Solving √(w²-20) - 2 = 2 A Step-by-Step Solution

In this article, we will delve into the intricacies of solving the equation √w²-20 - 2 = 2. This equation, which involves a square root and a variable, requires a systematic approach to find its solutions. We will break down each step, providing clear explanations and justifications along the way. By the end of this guide, you will not only be able to solve this specific equation but also gain a deeper understanding of how to tackle similar algebraic problems. Our focus will be on ensuring clarity and accuracy, making this a valuable resource for students and anyone interested in mathematics.

Understanding the Equation Structure

The equation √w²-20 - 2 = 2 is a radical equation, specifically involving a square root. The variable w is squared and then has 20 subtracted from it, the square root of the result is taken, and finally, 2 is subtracted from that. The entire expression is set equal to 2. To solve this, we need to isolate the square root term and then eliminate it by squaring both sides of the equation. It is crucial to remember that squaring both sides can introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original one. Therefore, we must check all potential solutions in the original equation.

The presence of the square root and the squared variable introduces complexities that require careful handling. The term inside the square root, w² - 20, must be non-negative for the square root to be real. This condition implies that w² ≥ 20, which means w must be greater than or equal to the square root of 20 or less than or equal to the negative square root of 20. This preliminary analysis gives us a sense of the possible range of solutions. The constant terms, -2 and 2, play a role in shifting the equation and need to be considered when isolating the radical. The interplay between these components is what makes solving the equation an interesting mathematical exercise.

Step-by-Step Solution

Step 1: Isolate the Square Root

The first step in solving the equation √w²-20 - 2 = 2 is to isolate the square root term. This means we want to get the term √w²-20 by itself on one side of the equation. To do this, we add 2 to both sides of the equation:

√w²-20 - 2 + 2 = 2 + 2

This simplifies to:

√w²-20 = 4

Now, we have the square root term isolated, which allows us to proceed to the next step of eliminating the square root.

Isolating the square root is a fundamental technique in solving radical equations. It allows us to apply the inverse operation (squaring) to eliminate the radical. By adding 2 to both sides, we maintain the balance of the equation, ensuring that the solutions remain valid. This step is crucial because it sets up the equation for the next operation, which involves squaring both sides. The simplicity of this step is deceptive; it is a critical move that simplifies the equation and allows us to move closer to the solution. Without isolating the square root, we would not be able to effectively eliminate it and solve for the variable.

Step 2: Eliminate the Square Root

To eliminate the square root in the equation √w²-20 = 4, we square both sides of the equation. Squaring both sides is the inverse operation of taking the square root, and it will remove the radical sign. This gives us:

(√w²-20)² = 4²

Simplifying this, we get:

w² - 20 = 16

Now, we have a quadratic equation without the square root, which is much easier to solve.

Eliminating the square root by squaring both sides is a powerful technique. However, it is also a step where extraneous solutions can be introduced. This happens because squaring both sides can make a false statement true. For example, -2 ≠ 2, but (-2)² = 2². Therefore, it is absolutely essential to check the solutions obtained after this step in the original equation. The resulting quadratic equation, w² - 20 = 16, is a standard form that we can solve using various methods, such as factoring, completing the square, or using the quadratic formula. The simplification from a radical equation to a quadratic equation is a significant step forward in the solution process.

Step 3: Solve the Quadratic Equation

Now we have the quadratic equation w² - 20 = 16. To solve this, we first add 20 to both sides to isolate the w² term:

w² - 20 + 20 = 16 + 20

This simplifies to:

w² = 36

To find the values of w, we take the square root of both sides:

√w² = ±√36

This gives us two possible solutions:

w = 6 and w = -6

Solving the quadratic equation involves isolating the squared term and then taking the square root of both sides. It is crucial to remember that taking the square root of a number yields both positive and negative solutions, as both the positive and negative values, when squared, will result in the same positive number. In this case, both 6 and -6, when squared, equal 36. This step is a critical transition from an equation involving a square to finding the individual values of the variable. The two potential solutions highlight the importance of the final step, which is to check these solutions in the original equation.

Step 4: Check for Extraneous Solutions

It is crucial to check our potential solutions, w = 6 and w = -6, in the original equation √w²-20 - 2 = 2 to ensure they are not extraneous.

Checking w = 6

Substitute w = 6 into the original equation:

√6²-20 - 2 = √36-20 - 2 = √16 - 2 = 4 - 2 = 2

Since this is true, w = 6 is a valid solution.

Checking w = -6

Substitute w = -6 into the original equation:

√(-6)²-20 - 2 = √36-20 - 2 = √16 - 2 = 4 - 2 = 2

Since this is also true, w = -6 is also a valid solution.

Checking for extraneous solutions is a vital step in solving radical equations. Squaring both sides of an equation can introduce solutions that do not satisfy the original equation. This is because the squaring operation can make negative values positive, potentially creating solutions that do not exist in the context of the original equation. By substituting each potential solution back into the original equation, we can verify whether it is a valid solution or an extraneous one. In this case, both 6 and -6 satisfy the original equation, making them both valid solutions. This step underscores the importance of rigor in mathematical problem-solving and ensures that we arrive at the correct answer.

Final Answer

Therefore, the solutions to the equation √w²-20 - 2 = 2 are:

• w = 6
• w = -6

Thus, the correct answers from the given options are:

• A. -6
• D. 6

We have successfully solved the equation and identified the correct solutions. The process involved isolating the square root, squaring both sides, solving the resulting quadratic equation, and, most importantly, checking for extraneous solutions. This comprehensive approach ensures that we have a complete and accurate solution. The final answers, w = 6 and w = -6, are the only values that satisfy the original equation. This exercise reinforces the importance of following a systematic method when solving radical equations and highlights the potential pitfalls of neglecting any of the steps.

Conclusion

In this comprehensive guide, we have successfully solved the equation √w²-20 - 2 = 2. We began by understanding the structure of the equation and then proceeded through a step-by-step solution process. We first isolated the square root term, then eliminated it by squaring both sides. This led us to a quadratic equation, which we solved to find two potential solutions: w = 6 and w = -6. Crucially, we checked both of these solutions in the original equation to confirm that they were not extraneous. This process demonstrated the importance of verifying solutions in radical equations.

Solving radical equations like this one requires a methodical approach. Each step, from isolating the radical to checking for extraneous solutions, is essential for obtaining the correct answer. The presence of a square root introduces the possibility of extraneous solutions, making the verification step non-negotiable. The transition from a radical equation to a quadratic equation is a common technique in solving these types of problems. The skills and techniques used in this problem are applicable to a wide range of algebraic problems, making this exercise a valuable learning experience. By mastering these concepts, you can confidently tackle similar equations and enhance your mathematical problem-solving abilities.