Dichloroethanoic acid has pKa = 1.29. A solution has a concentration of 0.10 mol/L…

Step 1. $K_a = 10^{-1.29} = 0.0513$.

Step 2. Approximation. $\sqrt{0.0513 \times 0.10} = 0.0716\ \text{mol/L}$, which is larger than it should be and gives $\text{pH} = 1.15$.

Step 3. Exact. The square root is $\sqrt{0.0513^2 + 4 \times 0.0513 \times 0.10} = 0.1522$. Then $x = \dfrac{-0.0513 + 0.1522}{2} = 0.0504\ \text{mol/L}$, and $\text{pH} = 1.30$.

Step 4. Ionised. $\dfrac{0.0504}{0.10} \times 100 = 50.4\ \%$.

Why the approximation fails. It takes $c - x$ as $c$. Here half the acid is ionised, so $c - x$ is only half of $c$.