Final answer:
The quadratic sequence with a second constant difference of 4 is solved to find x = 12 and y = 21.
Step-by-step explanation:
The student asked to calculate the values of x and y in a quadratic sequence 5; x; y; 29; where the second constant difference is equal to 4. A quadratic sequence has the form t_n = an^2 + bn + c, where t_n represents the nth term of the sequence, and a, b, and c are constants. The second difference being constant and equal to 4 indicates that 2a = 4, hence a = 2.
Since the first term is 5, we have c = 5. With this information, we have the sequence 5; 5 + 2 + b; 5 + 8 + 3b; 29. Solving these equations for b and the subsequent terms for x and y, we get that x is the second term and y is the third term of the sequence.
The value of b comes out to be 1, leading to the sequence being of the form t_n = 2n^2 + n + 2. Thus, x = 12 and y = 21.