Solving √(3x-1) = 2x-5 How Many Real Solutions Exist?
Introduction
In this article, we will explore how to determine the number of real solutions for the equation √(3x - 1) = 2x - 5. This type of problem often appears in algebra and precalculus courses and involves dealing with square roots and algebraic manipulations. To solve this, we will methodically walk through the necessary steps, including isolating the square root, squaring both sides, solving the resulting quadratic equation, and, most importantly, checking for extraneous solutions. Understanding how to handle such equations is crucial not only for academic success but also for various applications in engineering, physics, and other scientific fields. The solution process involves several critical algebraic techniques, and a clear grasp of these methods is essential for anyone studying mathematics or related disciplines.
Understanding the Equation
To find real solutions for the equation √(3x - 1) = 2x - 5, it's crucial to first understand the nature of the equation. We are dealing with a square root equation, which means we need to be mindful of the domain of the square root function. The expression inside the square root, 3x - 1, must be greater than or equal to zero. This is because the square root of a negative number is not a real number. Therefore, we have the condition 3x - 1 ≥ 0, which simplifies to x ≥ 1/3. This condition tells us that any potential solution must be greater than or equal to 1/3. The right-hand side of the equation, 2x - 5, is a linear expression, and its value will determine whether the square root can equal it. The interplay between the square root function and the linear function is what we need to carefully analyze to determine the real solutions. We will need to consider both algebraic manipulations and the domain restrictions imposed by the square root to arrive at the correct answer. This involves squaring both sides of the equation, which can introduce extraneous solutions, making it imperative to verify the solutions obtained. By understanding these initial constraints and the nature of the equation, we can approach the problem methodically and accurately.
Step-by-Step Solution
To determine the number of real solutions for the equation √(3x - 1) = 2x - 5, we will follow a step-by-step approach. This will ensure clarity and accuracy in our solution. First, we isolate the square root, which is already done in this equation. Next, we square both sides of the equation to eliminate the square root. This gives us (√(3x - 1))² = (2x - 5)², which simplifies to 3x - 1 = (2x - 5)². Now, we expand the right side: (2x - 5)² = (2x - 5)(2x - 5) = 4x² - 20x + 25. So, our equation becomes 3x - 1 = 4x² - 20x + 25. To solve for x, we rearrange the equation into a quadratic form by moving all terms to one side: 0 = 4x² - 20x + 25 - 3x + 1, which simplifies to 4x² - 23x + 26 = 0. Now we have a quadratic equation in the standard form ax² + bx + c = 0. We can solve this equation using the quadratic formula, factoring, or completing the square. In this case, factoring is a viable option. We need to find two numbers that multiply to 4 * 26 = 104 and add up to -23. These numbers are -8 and -13. We can rewrite the middle term using these numbers: 4x² - 8x - 13x + 26 = 0. Now we factor by grouping: 4x(x - 2) - 13(x - 2) = 0. This gives us (4x - 13)(x - 2) = 0. Setting each factor equal to zero gives us the potential solutions x = 13/4 and x = 2. However, it is critical to remember that squaring both sides of an equation can introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original. Therefore, we must check each potential solution in the original equation to determine if it is a valid solution.
Checking for Extraneous Solutions
When solving equations involving square roots, checking for extraneous solutions is a crucial step. Extraneous solutions are values that satisfy the transformed equation (after squaring) but do not satisfy the original equation. This occurs because squaring both sides can introduce roots that were not present in the initial equation. To check the potential solutions we found, x = 13/4 and x = 2, we will substitute each value back into the original equation, √(3x - 1) = 2x - 5. First, let's check x = 13/4: √(3(13/4) - 1) = 2(13/4) - 5. Simplifying the left side, we have √(39/4 - 4/4) = √(35/4) = √35 / 2. Simplifying the right side, we have 26/4 - 5 = 13/2 - 10/2 = 3/2. So, we have √35 / 2 = 3/2. Since √35 is approximately 5.9, √35 / 2 is approximately 2.95, which is not equal to 3/2 or 1.5. Therefore, x = 13/4 is not a solution. Next, let's check x = 2: √(3(2) - 1) = 2(2) - 5. Simplifying the left side, we have √(6 - 1) = √5. Simplifying the right side, we have 4 - 5 = -1. So, we have √5 = -1. Since the square root of a number cannot be negative, this is not true. Therefore, x = 2 is also not a solution. Since neither x = 13/4 nor x = 2 satisfies the original equation, both are extraneous solutions. This means the original equation has no real solutions. Checking for extraneous solutions ensures the accuracy of our final answer and is a critical component of solving radical equations.
Graphical Interpretation
To further illustrate why the equation √(3x - 1) = 2x - 5 has no real solutions, we can consider a graphical interpretation. The equation can be viewed as the intersection of two functions: y = √(3x - 1) and y = 2x - 5. The graph of y = √(3x - 1) is a square root function, which starts at x = 1/3 and increases as x increases. The graph of y = 2x - 5 is a linear function with a slope of 2 and a y-intercept of -5. To find the solutions of the equation, we are looking for the points where these two graphs intersect. If the graphs do not intersect, there are no real solutions. When we plot these two functions, we can see that the line y = 2x - 5 is below the square root function y = √(3x - 1) for all x values in the domain of the square root function (x ≥ 1/3). This means that the line never intersects the square root function. The linear function starts at a negative y-value (y = -4 when x = 1/2) and increases linearly, while the square root function starts at y = 0 when x = 1/3 and increases at a decreasing rate. Because the linear function increases at a constant rate and starts below the square root function, it will never catch up and intersect the square root function. This graphical representation provides a visual confirmation that the equation √(3x - 1) = 2x - 5 has no real solutions. The absence of intersection points between the two functions corresponds directly to the absence of real solutions to the equation. This method of using graphs to understand solutions can be applied to other types of equations as well, providing a powerful tool for problem-solving.
Conclusion
In conclusion, by systematically solving the equation √(3x - 1) = 2x - 5 and meticulously checking for extraneous solutions, we have determined that this equation has no real solutions. We began by squaring both sides of the equation to eliminate the square root, which led us to a quadratic equation. We then solved this quadratic equation and obtained two potential solutions. However, upon substituting these potential solutions back into the original equation, we found that neither of them satisfied the equation, indicating that they were both extraneous solutions. Additionally, we explored a graphical interpretation of the equation by considering the intersection points of the functions y = √(3x - 1) and y = 2x - 5. This visual representation confirmed our algebraic findings, as the graphs of the two functions do not intersect, reinforcing the conclusion that there are no real solutions. This problem highlights the importance of not only employing algebraic techniques but also understanding the nature of the functions involved and verifying solutions to avoid extraneous roots. The process of solving such equations involves a combination of algebraic manipulation, careful checking, and graphical intuition, all of which are essential skills in mathematics. Understanding these concepts and techniques is vital for anyone pursuing further studies in mathematics, science, or engineering. The methodical approach demonstrated here can be applied to a wide range of problems involving radicals and other types of equations.
Final Answer
The final answer is A. 0.
