Root finding using the Bisection Method - exact solutions

As a final blog on this program, I want to cover the situation where an exact solution of f(x)=0 is found by the bisection method, i.e. where the program finds α where f(α)=0 and not just an approximation to α.

Consider the quadratic equation 160x²-428x-645=0. This equation has a root around x=-1 because f(-1.1)=19.4 and f(-0.9)=-130.2 (where f(x)=160x²-428x-645), so f changes sign between -1.1 and -0.9. If we now edit the program to have 160C²-428C-645 instead of C-COS(C) in the "subroutine" then on running the program with A being -1.1, B being -0.9 and D set to 1x10^-5 we obtain the root -1.075 in 3 bisections. This is the exact value for the root because the quadractic equation can be factorised to (x-3.75)(x+1.075)=0, so the roots are x=-1.075 and x=3.75.

This shows that the program can cope with the situation where it finds the root exactly. Of course, this doesn't always happen even if an exact value could be found. For example, rerunning the program with A set to -1.5, B set to -0.5 and D as before, we obtain C as -1.075004578 in 17 iterations.