Solving (x-6)^2 = 25 Step-by-Step Guide

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In the realm of algebraic equations, the quest to find the unknown variable, often represented by 'x', is a fundamental pursuit. This article delves into the intricacies of solving the equation (x-6)^2 = 25, providing a comprehensive step-by-step guide to unravel its solutions. We will explore the underlying principles, employ strategic techniques, and present the answers in a clear and concise manner.

Understanding the Equation's Structure

At its core, the equation (x-6)^2 = 25 embodies a squared expression, (x-6)^2, equated to a constant value, 25. The expression (x-6)^2 signifies the square of the difference between 'x' and 6. To solve for 'x', we must reverse the squaring operation and isolate 'x' on one side of the equation.

Method 1: Utilizing the Square Root Property

The square root property states that if a^2 = b, then a = ±√b. Applying this property to our equation, we take the square root of both sides, introducing the ± sign to account for both positive and negative roots.

√(x-6)^2 = ±√25

This simplifies to:

x - 6 = ±5

Now, we have two separate equations to solve:

  1. x - 6 = 5
  2. x - 6 = -5

Solving the first equation, we add 6 to both sides:

x = 5 + 6

x = 11

Solving the second equation, we again add 6 to both sides:

x = -5 + 6

x = 1

Therefore, the solutions to the equation (x-6)^2 = 25 are x = 11 and x = 1.

Method 2: Expanding and Factoring

Alternatively, we can solve the equation by expanding the squared expression, rearranging the terms, and factoring the resulting quadratic equation. Expanding (x-6)^2, we get:

x^2 - 12x + 36 = 25

Subtracting 25 from both sides, we obtain:

x^2 - 12x + 11 = 0

This is a quadratic equation in the standard form ax^2 + bx + c = 0. To solve it, we can factor the quadratic expression. We seek two numbers that multiply to 11 and add up to -12. These numbers are -11 and -1.

Thus, we can factor the equation as:

(x - 11)(x - 1) = 0

Setting each factor equal to zero, we get:

  1. x - 11 = 0
  2. x - 1 = 0

Solving these equations, we find:

  1. x = 11
  2. x = 1

Again, we arrive at the same solutions: x = 11 and x = 1.

Verifying the Solutions

To ensure the accuracy of our solutions, we can substitute them back into the original equation and check if they hold true.

For x = 11:

(11 - 6)^2 = 5^2 = 25

For x = 1:

(1 - 6)^2 = (-5)^2 = 25

Both solutions satisfy the equation, confirming their validity.

Expressing the Answer

As requested, we list the solutions separated by a comma: 11,1

This detailed guide provides a comprehensive exploration of solving the algebraic equation (x-6)^2 = 25. We've covered two distinct methods: employing the square root property and expanding and factoring. Both methods lead to the same solutions, x = 11 and x = 1, which we verified by substituting them back into the original equation. This methodical approach ensures accuracy and enhances understanding of algebraic problem-solving techniques. Let's delve deeper into each method, providing further clarity and examples.

Method 1: The Elegance of the Square Root Property

The square root property is a powerful tool for solving equations where a squared expression is isolated on one side. It elegantly undoes the squaring operation, allowing us to directly solve for the variable. The core principle is that if a^2 = b, then 'a' must be either the positive or negative square root of 'b', represented as a = ±√b. This ± sign is crucial because both the positive and negative square roots, when squared, yield the same positive result.

Applying the Property to (x-6)^2 = 25

  1. Take the square root of both sides: This is the foundational step. Applying the square root to both sides maintains the equation's balance and introduces the ± sign.

    √(x-6)^2 = ±√25

  2. Simplify: The square root of (x-6)^2 simplifies to (x-6), and the square root of 25 is 5.

    x - 6 = ±5

  3. Separate into two equations: The ± sign now branches the problem into two distinct equations:

    • x - 6 = 5
    • x - 6 = -5

  4. Solve each equation: Isolate 'x' in each equation by adding 6 to both sides.

    • x = 5 + 6 => x = 11
    • x = -5 + 6 => x = 1

Advantages of the Square Root Property

  • Direct and Efficient: When applicable, this method is often the quickest route to the solution.
  • Conceptual Clarity: It directly addresses the inverse relationship between squaring and taking the square root.

Example Scenarios

Let's consider a similar equation: (y + 3)^2 = 16

  1. √(y + 3)^2 = ±√16
  2. y + 3 = ±4
    • y + 3 = 4 => y = 1
    • y + 3 = -4 => y = -7

The solutions are y = 1 and y = -7.

Another example: (2z - 1)^2 = 9

  1. √(2z - 1)^2 = ±√9
  2. 2z - 1 = ±3
    • 2z - 1 = 3 => 2z = 4 => z = 2
    • 2z - 1 = -3 => 2z = -2 => z = -1

The solutions are z = 2 and z = -1.

