Extraneous Solutions Explained Solving $\sqrt{-3x-2} = X+2$

When solving radical equations, it's crucial to understand the concept of extraneous solutions. These are solutions that arise during the solving process but do not satisfy the original equation. In this comprehensive guide, we will delve into extraneous solutions, focusing specifically on the equation $\sqrt{-3x-2} = x+2$. We'll explore the steps to solve the equation, identify potential extraneous solutions, and verify our answers to ensure accuracy. Understanding this concept is vital for anyone studying algebra and precalculus, as it helps in avoiding common mistakes and ensuring correct solutions. Let's embark on this mathematical journey to master extraneous solutions.

What are Extraneous Solutions?

Extraneous solutions are essentially 'false positives' in the world of equation-solving. They appear as valid answers when you follow the algebraic steps to solve an equation, but when you plug them back into the original equation, they don't hold true. This phenomenon commonly occurs when dealing with radical equations, which involve square roots, cube roots, and other radicals. The process of squaring both sides of an equation, for example, can introduce extraneous solutions because it can turn negative values into positive ones, thereby altering the equation's fundamental constraints.

To truly grasp this, consider a simple equation like $\sqrt{x} = -2$. Intuitively, we know that the square root of a number cannot be negative. However, if we were to square both sides, we'd get $x = 4$. Plugging this back into the original equation, we find $\sqrt{4} = 2$, which is not equal to $-2$. Thus, $x = 4$ is an extraneous solution. This example underscores the importance of always checking your solutions in the original equation, especially when dealing with radicals.

Extraneous solutions arise because certain algebraic operations, while valid in themselves, can expand the domain of possible solutions beyond what the original equation allows. In the context of radical equations, the domain is often restricted by the fact that you cannot take an even root (square root, fourth root, etc.) of a negative number within the realm of real numbers. Thus, any solution that leads to taking the square root of a negative number is immediately suspect and must be carefully scrutinized. Recognizing and eliminating extraneous solutions is a critical skill in algebra, ensuring that you arrive at the correct and meaningful answers.

Solving the Equation $\sqrt{-3x-2} = x+2$

To tackle the equation $\sqrt{-3x-2} = x+2$, we'll follow a systematic approach, emphasizing each step to ensure clarity and accuracy. This process involves isolating the radical, squaring both sides, solving the resulting quadratic equation, and, crucially, checking for extraneous solutions. By meticulously working through each step, we can confidently determine the true solutions to the equation.

Step 1: Isolate the Radical

The first step in solving any radical equation is to isolate the radical term. In this case, the square root term, $\sqrt{-3x-2}$, is already isolated on one side of the equation. This simplifies our initial setup, allowing us to move directly to the next step. The isolation of the radical is crucial because it sets the stage for eliminating the square root by squaring both sides, a technique that will help us transform the equation into a more manageable form.

Step 2: Square Both Sides

To eliminate the square root, we square both sides of the equation. Squaring both sides is a fundamental algebraic operation that maintains the equality of the equation while removing the radical. This gives us: $(\sqrt{-3x-2})^2 = (x+2)^2$. When we simplify this, we get $-3x-2 = (x+2)^2$. Squaring both sides is a key step, but it's also where the potential for introducing extraneous solutions arises. This is because the squaring operation can make two unequal quantities equal (e.g., $-2$ and $2$ both become $4$ when squared). Therefore, it's imperative to check our solutions later in the original equation.

Step 3: Expand and Simplify

Next, we expand the right side of the equation and simplify: $-3x-2 = x^2 + 4x + 4$. To solve this equation, we need to set it to zero, so we move all terms to one side to obtain a standard quadratic equation form. Adding $3x$ and $2$ to both sides gives us $0 = x^2 + 7x + 6$. This quadratic equation is now ready to be solved, and we can use factoring, completing the square, or the quadratic formula to find the roots.

Step 4: Solve the Quadratic Equation

Now we solve the quadratic equation $x^2 + 7x + 6 = 0$. This equation can be factored easily. We look for two numbers that multiply to $6$ and add to $7$. These numbers are $1$ and $6$. Thus, we can factor the quadratic as $(x+1)(x+6) = 0$. Setting each factor equal to zero gives us two potential solutions: $x+1 = 0$ and $x+6 = 0$. Solving these, we find $x = -1$ and $x = -6$. These are our candidate solutions, but we must remember that we squared the equation, so we need to check for extraneous solutions.

