Finding Values For C And D To Validate An Equation

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In this article, we will walk you through the process of determining the values of c and d that make the following equation true:

162x6y53=3x2y(cyd3)\sqrt[3]{162 x^6 y^5}=3 x^2 y(\sqrt[3]{c y^d})

We'll break down the steps, providing a detailed explanation to ensure you understand the underlying concepts. So, let's dive in and solve this mathematical puzzle together!

Simplifying the Left-Hand Side of the Equation

To find the values of c and d, our initial step involves simplifying the left-hand side (LHS) of the given equation. This process will make it easier to compare both sides and identify the correct values. We begin with the expression:

162x6y53\sqrt[3]{162 x^6 y^5}

Our goal is to extract any perfect cubes from within the cube root. Let's start by factoring 162. The prime factorization of 162 is 2 × 3 × 3 × 3 × 3, which can be written as 2 × 3⁴. We can rewrite this as 2 × 3³ × 3.

Now, let's look at the variables. We have x⁶ and y⁵. For x⁶, we can express it as (x²)³, which is a perfect cube. For y⁵, we can rewrite it as y³ × y², where y³ is a perfect cube.

Substituting these factorizations back into the cube root, we get:

2×33×3×x6×y3×y23\sqrt[3]{2 \times 3^3 \times 3 \times x^6 \times y^3 \times y^2}

We can now separate the perfect cubes from the remaining terms:

33×(x2)3×y3×2×3×y23\sqrt[3]{3^3 \times (x^2)^3 \times y^3 \times 2 \times 3 \times y^2}

Taking the cube root of the perfect cubes, we have:

3x2y2×3×y233 x^2 y \sqrt[3]{2 \times 3 \times y^2}

Simplifying the terms inside the cube root, we get:

3x2y6y233 x^2 y \sqrt[3]{6 y^2}

Thus, the simplified form of the left-hand side of the equation is 3x²y∛(6y²). This simplification is a crucial step towards matching it with the right-hand side and determining the values of c and d.

Matching the Simplified LHS with the RHS

Now that we have simplified the left-hand side (LHS) of the equation, we can move on to comparing it with the right-hand side (RHS) to determine the values of c and d. The simplified LHS is:

3x2y6y233 x^2 y \sqrt[3]{6 y^2}

The given RHS is:

3x2y(cyd3)3 x^2 y (\sqrt[3]{c y^d})

To find the values of c and d, we need to match the expressions inside the cube root on both sides of the equation. By comparing the two expressions, we can see that:

6y23=cyd3\sqrt[3]{6 y^2} = \sqrt[3]{c y^d}

This implies that the terms inside the cube root must be equal. Therefore, we can equate the expressions:

6y2=cyd6 y^2 = c y^d

This equation allows us to directly identify the values of c and d. By inspection, it is clear that:

  • c = 6
  • d = 2

This matching process is a fundamental step in solving equations of this nature. By simplifying and comparing both sides, we can isolate the unknown variables and determine their values. In this case, we found that c is 6 and d is 2. This methodical approach ensures accuracy and understanding in solving mathematical problems.

Identifying the Correct Option

After finding the values of c and d, which are c = 6 and d = 2, we now need to identify the correct option from the given choices. The options provided are:

A. c = 2, d = 2 B. c = 2, d = 4 C. c = 6, d = 2 D. c = 6, d = 4

By comparing our calculated values with the options, we can see that:

  • Option A has c = 2 and d = 2, which does not match our result.
  • Option B has c = 2 and d = 4, which also does not match our result.
  • Option C has c = 6 and d = 2, which exactly matches our calculated values.
  • Option D has c = 6 and d = 4, which does not match our result.

Therefore, the correct option is C. c = 6, d = 2. This step is crucial in any problem-solving scenario to ensure that the final answer aligns with the available choices. By systematically comparing our results with each option, we can confidently select the correct answer and avoid errors. This process reinforces the importance of accuracy and attention to detail in mathematical problem-solving.

