Final answer:
Complex roots for the equation 2x^2 + 5 = 0 are ±i√(5/2), no solutions for √(x + 3) = -2, the solutions for the system {2x - y = 4, x + 3y = 5} are x = 2 and y = 0, and the roots of 3x^2 - 6x + 2 = 0 are (3 ± √3)/3.
Step-by-step explanation:
To solve the equation 2x^2 + 5 = 0 for complex roots, we move 5 to the other side of the equation to get 2x^2 = -5, and then divide by 2 to isolate x^2, resulting in x^2 = -5/2. Taking the square root of both sides gives us x = ±√(-5/2), which means the complex roots are ±i√(5/2).
For the equation √(x + 3) = -2, no real solutions exist because the square root of a real number cannot be negative. Therefore, there are no real or complex solutions to this equation, since the square root function outputs only nonnegative results for real inputs.
To solve the system of equations {2x - y = 4, x + 3y = 5} for complex values, we can use substitution or elimination method. Let's use elimination by multiplying the second equation by 2 and adding to the first equation. We get 4x + 6y = 10, and adding this to 2x - y = 4, gives us 6x + 5y = 14. Solving these equations simultaneously, we find the solutions x = 2 and y = 0.
To determine the roots of the equation 3x^2 - 6x + 2 = 0, we can use the quadratic formula, x = [-b ± √(b^2 - 4ac)]/(2a). Plugging in the values gives us x = [6 ± √(36 - 24)]/6, which simplifies to x = [6 ± √12]/6. Therefore, the roots are (3 ± √3)/3.