Calculation of Degree of Dissociation from Ostwald's Dilution Law
The OP asked a simple question to which I gave a complex answer. My daughter called such answers "Dad answers". 

The OP essentially asked "If the dilute solution approximation for $\alpha$ is used when calculating, how do you know how good the approximation is?"

In his answer user porphyrin focused on the pertinent idea. 

Dilute Solution Approximation

Now if we assume that $(1-\alpha) \approx 1$ meaning that very little of the compound dissociated, then we get the standard approximation, $\alpha'$:

$$  \alpha^{'} \approx \sqrt{\dfrac{K_d}{c_{0}}}\tag{1}$$

Exact Solution

The equation below is an exact solution even though it doesn't allow $\alpha$ to be calculated directly. You would have to guess at $\alpha$ values and iterate until you found a solution. 

$$\dfrac{\alpha}{\sqrt{1 - \alpha}} = \sqrt{\dfrac{K_d}{c_{0}}}\tag{2}$$

Estimated % Error

Now let's calculate an estimated % error as:

$$\pu{Est \% Error} = 100 \times \dfrac{\dfrac{\alpha'}{\sqrt{1 - \alpha'}} - \alpha'}{\alpha'}\tag{3}$$

Quadratic Solution

The exact value of $\alpha$ can be calculated directly using a quadratic equation. The solution is given below. 

$$\alpha = - \dfrac{K_d}{2\cdot c_0} + \sqrt{\left(\dfrac{K_d}{2\cdot c_0}\right)^2 + \dfrac{K_d}{c_0}}\tag{4}$$

Data Table

\begin{array}{|c|c|c|c|c|} \hline
(\alpha^{'})^2=K_d/c_0 &amp; \alpha (quadratic) &amp; approx &amp; true\ \% \ error &amp; est.\ \% \ error \\ \hline
1.00000000 &amp; 0.61803399 &amp; 1.00000000 &amp; 61.80339887 &amp;  \\ \hline
0.81000000 &amp; 0.58192705 &amp; 0.90000000 &amp; 54.65856100 &amp; 216.22776602 \\ \hline
0.64000000 &amp; 0.54162637 &amp; 0.80000000 &amp; 47.70329614 &amp; 123.60679775 \\ \hline
0.49000000 &amp; 0.49663670 &amp; 0.70000000 &amp; 40.94810050 &amp; 82.57418584 \\ \hline
0.36000000 &amp; 0.44641839 &amp; 0.60000000 &amp; 34.40306509 &amp; 58.11388301 \\ \hline
0.25000000 &amp; 0.39038820 &amp; 0.50000000 &amp; 28.07764064 &amp; 41.42135624 \\ \hline
0.16000000 &amp; 0.32792156 &amp; 0.40000000 &amp; 21.98039027 &amp; 29.09944487 \\ \hline
0.09000000 &amp; 0.25835623 &amp; 0.30000000 &amp; 16.11874208 &amp; 19.52286093 \\ \hline
0.04000000 &amp; 0.18099751 &amp; 0.20000000 &amp; 10.49875621 &amp; 11.80339887 \\ \hline
0.01000000 &amp; 0.09512492 &amp; 0.10000000 &amp; 5.12492197 &amp; 5.40925534 \\ \hline
0.00810000 &amp; 0.08604108 &amp; 0.09000000 &amp; 4.60119879 &amp; 4.82848367 \\ \hline
0.00640000 &amp; 0.07686397 &amp; 0.08000000 &amp; 4.07996803 &amp; 4.25720703 \\ \hline
0.00490000 &amp; 0.06759286 &amp; 0.07000000 &amp; 3.56123125 &amp; 3.69516947 \\ \hline
0.00360000 &amp; 0.05822699 &amp; 0.06000000 &amp; 3.04498988 &amp; 3.14212463 \\ \hline
0.00250000 &amp; 0.04876562 &amp; 0.05000000 &amp; 2.53124512 &amp; 2.59783521 \\ \hline
0.00160000 &amp; 0.03920800 &amp; 0.04000000 &amp; 2.01999800 &amp; 2.06207262 \\ \hline
0.00090000 &amp; 0.02955337 &amp; 0.03000000 &amp; 1.51124937 &amp; 1.53461651 \\ \hline
0.00040000 &amp; 0.01980100 &amp; 0.02000000 &amp; 1.00499988 &amp; 1.01525446 \\ \hline
0.00010000 &amp; 0.00995012 &amp; 0.01000000 &amp; 0.50124999 &amp; 0.50378153 \\ \hline
0.000001 &amp; 0.0009995 &amp; 0.00100000 &amp; 0.05001250 &amp; 0.05003753 \\ \hline
0.00000001 &amp; 9.9995E-05 &amp; 0.00010000 &amp; 0.00500012 &amp; 0.00500038 \\ \hline
1E-10 &amp; 9.99995E-06 &amp; 0.00001000 &amp; 0.00050000 &amp; 0.00050000 \\ \hline
\end{array}

So the Est % Error from equation (3) works quite well. 


