Quadratic Equation

This is a discussion on Quadratic Equation within the C Programming forums, part of the General Programming Boards category; so the exceptions i came up with is. Exceptions: 1.if a=0… there are infinite solutions 2.if b^2-(4ac)<0… YOu get an ...

  1. #16
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    so the exceptions i came up with is.
    Exceptions:
    1.if a=0… there are infinite solutions
    2.if b^2-(4ac)<0… YOu get an imaginary number
    Is that the only exceptions i need to modify my program for? That i wont get regular integers. Its just the math part of it im confused on.

  2. #17
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    If a is zero then you are just doing a regular linear function, correct? And read my last post. Use the complex data type that your compiler may supply you.

  3. #18
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    Besides, you are also forgetting that if "a" is zero, then invariably you end up dividing by zero. Which isn't technically a limitation of the quadratic formula since it is designed to only deal with quadtratics and not linear functions.

  4. #19
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    YAY

    Code:
    #include <FPT.h>
    
    
    int main()
    {
      double a,b,c,d,x1,x2,rp,ip ;
    
      a = inD() ;
      b = inD() ;
      c = inD() ;
    
    
      if (a == 0.0) {
    
        // linear equation?
        if ((b == 0.0) && (c == 0.0)) {
            outS("Any number is a solution.\n") ;
        } else if (b == 0.0) {
            outS("There are no solutions.\n") ;
        } else {
            outS("There is a single real solution, ") ;
            x1 = -c/b ;
            outD(x1) ;
            outS("\n") ;
        }
    
    
      } else 
      {
    
        d = b*b - 4*a*c ;
    
        if (d == 0) {
           x1 = -b/(2*a) ;
           outS("There are two identical real solutions, ") ;
           outD(x1) ;
           outS(" and ") ;
           outD(x1) ;
           outS("\n") ;
        } else if (d > 0) {
           d = sqrt(d) ;
           x1 = (-b + d)/(2*a) ;
           x2 = (-b - d)/(2*a) ;
           outS("There are two real solutions, ") ;
           outD(x1) ;
           outS(" and ") ;
           outD(x2) ;
           outS("\n") ;
         } else {
           d = sqrt(-d) ;
           rp = -b/(2*a) ;
           ip =  d/(2*a) ;
           outS("There are two complex answers, ") ;
    
           outD(rp) ;
           if (ip >= 0) {
    	 outS(" + ") ;
             outD(ip) ;
             outS("i") ;
           } else {
    	 outS(" - ") ;
             outD(-ip) ;
             outS("i") ;
           }
    
           outS(" and ") ;
    
           ip = -ip ;
           outD(rp) ;
           if (ip >= 0) {
    	 outS(" + ") ;
             outD(ip) ;
             outS("i") ;
           } else {
    	 outS(" - ") ;
             outD(-ip) ;
             outS("i") ;
           }
    
           outS("\n") ;
         }
    
      }
    
    
    
    }
    I believe that i got a working program! Someone shoot me some numbers so that i know it solves all numbers!

  5. #20
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    I believe your code looks good. What happens if you pass in NaN or positive (or negative) infinity? Your program should at least test for those values and throw them out if they are in use. It should actually just treat all of those as NaN and spit out an IO error.

  6. #21
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    sorry im confused as to what you're stating. Please Explain (its been a long day of classes)

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