Question

what is the square root of 3.1

Answer 1

You can simply use a calculator, or do a Taylor approximation
of some Nth degree. The first degree approximation, otherwise
known as linearization, using \(f(x)=\sqrt{x}\) where \(a=4\), you get:

\(\large\color{black}{ \displaystyle {\rm L}(3.1)=f(a)+f'(a)\left(x-a\right)}\)

\(\large\color{black}{ \displaystyle {\rm L}(3.1)=\sqrt{4}+\frac{1}{2 \sqrt{4}} \left(3.1-4\right)}\)

\(\large\color{black}{ \displaystyle {\rm L}(3.1)=2+\frac{1}{4} \left(-0.9\right)}\)

\(\large\color{black}{ \displaystyle \sqrt{3.1}\approx2-0.225=1.775}\)