Bunuel
If \(x\neq{0}\), then \(\frac{\sqrt{x^2}}{x}=\)
A. -1
B. 0
C. 1
D. x
E. \(\frac{|x|}{x}\)
General rule: \(\sqrt{x^2}=|x|\).
When we see the equation of a type: \(y=\sqrt{x^2}\) then \(y=|x|\), which means that \(y\) cannot be negative but \(x\) can.
\(y\) cannot be negative as \(y=\sqrt{some \ expression}\), and even root from the expression (some value) is never negative (as for GMAT we are dealing only with real numbers).
When the GMAT provides the square root sign for an even root, such as a square root, then the only accepted answer is the positive root.
That is, \(\sqrt{16} = 4\), NOT +4 or -4. In contrast, the equation x^2 = 16 has TWO solutions, +4 and -4.
Even roots have only a positive value on the GMAT.
Odd roots will have the same sign as the base of the root. For example, CBRT(64) = 4, CBRT(-27) = -3.
In our original question we have: \(\frac{\sqrt{x^2}}{x}=\frac{|x|}{x}\), if we knew that \(x\) is positive then the answer would be 1, if we knew that \(x\) is negative the answer would be -1. BUT we don't know the sign of x, hence we cannot simplify expression \(\frac{|x|}{x}\) further.
Hope it's clear.
Experts -
To clarify, for the general rule mentioned above: x2−−√=|x|x2=|x|
X can be - or +, correct? When I saw just the absolute value bars, I was not sure if you should only be taking the positive solution or if the absolute value bars mean you need to then solve for the absolute value by having one positive and one negative case for x.
Many thanks!