A solution of quadratic equation in mathematics (the other two are formula method and decomposition method)

The specific process is as follows:

1. Transform this unary quadratic equation into the form of AX 2+BX+C = 0 (this unary quadratic equation satisfies the real root).

2. Convert the quadratic coefficient into 1.

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

4. At the same time, add half the square of the first coefficient to the left and right sides of the equal sign.

5. Write the algebraic expression on the left side of the equal sign as a complete square.

6. Simultaneously Square

7. The root of the original equation can be obtained by sorting.

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

1.2x^2-6x+4=0

2.x^2-3x+2=0

3.x^2-3x=-2

4.x 2-3x+2.25 = 0.25 (+2.25: plus 3.5 square, -2 should also add 3.5 square to make both sides of the equation equal).

5. (x-1.5) 2 = 0.25 (a 2+2b+1= 0, that is, (A+ 1) 2 = 0).

6.x- 1.5= 0.5

7.x 1=2

x2= 1

Quadratic function collocation skills;

Y = ax 2-bx+c is converted into y = a (x+h) 2+k h) 2+k.

=a(x+b/2a)^2+(c-b^2/4a)

Cross multiplication

Cross multiplication can decompose some quadratic trinomials. The key of this method is to decompose the quadratic coefficient A into the product of two factors a 1 and A2 A 1. 6? 1a2, decompose the constant term c into the product of two factors, c 1 and C2? 6? 1c2, and make a 1c2+a2c 1 just be a linear term b, then the result can be written directly: when factorizing factors in this way, we should pay attention to observation, try and realize that it is actually the inverse process of binomial multiplication. When the first coefficient is not 1, it often needs to be tested many times, so be sure to pay attention to the sign of each coefficient.

This example puts 2x^2；; -7x+3 factorization factor.

Analysis: first decompose the quadratic coefficient and write it in the upper left corner and lower left corner of the crosshair, then decompose the constant term and divide it into two parts.

Don't write it in the upper right corner and lower right corner of the crosshair, and then cross multiply to find the algebraic sum to make it equal to the coefficient of the first term.

Quadratic coefficient decomposition (positive factor only):

2= 1×2=2× 1;

Decomposition of constant term:

3= 1×3=3× 1=(-3)×(- 1)=(- 1)×(-3).

Draw a cross line to represent the following four situations:

1 1

╳

2 3

1×3+2× 1

=5

1 3

╳

2 1

1× 1+2×3

=7

1 - 1

╳

2 -3

1×(-3)+2×(- 1)

=-5

1 -3

╳

2 - 1

1×(- 1)+2×(-3)

=-7

After observation, the fourth case is correct, because after cross multiplication, the algebraic sum of the two terms is exactly equal to the coefficient of the first term -7.

Solution 2x^2；; -7x+3=(x-3)(2x- 1)。

Generally speaking, for the quadratic trinomial ax2+bx+c(a≠0), if the quadratic term coefficient A can be decomposed into the product of two factors, that is, a=a 1a2, the constant term C can be decomposed into the product of two factors, that is, c=c 1c2, and A/KLOC-.

a 1 c 1

? ╳

a2 c2

a 1c2+a2c 1

Cross-multiply diagonally, and then add to get a 1c2+a2c 1. If it is exactly equal to the first term coefficient b of the quadratic trinomial ax2+bx+c, that is, a 1c2+a2c 1=b, the quadratic trinomial can be decomposed into two factors a65438+.

ax2+bx+c =(a 1x+c 1)(a2x+C2)。

The way to help us decompose the quadratic trinomial by drawing cross lines like this is usually called cross multiplication.