Each of the equations $3x^2-2=25, (2x-1)^2=(x-1)^2, \sqrt{x^2-7}=\sqrt{x-1}$ has: 
$\textbf{(A)}\ \text{two integral roots}\qquad\textbf{(B)}\ \text{no root greater than 3}\qquad\textbf{(C)}\ \text{no root zero}\\ \textbf{(D)}\ \text{only one root}\qquad\textbf{(E)}\ \text{one negative root and one positive root}$
Solution
Since the question asks us about the unifying characteristic of all three equations' roots, we have to first determine them.
$3x^2-2 = 25$ can be rewritten as $3x^2 - 27 = 0$, which gives the following roots $+3$ and $-3$.
$(2x-1)^2 = (x-1)^2$ can be expanded to $4x^2-4x+1=x^2-2x+1$, which in turn leads to $3x^2-2x=0$. The roots here are $0$ and $\frac{2}{3}$.
$\sqrt{x^2-7}=\sqrt{x-1}$, when squared, also turns into a quadratic equation: $x^2 - x - 6 = 0$. Binomial factoring gives us the roots $-2$ and $3$.
We can clearly see that, between all of the equations, there is $\boxed{\textbf{(B)} \text{no root greater than 3}}$.
Note: There are probably extraneous solutions somewhere, but that does not affect the solution.