x^3-70x^2+1601.375x-12285.5=0

Question

find the roots of the equation.

Answer 1

This cubic has only one real root. A graph reveals the root is near to x=30.

Newton's Method can be used to find the root using calculus.

f(x)=x³-70x²+1601.375x-12285.5; f'(x)=3x²-140x+1601.375.

x_n+1=x_n-f(x_n)/f'(x_n) where x₀=30.

x₁=32.4093..., x₂=31.8047..., x₃=31.7516..., x₄=31.7512..., x₅=31.75129779, x₆=31.75129779, so we take this as a fairly accurate value for the real root.

To find the complex roots we need the quadratic, which can be found using synthetic division by the real root:

31.75129779 | 1  -70                   1601.375        -12285.5

                        1   31.75129779 -1214.445634 | 12285.5

                        1  -38.24870221   386.929066 |       0       =

x²-38.24870221x+386.929066

We can now find the complex roots:

x=(38.24870221±i√84.75304352)/2=19.12435±4.60307i approx.