To solve the problem, we can use algebraic equations. Let x be the smaller real number and y be the larger real number. We know that y = x + 4 and that 8y + x^2 = 48.
Substituting y = x + 4 into the second equation, we get:
8(x + 4) + x^2 = 48
Expanding and rearranging, we get:
x^2 + 8x - 16 = 0
Using the quadratic formula, we can solve for x:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
where a = 1, b = 8, and c = -16. Plugging in these values, we get:
x = (-8 ± sqrt(8^2 - 4(1)(-16))) / 2(1)
Simplifying, we get:
x = (-8 ± sqrt(96)) / 2
x = (-8 ± 4sqrt(6)) / 2
x = -4 ± 2sqrt(6)
Since x is a positive real number, we take the positive root:
x = -4 + 2sqrt(6)
Now that we have found x, we can use y = x + 4 to find y:
y = -4 + 2sqrt(6) + 4
y = 2 + 2sqrt(6)
Therefore, the two real numbers are approximately -4 + 2sqrt(6) and 2 + 2sqrt(6).