Solving √(v+7) - 4 = 3 Step-by-Step Guide
Introduction
In this article, we will embark on a step-by-step journey to solve for the variable v in the given equation: √(v+7) - 4 = 3. This equation involves a square root, which adds a layer of complexity, but with a systematic approach, we can unravel the solution. Our goal is to isolate v and determine its value(s) that satisfy the equation. We will also delve into the importance of checking our solution(s) to ensure they are valid within the context of the original equation. Equations like this are fundamental in various fields of mathematics and physics, making understanding the solution process crucial.
Understanding the Equation
Before we dive into the solution, let's break down the equation. The equation √(v+7) - 4 = 3 features a square root term, √(v+7), which means we are looking for a value of v such that when 7 is added to it, the square root of the result, diminished by 4, equals 3. The presence of the square root necessitates a careful approach, as we need to consider the domain of the square root function. The expression inside the square root, v+7, must be greater than or equal to 0, as we are dealing with real numbers. This constraint will play a crucial role in verifying the validity of our solution.
Step-by-Step Solution
To solve for v, we will follow a series of algebraic manipulations to isolate v on one side of the equation. First, we will isolate the square root term by adding 4 to both sides of the equation. This step simplifies the equation and brings us closer to isolating the variable. Next, to eliminate the square root, we will square both sides of the equation. Squaring both sides is a valid operation, but it is essential to remember that it can sometimes introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original equation. Therefore, we must check our solution(s) at the end. After squaring, we will have a simple linear equation in v, which we can solve by further algebraic manipulation. Finally, we will check our solution(s) by substituting them back into the original equation to ensure they are valid.
Solving the Equation √(v+7) - 4 = 3
Step 1: Isolate the Square Root Term
The first step in solving for v is to isolate the square root term. To achieve this, we add 4 to both sides of the equation:
√(v+7) - 4 + 4 = 3 + 4
This simplifies to:
√(v+7) = 7
Now, we have the square root term isolated on one side of the equation, which sets the stage for the next step.
Step 2: Eliminate the Square Root
To eliminate the square root, we square both sides of the equation:
(√(v+7))^2 = 7^2
This gives us:
v+7 = 49
By squaring both sides, we have transformed the equation into a simpler linear equation.
Step 3: Solve for v
Now, we solve for v by subtracting 7 from both sides of the equation:
v+7 - 7 = 49 - 7
This results in:
v = 42
So, we have found a potential solution for v. However, we need to verify if this solution is valid.
Step 4: Check the Solution
It is crucial to check the solution by substituting it back into the original equation to ensure it is not an extraneous solution. Substituting v = 42 into the original equation, we get:
√(42+7) - 4 = 3
Simplifying the expression inside the square root:
√(49) - 4 = 3
The square root of 49 is 7:
7 - 4 = 3
This simplifies to:
3 = 3
Since the equation holds true, v = 42 is a valid solution.
Final Answer
Therefore, the solution to the equation √(v+7) - 4 = 3 is:
v = 42
Importance of Checking Solutions
The importance of checking solutions, especially when dealing with radical equations, cannot be overstated. Squaring both sides of an equation, while a valid algebraic manipulation, can introduce extraneous solutions. These are solutions that emerge during the solving process but do not satisfy the original equation. In the context of our problem, if we had not checked our solution, we might have incorrectly concluded that v = 42 is the correct answer. Checking the solution ensures that we only accept values that truly satisfy the original equation and maintain the equality.
Extraneous Solutions Explained
Extraneous solutions arise because the squaring operation can mask the sign of a number. For example, both 3 and -3, when squared, result in 9. Therefore, when we square both sides of an equation, we are essentially saying that if a² = b², then a = b or a = -b. This introduces the possibility of solutions that satisfy the squared equation but not the original equation. To avoid accepting extraneous solutions, it is a mandatory practice to substitute the obtained values back into the original equation and verify their validity.
Real-World Implications
The concept of extraneous solutions is not just a mathematical curiosity; it has real-world implications. In various scientific and engineering applications, equations involving square roots or other radicals are used to model physical phenomena. For instance, in physics, equations describing projectile motion or wave behavior often involve square roots. In such scenarios, extraneous solutions can lead to incorrect predictions or designs. Therefore, understanding and addressing the possibility of extraneous solutions is crucial for accurate modeling and problem-solving.
Domain Considerations
When dealing with equations involving square roots, it is essential to consider the domain of the square root function. The domain is the set of all possible input values for which the function is defined. For the square root function, the expression inside the square root must be greater than or equal to zero. In our equation, √(v+7), the expression inside the square root is v+7. Therefore, we must have:
v+7 ≥ 0
Solving this inequality for v, we get:
v ≥ -7
This means that any solution we find for v must be greater than or equal to -7. This domain restriction is crucial because the square root of a negative number is not a real number. If we had obtained a solution less than -7, we would have immediately recognized it as an extraneous solution.
Impact on Solution Verification
The domain restriction reinforces the importance of checking solutions. Even if a solution satisfies the algebraic manipulations we perform, it may still be invalid if it falls outside the domain of the original equation. In our case, the solution v = 42 is well within the domain v ≥ -7, which provides further confidence in its validity. However, if we had obtained a solution like v = -8, we would have immediately rejected it because it violates the domain restriction.
Practical Applications of Domain Considerations
Domain considerations are not limited to mathematical exercises; they have practical applications in various fields. For example, in physics, the speed of an object cannot be negative, and in chemistry, concentrations of substances cannot be negative. When modeling such scenarios with equations, it is essential to consider these domain restrictions to ensure that the solutions obtained are physically meaningful. Neglecting domain considerations can lead to absurd or nonsensical results.
Conclusion
In this comprehensive guide, we have successfully solved for v in the equation √(v+7) - 4 = 3. We followed a systematic approach, isolating the square root term, eliminating the square root, and solving the resulting linear equation. We also emphasized the crucial step of checking our solution to ensure its validity and avoid extraneous solutions. Furthermore, we discussed the importance of domain considerations when dealing with square root equations. By understanding these concepts and techniques, you will be well-equipped to tackle similar equations and problems in mathematics and beyond. Remember to always check your solutions and consider the domain to ensure accurate and meaningful results.
