C语言编程求解一元二次方程 ax2 bx c=0

#include"stdio.h"#include"math.h"main(){floata,b,c,p,q,k,l;{printf("\n\n\n");printf(">>输入a,b,c\n\n>>

#include"stdio.h"#include"math.h"/*求ax*x+bx+c=0的解*/main(){floata,b,c,x1,x2,d;printf("请输入a：");scanf("

main(){floata,b,c;\x09floattemp;//计算b*b-4*a*c\x09floatnum;//开根号\x09floatx1,x2;//方程的根\x09printf("Ente

//只一处有错,还有一个注意输入格式.#include#includeintmain(){doublep,q,x1,x2,disc,a,b,c;scanf("%lf,%lf,%lf",&a,&b,&c

scanf("%f,%f,%f",a,b,c);a,b,c前加个&符号还有x1=(e-b)/2a,要x1=(e-b)/(2*a)

lf%错了,应该是%lf

很高兴为您解答.原代码中的scanf和printf中的%要放在d和lf的前面才对,改正后运算无误~#include#includevoidmain(){doublex1;//x1,x2分别为方程的2个

double改做float再问：yiyuanercifangcheng.cpp(25):warningC4244:'=':conversionfrom'int'to'float',possiblelo

#include#includeintmain(){doublea,b,c,disc,p,q,x1,x2;scanf("%lf%lf%lf",&a,&b,&c);disc=b*b-4*a*c;if(a

#includefloatf(float);voidmain(){floata,b,c,d,x1,x2,p,q;printf("a=");scanf("%f",&a);printf("b=");sca

disp('方程形式:a*x^2+b*x+c=0');a=input('a=');b=input('b=');c=input('c=');p=[abc];ans=roots(p)哥们,Mablab输出

#include#includeintmain(){inta,b,c,m;doublex1,x2,n;//解为double类型printf("请输入ax2+bx+c=0中的a,b,c:\n");sca

刚做了,不知道是否都是你问的,源程序如下#include#includeintmain(){floata,b,c,p,x1,x2;/*a,b,c为方程的系数,p用来存放b*b-4ac的值,x1,x2存

cleartext一元二次方程求解ax^2+bx+c=0endtextinput"请输入a的值："toainput"请输入b的值："tobinput"请输入c的值："tocm=b*b-4*a*cifm

可以用Scanner逻辑上应该没错误importjava.util.Scanner;publicclassTest2{publicstaticStringx(inta,intb,intc){intx=

#include#include#includevoidmain(){floata,b,c,x1,x2,delta;intflag;printf("a=");scanf("%f",&a);printf

#include#includevoidmain(){floata,b,c,disc,x1,x2,realpart,imagpart;scanf("%f,%f,%f",&a,&b,&c);disc=b

x=solve('a*x^2+b*x+c','x')x=-(b+(b^2-4*a*c)^(1/2))/(2*a)-(b-(b^2-4*a*c)^(1/2))/(2*a)

1#include#includevoidmain(){printf("输入二次项系数、一次项系数和常数项：");scanf("%f%f%f",a,b,c);floatd=b*b-4*a*c;

#include <stdio.h> #include <math.h>void b1 () { floa