Dimensional analysis

Treat units as algebra. Both sides of a true equation must reduce to the same combination of metres, kilograms and seconds, and any term you add to another must already match it.

Check $v^2 = v_0^2 + 2a(x - x_0)$. The left side is $\text{m}^2/\text{s}^2$. On the right, $a(x-x_0)$ is $(\text{m/s}^2)(\text{m}) = \text{m}^2/\text{s}^2$. Both sides agree, and the pure number 2 is invisible to the test.

Check $T_s = 2\pi\sqrt{m/k}$. A spring constant is in $\text{N/m}$, which unpacks to $\text{kg/s}^2$, so $m/k$ is in $\text{s}^2$ and the square root is in seconds. Correct.

On the multiple choice section this is a scoring technique rather than a study aid. When four symbolic options differ in structure, work out the units of each and cross off the ones that cannot be a force, an energy or a time. The AP Physics C: Electricity and Magnetism CED lists it among its instructional strategies, noting that dimensional analysis of units can be used to determine whether a given calculation will produce the desired answer.

Its limit is precision. Units cannot catch a missing factor of 2, a wrong sign, or a swapped sine and cosine, since all of those are dimensionless. Passing the check means an option survives, not that it is right.