Quadratic equation in python program, problem!?

Question

Okay, so i'm writing this program in python to solve quadratic equations, and I've found out that whenever all the variables (a, b, c) are positive, I get an error. Why is that so?

Source code:

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# its in estonian but the strings of text doesn't have anything to do with the
# problem.

# ax^2 + bx + c = 0

import time
import math
import sys
import os

a = input("sisestage ruutliikme vaartus (a) ") # input a
time.sleep(1)
print "..."
b = input ("sisestage lineaarliikme vaartus (b) ") # input b
time.sleep(1)
print "..."
c = input ('sisestage vabaliikme vaartus (c) ') # input c
time.sleep(1)
x1 = (-b + (math.sqrt (b*b -4*a*c))) / (2*a)
x2 = (-b - (math.sqrt (b*b -4*a*c))) / (2*a)
print "Antud funktsiooni nullkohad" # x1, x2 = ...
time.sleep(1)
print "..."
print "x1"
print x1
print "x2"
print x2
Xh = -b/2*a
Yh = a*Xh*Xh+b*Xh+c
print "Antud funktsiooni haripunkti koordinaadid" # Xh, Yh = ...
time.sleep(1)
print "..."
print "Xh"
print Xh
print "Yh"
print Yh
def restart_program(): # restart, yes, no?
python = sys.executable
os.execl(python, python, * sys.argv)

if __name__ == "__main__":
answer = raw_input("Do you want to restart this program ? ")
if answer.strip() in "y Y yes Yes YES".split():
restart_program()

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Example of the error I get when all variables are positive:

(a) 2
...
(b) 4
...
(c) 5

Traceback (most recent call last):
File "C:\Users\kasutaja\Documents\Programs\Ruutfunktsioon.py", line 19, in
x1 = (-b + (math.sqrt (b*b -4*a*c))) / (2*a)
ValueError: math domain error


Any help would be greatly appreciated!

Answer 1

Install the Python interpreter on your computer. This is the program that reads Python programs and carries out their instructions; you need it before you can do any Python programming.

Answer 2

Even if all variables are positives, the radicant MAY become negative...

a=1, b=1, c=1 ==> b² - 4ac = 1 - 4 = -3

a=1, b=2, c=3 ==> 4 - 12 = -8 etc...



I have not seen any check to see if the radicant (b²-4ac) is positive. If it is negative, your math function will return a "domain error" (imaginary root). You should also check if the radicant is not 0 (two equal roots), as well as if "a" is not zero (division by zero). In this case, you would not be solving a quadratic equation, but a simple linear, but your input system ALLOWS the user to do so! NEVER ASSUME the user to do what you expect!!!! LOL.



Sorry, I am not sure we call it "radicant". maybe "radical"? I confuse languages, sometimes... :-)

Answer 3

If x is a root of this equation, then 1/x will probably be a root of the related operate. So (1/x)^2 + 1/x -12 = zero, or 1 + x - 12x^2 = zero or 12x^2 -x -1 = zero has the reciprocals as roots.