Approximating the concentration of ions when a salt dissociates In my textbook written that:


  Given that $K_\mathrm{sp}(\ce{Ag2CrO4})=9.0\times 10^{-12}$, consider the solubility of $\ce{Ag2CrO4}$ in a $0.100\:\mathrm{M}$ solution of $\ce{AgNO3}$.
  
  Initial concentrations (before any $\ce{Ag2CrO4}$ dissolves) are,
  $[\ce{Ag+}]_0=0.100\:\mathrm{M}$ (from $\ce{AgNO3})$ and $[\ce{CrO4^{2-}}]_0=0\:\mathrm{M}$
  
  The system comes to equilibrium as $\ce{Ag2CrO4}$ dissolves according to the reaction:
  $$\ce{Ag2CrO4(s)\rightleftharpoons 2Ag+(aq) + CrO4^{2-}(aq)}$$
  $$K_\mathrm{sp}=[\ce{Ag+}]^{2}[\ce{CrO4^{2-}}]=9.0\times 10^{-12}$$
  
  We assume that $x\:\mathrm{M}$ of $\ce{Ag2CrO4}$ dissolves to reach equilibrium, which means that:
  $$x\:\mathrm{M\:\ce{Ag2CrO4}(s)}\rightarrow 2x\:\mathrm{M\:\ce{Ag+}(aq)}+x\:\mathrm{M\:\ce{CrO4^{2-}}\:(aq)}$$
  
  Now we can specify the equilibrium concentrations in terms of $x$:
  $$[\ce{Ag+}]=[\ce{Ag+}]_{0}+\Delta[\ce{Ag+}]=(0.100+2x)\:\mathrm{M}$$
  $$[\ce{CrO4^2-}]=[\ce{CrO4^2-}]_{0}+\Delta[\ce{CrO4^2-}]=(0+x)\:\mathrm{M}$$
  
  Giving:
  $$9.0\times 10^{-12}=[0.100+2x]^{2} \cdot [x]$$


Also, here is line I want to discuss (again, taken from the textbook):


  Since the $K_{\mathrm{sp}}$ value for $\ce{Ag2CrO4}$ is small (the position of the equilibrium lies far to the left), $x$ is expected to be small compared with $0.100\:\mathrm{M}$. Therefore:
  $${9.0\times 10^{-12}=[0.100+2x]^{2}[x] \approx [0.100}]^{2}[x]$$


When can this approximation be made and when can it not be made?
Really, your question seems to be a math question.  

You need to solve the cubic equation:

$$\mathrm{9.0\cdot10^{-12}=[0.100+}2x]^{2}[x]$$

Cubic equations are exactly solvable; however, it is common to use an approximate method.  

Whether or not it is appropriate to approximate 0.100 +2x as 0.100 depends upon how small x is and how precise an answer you need.

Once you get an approximate solution for x, you could recalculate 0.100 +2x and use that value to re-solve the problem. By repeating this process you can get arbitrary close to the correct value by iteration.