Solving The Equation √1-3x = X + 3 A Step-by-Step Guide
Introduction
In this article, we will delve into the step-by-step solution of the equation √1-3x = x + 3. This equation is a classic example of a radical equation, where the variable x is present under a square root. Solving radical equations requires careful attention to detail, as we need to eliminate the radical and solve the resulting algebraic equation. However, it is crucial to check for extraneous solutions, which are solutions obtained during the solving process but do not satisfy the original equation. This guide will not only provide the solution but also explain the underlying concepts and potential pitfalls, ensuring a thorough understanding of the process. By understanding the methodology involved in solving radical equations, you will enhance your problem-solving abilities in algebra and gain a deeper appreciation for mathematical precision.
Understanding Radical Equations
Before we dive into the solution, let's first understand what radical equations are and the basic principles involved in solving them. A radical equation is an equation in which a variable is under a radical, most commonly a square root. The key strategy for solving these equations is to isolate the radical and then eliminate it by raising both sides of the equation to the appropriate power. For a square root, this means squaring both sides. However, this process can sometimes introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original radical equation. These extraneous solutions arise because squaring both sides of an equation can turn a false statement into a true one. For instance, if we have -2 = 2 (which is false) and square both sides, we get 4 = 4 (which is true). Thus, it is essential to verify all potential solutions by substituting them back into the original equation to ensure they are valid. This step is not merely a formality; it is a critical part of the solution process that ensures accuracy. The concepts of isolating the radical, eliminating it by squaring, and checking for extraneous solutions are fundamental in dealing with radical equations. Applying these principles methodically will lead to accurate and reliable solutions, thereby solidifying your understanding of algebraic techniques.
Step-by-Step Solution of √1-3x = x + 3
Our primary goal is to solve the equation √1-3x = x + 3. The first step in solving this radical equation is to isolate the radical term, which in this case is √1-3x. Fortunately, the radical term is already isolated on the left side of the equation, making our task a bit easier. Now that the radical is isolated, we need to eliminate it to solve for x. To do this, we square both sides of the equation. Squaring both sides will remove the square root on the left side, allowing us to work with a more familiar algebraic form. Remember, what we do to one side of the equation, we must also do to the other side to maintain equality. So, we square both √1-3x and (x + 3). Squaring √1-3x gives us 1-3x. Squaring (x + 3) requires a bit more attention, as we need to use the binomial expansion formula (a + b)² = a² + 2ab + b². Applying this formula, we get (x + 3)² = x² + 2(x)(3) + 3² = x² + 6x + 9. So, our equation now becomes 1-3x = x² + 6x + 9. This is a quadratic equation, which we can solve by rearranging it into standard form and then factoring, using the quadratic formula, or completing the square. The next step is to bring all terms to one side to set the equation to zero. This allows us to solve the quadratic equation more easily. By systematically following these steps, we transform the radical equation into a manageable quadratic equation that we can solve using standard algebraic techniques. Understanding each step and the reasoning behind it is crucial for mastering the process of solving radical equations.
Transforming to a Quadratic Equation
After squaring both sides of the original equation √1-3x = x + 3, we arrived at the equation 1-3x = x² + 6x + 9. To transform this into a standard quadratic equation, we need to set one side of the equation to zero. This is achieved by moving all terms to one side. In this case, it’s convenient to move the terms on the left side to the right side to keep the x² term positive, which simplifies the factoring process, if applicable, or the use of the quadratic formula. So, we add 3x to both sides and subtract 1 from both sides. This gives us: 0 = x² + 6x + 3x + 9 - 1. Simplifying this, we get: 0 = x² + 9x + 8. Now we have a quadratic equation in the standard form ax² + bx + c = 0, where a = 1, b = 9, and c = 8. The standard form of a quadratic equation is essential because it allows us to apply various solution methods such as factoring, completing the square, or using the quadratic formula. In this case, the quadratic equation x² + 9x + 8 = 0 can be solved relatively easily by factoring. Recognizing the structure of a quadratic equation and knowing how to manipulate it into standard form is a fundamental skill in algebra. This skill not only helps in solving radical equations but also in various other mathematical contexts. The ability to transform equations into recognizable forms is a key aspect of mathematical problem-solving, enhancing both efficiency and accuracy.
