Solving $18r^3 + 33r^2 - 30r = 0$ A Step-by-Step Guide

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Introduction: Mastering Polynomial Equations

In the realm of mathematics, solving polynomial equations is a fundamental skill that underpins many advanced concepts. Polynomial equations, characterized by terms involving variables raised to non-negative integer powers, appear frequently in various scientific and engineering applications. This article delves into the process of solving the specific cubic equation 18r3+33r2βˆ’30r=018r^3 + 33r^2 - 30r = 0. We will explore the techniques involved in finding the roots of this equation and verifying the solutions to ensure accuracy. Understanding these methods provides a strong foundation for tackling more complex algebraic problems. Let's embark on this journey to enhance our mathematical prowess and conquer the challenges posed by polynomial equations. The ability to solve such equations efficiently and accurately is crucial for students, engineers, and professionals alike. This comprehensive guide aims to equip you with the necessary tools and knowledge to approach and solve similar polynomial equations with confidence. By breaking down the process into manageable steps, we aim to demystify the complexities often associated with higher-degree equations. So, whether you are a student grappling with algebra or a professional needing a refresher, this article provides a valuable resource for mastering polynomial equations. The equation 18r3+33r2βˆ’30r=018r^3 + 33r^2 - 30r = 0 serves as an excellent example to illustrate the typical steps involved in solving cubic equations. We will start by identifying common factors, simplifying the equation, and then applying appropriate methods to find the roots. Each step will be explained in detail to ensure a clear understanding of the underlying principles. By the end of this guide, you will not only know how to solve this particular equation but also gain a broader perspective on solving other polynomial equations. Remember, practice is key to mastering any mathematical skill. Therefore, working through examples like this one will significantly enhance your problem-solving abilities and build your confidence in tackling algebraic challenges.

Step-by-Step Solution: Unraveling 18r3+33r2βˆ’30r=018r^3 + 33r^2 - 30r = 0

1. Identifying the Common Factor: The First Key Step

The first step in solving the equation 18r3+33r2βˆ’30r=018r^3 + 33r^2 - 30r = 0 is to identify the greatest common factor (GCF) of all the terms. This simplifies the equation, making it easier to handle. In this case, we can observe that each term is divisible by both 3 and r. Thus, the GCF is 3r. Factoring out 3r from the equation, we get:

3r(6r2+11rβˆ’10)=03r(6r^2 + 11r - 10) = 0

This step is crucial because it reduces the degree of the polynomial within the parentheses, transforming the cubic equation into a simpler quadratic equation. By factoring out the common term, we have effectively separated one root of the equation, which is r = 0. This is because if 3r equals zero, then the entire expression equals zero. The remaining quadratic equation, 6r2+11rβˆ’10=06r^2 + 11r - 10 = 0, can be solved using various methods, such as factoring, completing the square, or applying the quadratic formula. Factoring out the greatest common factor is a fundamental technique in solving polynomial equations and should always be the first step in your problem-solving process. It not only simplifies the equation but also helps identify one or more roots directly. The identification of the GCF sets the stage for further analysis and ultimately leads to a complete solution. By reducing the complexity of the equation, we pave the way for applying more specific techniques to find the remaining roots. This systematic approach is a cornerstone of algebraic problem-solving and should be diligently practiced to achieve mastery. So, always remember to look for common factors first when faced with a polynomial equation.

2. Solving the Quadratic Equation: Tackling 6r2+11rβˆ’10=06r^2 + 11r - 10 = 0

Now we need to solve the quadratic equation 6r2+11rβˆ’10=06r^2 + 11r - 10 = 0. There are several methods to accomplish this, but factoring is often the most efficient if it's possible. We are looking for two binomials that, when multiplied, give us the quadratic expression. This involves finding two numbers that multiply to the product of the leading coefficient (6) and the constant term (-10), which is -60, and add up to the middle coefficient (11). The numbers 15 and -4 satisfy these conditions because 15 * -4 = -60 and 15 + (-4) = 11. Using these numbers, we can rewrite the middle term and factor by grouping:

