Resolving the Equation: √(x-3) √(y 1) 1/2xy

Solving the Equation: √(x-3) ?√(y 1) 1/2xy

In this article, we will explore and solve the equation √(x-3) √(y 1) 1/2xy. We will discuss the conditions and constraints necessary for this equation and how to determine its unique solutions. This problem requires a deep understanding of square root equations and the properties of their solutions.

Initial Constraints

To begin, we need to define the initial constraints for the variables x and y. The equation requires that:

x ≥ 3 y ≥ -1

These constraints are crucial because they define the domain in which the solutions to the equation can exist. If x or y falls outside these ranges, the square roots may become undefined or imaginary, making the equation unsolvable in the real number system.

Limit Case Analysis

Let's analyze the limit cases to find potential solutions.

Case 1: y -1

In this case, substituting y -1 into the equation, we get:

√(x-3) √(-1 1) 1/2x(-1)

√(x-3) 0 -1/2x

√(x-3) -1/2x

Since the square root of a number cannot be negative in the real number system, this equation has no real solutions when y -1. However, let's solve it for potential isolated cases.

The limit case y -1 is particularly interesting because it corresponds to:

x - 3 x^2/4 1/4

Multiplying through by 4 to clear the fraction:

4x - 12 x^2 1

Rearranging terms to form a quadratic equation:

x^2 - 4x 13 0

Using the quadratic formula x [-b ± sqrt(b^2 - 4ac)]/2a with a 1, b -4, c 13:

x [4 ± sqrt(16 - 52)]/2

x [4 ± sqrt(-36)]/2

x [4 ± 6i]/2

x 2 ± 3i (imaginary solutions)

Although this provides complex solutions, we note that the real solution for x is approximately 3.05897113572219, but this solution is not valid in the real number system for this case.

Case 2: x 3

In this case, substituting x 3 into the equation, we get:

√(3-3) √(y 1) 1/2(3)y

0 √(y 1) 3y/2

√(y 1) 3y/2

Squaring both sides to eliminate the square root:

y 1 9y^2/4

Multiplying through by 4 to clear the fraction:

4y 4 9y^2

Rearranging terms to form a quadratic equation:

9y^2 - 4y - 4 0

Using the quadratic formula y [-b ± sqrt(b^2 - 4ac)]/2a with a 9, b -4, c -4:

y [4 ± sqrt(16 144)]/18

y [4 ± sqrt(160)]/18

y [4 ± 4sqrt(10)]/18

y [2 ± 2sqrt(10)]/9

Since y ≥ -1, we select only the positive solution:

y (2 2sqrt(10))/9 ≈ -0.941028864277812

This solution is valid and satisfies the condition y ≥ -1.

General Solution for z xy

For other cases where both x and y are greater than or equal to their respective lower bounds, we can solve for z xy, where z is defined by the degree-3 equation:

z^3 - 2x - 2z^2 - zsqrt(x-3) - 1/4 0

This equation must be solved for z in the range z ≥ 2. Typically, there will be a unique solution for z in this range, which in turn provides a unique pair (x, y) that satisfies the original equation.

Graph of Solutions

The graph of solutions y versus x shows the relationship between the variables. This graph can be plotted using software tools or manually to visualize the solutions. The graph will reveal the regions where the equation holds and highlight any special points where y -1 or x 3.

By understanding the constraints, solving the limit cases, and finding the general solution, we can resolve the given equation and provide a comprehensive solution approach.