Step-by-Step Recap of the Solution

To ensure a clear understanding, let's recap the step-by-step solution to the problem of finding the values of c and d in the equation:

162x6y53=3x2y(cyd3)\sqrt[3]{162 x^6 y^5}=3 x^2 y(\sqrt[3]{c y^d})

  1. Simplify the Left-Hand Side (LHS):
    • We started by factoring 162 into its prime factors: 162 = 2 × 3⁴ = 2 × 3³ × 3.
    • We rewrote x⁶ as (x²)³ and y⁵ as y³ × y².
    • The LHS then became \sqrt[3]{2 \times 3^3 \times 3 \times x^6 \times y^3 \times y^2}.
    • We separated the perfect cubes: \sqrt[3]{3^3 \times (x2)3 \times y^3 \times 2 \times 3 \times y^2}.
    • Taking the cube root of the perfect cubes, we obtained 3x²y∛(6y²).
  2. Match the Simplified LHS with the Right-Hand Side (RHS):
    • The simplified LHS was 3x²y∛(6y²), and the RHS was 3x²y(∛(cyᵈ)).
    • We equated the expressions inside the cube root: ∛(6y²) = ∛(cyᵈ).
    • This implied that 6y² = cyᵈ.
  3. Determine the Values of c and d:
    • By comparing the coefficients and exponents, we found that c = 6 and d = 2.
  4. Identify the Correct Option:
    • The given options were:
      • A. c = 2, d = 2
      • B. c = 2, d = 4
      • C. c = 6, d = 2
      • D. c = 6, d = 4
    • We compared our calculated values (c = 6, d = 2) with the options.
    • Option C (c = 6, d = 2) matched our result.

Therefore, the final answer is C. c = 6, d = 2. This detailed recap ensures that each step is clear and reinforces the process of solving the equation, making it easier to understand and apply similar problem-solving techniques in the future.

Why Option C is the Correct Answer

To further emphasize why option C is the correct answer, let's delve into a more detailed explanation. We found that c = 6 and d = 2 by simplifying and matching the terms in the given equation:

162x6y53=3x2y(cyd3)\sqrt[3]{162 x^6 y^5}=3 x^2 y(\sqrt[3]{c y^d})

We simplified the left-hand side (LHS) to 3x²y∛(6y²). The right-hand side (RHS) is given as 3x²y(∛(cyᵈ)). For the equation to hold true, the expressions inside the cube root on both sides must be equal:

∛(6y2)=∛(cyd)∛(6y²) = ∛(cyᵈ)

This means:

6y2=cyd6y² = cyᵈ

Now, let's analyze each option:

  • Option A: c = 2, d = 2
    • If we substitute c = 2 and d = 2, we get 2y². This does not match 6y².
  • Option B: c = 2, d = 4
    • If we substitute c = 2 and d = 4, we get 2y⁴. This also does not match 6y².
  • Option C: c = 6, d = 2
    • If we substitute c = 6 and d = 2, we get 6y². This exactly matches the simplified LHS, 6y².
  • Option D: c = 6, d = 4
    • If we substitute c = 6 and d = 4, we get 6y⁴. This does not match 6y².

From this analysis, it is evident that only option C makes the equation true. By substituting c = 6 and d = 2 into the RHS, we get:

3x2y(6y23)3 x^2 y (\sqrt[3]{6 y^2})

This matches the simplified LHS that we derived earlier. Therefore, option C is the correct answer because it provides the values of c and d that make the equation valid. This step-by-step verification further solidifies our understanding and confirms the accuracy of our solution.

Practical Applications of Simplifying Radicals

Understanding how to simplify radicals, as we did in this problem, has several practical applications in various fields. Simplifying radicals makes mathematical expressions easier to work with, which is particularly useful in algebra, calculus, and other advanced mathematical areas. Let's explore some specific applications:

  1. Engineering:
    • In structural engineering, simplifying radicals is crucial for calculating stresses and strains in materials. For example, when dealing with geometric shapes and forces, engineers often encounter expressions involving square roots and cube roots. Simplifying these radicals helps in obtaining more manageable and accurate results.
    • Electrical engineering also uses simplified radicals in circuit analysis. Calculating impedances and currents in AC circuits often involves complex numbers and radicals. Simplifying these expressions ensures accurate calculations and efficient circuit designs.
  2. Physics:
    • Physics frequently involves simplifying radicals in various calculations. For instance, in mechanics, calculations related to energy, momentum, and velocity often include square roots. Simplifying these radicals allows physicists to solve problems more efficiently and understand the relationships between different physical quantities.
    • Optics and electromagnetism also utilize radicals in calculations involving wavelengths, frequencies, and intensities. Simplified radicals make these calculations easier to perform and interpret.
  3. Computer Graphics and Game Development:
    • In computer graphics, radicals are commonly used in transformations and calculations involving vectors and matrices. Simplifying these radicals helps in optimizing the performance of graphics rendering algorithms.
    • Game development involves numerous calculations related to physics simulations, collision detection, and character movements. Simplifying radicals in these calculations can significantly improve the efficiency and responsiveness of games.
  4. Financial Analysis:
    • Financial analysts use radicals in various models, such as calculating investment returns and risk assessments. For example, the standard deviation, a key measure of risk, involves square roots. Simplifying these radicals can make financial models easier to understand and apply.
    • Calculating compound interest and other financial metrics often involves radicals. Simplified radicals help in accurate and efficient financial planning and analysis.

