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You could find the numbers on either side of 5 that are perfect squares {4 & 9} then find their square roots and interpolate. So for y = f(x) = sqrt(x):

x | y

4 | 2

9 | 3

So you could interpolate using Δy/Δx = (3-2)/(9-4) = 1/5. So Δx between 4 and 5 is 1, so Δy = Δx * (Δy/Δx) = 1 * (1/5) = 1/5. Then add Δy to y,

which is 1/5 + 2 = 2 1/5. So it is closest to 2. Of course you could just look at it and see that 5 is closer to 4 than 9, so infer that 2 is closer to the square root of 5, than 3.

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