Solving (x-b)/√(x²-b²) + 4 = (b-x)/√(x²-b²) Expressing X In Terms Of B
This article provides a detailed solution for the equation (x-b)/√(x²-b²) + 4 = (b-x)/√(x²-b²), where b is a positive constant. We will explore the steps to isolate x and express it in terms of b, ensuring we consider the domain restrictions imposed by the square root and denominator. Understanding such equations is crucial in various mathematical contexts, including calculus and algebra, and this guide aims to provide a clear, step-by-step approach to solving it.
1. Introduction to the Problem
The problem at hand involves solving an equation with a radical term in the denominator. The equation is given by:
(x-b)/√(x²-b²) + 4 = (b-x)/√(x²-b²)
where b is a positive constant. Our goal is to find the value(s) of x that satisfy this equation, expressing x in terms of b. This requires careful manipulation of the equation while considering the domain restrictions imposed by the square root and the denominator. Specifically, we need to ensure that x² - b² > 0, as the square root of a negative number is not real, and the denominator cannot be zero.
Understanding the Context
Before diving into the algebraic manipulations, it’s essential to understand the context of the problem. Equations like this often appear in various fields of mathematics and physics. They can model physical phenomena, represent geometric relationships, or arise in calculus problems involving optimization or curve analysis. The presence of a square root term introduces a non-linear element, making the equation more complex and potentially leading to multiple solutions or restrictions on the domain. In practical terms, this equation might represent a relationship between distances, velocities, or other physical quantities, where x and b are parameters that define the system. Therefore, solving this equation not only provides a mathematical solution but also offers insights into the underlying relationship between the variables.
Initial Considerations and Restrictions
The first step in solving any equation involving radicals is to identify any domain restrictions. In this case, the term √(x² - b²) in the denominator imposes a significant restriction: x² - b² must be greater than zero. This is because the square root of a negative number is not a real number, and division by zero is undefined. Thus, we have the inequality:
x² - b² > 0
This inequality can be factored as:
(x - b)(x + b) > 0
Analyzing the inequality, we find that it holds true when:
- x > b
- x < -b
This means that the solution for x must lie outside the interval [-b, b]. This restriction is crucial because any solution we find must satisfy this condition. If a solution falls within this interval, it must be discarded as an extraneous solution. The presence of this restriction highlights the importance of checking solutions against the original equation, especially when dealing with radical equations. This initial analysis sets the stage for the algebraic manipulations that follow, ensuring that we only consider valid solutions.
2. Solving the Equation Step-by-Step
To solve the equation (x-b)/√(x²-b²) + 4 = (b-x)/√(x²-b²), we will follow a step-by-step approach. Our goal is to isolate x on one side of the equation. This involves combining like terms, eliminating the radical in the denominator, and solving the resulting algebraic equation. Each step is carefully explained to ensure clarity and accuracy.
Step 1: Combine Terms with the Radical
The first step is to combine the terms that have the square root in the denominator. We can do this by subtracting (b-x)/√(x²-b²) from both sides of the equation:
(x-b)/√(x²-b²) - (b-x)/√(x²-b²) + 4 = 0
Notice that (b-x) is the negative of (x-b), so we can rewrite the second term as:
(x-b)/√(x²-b²) + (x-b)/√(x²-b²) + 4 = 0
Now, we can combine the like terms:
2(x-b)/√(x²-b²) + 4 = 0
This simplifies the equation and prepares it for further manipulation.
Step 2: Isolate the Radical Term
Next, we want to isolate the term with the square root. To do this, subtract 4 from both sides of the equation:
2(x-b)/√(x²-b²) = -4
Now, divide both sides by 2:
(x-b)/√(x²-b²) = -2
This step brings us closer to eliminating the radical and solving for x.
Step 3: Eliminate the Square Root
To eliminate the square root, we first multiply both sides by √(x²-b²):
x - b = -2√(x² - b²)
Now, square both sides of the equation to get rid of the square root:
(x - b)² = (-2√(x² - b²))²
Expanding both sides, we get:
x² - 2bx + b² = 4(x² - b²)
This step is crucial as it transforms the equation into a quadratic equation, which we can solve using standard techniques.