6r2+15rβˆ’4rβˆ’10=06r^2 + 15r - 4r - 10 = 0

Now, we factor by grouping:

3r(2r+5)βˆ’2(2r+5)=03r(2r + 5) - 2(2r + 5) = 0

(3rβˆ’2)(2r+5)=0(3r - 2)(2r + 5) = 0

Setting each factor equal to zero, we get:

3rβˆ’2=03r - 2 = 0 or 2r+5=02r + 5 = 0

Solving for r in each equation gives us:

r=23r = \frac{2}{3} or r=βˆ’52r = -\frac{5}{2}

This method of factoring the quadratic equation allows us to find the roots without resorting to more complex techniques. However, if factoring proves difficult, the quadratic formula is always a reliable alternative. The quadratic formula is a universal method for solving quadratic equations of the form axΒ² + bx + c = 0 and is given by:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b Β± \sqrt{b^2 - 4ac}}{2a}

In our case, a = 6, b = 11, and c = -10. Plugging these values into the quadratic formula will yield the same roots, r = 2/3 and r = -5/2. Understanding both factoring and the quadratic formula equips you with a versatile toolkit for solving quadratic equations. The choice of method often depends on the specific equation and your personal preference. However, practicing both techniques ensures you are well-prepared for any quadratic equation you encounter.

3. Complete Solution Set: All Roots Revealed

Combining the root we found earlier by factoring out 3r, which is r = 0, with the two roots we found from the quadratic equation, r = 2/3 and r = -5/2, we obtain the complete solution set for the equation 18r3+33r2βˆ’30r=018r^3 + 33r^2 - 30r = 0. Therefore, the roots of the equation are:

r=0,r=23,r=βˆ’52r = 0, r = \frac{2}{3}, r = -\frac{5}{2}

This completes the process of solving the cubic equation. By systematically factoring out the common factor and then solving the resulting quadratic equation, we were able to find all three roots. A cubic equation, in general, will have up to three roots, which may be real or complex. In this case, all three roots are real numbers. The complete solution set provides a comprehensive understanding of the behavior of the equation. Each root represents a point where the polynomial function intersects the x-axis, also known as the zeros of the function. Knowing all the roots is essential for various applications, such as graphing the polynomial function, analyzing its properties, and solving related problems. The ability to find the complete solution set is a testament to a solid understanding of algebraic techniques and problem-solving strategies. It is a crucial skill for anyone working with polynomial equations and related mathematical concepts. Therefore, mastering the process of finding all roots is an important step in your mathematical journey.

Verification: Ensuring Solution Accuracy

1. Substituting the Roots: The Crucial Check

To ensure the accuracy of our solutions, it is essential to verify each root by substituting it back into the original equation. This process confirms that the values we found indeed satisfy the equation 18r3+33r2βˆ’30r=018r^3 + 33r^2 - 30r = 0. Let's start by verifying the root r = 0:

18(0)3+33(0)2βˆ’30(0)=0+0βˆ’0=018(0)^3 + 33(0)^2 - 30(0) = 0 + 0 - 0 = 0

This confirms that r = 0 is a valid solution. Next, let's verify r = 2/3:

18(23)3+33(23)2βˆ’30(23)=18(827)+33(49)βˆ’30(23)=163+443βˆ’20=603βˆ’20=20βˆ’20=018(\frac{2}{3})^3 + 33(\frac{2}{3})^2 - 30(\frac{2}{3}) = 18(\frac{8}{27}) + 33(\frac{4}{9}) - 30(\frac{2}{3}) = \frac{16}{3} + \frac{44}{3} - 20 = \frac{60}{3} - 20 = 20 - 20 = 0

This also confirms that r = 2/3 is a valid solution. Finally, let's verify r = -5/2:

