How to do the matching method of a one dimensional quadratic equation 80

First, the coefficient of the square is proposed, the coefficient is advanced, and then the formula is formulated, where the coefficient related to the unknown is divided by the coefficient of advance, then divided by 2, and finally the square of this number is a constant that we want to add, and finally subtract this constant by the coefficient of advance.

As. 6x^2+4x+5=6（x^2+2／3x+1／9)-2／3+5=（x-1／3)^2+13／3

x^2-2x-4=0

Solution: x 2-2x = 4

x 2-2x+(1) 2=4+(1) 2 (the left is a perfect flat) x-1) 2=5

x-1 = positive and negative root number 5 (open squared).

x1=6，x2=-4

Move the constant term to the right of the equation, the quadratic term and the primary term on the left side of the equation remain unchanged, and then multiply by the square of one side of the primary term, and the left side of the equal sign becomes a perfect flat method, and you will definitely do it.

Please, you must make me do it, my wealth is already zero! Please!!

The method of matching a quadratic equation is as follows:

The formula of the unary quadratic equation is ax +bx+c=0(a≠0). where ax is called the quadratic term, a is the quadratic coefficient, bx is called the primary term, b is the primary term, and c is called the constant term.

After simplification, an integer equation with only one unknown (one element) and the highest number of unknowns is 2 (quadratic) is called a quadratic equation with one unknown. The value of the unknowns that make the left and right sides of a quadratic equation equal is called the solution of a quadratic equation, also known as the root of a quadratic equation.

An analysis of the algebraic problems on ancient Babylonian tablets shows that as early as 2250 BC the ancient Babylonians had mastered the algebra related to solving unary quadratic equations and applied them to solving problems related to the area and edges of rectangles. The algorithm can be traced back to the Third Dynasty of your.

In two ancient Egyptian papyri found in Kahun, the problem of solving quadratic equations by the test position method also arises.

Around 300 B.C., the contents of Propositions 5 and 6 and Propositions 12 and 13 of Propositions 12 and 13 in Books II, Propositions 5 and 6 and Propositions 12 and 13 of Euclid Elements, a mathematician who was active in Alexandria, the center of ancient Greek culture, were equivalent to the geometric solutions of quadratic equations.

After Euclid, Diophantus, the representative of Alexander's second "** era" of mathematical development, published arithmetica. The book presents a number of problems with quadratic equations or that can be reduced to quadratic equations. This is enough to show that Diophantine is proficient in finding the roots of quadratic equations, but is still limited to positive rational roots.

But he always takes only one root, and if there are two positive roots, he takes the larger one. Ancient Chinese mathematics has long involved the problem of quadratic equations. The most important work of traditional Chinese mathematics, Nine Chapters of Arithmetic, has already covered these issues.

Therefore, it is certain that quadratic equations and their solutions have been known since the Eastern Han Dynasty.

The steps to solve a quadratic equation by the matching method are as follows:

The fitting method solves the quadratic equation step.

If there is only one unknown x, the highest number of unknowns is 2, and the coefficient is not 0, such an equation is called in the worldUnary quadratic equations

The general form of a quadratic equation is: ax 2 (2 is the degree, i.e. the square of x) + bx + c = 0, (a≠0), which is an integer equation with only one unknown and the highest order of the unknown is 2.

Therefore, a quadratic equation must meet the following 3 conditions:

Both sides of the equation are equations about unknowns.

There is only one unknown.

The highest number of unknowns is 2

For example, 2x - 4x 3=0 and 3x =5 are unary quadratic equations.

Matching method:This method can be used when a quadratic equation cannot be solved by direct slicing and factorization after it has been reduced to a general formula.

Solution steps: If the coefficient of the quadratic limb term of the equation is not 1, the terms in the equation are divided by the coefficient of the quadratic term, so that the coefficient of the quadratic term is 1;

Move the constant term to the right of the equal sign;

On both sides of the equation add the square of half of the coefficient of the subterm;

The left side of the equation becomes a perfect clear and fully flat method, and the right side merges similar terms into a real number;

Both sides of the equation are squared at the same time, so as to find the two roots of the equation;

Matching method: The unary quadratic equation is arranged into the form of (x+m) 2=n, and then the method of solving it by the direct open-level method is used.

Reducing the original equation to a general form;

Divide both sides of the equation by the quadratic coefficient to make the quadratic coefficient 1, and move the constant term to the right of the equation;

Add half of the square of the coefficient of the primary term to both sides of the equation;

The left side is matched into a perfectly flat method, and the right side is turned into a constant;

Further, the solution of the equation is obtained by the direct open-level method, and if the right side is a non-negative number, the equation has two real roots; If the right side is a negative number, then the equation has a pair of conjugate imaginary roots.

y=ax²+bx+c

a(x +bx a)+c plus half the square of b a and subtract this number.

a(x²+bx/a+b²/4a²--b²/4a²)+c=a(x+b/2a)²-a*b²/4a²+c=a(x+b/2a)²-b²/4a+c

For example, x +6x

x² +6x +3² -3²

x+3)² 9

Generally, it is half the square of the coefficient of the primary term.

The first year of junior high school is a summary of the knowledge of primary school, of course, as the grade level increases, the knowledge will also be difficult!

No way! The factorization and matching methods are both junior high, ok? Don't pretend to understand.

ax^2+bx+c=0

a(x^2+bx/a)+c=0

a[x+b (2a)] 2+c-b 2 (4a)=0 The trick is to divide the coefficient of the primary term by the coefficient of the quadratic term and divide it by 2

2(y²-3y)+1

2(y²-3y+9/4-9/4)+1

2(y²-3y+9/4)-9/2+1

2(y-3/2)²-7/2

Can you read it? It's not good to do fractions and squares.

The specific process of solving a quadratic equation by the matching method is as follows:

1.This unary quadratic equation is reduced to the form ax 2+bx+c=0 (this unary quadratic equation satisfies the real root).

2.The quadratic coefficient is reduced to 1

3.Move the constant term to the right of the equal sign.

4.The left and right sides of the equal sign are simultaneously added to the square of half of the coefficient of the primary term 5Write the algebraic formula to the left of the equal sign in perfectly squared form.

6.The left and right sides are squared at the same time.

7.Sorting out can get the root of the original equation.

Example: Solve the equation 2x 2+4=6x

5.(x2=1

1.Transformation: Transform this unary quadratic equation into the form ax 2+bx+c=0 (i.e., the general form of the unary quadratic equation) into the general form 2Shift: The constant term is moved to the right of the equation.

You can get the root of the original equation) Algebraic notation: Note ( 2 is the meaning of squared. )

ax^2+bx+c=a(x+b/2a)^2+（4ac-b^2）/4a=a[(x+m)^2-n^2]=a(x+m+n)*(x+m-n)

Example: Solve the equation 2x 2+4=6x

: Add 3 and a half squared, and -2 also add 3 and a half squared to make both sides of the equation equal)

5.( a 2+2b+1=0 i.e. (a+1) 2=0).

x2=1 (a quadratic equation usually has two solutions, x1 x2).