You are mistaken, there is no square root.
You have a degree 6 polynomial that you multiply by x, giving a degree 7 polynomial. You integrate that, giving a degree 8 polynomial.
You have a degree 6 polynomial that you integrate, giving a degree 7 polynomial.
You have the ratio of those two equaling a constant. Multiply through by the degree 7 polynomial to get degree 7 on one side and degree 8 on the other. Bring all the terms to the same side and you have a degree 8 polynomial. That has 7 roots.
No square roots involved.
No particular reason to favor one over another if they are real-valued (and perhaps favour real and positive)
syms x
Q = @(v) sym(v);
y=(-Q(167345)*x^6+Q(200645)*x^5-Q(90447)*x^4+Q(18683)*x^3-Q(1852.8)*x^2+Q(58.263)*x+Q(48.285))
part1 = int(x*y, [x Q(0.420305)])
part2 = int(y,[x Q(0.420305)])
z = Q(0.32446) == part1 / part2
k = solve(z,x)
vk = vpa(k)
real_positive = vk(imag(vk) == 0 & vk > 0)
