编写程序求一元二次方程的根,考虑不同的根情况
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PrivateSubCommand1_Click()DimaAsSingle,bAsSingle,cAsSingleDimdAsSingle,x1AsSingle,x2AsSinglea=InputB
double x1 = 0;//解1double x2 = 0;//解2Console.WriteLine("求 ax^
//Equation.h#ifndef_Equation_h#define_Equation_hclassEquation{private:doublea;doubleb;doublec;voidSh
#include"stdio.h"#include"math.h"doublex1,x2,p;floatfile1(floata,floatb){x1=(-b+sqrt(p))/2*a;x2=(-b-
PublicClassForm1PrivateSubButton1_Click(ByValsenderAsSystem.Object,ByValeAsSystem.EventArgs)HandlesB
#include#includevoidm(floata,floatb,floatc){\x09doublex1,x2;\x09x1=(-b+sqrt(b*b-4*a*c))/(2*a);\x09x2
#include#includeintmain(){doublea,b,c,disc,x1,x2,p,q,x;scanf("%lf%lf%lf",&a,&b,&c);disc=b*b-4*a*c;if
我也刚学C,费了好几个小时,终于把这个问题搞定了!已经运行过了,结果跟谭版结果一样,敬请放心使用.#include"stdio.h"#include"math.h"voidmain(){doublea
对于形如a*x^2+b*x+c=0的方程可以使用下面的程序求根x=roots([abc])例如4*x^2-5*x+1=0x=roots([4-51])x=1.00000.2500祝你学习愉快!再问:是
#include"stdio.h"#include"math.h"voidmain(){floata,b,c;floatdelta;printf("inputa:");scanf("%f",&a);p
a=-10;b=10;n=0;whileb-a>epst=(a+b)/2;n=n+1;if4*t^2+3*t-6==0break;elseif(4*a^2+3*a-6)*(4*t^2+3*t-6)>0
第二题:#includevoidmain(){inti,g,s,b;for(i=100;i
C++的代码:#include#includevoidmain(void){doublea,b,c,d;charch('y');do{coutb>>c;if(-0.0001
#include#include
#include#includevoidb1(){floatl,s,k;inta,b,c,h;printf("\n");printf("\n");printf("输入a,b,c的值\n");print
dimaasdouble,basdouble,casdoubledimx1asdouble,x2asdoublea=val(inputbox(""))b=val(inputbox(""))c=val(
以下是画一元二次方程的图的代码:(假设y=a*x^2+b*x+c,将窗口的大小调为4800×4800)PrivateSubCommand1_Click()ClsLine(0,2400)-(4800,2
PrivateSubCommand1_Click()Dima#,b#,c#,d#,x1#,x2#a=Val(InputBox("a=","数据输入框",1))b=Val(InputBox("b=","
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;