x%5E2-(3-2i)y%3D2x-3y%2B5i.jpeg "(2+3i)x^2-(3-2i)y=2x-3y+5i")
Solving Complex Quadratic Equations: (2 + 3i)x^2 - (3 - 2i)y = 2x - 3y + 5i
This article will explore how to solve a complex quadratic equation involving both real and imaginary variables. Let's consider the equation:
(2 + 3i)x^2 - (3 - 2i)y = 2x - 3y + 5i
To tackle this problem, we need to separate the real and imaginary parts of the equation.
Isolating Real and Imaginary Components
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Expand the equation: (2 + 3i)x^2 - (3 - 2i)y = 2x - 3y + 5i 2x^2 + 3ix^2 - 3y + 2iy = 2x - 3y + 5i
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Group the real and imaginary terms: (2x^2 - 3y) + (3x^2 + 2y)i = (2x - 3y) + 5i
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Equate the real and imaginary parts: Real Part: 2x^2 - 3y = 2x - 3y Imaginary Part: 3x^2 + 2y = 5
Solving the System of Equations
We now have a system of two equations with two unknowns:
- 2x^2 - 2x = 0
- 3x^2 + 2y = 5
Let's solve for x and y:
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Solving for x:
- Factor out 2x from the first equation: 2x(x - 1) = 0
- This gives us two possible solutions: x = 0 and x = 1
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Solving for y:
- Substitute each value of x into the second equation:
- If x = 0: 3(0)^2 + 2y = 5 => y = 5/2
- If x = 1: 3(1)^2 + 2y = 5 => y = 1
- Substitute each value of x into the second equation:
The Solutions
Therefore, the solutions to the complex quadratic equation (2 + 3i)x^2 - (3 - 2i)y = 2x - 3y + 5i are:
- x = 0, y = 5/2
- x = 1, y = 1
Conclusion
Solving complex quadratic equations requires careful separation of real and imaginary components, leading to a system of equations that can be solved to find the solutions. This approach allows us to handle complex variables and find solutions for both x and y in the equation.