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A simple program to solve quadratic equations with

, 8 Nov 2010 CPOL
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Simple and prints imaginary roots too!float a,b,c,x1,x2,d,dsq;printf("ax^2 + bx + c = 0");printf("\nEnter a,b,c separated by commas : \n");scanf("%f,%f,%f",&a,&b,&c);d = b*b-(4*a*c);if(d>=0){dsq=sqrt(d);x1 = (-b+dsq)/(2*a);x2 = (-b-(dsq))/(2*a);printf("\nRoot 1 : %f\nRoot 2...
Simple and prints imaginary roots too!
float a,b,c,x1,x2,d,dsq;
printf("ax^2 + bx + c = 0");
printf("\nEnter a,b,c separated by commas : \n");
d = b*b-(4*a*c);
x1 = (-b+dsq)/(2*a);
x2 = (-b-(dsq))/(2*a);
printf("\nRoot 1 : %f\nRoot 2 : %f",x1,x2);
d = ((4*a*c)-pow(b,2))/(2*a);
printf("\nRoot 1 : %f+%fi",((-b)/(2*a)),d);
printf("\nRoot 2 : %f-%fi",((-b)/(2*a)),d);}


This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)


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Anshul R
India India
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Comments and Discussions

GeneralReason for my vote of 2 Can crash too ! PinmemberYvesDaoust9-Nov-10 0:27 
Generalthats the most basic rule in the book PinmemberJFergulbops8-Nov-10 23:38 
General1) It doesn't matter what the real world quadratic equations... PinmemberSilic0re098-Nov-10 5:34 
1) It doesn't matter what the real world quadratic equations say, if I run your program, I have the freedom to (stupidly or not), put a zero where asked for "a". The program should recognize this and ask for a different number, not create a divide by zero exception. One great axiom of programming is to ALWAYS ASSUME your input data is dirty, incorrect, and malicious. You need to protect your program from the user.
GeneralThank you for the suggestion.. I do not wish to be rude. If... Pinmemberpranav958-Nov-10 1:13 
GeneralReason for my vote of 2 Inefficient and does not handle divi... PinmemberAndrew Phillips7-Nov-10 16:19 
GeneralThis is better than the others as you avoid the domain error... PinmemberAndrew Phillips7-Nov-10 16:14 

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