Solving the Equation: (x-1)² = 4√(x-4)²
This article will guide you through the process of solving the equation (x-1)² = 4√(x-4)². We'll explore the steps involved and discuss the importance of checking for extraneous solutions.
Understanding the Equation
First, let's analyze the given equation:
- (x-1)²: Represents the square of the expression (x-1).
- 4√(x-4)²: Represents 4 times the square root of the square of the expression (x-4).
Solving the Equation
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Simplify the square root: Since we're dealing with the square root of a square, we can simplify the right side of the equation: √(x-4)² = |x-4|
The absolute value is essential because the square root of a square always results in a non-negative value.
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Rewrite the equation: The equation now becomes: (x-1)² = 4|x-4|
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Consider cases: Due to the absolute value, we need to consider two cases:
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Case 1: x-4 ≥ 0 (or x ≥ 4) In this case, |x-4| = x-4. Our equation becomes: (x-1)² = 4(x-4)
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Case 2: x-4 < 0 (or x < 4) In this case, |x-4| = -(x-4). Our equation becomes: (x-1)² = -4(x-4)
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Solve each case:
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Case 1: (x-1)² = 4(x-4) Expand the squares and simplify: x² - 2x + 1 = 4x - 16 x² - 6x + 17 = 0 This quadratic equation does not factor easily. We can use the quadratic formula to find the solutions: x = (6 ± √(6² - 4 * 1 * 17)) / (2 * 1) x = (6 ± √(-32)) / 2 x = (6 ± 4i√2) / 2 x = 3 ± 2i√2
These solutions are complex numbers and are valid solutions for the original equation.
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Case 2: (x-1)² = -4(x-4) Expand the squares and simplify: x² - 2x + 1 = -4x + 16 x² + 2x - 15 = 0 Factor the quadratic: (x+5)(x-3) = 0 Therefore, x = -5 or x = 3.
However, both of these solutions fall outside the condition for Case 2 (x < 4), which means they are extraneous solutions.
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Checking for Extraneous Solutions
It's crucial to check our solutions against the original equation to ensure they are valid. We already determined that the solutions for Case 1 are complex numbers and are valid. However, the solutions for Case 2, x = -5 and x = 3, need to be checked.
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For x = -5: (-5-1)² = 4√(-5-4)² 36 = 4√81 36 = 36 This solution is valid.
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For x = 3: (3-1)² = 4√(3-4)² 4 = 4√1 4 = 4 This solution is valid.
Final Solution
Therefore, the final solutions for the equation (x-1)² = 4√(x-4)² are:
- x = 3 ± 2i√2 (Complex solutions from Case 1)
- x = -5 (Valid solution from Case 2)
- x = 3 (Valid solution from Case 2)