Solving the Equation: (x1)² = 4√(x4)²
This article will guide you through the process of solving the equation (x1)² = 4√(x4)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:
 (x1)²: Represents the square of the expression (x1).
 4√(x4)²: Represents 4 times the square root of the square of the expression (x4).
Solving the Equation

Simplify the square root: Since we're dealing with the square root of a square, we can simplify the right side of the equation: √(x4)² = x4
The absolute value is essential because the square root of a square always results in a nonnegative value.

Rewrite the equation: The equation now becomes: (x1)² = 4x4

Consider cases: Due to the absolute value, we need to consider two cases:

Case 1: x4 ≥ 0 (or x ≥ 4) In this case, x4 = x4. Our equation becomes: (x1)² = 4(x4)

Case 2: x4 < 0 (or x < 4) In this case, x4 = (x4). Our equation becomes: (x1)² = 4(x4)


Solve each case:

Case 1: (x1)² = 4(x4) 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.

Case 2: (x1)² = 4(x4) Expand the squares and simplify: x²  2x + 1 = 4x + 16 x² + 2x  15 = 0 Factor the quadratic: (x+5)(x3) = 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.

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.

For x = 5: (51)² = 4√(54)² 36 = 4√81 36 = 36 This solution is valid.

For x = 3: (31)² = 4√(34)² 4 = 4√1 4 = 4 This solution is valid.
Final Solution
Therefore, the final solutions for the equation (x1)² = 4√(x4)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)