<s>[INST] Is there any simple method to calculate $\sqrt x$ without using logarithm Suppose that we don't know logarithm, then how we would able to calculate $\sqrt x$, where $x$ is a real number? More generally, is there any algorithm to calculate $\sqrt [ n ]{ x } $ without using logarithm? More simple techniques would be nice.
Here is a simple technique used to approximate square roots by Persian author Hassan be al-Hossein:
For example: $\sqrt {78}\approx 8\frac { 14 }{ 17 } $ , where $8$ is the nearest integer root of $78$, $14 = 78 - 8^2$, $17 = 2 \times 8 + 1$.
if $n=2^k$ we can use the method above. 
For example, for $k=2$ Lets calculate $\sqrt [ 4 ]{ 136 } $: $$\sqrt [ 4 ]{ 136 } =\sqrt { \sqrt { 136 }  } \approx \sqrt { 11\frac { 136-{ 11 }^{ 2 } }{ 11\times 2+1 }  } =\sqrt { 11\frac { 15 }{ 23 }  } \\ \sqrt { 11\frac { 15 }{ 23 }  } \approx 3\frac { 11\frac { 15 }{ 23 } -{ 3 }^{ 2 } }{ 3\times 2+1 } =\frac { 544 }{ 161 } =3.38\\$$ The exact result is$$ \sqrt [ 4 ]{ 136 } =3.4149\cdots$$ The method approximates well, but it is working for only $n=2^k$ as I know. 
 [/INST] Every positive number $x$ can be always written as the sum of two other numbers (or the difference).
Say that one of the two is a perfect square $n^2$, then we can always say that
$$x = n^2 + q$$
Where $q$ is the obvious remainder. Since we can also adopt the difference convention, it's better to write 
$$x = n^2 \pm q$$
When to choose the plus or the minus? Well the rule is that the smaller is $q$, the better.
After that, we can use the following approximation:
$$\sqrt{x} = \sqrt{n^2 \pm q} \approx n \pm \frac{q}{2n}$$
Example
Let's calculate $\sqrt{40}$. Either we choose $40 = 36 + 4$ or $40 = 49 - 9$. The first one is better of course, hence
$$\sqrt{40} = \sqrt{36 + 4} \approx 6 + \frac{4}{12} = 6 + \frac{1}{3} = 6.\bar 3$$
Notice that the real value is
$$\sqrt{40} = 6.324(...)$$
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