Find the solution of a quadratic equation by programming!

vb wants to control oh add it yourself.

**：private sub command1_click()dim a, b, c, delta, x1, x2, x as double

a = val(

b = val(

c = val(

if a = 0 then

The quadratic coefficient is 0 and is not a quadratic equation"

end if

delta = b * b - 4 * a * c ‘b^2-4ac= "b^2-4ac=" & delta 'This sentence can be left out.

if delta = 0 then

x = (-b) / 2 * a

The equation has two identical real roots x=" & xend if

if delta > 0 then

y = sqr(delta) 'b 2-4acx1 under root number = (y - b) 2 * a

x2 = (-b + y)) / 2 * a= x1 & x2

end if

if delta < 0 then

Equations have no real roots"

end if

end sub

Microsoft C++.

#include

#include

using namespace std;

int main()

double a,b,c;

double delta,x1,x2;

int sign;

cout<<"The unary quadratic equation a*x*x+b*x+c=0";

cout<<"Enter three coefficients a(a!).=0),b,c;"<>a>>b>>c;

cout<<"a="<0)

sign=1;

else sign=0;

delta=sqrt(fabs(delta));

x1=-b/(2*a);

x2=delta/(2*a);

if(sign)

cout<<"The equation has two different real roots:"

Anonymous users2024-02-04

The general method of solving a quadratic equation is of course to solve it directly by the formula method, which is applicable to all quadratic equations, but it is not the easiest method.

In fact, when we solve a quadratic equation, we should adopt the corresponding method according to the characteristics of the equation, so as to improve the efficiency of solving the problem

1) Root-finding formula method, applicable to all one-dimensional quadratic equations;

2) Factorization, the equation is deformed into the form of (x-x1)(x-x2)=0, so that x-x1=0 and x-x2=0 obtain the two roots of the equation.

3) Matching method: The matching method is also suitable for solving all one-dimensional quadratic equations. The method is not described separately.

4) Direct opening and leveling method: When both sides of the equation are perfectly squared, you can directly square both sides of the equation and take positive and negative.

According to the characteristics of different equations, choose different methods to achieve twice the result with half the effort!

There is no one fixed method, only flexibility.

There are generally two common solutions to binary equations:

1.Substituting the elimination method: 2, addition and subtraction elimination method.

1.Substitution of the elimination method.

Substitution elimination method: Turn the coefficient of an unknown number of one of the equations into 1 and substitute it into another equation.

For example: 2x+y=9

5x+3y=21②

Solution: Derive: y=9-2x

Substituting "gets: 5x+3(9-2x)=21".

5x+27-6x =21

5x-6x = 21-27

x = 6x =6

Substituting x=6 yields: y=-3

The solution of the system of equations is x=6 y=-3

2.Addition, subtraction, and elimination method uses the properties of the equation to make the absolute value of the coefficient before one of the two unknowns in the equation system equal, and then add (or subtract) the two equations to eliminate the unknown, so that the equation contains only one unknown and can be solved.

3x+2y=7 ①

5x-2y=1 ②

Solution: 3x+5x)+2y+(-2y))=7+1)8x=8

x=1 substitute x : 3x+2y=7

3×1+2y=7

2y=4 y=2x=1

Binary equations, if you really don't know how to solve them, then use the root formula, but sometimes it can be very troublesome and complicated. After all, you can't use a calculator during the exam. Find the memory of the root formula, if you use it too much, you will naturally remember it.

But it's best to try your best to learn how to multiply crosses! This would be convenient, if applicable. There is also learning how to trim, which is necessary!

If you really don't understand, then dissect the example problem!

All the formulas can be solved, which is -b+- root number... As for memory, it's OK to use it a few more times.

Listen to the teacher and do a few simple ways to find it.

First of all, this is not a quadratic equation, but a quadratic inequality!

Solution: f'(x)=3(x^2)-6x-9

3[(x 2)-2x-3] (Factorization: Extracting the common factor 3) = 3(x+1)(x-3) (Factoring: Cross multiplication) 1, by f'(x) 0, there is:

x+1)(x-3) 0 i.e.: x+1 0, x-3 0.........1) Or: x+1 0, x-3 0.........2) By the group of inequalities (1), there are:

x -1, x 3, get: x 3 by the inequality group (2), have: x -1, x 3, get:

x -1 is combined to obtain: x 3, or: x -1;

2. By f'(x) 0, there is: (x+1)(x-3) 0 can be seen: x+1 and x-3 must have different signs, i.e.: x+1 and x-3 must be one positive and one negative).

Obviously: x-3 x+1

So: x+1 0, x-3 0

Solution: x -1, x 3

i.e.: -1 x 3.

Do you understand?

In the end, there are two ways. 1, using the root finding formula 2, multiply the crosses.

First of all, factorization! Then let each term be equal to 0 to get x.

3x-5y=-2

2x+7y=40

6x-10y=4 1

6x+21y=120 2

2-1 Laugh Wood.

31y=116 y=116 Excavation section 31

Substituting 1 x = 642 93

There are generally two ways to do this.

For example, solution x+y=8 1).

3x+y=12 2）

Method 1: Substitution equation method.

Obtained from (1).

y=8-x 3）

Replace (3) with (2).

3x+(8-x)=12

x=2 and then replace x=2 back to (1).

Get 2+y=8

y=6 Method 2: Addition and subtraction equation method.

2) Formula - (1) Formula.

Get 2x=4x=2 and substitute x=2 back to (1).

Get 2+y=8

y=6 In general, the first type is used.