From Algebra, the statement is equivalent to say that $(x^2− 11x + 22)_{r}$ = $(x − 3)_{r} \cdot (x − 6)_{r}$. Doing operations we arrive at $3 + 6 = 11_{r} = r + 1$, and $(3)(6) = 22_{r} = 2 \cdot 11_{r}$. In any case, $r = 8$.
This is the solution to the problem, but how do I yield to the conclusion that $3 + 6 = 11_{r} = r + 1$?
If the roots are $3$ and $6$, the equation is $(x-3)(x-6)=x^2-(3+6)x+3\cdot 6=0$ So $11_r=3+6$ and $11$ in base $r$ is $r+1$ because the leading digit is $r$. Similarly we have $3\cdot 6 = 22_r=2r+2$ Each of these gives us $r=8$
Multiply through $(x-3)(x-6)=x^2-9x+18$ (the coefficients are in base $10$).
Now the coefficients in the original equation are $11_r =r+1$ and $22_r=2r+2$ because $ab_r=ar+b$ where the suffix $r$ indicates base $r$.
So the original equation is $x^2-(r+1)x+(2r+2)=0$ and that has to be the same as the equation converted to base $10$.
Hence, equating coefficients, $r+1=9$ and $2r+2=18$ whence $r=8$ and the question is well-posed because the equations are consistent.
The rule stating that in the equation $x^2 - sx + p^2 = 0$ the sum of the roots is $s$ and their product is $p$ is true whatever the basis is. Just write the equation as $(x-r_1)(x-r_2)$ where $r_1$ and $r_2$ are the roots and do the multiplication.
