Question

If 251 base x =100, find x​

Answer 1

If $251_x\equiv100_{10}$, then this translates to the equation

$251_x = 2x^2 + 5x + 1 = 100$

Solve for $x$.

$2x^2 + 5x = 99$

$2\left(x^2 + \dfrac52 x\right) = 99$

$2 \left(x^2 + \dfrac52 x + \dfrac{25}{16}\right) - \dfrac{25}8 = 99$

$2 \left(x + \dfrac54\right)^2 = \dfrac{817}8$

$\left(x + \dfrac54\right)^2 = \dfrac{817}{16}$

$x + \dfrac54 = \pm \dfrac{\sqrt{817}}4$

$x = \dfrac{-5\pm\sqrt{817}}4$

though it is a bit unusual (but not entirely out of the question) to have an irrational base number system.

In case you meant $100_2$ on the right side, then $100_2 = 2^2 = 4_{10}$, so that

$2x^2 + 5x + 1 = 4 \\\\ 2x^2 + 5x - 3 = 0 \\\\ (x+3) (2x-1) = 0 \\\\ x=-3 \text{ or } x = \dfrac12$

which at first glance seem like more reasonable choices of base.