# The square root of 1326

So, there I was, happily working out that the square root of 5,100 is very close to $70 + \sqrt{2}$…

OK, OK, OK, big aside here. Why is that? I’ll tell you why that is. It’s due to the binomial expansion, in a slightly surprising way:

- $\left( 5000 + x\right)^{1/2} \approx 50\sqrt{2} + \frac{x}{100\sqrt{2}}$
- $\left( 5000 - x\right)^{1/2} \approx 50\sqrt{2} - \frac{x}{100\sqrt{2}}$
- So $\sqrt{5100} - \sqrt{4900} \approx \frac{100}{100\sqrt{2}} + \frac{100}{100\sqrt{2}}$
- Or $\sqrt{5100} - 70 \approx \sqrt{2}$.

Lovely stuff. But then in stepped in @shalock, who noted that it’s not even the best such approximation. That would be $\sqrt{1326} \approx 35 + \sqrt{2}$.

Now *that* – that just seems wrong. We can do something sort of similar, based around 1275.125 – the square of $\frac{101\sqrt{2}}{4}$:

- $\left( 1275.125 + x\right)^{1/2} \approx \frac{101\sqrt{2}}{4} + \frac{x\sqrt{2}}{101}$
- $\left( 1275.125 - y\right)^{1/2} \approx \frac{101\sqrt{2}}{4} - \frac{y\sqrt{2}}{101}$
- So $\sqrt{1326} - \sqrt{1225} \approx \frac{407\sqrt{2}}{808} + \frac{401\sqrt{2}}{808}$
- Which is, again, approximately $\sqrt{2}$.

I haven’t worked through the higher-order terms, on the grounds that they’re awful. The $x^2$ terms in the expansions around 5000 cancel out; the $x^2$ terms around 1275.125 don’t cancel (because the $x$s are different), but I suspect they might balance out with the $x^3$ terms to make a closer approximation.

As for why $\sqrt{1326}$ and $\sqrt{5100}$ are very close to 35 apart, I don’t know. Any ideas?