In summary, the ability to simplify radicals is a fundamental skill with far-reaching applications. Whether in engineering, physics, computer science, or finance, this skill helps in making complex calculations manageable and accurate. Mastering this technique not only aids in solving mathematical problems but also in applying mathematical principles to real-world scenarios.

Common Mistakes to Avoid When Simplifying Radicals

When simplifying radicals, it's essential to be aware of common mistakes that can lead to incorrect answers. Avoiding these pitfalls ensures accuracy and a better understanding of the process. Here are some frequent errors to watch out for:

  1. Incorrectly Factoring the Radicand:
    • A common mistake is not completely factoring the radicand (the number inside the radical) into its prime factors. For example, when simplifying \√18, one might stop at \√(9 × 2) without recognizing that 9 can be further factored into 3 × 3. The correct factorization is \√(3 × 3 × 2), which simplifies to 3\√2. Always ensure that you break down the radicand into its prime factors to identify all perfect squares, cubes, etc.
  2. Forgetting to Simplify the Coefficient:
    • Sometimes, students focus solely on simplifying the radical part and overlook the coefficient (the number outside the radical). For example, in the expression 2\√(4x²), it’s crucial to simplify both the radical and the coefficient. The \√(4x²) simplifies to 2x, and then you must multiply this by the existing coefficient 2, resulting in 4x. Failing to do so will lead to an incomplete simplification.
  3. Misapplying the Product and Quotient Rules:
    • The product rule states that \√(ab) = \√a × \√b, and the quotient rule states that \√(a/b) = \√a / \√b. These rules are powerful tools but can be misapplied. For instance, you cannot apply these rules directly to addition or subtraction inside the radical. \√(a + b) is not equal to \√a + \√b. Ensure you only use these rules when dealing with multiplication and division.
  4. Incorrectly Handling Variables:
    • When simplifying radicals involving variables, it’s essential to correctly apply the rules of exponents. For example, \√(x⁴) simplifies to x², but \√(x³) simplifies to x\√x. Pay close attention to whether the exponent is even or odd and how it affects the simplified form.
  5. Not Rationalizing the Denominator:
    • Expressions with radicals in the denominator are generally not considered fully simplified. For example, 1/\√2 should be rationalized by multiplying both the numerator and the denominator by \√2, resulting in \√2/2. Forgetting this step leaves the expression in an unsimplified form.
  6. Making Arithmetic Errors:
    • Simple arithmetic errors, such as incorrect multiplication or division, can derail the entire simplification process. Always double-check your calculations to ensure accuracy. Mistakes in basic arithmetic can easily lead to incorrect final answers.

By being mindful of these common mistakes, you can improve your accuracy and confidence when simplifying radicals. Practice and attention to detail are key to mastering this fundamental mathematical skill.

Conclusion

In conclusion, finding the correct values of c and d that satisfy the equation \sqrt[3]{162 x^6 y^5}=3 x^2 y(\sqrt[3]{c y^d}) involves a methodical approach that includes simplifying radicals and comparing terms. By breaking down the problem into manageable steps, we were able to simplify the left-hand side of the equation, match it with the right-hand side, and accurately determine that c = 6 and d = 2. This process led us to identify option C as the correct answer.

Throughout this article, we emphasized the importance of careful simplification, accurate factoring, and attention to detail. We also highlighted the practical applications of simplifying radicals in various fields, such as engineering, physics, computer graphics, and finance. Additionally, we discussed common mistakes to avoid when simplifying radicals, ensuring a more robust understanding of the topic.

Mastering the simplification of radicals is a valuable skill that not only helps in solving mathematical problems but also in applying mathematical concepts to real-world scenarios. By understanding the underlying principles and practicing consistently, you can confidently tackle similar problems and enhance your mathematical proficiency. Remember, the key to success in mathematics is a combination of knowledge, practice, and a methodical approach.