Solving the Quadratic Equation
Now that we have the quadratic equation x² + 9x + 8 = 0, we need to solve for x. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. In this case, factoring is the most straightforward approach. Factoring involves finding two numbers that multiply to give the constant term (8) and add up to the coefficient of the linear term (9). These numbers are 8 and 1, since 8 * 1 = 8 and 8 + 1 = 9. Therefore, we can rewrite the quadratic equation as: (x + 8)(x + 1) = 0. To find the solutions for x, we set each factor equal to zero: x + 8 = 0 or x + 1 = 0. Solving these linear equations gives us two potential solutions: x = -8 and x = -1. These values are the roots of the quadratic equation and represent the x-values where the parabola intersects the x-axis. While factoring is a powerful method for solving quadratic equations, it's not always feasible, especially when the roots are not integers or the equation is more complex. In such cases, the quadratic formula is a reliable alternative. However, when factoring is possible, it is often quicker and less prone to errors. Understanding and being proficient in various methods for solving quadratic equations is crucial for mathematical problem-solving, providing flexibility and efficiency in tackling different types of problems. Recognizing when factoring is an efficient approach and applying it correctly can significantly streamline the solution process.
Checking for Extraneous Solutions
After solving the quadratic equation, we obtained two potential solutions: x = -8 and x = -1. However, because we squared both sides of the original radical equation, it is crucial to check for extraneous solutions. Extraneous solutions are values that satisfy the transformed equation (in this case, the quadratic equation) but do not satisfy the original equation. These solutions arise because squaring both sides can introduce solutions that were not present in the original equation. To check for extraneous solutions, we substitute each potential solution back into the original equation, √1-3x = x + 3, and see if the equation holds true. First, let's check x = -8: √1 - 3(-8) = -8 + 3. √1 + 24 = -5. √25 = -5. 5 = -5. This is not true, so x = -8 is an extraneous solution. Now, let's check x = -1: √1 - 3(-1) = -1 + 3. √1 + 3 = 2. √4 = 2. 2 = 2. This is true, so x = -1 is a valid solution. By performing this check, we confirm that x = -8 does not satisfy the original equation and is therefore an extraneous solution. This process of verifying solutions is a critical step in solving radical equations. It ensures that we only accept valid solutions and discard any that were introduced by the algebraic manipulation. Neglecting to check for extraneous solutions can lead to incorrect answers, so it is an essential part of the problem-solving strategy.
Final Solution
After solving the equation √1-3x = x + 3, we found two potential solutions: x = -8 and x = -1. However, after rigorously checking for extraneous solutions by substituting these values back into the original equation, we determined that x = -8 is an extraneous solution and does not satisfy the original equation. On the other hand, x = -1 does satisfy the original equation. Therefore, the final solution to the equation √1-3x = x + 3 is x = -1. This comprehensive solution process highlights the importance of not only solving the equation but also verifying the solutions to ensure they are valid. Extraneous solutions can often arise in radical equations due to the process of squaring both sides, which can introduce solutions that do not actually fit the original equation. By meticulously checking each potential solution, we can confidently arrive at the correct answer. This careful approach is crucial in mathematics, where precision and accuracy are paramount. The process of solving radical equations reinforces the understanding of algebraic manipulations and the necessity of solution verification, which are valuable skills applicable in various mathematical contexts.
Conclusion
In conclusion, we have successfully solved the radical equation √1-3x = x + 3. The step-by-step process involved isolating the radical, squaring both sides, transforming the resulting equation into a quadratic equation, solving the quadratic equation, and, most importantly, checking for extraneous solutions. This thorough approach led us to the final and correct solution, which is x = -1. Solving radical equations requires a blend of algebraic skills and a keen eye for detail. The potential for extraneous solutions necessitates a rigorous verification process to ensure accuracy. This exercise demonstrates the importance of understanding not only the mechanics of solving equations but also the underlying principles that govern the validity of solutions. Mastering these techniques is essential for success in algebra and beyond. The ability to solve radical equations effectively builds confidence and competence in mathematical problem-solving, preparing students for more advanced topics and real-world applications. By understanding the nuances of radical equations, one can develop a deeper appreciation for the elegance and precision of mathematics.