Method 2: Expanding and Factoring – A Quadratic Journey

This method transforms the original equation into a quadratic equation in standard form (ax^2 + bx + c = 0), which can then be solved by factoring. Factoring involves expressing the quadratic expression as a product of two binomials. This approach relies on recognizing patterns and applying factoring techniques.

Transforming (x-6)^2 = 25 into a Quadratic Equation

  1. Expand the squared expression: Use the FOIL (First, Outer, Inner, Last) method or the binomial theorem to expand (x-6)^2.

    (x - 6)^2 = (x - 6)(x - 6) = x^2 - 6x - 6x + 36 = x^2 - 12x + 36

  2. Rewrite the equation: Substitute the expanded expression back into the original equation.

    x^2 - 12x + 36 = 25

  3. Set the equation to zero: Subtract 25 from both sides to obtain the standard quadratic form.

    x^2 - 12x + 36 - 25 = 0 x^2 - 12x + 11 = 0

Factoring the Quadratic Expression

  1. Identify factors: Find two numbers that multiply to 'c' (11 in this case) and add up to 'b' (-12 in this case). These numbers are -11 and -1.

  2. Write the factored form: Use the identified factors to write the quadratic expression as a product of two binomials.

    (x - 11)(x - 1) = 0

Solving for x

  1. Set each factor to zero: Apply the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero.

    • x - 11 = 0
    • x - 1 = 0

  2. Solve each equation: Isolate 'x' in each equation.

    • x = 11
    • x = 1

Advantages of Expanding and Factoring

  • Versatility: This method is applicable to a wider range of quadratic equations.
  • Reinforces Factoring Skills: It provides practice in factoring, a crucial skill in algebra.

Example Scenarios

Consider the equation (y + 2)^2 = 9

  1. Expand: (y + 2)^2 = y^2 + 4y + 4
  2. Rewrite: y^2 + 4y + 4 = 9
  3. Set to zero: y^2 + 4y - 5 = 0
  4. Factor: (y + 5)(y - 1) = 0
  5. Solve: y = -5, y = 1

Another example: (3z - 2)^2 = 16

  1. Expand: (3z - 2)^2 = 9z^2 - 12z + 4
  2. Rewrite: 9z^2 - 12z + 4 = 16
  3. Set to zero: 9z^2 - 12z - 12 = 0
  4. Simplify (divide by 3): 3z^2 - 4z - 4 = 0
  5. Factor: (3z + 2)(z - 2) = 0
  6. Solve: z = -2/3, z = 2

Both methods, the square root property and expanding and factoring, consistently yield the solutions x = 11 and x = 1. As per the instructions, we present these solutions separated by a comma: 11,1

This article has meticulously dissected the process of solving the equation (x-6)^2 = 25, showcasing two distinct yet equally effective methods. The square root property offers a direct and elegant approach, while expanding and factoring provides a versatile technique applicable to a broader range of quadratic equations. The key takeaway is that understanding the underlying principles and mastering these methods empowers you to confidently tackle algebraic challenges.

Key Concepts Revisited

Square Root Property

  • If a^2 = b, then a = ±√b
  • Crucial for equations with isolated squared expressions.
  • Remember to consider both positive and negative roots.

Expanding and Factoring

  • Transform the equation into the standard quadratic form (ax^2 + bx + c = 0).
  • Factor the quadratic expression into two binomials.
  • Apply the zero-product property to solve for the variable.

Problem-Solving Strategies

  1. Understand the Problem: Carefully analyze the equation and identify its structure.
  2. Choose the Appropriate Method: Select the method that best suits the equation's form. The square root property is ideal for isolated squared expressions, while expanding and factoring is more versatile for general quadratic equations.
  3. Execute the Steps: Meticulously follow the steps of the chosen method, ensuring accuracy in each step.
  4. Verify the Solutions: Substitute the solutions back into the original equation to check their validity.
  5. Present the Answer: Express the solutions clearly and concisely, following the given instructions.

Beyond this Equation: Expanding your Algebraic Toolkit

The techniques learned in solving (x-6)^2 = 25 are applicable to a wide array of algebraic equations. As you delve deeper into algebra, you will encounter more complex equations that may require a combination of these methods or the introduction of new techniques, such as the quadratic formula or completing the square. However, the fundamental principles of isolating the variable and applying inverse operations remain constant.

The Importance of Practice

The key to mastering algebraic problem-solving is consistent practice. By working through numerous examples and diverse problem types, you will develop a strong intuition for choosing the right method and executing the steps efficiently. Embrace the challenges, learn from your mistakes, and celebrate your successes. With dedication and perseverance, you can unlock the power of algebra and its applications in various fields.

Concluding Thoughts

Solving the equation (x-6)^2 = 25 is more than just finding the answers; it's a journey into the heart of algebraic thinking. By understanding the underlying concepts, mastering the techniques, and practicing consistently, you can build a solid foundation for further exploration of mathematics and its applications in the world around us. The solutions, 11 and 1, are not just numbers; they are the culmination of a logical process, a testament to the power of algebraic problem-solving.