(This is a post aimed more at high school students or young college students.)

This is a neat mental math trick I learned from television. Let’s start easy and have one of your friends find a 4-digit perfect square (i.e. square any integer between 32 and 99, inclusive). This trick will allow you to find the square root almost immediately. For example, let’s use 3136. Call it S.

- Estimate the square root to the nearest ten. 50 squared is 2500 and 60 squared is 3600, so the square root is between 50 and 60. I’m going to pick 50 because it’s easier to divide by. Let X = 50.
- Divide S by X, ignoring the remainder. 3136 / 50 = 62-point-something. Let W = 62.
- Average X and W. That is the square root! So for our example, the average of 50 and 62 is 56. You can check that this is right.

However, your friends who are on the ball might not be that impressed since, because it is known that 3136 is a perfect square, we would have guessed that the units digits is either a 4 or a 6 since that is the only way to get a 6 in the units’ digit of 3136. So let’s up the ante: we will now compute the square root of any 4-digit number to one decimal point. For this, I asked a random number generator and got 5389.

- 70 squared is 4900, 80 squared is 6400, so we’re somewhere between there. Let X = 70, since 4900 is closer, and like most iterative algorithms, Newton’s Method works better when we start closer to the endpoint.
- S / X is almost exactly 77. Let W = 77.
- The average of X and W is 73.5. Because this method tends to overestimate, let’s call this 73.

Now let us repeat steps 2 and 3, but with X equal to the number found in step 3.

- (again): S / X is a bit trickier with X = 73. I have a mental math trick for this (which I’ll explain later): the answer is 73 + (5389 – 4900 – 420 – 9) / 73 = 73 + (489 – 420 – 9) / 73 = 73 + (69 – 9) / 73 = 73 + 60 / 73, or about 73.8.
- (again): The average of 73 and 73.8 is 73.4. To check this, the square root of 5389 is 73.4098086…

Before I explain how it works, I find the student learns more when they try it for themselves. So here are some questions:

- Show how this algorithm is derived from Newton’s method. Hint: use y(x) = x^2 – S. Why does this algorithm tend to overestimate?
- Explain the mental math trick used in step 2 (again). Hint: use algebra.

WARNING: Try the problems posted in part 1 before reading this post. It took me some minutes, but mostly because I had to rederive Newton’s Method and tried the wrong y(x) first. I find that students do not really learn when answers are just handed to them; you have do it for skills to sink in. Do you think basketball stars got that good by listening or reading about how to play? They go onto the court and practice their skills. Math is the same.

1) I have a confession: I have a really bad memory for formulas. However, this is not a problem in math if you know what you’re doing. This is the formula for Newton’s Method that is easiest for me to remember:

y(x_i) = (x_i – x_{i+1}) [dy/dx evaluated at x_i]