题目

c语言程序设计题目:计算一元二次方程的根
【程序填空】
#include
//
else if(【?】)
{x1=(-b+sqrt(disc))/(2*a);
x2=(-b-sqrt(disc))/(2*a);
printf("has distinct real roots:%8.4f and %.4f ",x1,x2);
}
else
{realpart=-b/(2*a);
imagpart=sqrt(-disc)/(2*a);
printf("has complex roots: ");
printf("%8.4f=%.4fi ",realpart,imagpart);
printf("%8.4f-%.4fi ",realpart,imagpart);
}
}

提问时间：2020-11-07

答案

#include
//
else if(disc>0)
{
x1=(-b+sqrt(disc))/(2*a);
x2=(-b-sqrt(disc))/(2*a);
printf("has distinct real roots:%8.4f and %.4f ",x1,x2);
}
else
{
realpart=-b/(2*a);
imagpart=sqrt(-disc)/(2*a);
printf("has complex roots: ");
printf("%8.4f+%.4fi ",realpart,imagpart);
printf("%8.4f-%.4fi ",realpart,imagpart);
}
}
}
2 -3 -5
the equationhas distinct real roots:2.5000 and -1.0000
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