How do I find the appoximation of square roots that are not perfect squares WITHOUT a calculator?
SAMPLES:
(1) sqrt{21} =
(2) sqrt{79} =
DO NOT USE A CALCULATOR.
Thanks!
$\displaystyle \sqrt{16}<\sqrt{21}<\sqrt{25}$
Now notice that 21 is almost right inbetween 16 and 25, so we'll say that: $\displaystyle \sqrt{21}\approx 4.5$
and a calculator gives: $\displaystyle \sqrt{21}\approx 4.58$
So I was off by an eight-hundredth, still pretty close.
$\displaystyle \sqrt{64}<\sqrt{79}<\sqrt{81}$
Notice that 79 is a lot closer to 81 than 64, so I'll estimate that $\displaystyle \sqrt{79}\approx 8.9$
And the calculator gives: $\displaystyle \sqrt{79}\approx 8.89$
So I was off by about 1-hundredth, which is very close.
Generally you would use Taylor Polynomials, which involves calculus to find such without a calculator. It's far more accurate, but Quick's method suffices.
Sqrt(81) = 9
Sqrt(64) = 8
Thus, sqrt(79) is between 8 and 9. Note that sqrt(79) is quite close to sqrt(76), which is 2*sqrt(19) (gotten from sqrt(4)*sqrt(19) = sqrt(76).
And you can do the same deal with sqrt(19) as I did before to try and get a better result. Obviously the result will be in the high 8's, since sqrt(79) is close to sqrt(81).
My study book for June state math exam takes me through the basics of high school mathematics going through a few calculus 1 and 2 topics.
I will cross that bridge when I get there.
In terms of Taylor Polynomials, let's not go there just yet.
Thanks!