Solving Quadratic Equations Using The Square Root Property: A Step-by-Step Guide

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In the realm of algebra, quadratic equations hold a significant place, and mastering their solutions is crucial for mathematical proficiency. Among the various methods to solve these equations, the square root property stands out as a direct and efficient technique, especially when dealing with equations in a specific form. This article delves into the application of the square root property to solve quadratic equations, providing a comprehensive understanding through explanations, examples, and a step-by-step approach.

Understanding Quadratic Equations

Before diving into the square root property, it's essential to grasp the basics of quadratic equations. A quadratic equation is a polynomial equation of the second degree, generally expressed in the form ax² + bx + c = 0, where a, b, and c are constants, and a is not equal to zero. These equations can have two, one, or no real solutions, depending on the discriminant (b² - 4ac). The solutions, also known as roots or zeros, represent the values of x that satisfy the equation.

Methods for Solving Quadratic Equations

Several methods exist for solving quadratic equations, each with its strengths and applicability:

  • Factoring: This method involves expressing the quadratic equation as a product of two linear factors. It is efficient when the factors are easily identifiable.
  • Completing the Square: This technique transforms the quadratic equation into a perfect square trinomial, allowing for a straightforward solution.
  • Quadratic Formula: This formula provides a universal solution for any quadratic equation, regardless of its factorability. The quadratic formula is given by: x = (-b ± √(b² - 4ac)) / 2a.
  • Square Root Property: This method is particularly useful when the quadratic equation is in the form (x - h)² = k, where h and k are constants.

The Square Root Property: A Direct Approach

The square root property offers a direct method for solving quadratic equations in the form (x - h)² = k. This property states that if x² = k, then x = ±√k. In simpler terms, if a squared term equals a constant, we can take the square root of both sides to find the solutions, remembering to consider both the positive and negative roots.

Applying the Square Root Property: A Step-by-Step Guide

To effectively utilize the square root property, follow these steps:

  1. Isolate the Squared Term: Begin by isolating the squared term on one side of the equation. This involves performing algebraic operations to ensure that the term (x - h)² stands alone.
  2. Take the Square Root of Both Sides: Once the squared term is isolated, take the square root of both sides of the equation. Remember to include both the positive and negative square roots, as both will satisfy the equation.
  3. Solve for x: After taking the square root, you'll have two separate equations. Solve each equation for x to find the two possible solutions.
  4. Simplify the Solutions: If necessary, simplify the solutions by reducing radicals or combining like terms.
  5. Verify the Solutions: To ensure accuracy, substitute each solution back into the original equation to verify that it holds true.

Illustrative Examples

Let's explore some examples to solidify the application of the square root property.

Example 1: Solve the equation (x - 2)² = 9.

  1. The squared term is already isolated.
  2. Take the square root of both sides: √(x - 2)² = ±√9, which simplifies to x - 2 = ±3.
  3. Solve for x:
    • x - 2 = 3 => x = 5
    • x - 2 = -3 => x = -1
  4. The solutions are x = 5 and x = -1.
  5. Verify the solutions:
    • For x = 5: (5 - 2)² = 3² = 9 (True)
    • For x = -1: (-1 - 2)² = (-3)² = 9 (True)

Example 2: Solve the equation (x + 1)² = 50.

  1. The squared term is already isolated.
  2. Take the square root of both sides: √(x + 1)² = ±√50, which simplifies to x + 1 = ±5√2.
  3. Solve for x:
    • x + 1 = 5√2 => x = -1 + 5√2
    • x + 1 = -5√2 => x = -1 - 5√2
  4. The solutions are x = -1 + 5√2 and x = -1 - 5√2.
  5. Verify the solutions (verification is left as an exercise for the reader).

Solving the Given Problem: (x - 1)² = 50

Now, let's apply the square root property to the given equation: (x - 1)² = 50. This problem exemplifies the power and elegance of this method.

1. Isolate the Squared Term: In this case, the squared term, (x - 1)², is already isolated on the left side of the equation. This initial setup makes the application of the square root property particularly straightforward.

