Added exercise 7

ex_07.py

@@ -0,0 +1,74 @@
+#!/usr/bin/python3
+
+from math import sqrt
+
+from util.io import readvalue
+
+def solution_count(a, b, c):
+	"""
+	Calculates the number of real solutions
+	for a*x**2 + b*x + c = 0.
+
+	Returns: 0 or 1 or 2
+
+	Method:
+
+	After doing some fiddling you can get the equation
+
+	(x + b/(2*a))**2 = (b/(2*a))**2 - c/a
+
+	You can derive that the equation has 
+	no real solution, if (b/(2*a))**2 - c/a < 0
+	one real solution, if (b/(2*a))**2 - c/a = 0,
+	two real solutions, if (b/(2*a))**2 - c/a > 0.
+	"""
+
+	rterm = (b/(2*a))**2 - c/a
+
+	if(rterm < 0):
+		return 0
+	if(rterm == 0):
+		return 1
+	if(rterm > 0):
+		return 2
+
+def solutions(a, b, c):
+	"""
+	Returns the solution(s) of a*x**2 + b*x + c = 0.
+
+	If the equation has no solution, it returns None,
+	a tuple of the soltions, otherwise.
+	"""
+	# Get the number of solutions:
+	number_of_solutions = solution_count(a, b, c)
+
+	if(not number_of_solutions):
+		return
+
+
+	if(number_of_solutions == 1):
+		return (-b / (2*a), )
+	else:
+		rterm = sqrt((b/(2*a))**2 - c/a)
+		return (-b / (2*a) + rterm, -b / (2*a) - rterm)
+
+
+
+if( __name__ == "__main__"):
+	print("Program to find the solution(s) for a*x**2 + b*x + c = 0")
+	a = readvalue("a > ", float)
+	b = readvalue("b > ", float)
+	c = readvalue("c > ", float)
+
+	number_of_solutions = solution_count(a, b, c)
+	
+	messages = ["There are no real solutions."
+			, "There is exactly one real solution."
+			, "There are exactly two real solutions."]
+	print(messages[number_of_solutions])
+
+	if(number_of_solutions == 1):
+		print("The solution is:", solutions(a, b, c)[0])
+	if(number_of_solutions == 2):
+		print("The first solution is:", solutions(a, b, c)[0])
+		print("The second solution is:", solutions(a, b, c)[1])