Respuesta :
If [tex]251_x\equiv100_{10}[/tex], then this translates to the equation
[tex]251_x = 2x^2 + 5x + 1 = 100[/tex]
Solve for [tex]x[/tex].
[tex]2x^2 + 5x = 99[/tex]
[tex]2\left(x^2 + \dfrac52 x\right) = 99[/tex]
[tex]2 \left(x^2 + \dfrac52 x + \dfrac{25}{16}\right) - \dfrac{25}8 = 99[/tex]
[tex]2 \left(x + \dfrac54\right)^2 = \dfrac{817}8[/tex]
[tex]\left(x + \dfrac54\right)^2 = \dfrac{817}{16}[/tex]
[tex]x + \dfrac54 = \pm \dfrac{\sqrt{817}}4[/tex]
[tex]x = \dfrac{-5\pm\sqrt{817}}4[/tex]
though it is a bit unusual (but not entirely out of the question) to have an irrational base number system.
In case you meant [tex]100_2[/tex] on the right side, then [tex]100_2 = 2^2 = 4_{10}[/tex], so that
[tex]2x^2 + 5x + 1 = 4 \\\\ 2x^2 + 5x - 3 = 0 \\\\ (x+3) (2x-1) = 0 \\\\ x=-3 \text{ or } x = \dfrac12[/tex]
which at first glance seem like more reasonable choices of base.
