Finding the decimal estimate of the square root of a number (Pre-Algebra)

While it is easy to find the whole number estimate for the square root of a number, sometimes students are asked to find a closer estimate than just the whole number. As students progressed into higher level math, they will be asked to find a closer estimate than just the whole number.  

For example, in order to estimate the square root of 40, students find 2 perfect squares that 40 is between.  36 is the perfect square of 6, and 49 is the perfect square of 7.  Since 40 is closer to 36 than 40 is to 49, students can initially estimate that the square root of 40 is 6.  There is a way to find a closer estimate for the square root of 40.

We would start out just like we have when estimating for the whole number with one caveat that we do not round up.  A better example would be finding the square root of 48.  In this example, we know that square root of 48 is between square root 36 and square root of 49.  Since we know that 48 is closer to 36, we will "assume" that square root of 48 is a closer estimate to square root 49, which happens to be 7.  So for younger students, square root of 48 can be estimated to be 7.  But if you were to take the perfect square of 7, you will know that the number is 49, which is definitely more than our initial question of 48.

So in this case, in order to find the closer estimate, we now know that square root of 48 is 6.#### (with #### being some decimal numbers).  Let us find the distance between the perfect squares of the 6 and 7.  The perfect square of 6 is 36, and the perfect square of 7 is 49.  The distance between 49 and 36 is 13.  And the distance between the 36 and the number in question of 48 is 12.  So if we do the long division of 12/13, you will get .92307..  So if add this decimal equivalent to the initial whole number estimate of 6 which gives us 6.92307, we should get a closer estimate for the actual answer to the question of square root of 48.  If you do the math, using a calculator or by hand, 6.92307 x 6.92307 = 47.9288982249, which is quite close to the actual number of 48 and much closer than the original estimate of 7.

In order to get more comfortable with this type of problem, I would suggest that the students look for more problems to solve.  In the meantime, you can try these examples on your own:

1. sqrt(101)
   
2. sqrt(75)