Find the base numeric system Find the numeric base we are using if $ x = 4 $ and $ x = 7 $ are zeros of $ 5x^2 - 50x + 118. $
When I plug in $ x = 4 $ and $ x = 7, $ I receive $ -2 $ and $ 13, $ respectively, but how do I proceed from that?
 A: Let $b$ be the base in question. Presumably $b>8$, since $8$ appears as a digit in $118$. We also know that $5x^2-50x+118$ is a multiple of $(x-4)(x-7)$, and since the $x^2$ term of $(x-4)(x-7)$ is simply $x^2$, we must have
$$5x^2-50x+118=5(x-4)(x-7)\;.$$
From this we deduce that $55$ in base ten is $50_b$. This clearly implies that $b=11$ (in base ten), and this is confirmed by the fact that $5\cdot4\cdot7=140$ in base ten is $118_b$: with $b=11$ this is $121+11+8=140$.
A: An easy way to do this is to convert to decimal from the unknown base $n$. The solutions are single digits, so they will remain unchanged in value. The equation itself will become:
$$5x^2 - 5nx + n^2 + n + 8 = 0$$
By putting $x=4$ and $x=7$ sucessively, you will find that the solutions to the resulting quadratics are $8,11$ and $11,23$ respectively. Verifying by testing these bases with the original equation shows that $11$ is the only admissible solution.
A: $$5_bx^2-50_bx+118_b=5_b(x-4_b)(x-7_b)$$
From $118_b$, we can assume that $b > 8$.
This implies $5_b = 5,\; 4_b = 4,\; 7_b = 7,\;$ and $\; 118_b = b^2 + b + 8$.
Hence
\begin{align}
   b^2 + b + 8 &= 5 \times 4 \times 7 \\
   b^2 + b - 132 &= 0 \\
   (b - 11)(b + 12) &= 0 \\
   b &= 11
\end{align}
CHECK:
\begin{align}
   5_{11}(x-4_{11})(x-7_{11})
   &= 5_{11}(x^2 - 10_{11}x + 26_{11}) \\
   &= 5_{11}x^2 - 50_{11}x +118_{11} \checkmark
\end{align}
A: Sum of the roots is $(5b + 0)/5 = b$ and hence $b=11$.
A: Here's a different way to do it.
$f (x)=5x^2 - 50_bx +118_b =$
$5x (x-b)+118_b $
$f (4)=20 (4-b)  +118_b$
$f (7)=35 (7-b)+118_b $ 
So $118_b= 20 (b-4)=35 (b-7) $
So $15b =7*35-80$
$b=11$.
Or we could note $35|118_b $ and $20|118_b $ so $140|118_b $ and $7|b-4$ and $4|b-7$.  If $140 =118_b $ then $b=11$ is a good guess.
And here's a third way:
$5*4^2 - 4*(5b+0) + (b^2+b+8)=0$
$b^2 -19b +88=0$
$b= \frac {19 \pm \sqrt {19^2- 4*88}}{2}= 8;11$
$5*7^2 -7*5b+(b^2+b+8)=0$
$b^2 -34b+253=0$
$b= \frac {34 \pm \sqrt {34^2- 4*253}}{2}= 11;23$
So $b=11$
Still my favorite way is $5 (x-4)(x-7) =5x^2-55x+140= 5x^2 -50_bx + 118_b$ So $50_b=5b=55$ and $118_b =b^2+b+8=140$ so $b=11$.