Step 4: Simplify and Rearrange the Equation
Now, we simplify and rearrange the equation to get it into standard quadratic form. Distribute the 4 on the right side:
x² - 2bx + b² = 4x² - 4b²
Move all terms to one side to set the equation to zero:
0 = 3x² + 2bx - 5b²
This is a quadratic equation in the form ax² + bx + c = 0, where a = 3, b = 2b, and c = -5b². The next step is to solve this quadratic equation for x.
3. Solving the Quadratic Equation
Now that we have the quadratic equation 3x² + 2bx - 5b² = 0, we need to find the values of x that satisfy it. 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 the Quadratic Equation
We look for two binomials that multiply to give the quadratic expression. The equation is:
3x² + 2bx - 5b² = 0
We need to find two numbers that multiply to -15 (3 * -5) and add to 2. These numbers are 5 and -3. So, we can rewrite the middle term as 5bx - 3bx:
3x² - 3bx + 5bx - 5b² = 0
Now, we factor by grouping:
3x(x - b) + 5b(x - b) = 0
Factor out the common term (x - b):
(3x + 5b)(x - b) = 0
This gives us two possible solutions for x:
- 3x + 5b = 0
- x - b = 0
Finding the Potential Solutions
Solving the first equation for x:
3x = -5b
x = -5b/3
Solving the second equation for x:
x = b
So, we have two potential solutions: x = -5b/3 and x = b. The next step is to check these solutions against the domain restrictions we identified earlier.
4. Checking for Extraneous Solutions
We found two potential solutions for x: x = -5b/3 and x = b. However, we need to check these solutions against the domain restrictions we identified earlier: x < -b or x > b. This is crucial because squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original equation.
Checking x = -5b/3
Since b is a positive constant, -5b/3 is a negative value. We need to check if -5b/3 < -b:
-5b/3 < -b
Multiply both sides by 3 (and remember to flip the inequality sign since we're multiplying by a negative number):
-5b < -3b
Add 5b to both sides:
0 < 2b
Since b is positive, this inequality holds true. Therefore, x = -5b/3 is a valid solution because it satisfies the domain restriction x < -b.
Checking x = b
Now, we check if x = b satisfies the domain restrictions. Recall that our restriction is x < -b or x > b. Clearly, x = b does not satisfy either of these conditions, as it falls exactly on the boundary x = b. Furthermore, if we substitute x = b into the original equation, we get:
(b - b)/√(b² - b²) + 4 = (b - b)/√(b² - b²)
This simplifies to:
0/0 + 4 = 0/0
which is undefined due to division by zero. Thus, x = b is an extraneous solution and must be discarded.
Final Solution
After checking for extraneous solutions, we find that only one solution is valid: x = -5b/3. This solution satisfies the original equation and the domain restrictions imposed by the square root and denominator. Therefore, the final solution to the equation is:
x = -5b/3
5. Conclusion: Expressing x in Terms of b
In this article, we successfully solved the equation (x-b)/√(x²-b²) + 4 = (b-x)/√(x²-b²) for x in terms of b. We began by identifying the domain restrictions imposed by the square root and denominator, which ensured that we only considered valid solutions. The step-by-step algebraic manipulations led us to a quadratic equation, which we solved by factoring. Finally, we checked for extraneous solutions and determined that the only valid solution is:
x = -5b/3
This solution expresses x directly in terms of b, satisfying the problem's requirement. The process of solving this equation highlights the importance of considering domain restrictions, carefully manipulating algebraic expressions, and checking for extraneous solutions. Understanding these steps is crucial for solving more complex equations and applying mathematical concepts in various fields. This exercise provides a solid foundation for tackling similar problems in algebra, calculus, and beyond.
The result demonstrates a comprehensive approach to solving equations with radicals, emphasizing the necessity of rigorous checks and a thorough understanding of mathematical principles. The ability to solve such equations is a valuable skill in various scientific and engineering disciplines, where mathematical models often involve complex relationships between variables.
