a polynomial equation. By using these formulas, we can solve problems involving the roots without explicitly finding them. This approach is particularly useful in problems where the relationship between the roots is given, as in this case. The ability to apply Vieta's formulas is a valuable skill in algebra, and it can significantly simplify the process of solving certain types of problems. This problem demonstrates the effectiveness of Vieta's formulas in finding the value of a parameter that satisfies a given condition on the roots of a quadratic equation.
Verification
To verify our solution, we substitute p = 3 back into the quadratic equation px² - 14x + 8 = 0, which gives us 3x² - 14x + 8 = 0. We can solve this equation by factoring or using the quadratic formula. Factoring, we look for two numbers that multiply to 3 * 8 = 24 and add up to -14. These numbers are -12 and -2. Rewriting the middle term, we get 3x² - 12x - 2x + 8 = 0. Factoring by grouping, we have 3x(x - 4) - 2(x - 4) = 0, which simplifies to (3x - 2)(x - 4) = 0. Setting each factor equal to zero, we find x = 2/3 and x = 4. Now, we check if one root is 6 times the other. Indeed, 4 = 6 * (2/3), so our solution p = 3 is correct. This verification step is crucial to ensure the accuracy of our solution and to catch any potential errors in our calculations. By substituting the value of p back into the original equation and solving for the roots, we can confirm that the given condition is satisfied. This process reinforces our understanding of the problem and the solution, and it highlights the importance of careful and thorough problem-solving in mathematics. The ability to verify solutions is a valuable skill that builds confidence and accuracy in mathematical work.
Solution
Thus, the value of p for which one root of the quadratic equation px² - 14x + 8 = 0 is 6 times the other is p = 3. We arrived at this solution by applying Vieta's formulas to relate the sum and product of the roots to the coefficients of the quadratic equation. This approach allowed us to set up a system of equations and solve for p. We then verified our solution by substituting p = 3 back into the original equation and confirming that the roots satisfy the given condition. This problem illustrates the power of Vieta's formulas in solving problems related to the roots of quadratic equations, and it highlights the importance of verification in ensuring the accuracy of solutions. Understanding and applying Vieta's formulas is a valuable skill in algebra, and it can significantly simplify the process of solving certain types of problems. The combination of analytical problem-solving and verification ensures a solid understanding and mastery of the concepts involved.
In conclusion, solving quadratic equations is a fundamental skill in mathematics with applications across various fields. We have explored two problems that demonstrate different aspects of solving quadratic equations. The first problem involved finding the roots of a quadratic equation using both factoring and the quadratic formula. We showed how to identify the coefficients, apply the appropriate methods, and verify the solutions. The second problem focused on finding a parameter value that satisfies a given condition on the roots of a quadratic equation. We utilized Vieta's formulas to relate the roots and coefficients, set up a system of equations, and solve for the parameter. We also emphasized the importance of verifying the solution to ensure its accuracy. These problems highlight the versatility of quadratic equations and the different techniques available for solving them. Mastering these techniques requires practice and a deep understanding of the underlying principles. Quadratic equations are not just abstract mathematical concepts; they are powerful tools that can be applied to solve real-world problems in various disciplines. By developing a strong foundation in solving quadratic equations, you are equipping yourself with a valuable skill that will serve you well in your academic and professional pursuits. Remember, the key to success in mathematics is consistent practice and a willingness to explore different approaches to problem-solving.
