Programming Secant Method into Python - python-3.x

Question

The sum of two numbers is 20. If each number is added to its square root, the product of the two sums is 155.55. Use Secant Method to approximate, to within 10^(-4), the value of the two numbers.

Answer 1

Based on http://campus.murraystate.edu/academic/faculty/wlyle/420/Secant.htm
#inital guess
x1 = 10
x2 = 50
Epsilon = 1e-4
#given function
def func(x):
return abs(x)**0.5 * (abs(x)+20)**0.5 - 155.55
y1 = func(x1)
y2 = func(x2)
#loop max 20 times
for i in range(20):
ans = x2 - y2 * (x2-x1)/(y2-y1)
y3 = func(ans)
print("Try:{}\tx1:{:0.3f}\tx2:{:0.3f}\ty3:{:0.3f}".format(i,x1, x2, y3))
if (abs(y3) < Epsilon):
break
x1, x2 = x2, ans
y1, y2 = y2, y3
print("\n\nThe numbers are: {:0.3f} and {:0.3f}".format(ans, ans+20))

Answer 2

Based on Your Title
This code works well in most of the cases. Taken from Secant Method Using Python (Output Included)
# Defining Function
def f(x):
return x**3 - 5*x - 9
# Implementing Secant Method
def secant(x0,x1,e,N):
print('\n\n*** SECANT METHOD IMPLEMENTATION ***')
step = 1
condition = True
while condition:
if f(x0) == f(x1):
print('Divide by zero error!')
break
x2 = x0 - (x1-x0)*f(x0)/( f(x1) - f(x0) )
print('Iteration-%d, x2 = %0.6f and f(x2) = %0.6f' % (step, x2, f(x2)))
x0 = x1
x1 = x2
step = step + 1
if step > N:
print('Not Convergent!')
break
condition = abs(f(x2)) > e
print('\n Required root is: %0.8f' % x2)
# Input Section
x0 = input('Enter First Guess: ')
x1 = input('Enter Second Guess: ')
e = input('Tolerable Error: ')
N = input('Maximum Step: ')
# Converting x0 and e to float
x0 = float(x0)
x1 = float(x1)
e = float(e)
# Converting N to integer
N = int(N)
#Note: You can combine above three section like this
# x0 = float(input('Enter First Guess: '))
# x1 = float(input('Enter Second Guess: '))
# e = float(input('Tolerable Error: '))
# N = int(input('Maximum Step: '))
# Starting Secant Method
secant(x0,x1,e,N)