Use a linear approximation (or differentials) to estimate the given number.
square root of 100.6

Would use the algorithm for solving square root.

For square root,  √n

x₁ = 0.5(x₀ + n/x₀)        

(This formula is known and for square root, and can be derived using Newton-Raphson's approximation equation)

Where x₀ is the initial guess.  x₁ becomes the new guess.

For √100.6  let our initial guess be 10,  x₀ = 10,  n = 100.6

Our approximation shall be to 3 decimal places. Once we get the same answer twice we stop the algorithm.

x₀ = 10,  x₁ = 0.5(x₀ + n/x₀),    x₁ = 0.5(10 + 100.6/10)  = 10.030, x₁ = 10.030

x₂ = 0.5(x₁ + n/x₁),    x = 0.5(10.030 + 100.6/10.030)  ≈10.015,  x₂ ≈ 10.030 (to 3 decimal places)

Since x₂≈ x₁, the algorithm stops.

So the √100.6 is ≈ 10.030 to 3 decimal places.

I hope this helps.

Main idea on left, details on right