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C語言 求解ax^2+bx+2=0

#include<stdio.h>

#include

<math.h>

main()

{float

a,b,c,d,disc,x1,x2,realpart,imagpart;

scanf("%f,%f,%f",&a,&b,&c);

輸入系數

printf("the

equation");

屏幕打印“the

equation”

if(fabs(a)

<=le-6)-------------------1e-6是10的-6次方的意思,a為0,則為壹次方程

printf("is

not

a

quaratic");

//屏幕打印“is

not

a

quaratic”,即如果上面條件成立,屏幕上出現

//the

equation

is

not

a

quaratic;

else

{disc=b*b-4*a*c;

//根的判別式b^2-4ac

if(fabs(disc)

<=le-6

//如果等於0(接近0,因為float型的數值不會嚴格相等);

printf("has

tow

equal

roots:%8.4\n",-b/(*a));

//屏幕打印"has

tow

equal

roots

:"和(-b/a)的值;四位小數

else

if(disc>le-6)

//如果判別式值大於0,則

{x1=(-b+sqrt(disc))/(2*a);

//根據公式算出x1,x2;

{x2=(-b-sqrt(disc))/(2*a);

printf("has

distinct

real

roots:%8.4f

and%8.4f\n",x1,x2);//屏幕打印“has

distinct

real

roots”,"x1,x2

//的值"格式同上,

//接下來應該判別disc小於0的情況;

if

(disc<le-6)

{printf("has

no

real

roots");

realpart=-b/(2*a);

//實根

imagpart=sqrt(-disc)/(2*a);//虛根的模

printf("but

tow

complex

roots");

printf("the

real

part

is

%8.4f

and

the

imagpart

is

%8.4f\n

i",realpart,imagpart);//第壹個根的實部和虛部

printf("the

real

part

is

%8.4f

and

the

imagpart

is

-

%8.4f\n

i",realpart,imagpart);//第二個根的實部和虛部

}

}

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