This is a neat mental math trick I learned in middle school. 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.
Step 1: 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 for reasons I'll explain below. Call it X.
Step 2: Divide S by X, ignoring the remainder. (The reason I chose X = 50 instead of 60 is because it's really easy to divide by 50. But, honestly, dividing by 60 is just as easy.) 3136 / 50 = 62-point-something. Let W = 62.
Step 3: 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'm going to use 5389.
Step 1: 70 squared is 4900, 80 squared is 6400, so we're somewhere between there. Let X = 70, since 4900 is a bit closer.
Step 2: S / X is almost 77. Let W = 77.
Step 3: 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.
Step 2 (again): S / X is a bit trickier with X = 73. I have a mental math trick for this (which I'll explain using algebra in the next post): 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.
Step 3 (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:
1) Show how this algorithm is derived from Newton's method. Hint: use y(x) = x^2 - S. Why does this algorithm tend to overestimate?
2) Explain the mental math trick used in Step 2 (again). Hint: use algebra.