Solving √(3x + 12) = 9 A Step-by-Step Guide With Extraneous Solutions

Introduction

In the realm of algebra, solving equations is a fundamental skill. This article will delve into the process of solving a radical equation, specifically $\sqrt{3x + 12} = 9$. We will meticulously walk through each step, ensuring clarity and understanding. Furthermore, we will explore the crucial concept of extraneous solutions – those seemingly valid answers that, upon verification, do not satisfy the original equation. By the end of this guide, you will be equipped with the knowledge and skills to confidently solve radical equations and identify any extraneous solutions that may arise. This comprehensive approach ensures a solid grasp of the topic, crucial for success in algebra and beyond. So, let's embark on this journey of algebraic exploration and unravel the mysteries of radical equations together.

Solving the Equation $\sqrt{3x + 12} = 9$

Our primary objective is to isolate the variable x and determine its value. The given equation, $\sqrt{3x + 12} = 9$, involves a square root, which necessitates a specific approach to eliminate it. The initial step in solving this radical equation involves squaring both sides. Squaring both sides of an equation is a fundamental algebraic manipulation that preserves equality while allowing us to eliminate the radical. When we square both sides of $\sqrt{3x + 12} = 9$, we obtain $(\sqrt{3x + 12})^2 = 9^2$. This simplifies to $3x + 12 = 81$. Now that we've eliminated the square root, the equation transforms into a linear equation, which is significantly easier to solve. The next step is to isolate the term containing x. We can achieve this by subtracting 12 from both sides of the equation. This operation maintains the balance of the equation and brings us closer to isolating x. Subtracting 12 from both sides of $3x + 12 = 81$ yields $3x + 12 - 12 = 81 - 12$, which simplifies to $3x = 69$. Finally, to isolate x, we divide both sides of the equation by 3. This is the last step in isolating the variable, and it will give us the potential solution for x. Dividing both sides of $3x = 69$ by 3 gives us $x = 23$. Therefore, our potential solution to the equation $\sqrt{3x + 12} = 9$ is $x = 23$. However, it's essential to remember that this is just a potential solution. We must verify this solution by substituting it back into the original equation to ensure it holds true. This verification process is crucial in identifying extraneous solutions, which we will discuss in the next section.

Checking for Extraneous Solutions

Now that we have found a potential solution, $x = 23$, it's crucial to verify whether this solution is valid or extraneous. Extraneous solutions are values that arise during the solving process but do not satisfy the original equation. This often happens when dealing with radical equations because the process of squaring both sides can introduce solutions that were not present in the original equation. To check if $x = 23$ is an extraneous solution, we substitute it back into the original equation, $\sqrt{3x + 12} = 9$. Substituting $x = 23$ into the equation gives us $\sqrt{3(23) + 12} = 9$. Now we simplify the expression under the square root. Multiplying 3 by 23 gives us 69, so the equation becomes $\sqrt{69 + 12} = 9$. Adding 69 and 12 gives us 81, so the equation further simplifies to $\sqrt{81} = 9$. The square root of 81 is 9, so we have $9 = 9$. This equality holds true, which means that $x = 23$ is indeed a valid solution and not an extraneous one. The importance of this verification step cannot be overstated. Without it, we might mistakenly accept an extraneous solution as a valid answer, leading to incorrect results. By substituting the potential solution back into the original equation and verifying that it satisfies the equation, we can confidently determine whether the solution is valid or extraneous. In this case, since $x = 23$ satisfies the original equation, we can conclude that it is a valid solution. Therefore, we have successfully solved the radical equation and confirmed that our solution is not extraneous.

