problem: Sviatoslav solved the quadratic equation $x^2-x-1=0$ by completing the square. In the process, he came up with the equivalent equation $$(x+a)^2 = b,$$where $a$ and $b$ are constants.
  
  What is $b$?
solution: The square which agrees with $x^2-x-1$ except for the constant term is $\left(x-\frac 12\right)^2$, which is equal to $x^2-x+\frac 14$, and thus to $(x^2-x-1) + \frac 54$.
  
  Therefore, by adding $\frac 54$ to each side, Sviatoslav rewrote the equation $x^2-x-1 = 0$ as $$\left(x-\frac 12\right)^2 = \frac 54.$$We have $a=-\frac 12$ and $b=\boxed{\frac 54}$.