Solving The Radical Equation √3v+12 + 4 = 1 A Step-by-Step Guide

Understanding the Problem

In this article, we will delve into solving the equation $\sqrt{3v + 12} + 4 = 1$ for the real number $v$. This problem falls under the category of algebraic equations, specifically those involving square roots. To find the solution, we will employ a series of algebraic manipulations to isolate the variable $v$. The key to solving equations with square roots is to first isolate the radical term and then square both sides of the equation to eliminate the square root. However, it is crucial to check the solutions obtained after squaring both sides, as this process can sometimes introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original equation. We will also explore the domain of the variable $v$ to ensure that the solutions we find are valid within the context of the original equation. This involves considering the expression inside the square root, which must be non-negative to yield real number solutions. Solving radical equations often requires a meticulous approach to ensure accuracy and avoid pitfalls such as extraneous solutions. Our discussion will cover each step in detail, providing a clear understanding of the methods used and the reasoning behind them. This will enable readers to tackle similar problems with confidence and a solid grasp of the underlying principles. The goal is not just to find the numerical solution, but also to understand the process and the conditions under which solutions are valid.

Step-by-Step Solution

To solve the equation $\sqrt{3v + 12} + 4 = 1$, we will follow these steps:

1. Isolate the square root term: The first step is to isolate the square root term on one side of the equation. We achieve this by subtracting 4 from both sides of the equation: $\sqrt{3v + 12} + 4 - 4 = 1 - 4$ $\sqrt{3v + 12} = -3$

2. Analyze the equation: Before proceeding with squaring both sides, it is crucial to analyze the equation. We have $\sqrt{3v + 12} = -3$. Since the square root of any real number cannot be negative, this equation implies that the square root term must equal a negative number, which is impossible in the realm of real numbers. The square root of a real number is always non-negative. This observation leads us to a critical conclusion about the existence of solutions. The square root function always returns a non-negative value. Therefore, if we have an equation where the square root is equated to a negative number, we can immediately conclude that there are no real solutions.

3. Conclusion: Since the square root term cannot be negative, there is no real solution to the equation $\sqrt{3v + 12} = -3$. This is because the principal square root of any real number is defined to be non-negative. If we were to continue with the process of squaring both sides, we would be introducing a step that is not valid in the context of real numbers. Therefore, we can stop at this point and state the solution. Understanding the properties of the square root function is crucial in solving equations involving radicals. This knowledge allows us to identify cases where solutions are not possible without having to go through unnecessary algebraic manipulations. In this instance, recognizing that a square root cannot be negative is a key insight that saves us time and prevents us from making errors. The nature of the square root being non-negative is a fundamental concept in algebra.

Verification (Optional, but Recommended)

Even though we have determined that there is no solution based on the nature of the square root, let's briefly consider what would happen if we proceeded with the algebraic steps to illustrate why this can lead to extraneous solutions.

If we were to square both sides of the equation $\sqrt{3v + 12} = -3$, we would get:

$(\sqrt{3v + 12})^2 = (-3)^2$

$3v + 12 = 9$

Now, we solve for $v$:

$3v = 9 - 12$

$3v = -3$

$v = -1$

However, if we substitute $v = -1$ back into the original equation, we get:

$\sqrt{3(-1) + 12} + 4 = \sqrt{-3 + 12} + 4 = \sqrt{9} + 4 = 3 + 4 = 7$

Since $7 \neq 1$, the solution $v = -1$ is an extraneous solution. This demonstrates why it is essential to check solutions obtained after squaring both sides of an equation. Squaring both sides can introduce solutions that do not satisfy the original equation, and these are called extraneous solutions. The process of verifying solutions helps us to filter out these extraneous solutions and ensure that we only accept the valid solutions. In this case, the verification step clearly shows that $v = -1$ is not a solution to the original equation. This reinforces the importance of understanding the domain of the variable and the properties of the square root function. Checking for extraneous solutions is a critical step in solving radical equations.

Domain Considerations

Before we definitively conclude that there is no solution, let's consider the domain of the variable $v$ in the equation $\sqrt{3v + 12} + 4 = 1$. The expression inside the square root, $3v + 12$, must be greater than or equal to zero for the square root to be a real number:

$3v + 12 \geq 0$

Subtracting 12 from both sides:

$3v \geq -12$

Dividing by 3:

$v \geq -4$

So, the domain of $v$ is $v \geq -4$. This means that any solution we find must be greater than or equal to -4. However, as we have already shown, the equation $\sqrt{3v + 12} = -3$ has no solution because the square root cannot be negative. The domain restriction further reinforces that there is no real solution. Even if we found a value of $v$ within the domain $v \geq -4$ after squaring both sides, it would still need to satisfy the original equation, which it cannot. The domain of the variable plays a crucial role in determining the validity of solutions in radical equations. By considering the domain, we can avoid unnecessary algebraic manipulations and identify cases where solutions are not possible. The combination of understanding the properties of the square root function and considering the domain of the variable allows us to efficiently and accurately solve radical equations.

Conclusion

In summary, after isolating the square root term in the equation $\sqrt{3v + 12} + 4 = 1$, we arrived at $\sqrt{3v + 12} = -3$. Since the square root of a real number cannot be negative, we conclude that there is no real solution to this equation. This determination is further supported by considering the domain of the variable $v$, which is $v \geq -4$. Even if we were to proceed with squaring both sides and solving for $v$, we would obtain an extraneous solution, highlighting the importance of checking solutions in radical equations. The key takeaway from this problem is the importance of understanding the properties of the square root function and the potential for extraneous solutions when solving equations involving radicals. The absence of a real solution in this case emphasizes the significance of these concepts in algebra. By carefully analyzing the equation and considering the domain, we can efficiently solve radical equations and avoid common pitfalls. This comprehensive approach ensures that we arrive at the correct solution and understand the underlying mathematical principles.