Identifying Potential Extraneous Solutions

After solving the equation, we arrived at two potential solutions: $x = -1$ and $x = -6$. However, because we squared both sides of the equation during the solving process, it is essential to verify whether these solutions are valid or extraneous. Extraneous solutions can arise due to the squaring operation, which can introduce roots that do not satisfy the original equation. The process of identifying extraneous solutions involves substituting each potential solution back into the original equation and checking for consistency. This step is not just a formality; it's a critical part of the solution process, ensuring that we only accept the true solutions.

Checking $x = -1$

To check if $x = -1$ is a valid solution, we substitute it into the original equation, $\sqrt{-3x-2} = x+2$: $\sqrt{-3(-1)-2} = -1+2$. This simplifies to $\sqrt{3-2} = 1$, which further simplifies to $\sqrt{1} = 1$. Since $1 = 1$, the solution $x = -1$ satisfies the original equation and is therefore a valid solution.

Checking $x = -6$

Next, we check the solution $x = -6$ by substituting it into the original equation: $\sqrt{-3(-6)-2} = -6+2$. This simplifies to $\sqrt{18-2} = -4$, which further simplifies to $\sqrt{16} = -4$. However, $\sqrt{16}$ is $4$, not $-4$. Therefore, $4 \neq -4$, and $x = -6$ does not satisfy the original equation. This indicates that $x = -6$ is an extraneous solution.

Verifying the Solutions

Verifying solutions is the final and crucial step in solving radical equations, especially after squaring both sides. It's where we confirm whether our potential solutions are genuine or extraneous. This process involves substituting each potential solution back into the original equation and evaluating whether the equation holds true. This step is essential because squaring both sides of an equation can introduce solutions that don't actually work in the original equation, and verifying helps us weed out these false solutions.

In our case, we have two potential solutions, $x = -1$ and $x = -6$. We've already performed the checks, but let's recap the process to emphasize the importance of verification.

Verification of $x = -1$

Substituting $x = -1$ into the original equation $\sqrt{-3x-2} = x+2$ gives us:

$$\sqrt{-3(-1)-2} = -1+2$$

$$\sqrt{3-2} = 1$$

$$\sqrt{1} = 1$$

$$1 = 1$$

Since the equation holds true, $x = -1$ is a valid solution.

Verification of $x = -6$

Now, let's substitute $x = -6$ into the original equation:

$$\sqrt{-3(-6)-2} = -6+2$$

$$\sqrt{18-2} = -4$$

$$\sqrt{16} = -4$$

$$4 = -4$$

This statement is false, which confirms that $x = -6$ is an extraneous solution. It arose during the solving process but does not satisfy the original equation.

Through this verification process, we've definitively shown that $x = -1$ is the only valid solution to the equation $\sqrt{-3x-2} = x+2$, while $x = -6$ is an extraneous solution.

Final Answer and Conclusion

After carefully solving the equation $\sqrt{-3x-2} = x+2$ and verifying our solutions, we have determined that $x = -1$ is the valid solution, while $x = -6$ is an extraneous solution. Extraneous solutions, as we've seen, are potential solutions that emerge during the solving process but do not satisfy the original equation. They often arise in radical equations due to the squaring of both sides, which can introduce roots that were not initially present.

Therefore, the extraneous solution of the equation $\sqrt{-3x-2} = x+2$ is $x = -6$. This underscores the importance of always checking your solutions when dealing with radical equations to ensure that they are valid and not extraneous. Understanding and identifying extraneous solutions is a critical skill in algebra, helping students to achieve accurate and meaningful results.

In conclusion, when solving radical equations, remember to isolate the radical, square both sides (if necessary), solve the resulting equation, and most importantly, check your solutions in the original equation. This process will help you avoid the common pitfall of accepting extraneous solutions and ensure that you arrive at the correct answer. The correct answer is A. $x=-6$

Choose the correct answer

A. $x=-6$ B. $x=-1$ C. $x=1$ D. $x=6