Why Extraneous Solutions Occur

Understanding why extraneous solutions arise is crucial for mastering the process of solving radical equations. Extraneous solutions are, in essence, solutions that emerge during the algebraic manipulation process but do not satisfy the original equation. The primary reason for their occurrence lies in the nature of squaring both sides of an equation. Squaring both sides is a powerful technique for eliminating square roots, but it can also introduce new solutions that were not present in the original equation. To illustrate this, consider the simple equation $x = 2$. Squaring both sides gives us $x^2 = 4$. While $x = 2$ is a valid solution to this new equation, $x = -2$ is also a solution because $(-2)^2 = 4$. However, $x = -2$ is not a solution to the original equation, $x = 2$. This demonstrates how squaring both sides can introduce additional solutions. In the context of radical equations, this phenomenon is particularly relevant because the squaring operation can eliminate the radical, but it can also create a situation where the squared equation has solutions that do not satisfy the original radical equation. For example, if we have an equation like $\sqrt{x} = -3$, we know that the square root of a number cannot be negative. However, if we square both sides, we get $x = 9$, which seems like a solution. But if we substitute $x = 9$ back into the original equation, we get $\sqrt{9} = -3$, which is false. This is a classic example of an extraneous solution. Therefore, it is essential to always check potential solutions by substituting them back into the original equation to ensure they are valid and not extraneous. This step is crucial for accurately solving radical equations and avoiding errors.

Summary and Key Takeaways

In this comprehensive guide, we have meticulously walked through the process of solving the radical equation $\sqrt{3x + 12} = 9$. We began by squaring both sides of the equation to eliminate the square root, which transformed the equation into a linear form. We then isolated x through a series of algebraic manipulations, ultimately arriving at the potential solution $x = 23$. However, finding a potential solution is only part of the process. The crucial next step is to check for extraneous solutions. Extraneous solutions are values that satisfy the transformed equation but not the original radical equation. These solutions often arise due to the squaring operation, which can introduce new solutions that were not initially present. To verify our potential solution, we substituted $x = 23$ back into the original equation. Upon simplification, we found that $\sqrt{3(23) + 12} = 9$, which simplifies to $9 = 9$. This confirmed that $x = 23$ is a valid solution and not an extraneous one. The key takeaway from this exercise is the importance of verifying solutions in radical equations. Always substitute your potential solutions back into the original equation to ensure they are valid. This step is essential for avoiding errors and accurately solving radical equations. Understanding the concept of extraneous solutions and the reasons for their occurrence is also crucial for mastering this topic. By following these steps and guidelines, you can confidently solve radical equations and identify any extraneous solutions that may arise. This skill is fundamental in algebra and will serve you well in more advanced mathematical studies.

Practice Problems

To solidify your understanding of solving radical equations and identifying extraneous solutions, let's work through a few practice problems. These problems will allow you to apply the steps and concepts we've discussed and further develop your skills.

1. Solve for $x$: $\sqrt{2x - 1} = 5$
2. Solve for $x$: $\sqrt{x + 4} = x - 2$
3. Solve for $x$: $\sqrt{5x + 10} = 3\sqrt{x}$

For each problem, remember to follow these steps:

• Isolate the radical.
• Square both sides of the equation.
• Solve the resulting equation.
• Crucially, check for extraneous solutions by substituting your potential solutions back into the original equation.

Working through these practice problems will not only reinforce your understanding but also help you develop confidence in solving radical equations. As you solve these problems, pay close attention to the verification step. Identifying extraneous solutions is a critical part of the process, and consistent practice will make you more adept at recognizing them. Remember, the key to mastering any mathematical concept is practice, so take the time to work through these problems and solidify your understanding.

Conclusion

In conclusion, solving radical equations is a fundamental skill in algebra that requires a systematic approach and a keen eye for detail. We've explored the step-by-step process of solving such equations, emphasizing the critical importance of checking for extraneous solutions. The technique of squaring both sides, while effective in eliminating radicals, can inadvertently introduce solutions that do not satisfy the original equation. Therefore, the verification step is not merely a formality but an essential safeguard against accepting incorrect answers. By substituting potential solutions back into the original equation, we can confidently distinguish between valid solutions and extraneous ones. The ability to solve radical equations accurately is a valuable asset in mathematics and related fields. It demonstrates a solid understanding of algebraic principles and the importance of rigorous verification. As you continue your mathematical journey, remember the lessons learned here and apply them to new challenges. With practice and diligence, you will master the art of solving radical equations and confidently navigate the world of algebra.