Final answer:
The question requires solving a problem involving Pythagorean triples and algebra to determine the ages of three children. The children are 8, 15, and 17 years old, corresponding to the Pythagorean triple (8, 15, 17).
Step-by-step explanation:
The question states that a man has three children whose ages form a Pythagorean triple and provides information about the relative differences in their ages. Specifically, it states that the oldest child is nine years older than the youngest, and the middle child is two years younger than the oldest.
To solve this problem, we can denote the age of the youngest child as 'y'. According to the given information, the ages of the three children are then y (youngest), y + 7 (middle), and y + 9 (oldest). Since their ages form a Pythagorean triple, we can set up the following equation:
y^2 + (y + 7)^2 = (y + 9)^2
Expanding the squares and simplifying gives:
y^2 + y^2 + 14y + 49 = y^2 + 18y + 81
Combining like terms and simplifying further:
2y^2 + 14y + 49 = y^2 + 18y + 81
y^2 - 4y - 32 = 0
Factoring the quadratic equation:
(y - 8)(y + 4) = 0
Since age cannot be negative, we take the positive solution:
y = 8
Thus, the ages of the three children are 8 (youngest), 15 (middle), and 17 (oldest) years old.