Binomial expansion approximation for $\sqrt{2}$

Question

I saw on another post the question:

Given the expansion $(4 - 5x)^5 = 2 + (5/4)x + (25/64)x^2$, use $x = 1/10$ to find an approximation of $\sqrt{2}$. The answer is given there, but I'm struggling on how to even approach this problem, even with the discussion on the other thread. Any help would be much appreciated.

Answer 1

You still have the wrong sign in the first expression. Correcting that gives you $4/9$ instead of $20/35$ which has square root $2/3$ (the numbers were chosen to make this rational). Feeding in $x = 1/10$ now gives $181/128$, as claimed in the linked post. You can check that that is an approximation of root 2 by calculating its square.

If you want you know why this works, you could start with https://en.wikipedia.org/wiki/Binomial_approximation , particularly the generalisation near the end.

Answer 2

Thanks all for the comments, that's very helpful. This is the answer I got, I still feel like I am missing something obvious

with $x=\frac{1}{10}$,

$\frac{20}{35}(4-\frac{5}{10}) = 2$, therefore $\frac{20}{35}^\frac{1}{2}(4-\frac{5}{10})^\frac{1}{2} = \frac{20}{35}^{\frac{1}{2}} \cdot 2 \cdot (1 - \frac{5}{40})^\frac{1}{2} = \sqrt{2}$

$\frac{20}{35}^{\frac{1}{2}} 2 [ 1 + \frac{5}{8}x - \frac{25}{128}x^2] = \sqrt{2}$

$\frac{20}{35}^{\frac{1}{2}} [ 2 + \frac{5}{40} - \frac{25}{6400}] = \sqrt{2}$

I've got a feeling this is not what they are after though...