18(βˆ’52)3+33(βˆ’52)2βˆ’30(βˆ’52)=18(βˆ’1258)+33(254)βˆ’30(βˆ’52)=βˆ’11254+8254+75=βˆ’3004+75=βˆ’75+75=018(-\frac{5}{2})^3 + 33(-\frac{5}{2})^2 - 30(-\frac{5}{2}) = 18(-\frac{125}{8}) + 33(\frac{25}{4}) - 30(-\frac{5}{2}) = -\frac{1125}{4} + \frac{825}{4} + 75 = -\frac{300}{4} + 75 = -75 + 75 = 0

Thus, r = -5/2 is also a valid solution. This step of verification is critical in mathematical problem-solving. It not only ensures that the solutions are correct but also helps in identifying any potential errors made during the solution process. Substituting the roots back into the original equation is a foolproof way to confirm their validity. This practice reinforces the understanding of the equation and the properties of its roots. The process of verification is a cornerstone of mathematical rigor and should be an integral part of your problem-solving strategy. It provides confidence in the accuracy of the solutions and enhances your overall understanding of the mathematical concepts involved. So, always remember to verify your solutions to ensure correctness and deepen your understanding.

2. Importance of Verification: The Final Seal of Approval

The importance of verifying solutions in mathematical problems cannot be overstated. It serves as a critical step in ensuring the accuracy and correctness of the results obtained. In the context of solving polynomial equations, verification involves substituting the calculated roots back into the original equation to confirm that they satisfy it. This process acts as a safeguard against errors that may have occurred during the solution process, such as mistakes in factoring, arithmetic errors, or incorrect application of formulas. By verifying the solutions, we gain confidence in their validity and eliminate the possibility of presenting incorrect answers. Furthermore, verification enhances our understanding of the relationship between the roots and the equation. It reinforces the concept that a root of an equation is a value that, when substituted for the variable, makes the equation true. This understanding is fundamental to mastering algebraic problem-solving and building a strong foundation in mathematics. The process of verification also promotes critical thinking and attention to detail. It requires careful substitution, accurate calculations, and a thorough check of the results. This level of rigor is essential for success in mathematics and other quantitative disciplines. In addition to its role in ensuring accuracy, verification can also serve as a learning opportunity. If a solution does not verify, it indicates that an error has been made somewhere in the solution process. Identifying and correcting these errors can deepen our understanding of the underlying concepts and improve our problem-solving skills. Therefore, verification should not be viewed as a mere formality but rather as an integral part of the problem-solving process. It is the final seal of approval that confirms the correctness of our solutions and solidifies our understanding of the mathematical principles involved. Always prioritize verification to ensure the accuracy and reliability of your mathematical work.

Conclusion: The Power of Problem-Solving

In conclusion, we have successfully solved the cubic equation 18r3+33r2βˆ’30r=018r^3 + 33r^2 - 30r = 0 by systematically factoring out the greatest common factor and then solving the resulting quadratic equation. We found the roots to be r = 0, r = 2/3, and r = -5/2. Furthermore, we verified each solution by substituting it back into the original equation, ensuring the accuracy of our results. This process demonstrates the power of algebraic techniques in solving complex equations. The ability to solve polynomial equations is a crucial skill in mathematics and has wide-ranging applications in various fields, including science, engineering, and economics. Mastering these techniques requires a combination of understanding the underlying concepts, practicing problem-solving strategies, and paying attention to detail. The steps involved in solving this particular equationβ€”identifying common factors, solving quadratic equations, and verifying solutionsβ€”are fundamental to solving a wide range of mathematical problems. By understanding and applying these principles, you can tackle more complex equations and develop a deeper appreciation for the beauty and power of mathematics. The journey of solving mathematical problems is not just about finding the right answers; it is also about developing critical thinking skills, building confidence in your abilities, and fostering a love for learning. Each problem solved is a step forward in your mathematical journey, and the skills you acquire along the way will serve you well in your academic and professional pursuits. So, embrace the challenges, persevere through difficulties, and celebrate your successes. The world of mathematics is vast and fascinating, and the power of problem-solving is the key to unlocking its secrets.