10-26 Exercise 3: Find roots of a quadratic equation

Question

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Write a program to find roots of a quadratic equation

A 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 Oct 26, 2017 in Chapter 5: Selection Statements by semicolon (5.2k points)
ID: 28934 - Available when: 2017-10-26 18:00 - Due to: Unlimited
reopened Oct 26, 2017 by thopd | 12k views

56 Answers

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112233 · Answer 1 · 2017-11-09 17:27:38

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answered Nov 9, 2017 by 112233 (-167 points)

112233 · Answer 2 · 2017-11-09 17:17:35

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answered Nov 9, 2017 by 112233 (-167 points)

garrylin2 · Answer 3 · 2017-11-02 19:54:55

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answered Nov 2, 2017 by garrylin2 (-204 points)

garrylin2 · Answer 4 · 2017-11-02 19:53:31

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answered Nov 2, 2017 by garrylin2 (-204 points)

rick · Answer 5 · 2017-11-02 19:34:07

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'
         printf("Roots of %.2fx^2 + %.2fx + %.2f = 0 are complex and different\nx1 = %.2f+%.2fi\nx2 = %.2f-%.2fi",a,b,c,x3,x4,x3,x4);
         ^~~~~~
prog.c:21:9: note: include '<stdio.h>' or provide a declaration of 'printf'

rick · Answer 6 · 2017-11-02 19:28:09

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