Quadratic equation solver

[joeyke]

Ask the user to enter the coefficients a, b and c. Have your program then solve the two (or possibly one) solution(s) for X. Be careful to make sure there is a solution by confirming that the square root is not executed on a negative number which is considered to be imaginary.

Be sure to #include<cmath> and call the pow() and sqrt() functions when making your math calculations. Also note, the coefficients a b and c are to be declared as a double, as well as your two solutions for X.
this is what i did so far:

#include <cmath>
#include <iostream>

using namespace std;

int main ()
{                       
int a;
int b;
int c;
int x;
double posx, negx; //declare positive x and negative y

cout<<"Enter the value of a: ";
cin>>a;
cout<<"Enter the value of b: ";
cin>>b;
cout<<"Enter the value of c: ";
cin>>c;

int check;
check=(b*b)-(4*a*c);      
if (check>0)
{
   posx=(-b+sqrt(check))/2*a;
   negx=(-b-sqrt(check))/2*a;

system ("pause");
return 0;

}

}

[grayMist]

I am assuming your program attempts to solve a quadratic equation.
please mention the issue you are facing in more details. From the problem description i could not make out any issues at all !!

Here are a few pointers:
1) You have not declared a,b,c as double.
2) Declare 'check' as double as well or sqrt() will complain of an ambiguous call.

Do this and then let us know exactly where you are getting stuck.

And , most important , don't forget to put your code snippets inside the code-tags.

[joeyke]

in my output it doesn't go beyond entering a value

i don't know how the output should look like because from what i get it shows this:
Last login: Thu Feb 25 21:57:40 on ttys000
/Volumes/JOANNEKEO/C++1.5/HWkeoAssignment 2-3A
joanne-keos-macbook:~ joanne$ /Volumes/JOANNEKEO/C++1.5/HWkeoAssignment 2-3A
Enter the value of a: 5
Enter the value of b: 6
Enter the value of c: 3
joanne-keos-macbook:~ joanne$

the picture is the quadratic equation

[grayMist]

I think you forgot to display the values of posx and negx .

that i think is quite important. ;)

[joeyke]

yes

oh yeah ! how do i go about that in my program? thank for you help btw!

[grayMist]

try cout<<" posx ="<<posx<<" negx = "<<negx;
that should do a neat job of printing the values.

[joeyke]

where do i put that? after this:
posx=(-b+sqrt(check))/2*a;
negx=(-b-sqrt(check))/2*a;

[grayMist]

When do you think the output should be displayed?
before the calculation or after it ? :)

[joeyke]

after i think. is it like this:

if (check>0)
{
cout<<" posx ="<<posx<<" negx = "<<negx;

cin>>posx, negx;

posx=(-b+sqrt(check))/2*a;

negx=(-b-sqrt(check))/2*a;


system ("pause");

return 0;

}

}

[Banfa]

Have you tried to compile that?
If it compiled what result did you get printed on the screen?
Was that the result you expected?

grayMist question "When do you think the output should be displayed? Before the calculation or after it ?" is still relevant but perhaps you should also answer the question

Which piece of my code actually does the calculation.

[joeyke]

i still get this in my output:
Enter the value of a: 1
Enter the value of b: 2
Enter the value of c: 3

should there a calculation done in my output? if so then why isn't it printing that?

[joeyke]

wait i get it!
Enter the value of a: 1
Enter the value of b: 4
Enter the value of c: 6
posx =-3.81582e-232 negx = 0

is this right?

[joeyke]

that's what prints now after i did
double check;
check=(b*b)-(4*a*c);
cout<<" posx ="<<posx<<" negx = "<<negx;
cin>>posx, negx;

[johny10151981]

you have a huge mistake in the below coldes

posx=(-b+sqrt(check))/2*a;

negx=(-b-sqrt(check))/2*a;



please rethink. you must get a result.

Regards,
Johny

[Banfa]

Enter the value of a: 1
Enter the value of b: 4
Enter the value of c: 6
posx =-3.81582e-232 negx = 0

is this right?

You should know if it is right the first time you run the program. You do this by using the co-efficients of a function that you know the roots of. For a quadratic (or any polynomial) this is relatively easy because many quadratics can be written in the form

(m.x + n).(o.x + p)

Set m and o to 1 to make things easy, set n to a small negative integer and p to a small postive integer and you get

a = 1
b = p + n
c = p * n

With roots -p and -n

So you have a known equation with known roots and you can plug those values into your program and you will know it it has output the right value.