Obtain a quadratic polynomial
p(x) = ax² + bx+c, where sum of zeros is
√2 and product of zeros is 1/3
a < 0.
If the roots are p and q, then
pq = 1/(3a)
p+q = -√2/a
Solving for p and q, we get
(±6/√6 √(3-2a) - 6/√2)/(6a)
If we go for the easy win, with a = -1, that gives us for the roots
p,q = (1/√2 ± 1/√6)
check:
(1/√2 - 1/√6)+(1/√2 + 1/√6) = 1/3
(1/√2 - 1/√6)+(1/√2 + 1/√6) = √6
(x-(1/√2-1/√6))(x-(1/√2+1/√6)) = x^2 - √2 x + 1/3