## anonymous one year ago what is the square root of 3.1

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: $$\large\color{black}{ \displaystyle {\rm L}(3.1)=f(a)+f'(a)\left(x-a\right)}$$ $$\large\color{black}{ \displaystyle {\rm L}(3.1)=\sqrt{4}+\frac{1}{2 \sqrt{4}} \left(3.1-4\right)}$$ $$\large\color{black}{ \displaystyle {\rm L}(3.1)=2+\frac{1}{4} \left(-0.9\right)}$$ $$\large\color{black}{ \displaystyle \sqrt{3.1}\approx2-0.225=1.775}$$