When the Est % Error is 5% or less, $\alpha' \le 0.1$, the Est % Error is quite good. 
So for the approximate equation when $\alpha^{'} = 0.1$ you get 1 significant figure, 0.01 gets two significant figures and so on. The improvement doesn't continue forever of course - this is chemistry not math. To get three significant figures or more you'd have to be a fastidious experimenter. 
For the approximate equation when $\alpha' = 0.7$ the estimated error is 82.6% which is much too large, but an 82.6% error still tells you that the  approximation $\alpha^{'}$ is no good. 
To make user M. Farooq's point in another way, by the time you calculate the approximation from equation (1) and the Est % Error from equation (3), you've pretty much done enough work to calculate the exact answer using equation (4). So why bother with the approximation?




EDIT A comment by user Buck Thorn "I never thought about the usefulness of the estimate to bracket the true value" made me think. 

It turns out that we can bracket the true value $\alpha$ using a simple calculation of the approximate value $\alpha'$: 

$$\alpha'\cdot(1-\alpha') \lt \alpha &lt; \alpha'\tag{5}$$

\begin{array}{|c|c|c|c|} \hline
(\alpha^{'})^2=K_d/c_0 &amp; \alpha (quadratic) &amp; \alpha'\cdot(1-\alpha') &amp; \alpha' \\ \hline
1.0000000000 &amp; 0.6180339887 &amp; 0.0000000000 &amp; 1.0000000000 \\ \hline
0.8100000000 &amp; 0.5819270490 &amp; 0.0900000000 &amp; 0.9000000000 \\ \hline
0.6400000000 &amp; 0.5416263691 &amp; 0.1600000000 &amp; 0.8000000000 \\ \hline
0.4900000000 &amp; 0.4966367035 &amp; 0.2100000000 &amp; 0.7000000000 \\ \hline
0.3600000000 &amp; 0.4464183905 &amp; 0.2400000000 &amp; 0.6000000000 \\ \hline
0.2500000000 &amp; 0.3903882032 &amp; 0.2500000000 &amp; 0.5000000000 \\ \hline
0.1600000000 &amp; 0.3279215611 &amp; 0.2400000000 &amp; 0.4000000000 \\ \hline
0.0900000000 &amp; 0.2583562262 &amp; 0.2100000000 &amp; 0.3000000000 \\ \hline
0.0400000000 &amp; 0.1809975124 &amp; 0.1600000000 &amp; 0.2000000000 \\ \hline
0.0100000000 &amp; 0.0951249220 &amp; 0.0900000000 &amp; 0.1000000000 \\ \hline
0.0081000000 &amp; 0.0860410789 &amp; 0.0819000000 &amp; 0.0900000000 \\ \hline
0.0064000000 &amp; 0.0768639744 &amp; 0.0736000000 &amp; 0.0800000000 \\ \hline
0.0049000000 &amp; 0.0675928619 &amp; 0.0651000000 &amp; 0.0700000000 \\ \hline
0.0036000000 &amp; 0.0582269939 &amp; 0.0564000000 &amp; 0.0600000000 \\ \hline
0.0025000000 &amp; 0.0487656226 &amp; 0.0475000000 &amp; 0.0500000000 \\ \hline
0.0016000000 &amp; 0.0392079992 &amp; 0.0384000000 &amp; 0.0400000000 \\ \hline
0.0009000000 &amp; 0.0295533748 &amp; 0.0291000000 &amp; 0.0300000000 \\ \hline
0.0004000000 &amp; 0.0198010000 &amp; 0.0196000000 &amp; 0.0200000000 \\ \hline
1.000000E-04 &amp; 0.0099501250 &amp; 0.0099000000 &amp; 0.0100000000 \\ \hline
1.000000E-06 &amp; 0.0009995001 &amp; 0.0009990000 &amp; 0.0010000000 \\ \hline
1.000000E-08 &amp; 0.0000999950 &amp; 0.0000999900 &amp; 0.0001000000 \\ \hline
1.000000E-10 &amp; 9.999950E-06 &amp; 9.999900E-06 &amp; 0.0000100000 \\ \hline
\end{array}