21. For certain values of $x$, the list $x, x+6$ and $x^{2}$ contains just two different numbers. How many such values of $x$ are there?
A 1
В 2
C 3
D 4
E 5
21. D Clearly, there are no values of $x$ for which $x=x+6$. So the possible values of $x$ will be solutions of the equation $x=x^{2}$ or the equation $x+6=x^{2}$.
Consider the first equation: $x=x^{2}$. This simplifies to $x(1-x)=0$. So $x=0$ or $x=1$.
The equation $x+6=x^{2}$ simplifies to $x^{2}-x-6=0$. So $(x-3)(x+2)=0$.
Therefore $x=3$ or $x=-2$. So there are four different values of $x$.