Finding The Value Of C For Extraneous Solution In A Radical Equation

In the realm of mathematical equations, extraneous solutions can sometimes emerge as deceptive results that satisfy a transformed equation but not the original one. These extraneous solutions often arise when dealing with radical equations, where the process of squaring both sides can introduce solutions that don't hold true in the initial equation. In this article, we delve into the problem of finding the value of the constant c that makes z = 4 an extraneous solution in the equation:

$$\sqrt{\frac{3}{2} z+10}=c z+8$$

To unravel this problem, we will embark on a step-by-step journey, exploring the concepts of extraneous solutions, radical equations, and the algebraic manipulations required to solve for the elusive value of c. By the end of this exploration, you will have a solid understanding of how to identify and eliminate extraneous solutions, ensuring the accuracy of your mathematical results.

Extraneous Solutions: A Deceptive Twist

Before we dive into the specifics of the problem, it's crucial to grasp the concept of extraneous solutions. These deceptive solutions arise when we perform operations on an equation that alter its original form, potentially introducing solutions that don't satisfy the initial equation. A common culprit in the emergence of extraneous solutions is the squaring of both sides of an equation, especially when dealing with square roots or other radicals.

Consider a simple example to illustrate this point. Suppose we have the equation:

$$\sqrt{x} = -3$$

Intuitively, we know that the square root of a number cannot be negative. However, if we square both sides of the equation, we get:

$$x = 9$$

Now, if we substitute x = 9 back into the original equation, we get:

$$\sqrt{9} = -3$$

This statement is clearly false, as the square root of 9 is 3, not -3. Therefore, x = 9 is an extraneous solution that emerged due to the squaring operation.

In the given problem, we have a radical equation involving a square root. To solve for c, we will need to square both sides of the equation, which might introduce extraneous solutions. Therefore, it's essential to check our final solution to ensure it satisfies the original equation.

Solving for c: A Step-by-Step Approach

Now that we understand the concept of extraneous solutions, let's embark on the journey of solving for c in the given equation:

$$\sqrt{\frac{3}{2} z+10}=c z+8$$

We are given that z = 4 is an extraneous solution. This means that when we substitute z = 4 into the equation, it will lead to a contradiction or a false statement. Our goal is to find the value of c that makes this happen.

Step 1: Substitute z = 4 into the equation

Let's substitute z = 4 into the equation:

$$\sqrt{\frac{3}{2} (4)+10}=c (4)+8$$

Step 2: Simplify the equation

Now, let's simplify the equation:

$$\sqrt{6+10}=4c+8$$

$$\sqrt{16}=4c+8$$

$$4=4c+8$$

Step 3: Isolate the term with c

Subtract 8 from both sides of the equation:

$$4-8=4c$$

$$-4=4c$$

Step 4: Solve for c

Divide both sides of the equation by 4:

$$\frac{-4}{4}=c$$

$$c=-1$$

Therefore, the value of c that makes z = 4 a solution to the transformed equation is -1. However, we need to verify if this value of c indeed makes z = 4 an extraneous solution to the original equation. This is a critical step in solving radical equations.

Verifying the Solution: The Extraneous Solution Test

To verify if c = -1 makes z = 4 an extraneous solution, we need to substitute both values back into the original equation:

$$\sqrt{\frac{3}{2} z+10}=c z+8$$

Substitute z = 4 and c = -1:

$$\sqrt{\frac{3}{2} (4)+10}=(-1)(4)+8$$

Simplify the equation:

$$\sqrt{6+10}=-4+8$$

$$\sqrt{16}=4$$

$$4=4$$

The equation holds true! This means that z = 4 is a solution to the equation when c = -1. However, this doesn't necessarily mean that z = 4 is an extraneous solution. To confirm this, we need to check if z = 4 satisfies the original equation before squaring both sides. If it doesn't, then it is indeed an extraneous solution.

Let's go back to the step where we substituted z = 4 and c = -1:

$$\sqrt{\frac{3}{2} (4)+10}=(-1)(4)+8$$

Before simplifying, we have:

$$\sqrt{16} = 4$$

As we saw earlier, this simplifies to:

$$4 = 4$$

This is a true statement. However, we must consider the fact that the square root function, by definition, yields only the non-negative root. In other words, $\sqrt{16}$ is defined as 4, and not -4. The equation we derived after squaring would also be satisfied if we had:

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

Which simplifies to:

$$-4 = -4$$

Thus, for z = 4 to be an extraneous solution, the right-hand side of the equation must be negative when z = 4 and c = -1 before squaring. Let's examine the right-hand side:

$$cz + 8 = (-1)(4) + 8 = -4 + 8 = 4$$

Since the right-hand side is positive, and equal to the square root which is also positive, z = 4 is not an extraneous solution for c = -1. We made an assumption that because it satisfies the simplified equation, but may not satisfy the original, that it is extraneous, but this is not the case here.

Let's think through the problem a bit differently. For z = 4 to be an extraneous solution, it must satisfy the equation after squaring, but not the original equation. Let's square both sides of the original equation:

$$\left(\sqrt{\frac{3}{2} z+10}\right)^2=(c z+8)^2$$

$$\frac{3}{2} z+10 = (c z+8)^2$$

Substituting z = 4, we get:

$$\frac{3}{2} (4)+10 = (4c+8)^2$$

$$6+10 = (4c+8)^2$$

$$16 = (4c+8)^2$$

Taking the square root of both sides, we get:

$$\pm 4 = 4c+8$$

We have two cases:

Case 1:

$$4 = 4c+8$$

$$-4 = 4c$$

$$c = -1$$

As we saw earlier, c = -1 leads to 4 = 4, so z = 4 is not an extraneous solution in this case.

Case 2:

$$-4 = 4c+8$$

$$-12 = 4c$$

$$c = -3$$

Now, let's check if z = 4 is an extraneous solution when c = -3:

Original equation:

$$\sqrt{\frac{3}{2} z+10}=c z+8$$

Substitute z = 4 and c = -3:

$$\sqrt{\frac{3}{2} (4)+10}=(-3)(4)+8$$

$$\sqrt{6+10}=-12+8$$

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

$$4=-4$$

This is a false statement. Therefore, z = 4 is an extraneous solution when c = -3.

Conclusion: The Value of c for Extraneous Solutions

In this exploration, we successfully navigated the realm of extraneous solutions and radical equations to determine the value of c that makes z = 4 an extraneous solution in the given equation. Through careful algebraic manipulation and verification, we discovered that the value of c is:

$$c = -3$$

This journey underscores the importance of checking solutions in radical equations to avoid the pitfall of extraneous solutions. By understanding the concept of extraneous solutions and employing rigorous verification techniques, we can ensure the accuracy and reliability of our mathematical results. In the realm of mathematical problem-solving, attention to detail and a thorough understanding of fundamental concepts are the keys to success. This exploration has highlighted these principles, providing valuable insights into the nature of equations and their solutions.