How to Estimate Square Roots in Mental Math

Calculating square roots, for example \(\sqrt{59.6} = 7.72010363…\), is quite difficult—but there are methods such as the mental square roots algorithm, for solving these with as much accuracy as you want.

This article describes an easier method for quickly estimating the value of a square root, to about 2–5 significant figures. I’ll also show you how to know how accurate your estimate is likely to be.

Useful Prerequisites

The following will enable to use this method more easily:

Be able to estimate divisions, especially by numbers up to 200—for example \(45 \div 138\) or \(28 \div 17\)
Know as many square numbers as possible—e.g. \(12^2 = 144\) or even \(77^2 = 5929\)

Estimation Method

As an example, I will estimate \(\sqrt{59.6}\):

Choose a nearby guess. For example, \(\sqrt{59.6} \approx 8\)
Calculate the difference between guess² and the number you want to square-root. For example, \(8^2 – 59.6 = 4.4\)
Divide it by twice your original guess. Here, \(4.4 \div (2 \times 8) = 4.4 \div 16 = 0.275\). This is your correction.
Add or subtract the correction from the guess to get the estimate of \(8 – 0.275 = 7.725\)

By comparing this to the calculated value, \(\sqrt{59.6} = 7.72010363…\), we see that our estimate was correct to 3 significant figures.

Improvements to Accuracy

Assuming you do not make any mistakes, the most important detail that affects accuracy is your choice of original guess: it should be as close as possible to the true square root value. If my original guess is 10× closer to the true value than yours, my final estimate will be 100× closer than your final answer! [See proof later] Therefore:

if you need more than 2 significant figures, choose a more precise guess.
after doing the method once, you might choose to do it again with another guess, inspired by your first result. For example, after estimating \(\sqrt{59.6} \approx 7.725\), you might choose to repeat, using 7.7 or 7.8

Example with Low Precision

To illustrate this, let’s repeat the method taking 7 as our guess. This is a bad guess, since 59.6 is much closer to 8² than to 7².

\(59.6 – 49 = 10.6\)
\(\frac{10.6}{2 \times 7} = \frac{53}{70} = 0.57143…\)
\(7 + 0.57143… = 7.57143…\)

The result of this estimation is not even accurate to 2 significant figures.

Example with High Precision

Now let’s take 7.7 as our guess. This is a great choice, as 7.7² = 59.29 is really close to 59.6.

\(59.6 – 59.29 = 0.31\)
\(\frac{0.31}{2 \times 7.7} = 0.02013…\)
\(7.7 + 0.02013… = 7.72013…\)

The result of this new estimation is now accurate to 5 significant figures!

Practical Considerations for Mental Math

Note that the only difficult calculation here is the division by 7.7. You have to be good at divisions to use this method with high precision, but an approximation such as \(\frac{0.31}{2 \times 7.7} \approx \frac{0.3}{15} = 0.02\) is almost as good.
Note that it was helpful to know the squares 77² = 5929 and 78² = 6084, from memory, in order to quickly select the most appropriate 2-digit guess.

You should practice both of these aspects in order to apply this method more precisely.

Finally, note that this method always gives an overestimate, so you might choose to round your answer down.

The remaining sections are only important if you want to know more about the Mathematics behind the method.

Mathematics Behind the Method

This method is a modified one-step variation of the Newton-Raphson algorithm, used by computers to estimate the solution to difficult equations.

Why does this square roots variation work? Here are two explanations—one using algebra, and the other using calculus.

Using Algebra

If our guess is \(G\) (e.g., 7.7 in the example), then the true value is \(G + \epsilon\), where \(\epsilon\) is small compared to \(G\), although could be positive or negative.

Let \(C\) be the true square root (so e.g. \(C^2 = 59.6\) in the example). Then:

\((G + \epsilon)^2 = C^2\)

\(G^2 + 2 G \epsilon + \epsilon^2 = C^2\)

Assuming our first guess was reasonably good, we can approximate \(\epsilon^2 \approx 0\) to leave:

\(G^2 + 2 G \epsilon = C^2\)

Rearranging the equation to solve for \(\epsilon\) gives:

\(\epsilon = \frac{C^2 – G^2}{2G}\)

Therefore our new estimate should be \(G + \epsilon\) with this value of \(\epsilon\), in agreement with the method.

Using Calculus

The gradient of the curve \(y = x^2\) has derivative \(2x\), so the curve at the point where \(x = G\) can be approximated by the straight line of gradient \(2G\).

Let our guess be \(G\). Then as \(x\) increases to \(G + \epsilon\), for any small (positive or negative) value of \(\epsilon\), its square increases by approximately gradient × \(\epsilon\) = \(2G \epsilon\).

Since in the method, we already calculated this increase (or decrease), we can simply divide it by \(2G\) to estimate \(\epsilon\)

Assessing Accuracy (Proof)

If the true correct value of the square root is \(C\), and our initial guess is \(G\), our estimate is given by:

\(G + \frac{C^2 – G^2}{2G}\)

The error between our estimate and the true value is therefore:

\(|G + \frac{C^2 – G^2}{2G} – C|\)

\(= \frac{1}{2G}|2G(G – C) – (G^2 – C^2)|\)

\(= \frac{1}{2G}|2G(G – C) – (G – C)(G + C)|\)

\(= \frac{1}{2G}|(G – C)[2G – G – C]|\)

\(= \frac{(G – C)^2}{2G}\)

\(\approx \frac{(G – C)^2}{2C}\)

As \((G – C)\) is the difference between our guess and the correct answer, this proves the assertion I made earlier that if you improve your guess to be 10× more precise, your estimate will be approximately ×10² = ×100 more accurate.