Let the set of all values of k R such that the equation z( z + 2 + i) + k(2 + 3i) = 0 , z C , has at least one solution, be the interval [ , ] . Then 9( + ) is equal to:

Step-by-step solution

Let z = x + iy , then z = x - iy . Substituting z into the given equation: (x + iy)(x - iy + 2 + i) + k(2 + 3i) = 0 x^2 + y^2 + 2x + ix + 2iy - y + 2k + 3ki = 0 Separating the real and imaginary parts, we get: Real part: x^2 + y^2 + 2x - y + 2k = 0 Imaginary part: x + 2y + 3k = 0 x = -2y - 3k Substituting x into the real part equation: (-2y - 3k)^2 + y^2 + 2(-2y - 3k) - y + 2k = 0 4y^2 + 12ky + 9k^2 + y^2 - 4y - 6k - y + 2k = 0 5y^2 + (12k - 5)y + 9k^2 - 4k = 0 For the equation to have at least one solution z C , t