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Write a program to find roots of a quadratic equation

quadratic equation is a second order equation having a single variable. Any quadratic equation can be represented as where a, b and c are constants ( a can't be 0) and x is unknown variable. 

寫一個找二次方程的求根程式

For Example

 is a quadratic equation where a, b and c are 2, 5 and 3 respectively.

To calculate the roots of quadratic equation we can use below formula. There are two solutions of a quadratic equation.

使用下列公式:

x = (-b + sqrt(D))/(2*a)
x = (-b - sqrt(D))/(2*a)

where, D = (b*b-4*a*c) is Discriminant (判別式), which differentiate the nature of the roots of quadratic equation.

For the complex result (複數根):

realPart = -b/(2*a);
imaginaryPart =sqrt(-D)/(2*a);

Note: We have used sqrt() function to find square root which is in math.h library.

 

Example input 1:

1 2 1

Example output 1:

Roots of 1.00x^2 + 2.00x + 1.00 = 0 are real and same
x1 = x2 = -1.00

 

Example input 2:

1 -3 2

Example output 2:

Roots of 1.00x^2 + -3.00x + 2.00 = 0 are real and different
x1 = 2.00
x2 = 1.00

 

Example input 3:

1 2 2

Example output 3:

Roots of 1.00x^2 + 2.00x + 2.00 = 0 are complex and different
x1 = -1.00+1.00i
x2 = -1.00-1.00i
[Exercise] Coding (C) - asked in Chapter 5: Selection Statements by (5.2k points)
ID: 28934 - Available when: 2017-10-26 18:00 - Due to: Unlimited

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Your code has newline at the end sometime. You learned this lesson many times!
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After i correct the newline in my code,there is still wrong output.However,I had run the program in Codeblocks.The output is still as well as the result. Seeking for help.
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56 Answers

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answered by (-167 points)
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answered by (-204 points)
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answered by (-255 points)
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prog.c: In function 'main':
prog.c:5:5: warning: implicit declaration of function 'scanf' [-Wimplicit-function-declaration]
     scanf("%f %f %f",&a,&b,&c);
     ^~~~~
prog.c:5:5: warning: incompatible implicit declaration of built-in function 'scanf'
prog.c:5:5: note: include '<stdio.h>' or provide a declaration of 'scanf'
prog.c:13:9: warning: implicit declaration of function 'printf' [-Wimplicit-function-declaration]
         printf("Roots of %.2fx^2 + %.2fx + %.2f = 0 are real and different\nx1=%.2f\nx2=%.2f\n",a,b,c,x1,x2);
         ^~~~~~
prog.c:13:9: warning: incompatible implicit declaration of built-in function 'printf'
prog.c:13:9: note: include '<stdio.h>' or provide a declaration of 'printf'
prog.c:17:9: warning: incompatible implicit declaration of built-in function 'printf'
         printf("Roots of %.2fx^2 + %.2fx + %.2f = 0 are real and same\nx1 = x2 = %.2f",a,b,c,x1);
         ^~~~~~
prog.c:17:9: note: include '<stdio.h>' or provide a declaration of 'printf'
prog.c:21:9: warning: incompatible implicit declaration of built-in function 'printf'
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         ^~~~~~
prog.c:21:9: note: include '<stdio.h>' or provide a declaration of 'printf'
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answered by (-255 points)
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answered by (-255 points)
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answered by (-255 points)
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prog.c: In function 'main':
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     ^~~~~~
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answered by (-255 points)
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prog.c: In function 'main':
prog.c:18:16: warning: too many arguments for format [-Wformat-extra-args]
         printf("Roots of %.2f x^2+%.2f x+%.2f=0 are real and same\n x1=x2=%.2f",a,b,c,x1,x2);
                ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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