0

## Estimating Square Roots: Improved Accuracy

Published on Sunday, September 10, 2017 in , , ,

Back about 5 years ago, I taught a simple method for estimating the square roots of non-perfect squares. I recently learned an improvement that makes this far more accurate, and even a little more impressive.

The new approach works well with the original approach I taught. I'll start with a review of the original method I taught, and then I'll delve into the details of the improved method.

Quick Refresher: As mentioned in the original post, you should be comfortable with mentally squaring 2-digit numbers, and being able to find the square roots of perfect squares. You'll also need to know the squares of the numbers from 1 to 31 off the top of your head, in order to handle the numbers from 1 to 1,000.

As an example number, we'll say you're given 130, and asked to find the square root. You start by partitioning the given number into the sum of the largest perfect square equal to or less than the number, plus any remaining amount. The largest perfect square equal to or less than 130 is 121, so we partition it into 121 + 9:

$\sqrt{130} = \sqrt{121 + 9}$

If you consider the sum now under the square root at this point to consist of numbers x and y (x = 121 and y = 9 in our example), then the original square root approximation is as follows:

$\sqrt{x +y} \approx \sqrt{x}+\frac{y}{2\sqrt{x}}$

This may seem harder than it is. Remember that x is a perfect square, so sqrt(x) is an integer, so it will be easy to work with. Applying this formula to our example, and recalling that sqrt(121) = 11, we get this approximation:

$\\ \sqrt{121 +9} \approx \sqrt{121}+\frac{9}{2\sqrt{121}} \\ \hphantom{spaces} \ \ \ \ \approx 11\frac{9}{22}$

If you square that estimation, you get roughly 130.167. That's not bad, but how can we do better?

Improved Version: This past Monday, UK-based math tutor Colin Beveridge wrote a post called The Mathematical Ninja takes a square root, which is the improved version I've been mentioning.

The whole part (11 in our above example) will not change, so we'll only focus on changing the fractional part. If you consider, as in Colin's post, the fractional part to be ab, then the improved estimation is as follows:

$\frac{ab}{b^{2}+a}$

So, for the new numerator (top number of the fraction) in our example, you'd multiply 9 × 22 to get 198. For the new denominator (bottom number of the fractions), you'd have to square the original bottom number, 22 squared = 484, and then add the original top number, 9, for a total of 493. Our new estimation for the square root of 13 becomes:

$\sqrt{130} \approx 11\frac{198}{493}$

If you square that estimate, you get roughly 129.997, which is much closer to 130 than 130.167!

If this approach interests you, do take the time to read Colin's original post, as it has plenty of nice touches. Most importantly, he suggests using the closest perfect square above or below the given number, which will improve the accuracy of your estimate as compared to always using the perfect square less than or equal to the given number.