The perfect squares on either side of 8 are

(8-4)/(9-4) = 4/5

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To get a closer estimate, divide 8 by this estimate, and average the result with this estimate. A couple of iterations like this will get you the square root good to 8 decimal places.

(2.8 + 8/2.8)/2 ≈ (2.8+2.8571)/2 ≈

This iteration method has been in use since Babylonian times. It is equivalent to Newton's iteration method for finding the root of the equation f(x) = x^2-8.

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From calculus, we know that the square root changes at a rate that is inversely proportional to itself. The constant of proportionality is 1/2. So, at 9, the square root is changing at a rate of about (1/2)*(1/3) = 1/6 per unit. Thus, we expect the square root of 10 (1 unit above 9) to be about 3 + 1/6, and the square root of 8 (1 unit below 9) to be about 3 - 1/6 = 2 5/6, or

**4**=2^2 and**9**=3^2. So we know the square root of 8 will be between 2 and 3. 8 is 4/5 of the distance between 4 and 9, so we expect the square root of 8 to be about 4/5 of the distance between 2 and 3. That value is 2 4/5 =**2.8**(8-4)/(9-4) = 4/5

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To get a closer estimate, divide 8 by this estimate, and average the result with this estimate. A couple of iterations like this will get you the square root good to 8 decimal places.

(2.8 + 8/2.8)/2 ≈ (2.8+2.8571)/2 ≈

**2.8286**This iteration method has been in use since Babylonian times. It is equivalent to Newton's iteration method for finding the root of the equation f(x) = x^2-8.

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From calculus, we know that the square root changes at a rate that is inversely proportional to itself. The constant of proportionality is 1/2. So, at 9, the square root is changing at a rate of about (1/2)*(1/3) = 1/6 per unit. Thus, we expect the square root of 10 (1 unit above 9) to be about 3 + 1/6, and the square root of 8 (1 unit below 9) to be about 3 - 1/6 = 2 5/6, or

**2.83**.