Let's denote the two consecutive positive even integers as x and x+2.
We know their squares sum up to 100, represented mathematically as:
x^2 + (x + 2)^2 = 100.
Expanding this equation, we get:
x^2 + x^2 + 4x + 4 = 100.
Combining like terms gives us:
2x^2 + 4x - 96 = 0.
This is a quadratic equation in the form of ax^2 + bx + c = 0, where a = 2, b = 4, and c = -96.
We can solve this quadratic equation for x using the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / 2a.
Substituting a, b, and c into this formula, we find that:
x = [-4 ± sqrt((4)^2 - 4*2*-96)] / 2*2,
x = [-4 ± sqrt(16 + 768)] / 4,
x = [-4 ± sqrt(784)] / 4,
x = [-4 ± 28] / 4.
So the solutions are x = (28 - 4) / 4 = 6 and x = (-28 - 4) / 4 = -8.
Since we are looking for positive even integers, we discard the negative solution.
Therefore, the two consecutive positive even integers are 6 and 6 + 2 = 8.