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I have always used a calculator for determining roots

But I think it would be useful to know how to do this on paper. Are there different procedures for calculating square roots then cubic roots or does it all use the same principles?

So my question is how does one calculate the root of a number by hand (so to speak)?

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A variety of techniques are shown in Wikipedia. Several of them generalize to cube roots. The choice depends on what accuracy you want and how much work you are willing to do. For low accuracy you can often use a table you may have memorized and a small correction. For example, to find $\sqrt{26}$ you can say $26=25\cdot 1.04$, so $\sqrt {26}=\sqrt{25}\sqrt {1.04}\approx 5\cdot 1.02 =5.1$ where we have used $\sqrt {1+x} \approx 1+\frac x2$ for small $x$. The same works for cube roots using $\sqrt[3]{1+x} \approx 1+\frac x3$

One practical way is to use Newton iteration (Wikipedia calls it the Babylonian method, check details there). For cubic roots:

$$

f(x) = x^3 – n = 0

$$

gives the Newton iteration:

$$

x_{k + 1} = x_k – \frac{f(x_k}{f'(x_k)} = \frac{1}{3} (2 x_k + n / x_k^2)

$$

Starting point can be cooked up just like for square roots. Let $n$ have $d + 1$ digits, then one can use:

$$

x_0 = \begin{cases}

1 \cdot 10^{d / 3} & d \bmod 3 = 0 \\

2 \cdot 10^{\lfloor d / 3 \rfloor} & d \bmod 3 \ne 0

\end{cases}

$$

A technique Newton himself used was to use the binomial expansion of $(1 + x)^{1/2}$ for small $\lvert x \rvert$, i.e., take the nearest square $s^2$ and:

$$

n^{1/2} = s \left(1 + \frac{n – s^2}{s^2}\right)^{1/2}

$$

There’s a paper-and pencil square root algorithm that works pretty well, based on the following idea: Suppose $n$ is the number we want to find the square root of, and $n’$ is the number you get by dropping its last two digits $\delta$. Then $n = 100n’ + \delta$.

Say we already have $g$, a good guess for $\sqrt{n’}$. We want to find a good guess for $\sqrt n$. $10g$ will be a good start, but we would like to adjust $10g$ upwards to take the last two digits, $\delta$, into account. So what $\epsilon$ can we add to $10g$ so that $(10g+\epsilon)^2 \approx 100n’ + \delta$?

We can try to solve for $\epsilon$ in:

$$\begin{align}

(10g+\epsilon)^2 & \approx 100n’ + \delta \\

100g^2+20g\epsilon+\epsilon^2 & \approx 100g^2+\delta \\

20g\epsilon+\epsilon^2 & \approx \delta\\

\epsilon & \approx \frac\delta{20g+\epsilon}

\end{align}

$$

so take $\epsilon$ so that $\epsilon(20g+\epsilon) \approx \delta$; then $10g+\epsilon$ is a guess for $\sqrt n$. We can repeat this process to find guesses for longer and longer numbers.

(This is the algorithm Brian mentioned in his comment.)

Here is an example. Let’s find the square root of 142857. To do that we need a guess for the square root of 1428, to do that we need a guess for the square root of 14. A good guess is 3. So we have $n’=14, g=3$. We know that $(30)^2\approx 1400\approx 1428$, and we want a better guess for $\sqrt{1428}$. So we’d like to find $\epsilon$ such that $$\begin{align}

(30+\epsilon)^2 & \approx 1428\\

900 + 60\epsilon + \epsilon^2 & \approx 1428 \\

\epsilon(60+\epsilon) & \approx 528

\end{align}

$$

Eyeballing this, it looks like $\epsilon=7$ is about right. So our new guess is that $\sqrt{1428}\approx 37$.

Now we want $\sqrt{142857}$ and we guess that $370$ is not too far off. We want $\epsilon$ so that

$$\begin{align}

(370+\epsilon)^2 & \approx 142857\\

136900 + 740\epsilon + \epsilon^2 & \approx 142857 \\

\epsilon(740+\epsilon) & \approx 5957

\end{align}

$$

$\epsilon=8$ is a just a shade too big, but much closer than $\epsilon=7$, so the answer is a shade under 378. If we wanted to continue, we could take $\epsilon=8$ and calculate the amount by which the square root should fall short of 378, or we could take $\epsilon=7$ and calculate the amount by which the square root should exceed 377.

It’s not hard to come up with a cube (or higher) root analog of this algorithm, but it’s not practical, because instead of trying to estimate an $\epsilon$ that makes $20g\epsilon+\epsilon^2\approx \delta$, which is a not-quite simple division, you have to estimate an $\epsilon$ that makes $300g^2\epsilon+20g\epsilon^2\epsilon^3\approx \delta$, which is too hard.

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