Simplifying and approximating square roots.

In class, we worked on simplifying square roots, which means taking sqrt(n) and turning it into a x sqrt(b), where b is a square free number. We used what I call the prison break method, where a pair of twins under the square root sign can make a break for it, but only one gets out and the other is shot trying to escape. (Thanks, Mrs. Kruger!)

sqrt(20) = sqrt( 2 x 2 x 5), so one of the pair of 2s can be brought out, while the 5 stays inside.

sqrt(20) = 2 x sqrt(5)

We can also go to our calculator and get an approximation of sqrt(20) = 4.472135955..., a decimal number that continues on forever and never gets into an infinitely repeating pattern because it is an irrational number. If we round, we get an approximation of sqrt(20), but the idea is that sqrt(20) = 20 exactly, and the approximations won't be exactly 20 when we square them.

sqrt(20) rounded to one place after the decimal = 4.5
4.5² = 20.25

sqrt(20) rounded to two places after the decimal = 4.47
4.47² = 19.9809

sqrt(20) rounded to three places after the decimal = 4.472
4.472² = 19.998784

Approximating to three places after the decimal is fairly standard, since the square of that approximation is usually off by less than a hundredth. For example,

sqrt(999) = 31.60696126... or 31.607 approximated to three places after the decimal
31.607² = 999.002449

Practice problems.

For the following square roots, simplify and approximate to three places after the decimal, then find the square of the approximation.

sqrt(72)

sqrt(73)

sqrt(74)

sqrt(75)

Answers in the comments.