# Calculating a Square Root Without Calculator

Hi all, I am trying to find a nice back-of-the-envelope way to numerically-calculate nasty things like square roots and stuff. For instance:
$\sqrt {4.51} = ?$

I tried to use a linearization:
$\sqrt x = \sqrt {n + \varepsilon } = \sqrt {\left[ {{\rm{integer}}} \right] + \left[ {{\rm{decimal part}}} \right]} = \sqrt n + L(\sqrt x ) \cdot \varepsilon$
…where you “linearly” count your way from $\sqrt {n + \varepsilon }$ to $\sqrt x$ by way of the linearization, $L(\sqrt x )$, namely, the derivative:
$L(\sqrt x ) = {\left. {\frac{d}{{dx}}\sqrt x } \right|_{x = n}} = \frac{n}{{2\sqrt n }}$

But that’s another square root to calculate, which would further-roughen my approximation. Are there any other numerical-approximations you’d suggest?

## The Attempt at a Solution

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berkeman
Mentor
Hi all, I am trying to find a nice back-of-the-envelope way to numerically-calculate nasty things like square roots and stuff. For instance:
$\sqrt {4.51} = ?$

I tried to use a linearization:
$\sqrt x = \sqrt {n + \varepsilon } = \sqrt {\left[ {{\rm{integer}}} \right] + \left[ {{\rm{decimal part}}} \right]} = \sqrt n + L(\sqrt x ) \cdot \varepsilon$
…where you “linearly” count your way from $\sqrt {n + \varepsilon }$ to $\sqrt x$ by way of the linearization, $L(\sqrt x )$, namely, the derivative:
$L(\sqrt x ) = {\left. {\frac{d}{{dx}}\sqrt x } \right|_{x = n}} = \frac{n}{{2\sqrt n }}$

But that’s another square root to calculate, which would further-roughen my approximation. Are there any other numerical-approximations you’d suggest?

## The Attempt at a Solution

I tried googling the title of your thread, and got what look to be lots of good hits:

I didn't click into any of them -- are they helpful?

marcusl
Gold Member
How accurate do you need? There are a lot of strategies if high accuracy is not the goal (and in any case, if I'm trying to estimate in my head I shoot for 5-10% accuracy). In the present case, I know sqrt(4) = 2 and sqrt(5) = 2.23 (which we had to memorize in junior high school). Taking the mean gives the estimate sqrt(4.51) = 2.12 (approx.).

OR, sqrt(4.5) = sqrt(9) * sqrt(0.5) = 3 * .707 = 2.12.

I'll often use logs, too. I wouldn't use them for this problem, but I'll do it to show how it works.
I've memorized four values, namely,
log10 of {2, 3, 5, 7} = {0.3, 0.48, 0.7, 0.85}.
I can figure out other simple ones from these: obviously log10(4)=0.6 and log10(9)=0.96, for instance.

Applied to present problem, x = sqrt(4.5)=10^(1/2*log10(4.5)).

First, 1/2 * log10(4.5) = 1/2 * log10(9*5/10) = 1/2 * (0.96 + 0.7 - 1) = 0.36

Second, find x=10^0.36. Looking at the values I know above, I find that 0.36 = 0.96 - 0.6
Using my known values in reverse (antilogs), x = 10^0.96 / 10^0.6 = 9 / 4 = 2.25. That's within 6%.

Logs are harder for this problem, but are easier for others (like finding cube roots, e.g.).

Last edited:
vela
Staff Emeritus
$$\sqrt{4.51} = \sqrt{4+0.51} = \sqrt{4}\sqrt{1+0.1275}$$