Question

I started yesterday the course and today tried to make a small program, but doesn't seem to work as I want
print "(a*(x**2))+(b*x)+c"
a=+3
b=-5
c=+2

D=(b**2-4*a*c)
x1=(-b+(D**(-1/2)))/(2*a)
x2=(-b-(D**(-1/2)))/(2*a)

if D>0 and (-b+(D**(-1/2)))/(2*a)>0 and (-b-(D**(-1/2)))/(2*a)>0:
print "x1=" x1
print "x2=" x2
elif D>0 and (-b+(D**(-1/2)))/(2*a)>0:
print "x1=" x1
elif D>0 and (-b-(D**(-1/2)))/(2*a)>0:
print "x2"= x2
else: print "x not found"

ok the part of printing the quadratic equation with a,b and c changed with numbers is the difficult.

Answer 1

but I also want to show the solution as x1= number. Can someone please explain what is wrong? I have only seen the first two video lectures, so I am not an expert... Thank you very much for your time!!!

Answer 2

What programming language are you using? Python?

Answer 3

oh. yes Python. sorry i didn't mention it.

Answer 4

It's ok :) What version of python, 2 or 3?

Answer 5

2

Answer 6

Actually, you will have to do this:


a=+3
b=-5
c=+2
D=(b**2-4*a*c)
x1=(-b+(D**(-1/2)))/(2*a)
x2=(-b-(D**(-1/2)))/(2*a)

print "(a*(x**2))+(b*x)+c"


if D>0 and (-b+(D**(-1/2)))/(2*a)>0 and (-b-(D**(-1/2)))/(2*a)>0:
print "x1=" x1
print "x2=" x2
elif D>0 and (-b+(D**(-1/2)))/(2*a)>0:
print "x1=" x1
elif D>0 and (-b-(D**(-1/2)))/(2*a)>0:
print "x2"= x2
else:
print "x not found"

Answer 7

I think you have to put the definitions BEFORE you tell it to print things.
Also, make sure you indent the else like this:
else:
print "Indent the else!"

Answer 8

it says, invalid syntax at: "x1=" x1
=( I tried to run it exactly as you said

Answer 9

Instead of print"x1=" x1,
do this:
print"x1=" + x1
print"x2=" + x2

Answer 10

a=+3
b=-5
c=+2
D=(b**2-4*a*c)
x1=(-b+(D**(-1/2)))/(2*a)
x2=(-b-(D**(-1/2)))/(2*a)

print "(a*(x**2))+(b*x)+c"


if D>0 and (-b+(D**(-1/2)))/(2*a)>0 and (-b-(D**(-1/2)))/(2*a)>0:
print "x1=" + x1
print "x2=" + x2
elif D>0 and (-b+(D**(-1/2)))/(2*a)>0:
print "x1=" + x1
elif D>0 and (-b-(D**(-1/2)))/(2*a)>0:
print "x2=" + x2
else:
print "x not found"

Answer 11

>>>
(a*(x**2))+(b*x)+c

Traceback (most recent call last):
File "C:/Users/Klaou/Desktop/hi.py", line 52, in <module>
print "x1=" + x1
TypeError: cannot concatenate 'str' and 'float' objects
>>>

Answer 12

Then do this:
a=+3
b=-5
c=+2
D=(b**2-4*a*c)
x1=str(-b+(D**(-1/2)))/(2*a)
x2=str(-b-(D**(-1/2)))/(2*a)

print "(a*(x**2))+(b*x)+c"


if D>0 and (-b+(D**(-1/2)))/(2*a)>0 and (-b-(D**(-1/2)))/(2*a)>0:
print "x1=" + x1
print "x2=" + x2
elif D>0 and (-b+(D**(-1/2)))/(2*a)>0:
print "x1=" + x1
elif D>0 and (-b-(D**(-1/2)))/(2*a)>0:
print "x2=" + x2
else:
print "x not found"

Answer 13

yes yes yes, basically you did the number string! so this is solved, but the result remains
(a*(x**2))+(b*x)+c
x1=1.0
x2=0.666666666667
I changed the code to something like that:

Answer 14

a,b,c = input("Enter the coefficients of a, b and c separated by commas: ")

print "(a*(x**2))+(b*x)+c"

D=(b**2-4*a*c)
x1=str((-b+(D**(-1/2)))/(2*a))
x2=str((-b-(D**(-1/2)))/(2*a))

if D>0 and (-b+(D**(-1/2)))/(2*a)>0 and (-b-(D**(-1/2)))/(2*a)>0:
print "x1=" + x1
print "x2=" + x2
elif D>0 and (-b+(D**(-1/2)))/(2*a)>0:
print "x1=" + x1
elif D>0 and (-b-(D**(-1/2)))/(2*a)>0:
print "x2=" + x2
else:
print "x not found"

Answer 15

Enter the coefficients of a, b and c separated by commas: 3,-5,2
(a*(x**2))+(b*x)+c
x1=1.0
x2=0.666666666667

Answer 16

so the a,b,c is still there :( want it to be 3,-5,2
But really thanks for the help!!!!

Answer 17

You're welcome!

Answer 18

Finally I've made it! :
a,b,c = input("Enter the coefficients of a, b and c separated by commas: ")
from math import sqrt

x="x"
print ((str(a) + x) + "^2")+( "[+]" + str(b) + x)+( "[+]" + str(c))

D=((b**2)-(4*a*c))
print "D=" + str(D)

if D>0:

sqrt_rootD=(sqrt(D))
print "sqrt_rootD=" + str(sqrt_rootD)

x1=((-b+(sqrt(D)))/(2*a))
print "x1=" + str(x1)

x2=((-b-(sqrt(D)))/(2*a))
print "x2=" + str(x2)

elif D==0:

x3=(-b/(2*a))
print "x3=" + str(x3)

else:
print "x not real number"