2. Take the Square Root of Both Sides: The next step involves taking the square root of both sides of the equation. Remember, when taking the square root, it's crucial to consider both the positive and negative roots, as both will yield valid solutions. This gives us: √(x - 1)² = ±√50. Simplifying the left side, we get x - 1 = ±√50. Now, let's focus on simplifying the square root of 50. We can express 50 as the product of its prime factors: 50 = 2 * 5 * 5 = 2 * 5². Therefore, √50 can be simplified as √(2 * 5²) = 5√2. Substituting this back into our equation, we have x - 1 = ±5√2.

3. Solve for x: To isolate x, we need to add 1 to both sides of the equation. This gives us two separate equations to solve:

  • x - 1 = 5√2 => x = 1 + 5√2
  • x - 1 = -5√2 => x = 1 - 5√2

Thus, we have found two potential solutions for x: x = 1 + 5√2 and x = 1 - 5√2.

4. Simplify the Solutions: In this particular problem, the solutions x = 1 + 5√2 and x = 1 - 5√2 are already in their simplest form. The radical term, √2, cannot be further reduced, and there are no like terms to combine.

5. Verify the Solutions: To ensure the accuracy of our solutions, it's essential to substitute each value of x back into the original equation, (x - 1)² = 50, and verify that the equation holds true.

  • For x = 1 + 5√2:

    Substitute x = 1 + 5√2 into the original equation: [(1 + 5√2) - 1]² = 50. Simplifying the expression inside the parentheses, we get (5√2)². Squaring this term, we have (5√2)² = 5² * (√2)² = 25 * 2 = 50. This confirms that x = 1 + 5√2 is a valid solution.

  • For x = 1 - 5√2:

    Substitute x = 1 - 5√2 into the original equation: [(1 - 5√2) - 1]² = 50. Simplifying the expression inside the parentheses, we get (-5√2)². Squaring this term, we have (-5√2)² = (-5)² * (√2)² = 25 * 2 = 50. This confirms that x = 1 - 5√2 is also a valid solution.

Therefore, by substituting each solution back into the original equation and verifying that it holds true, we can confidently conclude that both x = 1 + 5√2 and x = 1 - 5√2 are the correct solutions to the equation (x - 1)² = 50.

Analyzing the Options

Now, let's analyze the given options in light of our solutions:

A. x = -49 (Incorrect) B. x = 1 - 5√2 (Correct) C. x = 1 + 5√2 (Correct) D. x = 51 (Incorrect)

Therefore, the two correct answers are B. x = 1 - 5√2 and C. x = 1 + 5√2.

Advantages of the Square Root Property

The square root property offers several advantages when solving quadratic equations:

  • Directness: It provides a direct and efficient solution for equations in the form (x - h)² = k.
  • Simplicity: The steps involved are relatively simple and straightforward.
  • Efficiency: It avoids the need for factoring or using the quadratic formula in specific cases.

Limitations of the Square Root Property

While the square root property is a valuable tool, it's essential to recognize its limitations:

  • Specific Form: It is only applicable to equations that can be expressed in the form (x - h)² = k.
  • Not Universal: It cannot be used to solve all quadratic equations, particularly those that do not readily fit the required form.

Conclusion

The square root property is a powerful technique for solving quadratic equations in the form (x - h)² = k. By isolating the squared term, taking the square root of both sides, and solving for x, we can efficiently find the solutions. While not universally applicable to all quadratic equations, it provides a direct and simple approach in specific cases. Mastering this property enhances one's algebraic toolkit and strengthens the ability to solve a wider range of mathematical problems. In the context of the given problem, (x - 1)² = 50, we successfully applied the square root property to determine the solutions x = 1 + 5√2 and x = 1 - 5√2, demonstrating the effectiveness of this method.

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Solve the equation (x−1)2=50(x-1)^2 = 50 and find the values of xx. Choose two correct answers from the options provided.

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Solving Quadratic Equations Using the Square Root Property A Step-by-Step Guide