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!

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- Jan 23rd 2007, 05:10 PMsymmetryWithout Calculator
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! - Jan 23rd 2007, 05:27 PMQuick
$\displaystyle \sqrt{16}<\sqrt{21}<\sqrt{25}$

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. - Jan 23rd 2007, 05:39 PMsymmetryok
How would you do sqrt{79} THE SAME WAY?

- Jan 23rd 2007, 05:45 PMQuick
$\displaystyle \sqrt{64}<\sqrt{79}<\sqrt{81}$

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. - Jan 23rd 2007, 05:45 PMAfterShock
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). - Jan 23rd 2007, 05:50 PMsymmetryok
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!