Enter 3 integers abc and find the solution of a x x b x c 0 How to program it

#include

#include

using namespace std;

int main()

float a = 1,b = 0, c= 0;

cout<<"a:";

cin>>a;

cout<<"b:";

cin>>b;

cout<<"c:";

cin>>c;

if(a==0)

return 0;

float x1,x2 = 0;

float i;

i = b*b-4*a*c;

if(i<0)

cout<<"There is no real root in the equation! "<0)

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

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

cout<<"The equation has two unequal real roots:"<>a;

cout<<"b:";

cin>>b;

cout<<"c:";

cin>>c;

Enter a value for b c.

Isn't the control structure the if statement here?

Take a closer look and understand it.

This question is a must-have example or exercise in a C language book, hehe.

buddyand23 can write it.

The one on the first floor is commonly used, and both solutions are explained in the book.

The equation that requires an integer solution is: (a+b) c + a+c) b + b+c) a = 4

a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2) /abc) =4

The molecule is carried out to obtain:

a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 = 4abc

a^2(b+c) +b^2(a+c) +c^2(a+b) =4abc

Now let's try to find integer solutions. Since every variable in the equation is symmetrical, we can assume a b c.

When a = 1, the equation becomes:

b+c) +b^2 + c^2 = 4bc

We can try different combinations of b and c to verify that the equation is full. By trying, we found that the equation holds when b = 2 and c = 3.

So, the integer solution is a = 1, b = 2, c = 3.

Proceed with the equation and get:

Problem: (a+b) c+(a+c) b+(b+c) a=4 to find the integer solution.

The equation is obtained: (a+b) c+(a+c) b+(b+c) a=4

The solution is suspicious: ,

Write a program and enter abc to find the value of ax + bx + c = 0.

Should we also consider the case where a is sometimes 0?

Find the root of the equation x 3 + ax 2 + bx + c = 0 using the string truncation method: (where x1 and x2 are required to be entered by the user are the intervals of the root estimated by the user).

#include

#include

floata,b,c;

Find the function value of f(x)=x3+ax 2+bx+c=0. */floatf(float

x) *Find the abscissa of the intersection of the lines of (x1,f(x1)) and (x2,f(x2)) and the x-axis: x [x1 f(x2) x2 f(x1)].

f(x2)－f(x1)]*

floatxpoint(float

x1,float

x2) * Find the real root of the equation in the interval (x1, x2) * floatroot(float

x1,float

x2)while(fabs(f(x))>1e-4);

return(x);

main()

while(f(x1)*f(x2)>=0);

x=root(x1,x2);

printf("a

rootof

equation

is%.4f",x);

For example, a, b, and c enter -5 respectively

80, x1 and x2 are input 1 each

10, the result is 5

By the condition has: a+b+c=1

16a-4b=45

y'=3x^2+2ax+b

No extremes. Rule.

Discriminant formulas of derivatives.

4a^2-12b<0

Rule. 4（a-6)^2<9

Namely. 3/2

It is easy to know that a(x-a)(x-b)=0

ax -a(a+b)x+aab=0, i.e. ax +bx+c=0 obviously a=a, -a =b(1+a).

b=-a+1-1/a+1，x=1

b za + 1 = 1 a = 0 (round) or a = -2

When a=-2, b=4, c=16

Therefore a+b+c=18

Because ab=48, ac=96, =48 96, that is, b c=48 96, so b=48c 96, because bc=72, substituting gets, 48c 96=72, c=12, and then a=8, b=6, so abc=576