Consider the following statements : I. x + x+1=0 has two irrational...

For Statement I:
Let $y = \sqrt{x}$, where $y \ge 0$.
The equation becomes $y^2 + y + 1 = 0$.
The discriminant is $\Delta = 1^2 - 4(1)(1) = -3 < 0$.
Since the discriminant is negative, there are no real roots. Thus, Statement I is incorrect.

For Statement II:
Let $y = \sqrt{x}$, where $y \ge 0$.
The equation becomes $y^2 - 5y + 4 = 0$.
Factoring the quadratic gives $(y - 1)(y - 4) = 0 \Rightarrow y = 1$ or $y = 4$.
Substituting $y = \sqrt{x}$, we get $x = 1$ or $x = 16$.
Both $1$ and $16$ are rational numbers. Thus, Statement II is correct.

Therefore, only Statement II is correct.

Answer: II only