Question 766155
Let the numbers be x and y.
Given : 
x^2 + y^2 = 4           ....(a)
x*y=1                   ....(b)
Multiplying equation 'b' by 2 on L.H.S and R.H.S,
2*x*y = 2               ....(c)
adding (a) and (c),
x^2 + 2*x*y + y^2 = 6
(x + y)^2 = 6
x + y = sqrt(6) (Taking only the positive square root. if x+y is taken as  -sqrt(6), values of x and y would interchange.)

y = sqrt(6) - x
Thus, x*(sqrt(6) -x ) = 1 (since product of the numbers equals 1).
x*sqrt(6) - x^2  = 1
x^2 - sqrt(6)*x + 1 = 0
Roots of the quadratic equation of the type
a*x^2 + b*x + c = 0 is given by : 
x = ((-b +- sqrt(b^2 - 4*a*c))/(2*a))

Thus, x = (sqrt(6) + sqrt(2))/2 or (sqrt(6) - sqrt(2))/2
y = (sqrt(6) - sqrt(2))/2 or (sqrt(6) + sqrt(2))/2