I'm not sure if I've interpreted this question correctly. I'm comparing the nature of the roots with the nature of the equation. The definition of f contains irrationals, so are you asking if the roots can be rational? "Nature" has not been defined in this context so I'm using the term as related to rationality.
In this case the sum of the roots (zeroes) of f(x) are: sum of roots=6/√5 and their product is 1.
If the roots are a and b then: a+b=6/√5 and ab=1.
b=1/a, a+1/a=6/√5, a2+1=6a/√5, a2-6a√5+1=0 (that is, f(a)=0 and f(1/a)=0, as expected).
If f(x)=V then we have √5x2-6x+√5=V for roots p and q.
p+q=6/√5 (as before), but pq=1-V/√5.
If V=n√5 where n is a rational number, then pq=1-n (rational product).
q=(1-n)/p; p+(1-n)/p=6/√5, √5p2+6p+√5(1-n)=0. When n=1, for example, p=6/√5 and q=0. One of the roots is therefore irrational.
Since p+q=6/√5 for any value of V and V=n√5 (irrational), q is rational but p=6/√5-q which is always irrational.
If V is rational then 1-V/√5 is irrational and pq=1-V/√5 is also irrational and the product of two rational numbers cannot be irrational, therefore at least one of the roots must be irrational.
Any equation involving f(x) will therefore have at least one irrational root.