Algebraically they are the same. We’re not talking about equations/functions as there are no dependent variables, so why do graphical discontinuities apply? If I were to multiply quantity B by one as (x+2)/(x+2) I would arrive at the equation on the left and have the same singularity.
I’m having trouble understanding when “real math” applies to the GRE. This real math seems to arbitrarily apply whereas things like sqrt(2) = +2 only or x^4=16 only having real solutions ignore the “real math.”
Can you provide clarity for this specific question, but also broader guidance on when to overthink and when to be simple on the quant portion?
There’s a removable discontinuity at x = 0 for \frac{x^2}{x}, whereas the function x is continuous at x = 0.
\sqrt{x} is conventionally defined to refer to the principal square root. The principal square root is only concerned with the unique non-negative square root. The motivation for this is so that we have an operator that can “undo” the squaring operation uniquely.
This doesn’t disagree with the notion that \sqrt{16} = 4.
To solve the equation x^2 = 16, we can take the square root (principal square root is implicitly implied here) of both sides:
\sqrt{x^2} = \sqrt{16}
By definition, \sqrt{x^2} = |x|. Consequently, we’re left with the equation |x| = 4, which you know how to solve.
(I’ve unintentionally changed your question from an x^4 to an x^2 because i think this example is more instructive. With x^4, we would get \sqrt{x^4} = |x^2| = x^2, which doesn’t directly address the issue at hand.)
I’m still confused as to why we’re considering these equations, as there is nothing in the question that hints that they are. With no dependent variables, should I always assume that variables present imply equations? Your response makes sense for functions, but (dis)continuities don’t make any sense when we’re talking about strictly algebraic manipulation of variables.
