The concrete process of solving quadratic equation of one variable Hurry up, senior high school

The concrete process of solving quadratic equation of one variable Hurry up, senior high school

The solution of quadratic equation of one variable
1、 Knowledge points:
One variable quadratic equation and one variable linear equation are integral equations, which is an important content of junior high school mathematics, and also the basis of learning mathematics in the future
The general form of quadratic equation with one variable is ax ^ 2 (2 is the degree, that is, the square of x) + BX + C = 0, (a ≠ 0). It is an integral equation with only one unknown and the highest degree of the unknown is 2
The basic idea and method of solving quadratic equation with one variable is to change it into two quadratic equations with one variable by "reducing degree"
1. Methods: direct leveling method; 2. Matching method; 3. Formula method; 4. Factorization method
2、 Methods and examples:
1. Direct leveling method:
The direct flattening method is to solve the quadratic equation of one variable by direct square root. The direct flattening method is used to solve the equation of form (x-m) 2 = n (n ≥ 0), and the solution is x = ± M
Solve equation (1) (3x + 1) 2 = 7 (2) 9x2-24x + 16 = 11
Analysis: (1) the equation is obviously easy to be solved by the direct square method. (2) the left side of the equation is the complete square equation (3x-4) 2, and the right side = 11 > 0, so the equation can also be solved by the direct square method
(3x+1)2=7×
∴(3x+1)2=5
Ψ 3x + 1 = ± (be careful not to lose the solution)
∴x=
The solution of the original equation is X1 =, x2=
2. Collocation method: solve the equation AX2 + BX + C = 0 (a ≠ 0) by collocation method
First move the constant C to the right of the equation: AX2 + BX = - C
The quadratic coefficient is reduced to 1: x2 + X=-
On both sides of the equation, add the square of half of the coefficient of the first term: x2 + X + () 2 = - + () 2
The left side of the equation becomes a complete square: (x +) 2=
When b2-4ac ≥ 0, x + = ±
X = (this is the root formula)
Solving the equation 3x2-4x-2 = 0 by the collocation method
Move the constant term to the right of the equation 3x2-4x = 2
The quadratic coefficient is reduced to 1: x2-x=
Add the square of half of the coefficient of the first term to both sides of the equation: x2-x + () 2 = + () 2
Formula: (x -) 2=
Direct square root: X - = ±
∴x=
The solution of the original equation is X1 =, X2 =
3. Formula method: change the quadratic equation of one variable into a general form, and then calculate the value of the discriminant △ = b2-4ac. When b2-4ac ≥ 0, substitute the values of various coefficients a, B, C into the root formula x = [- B ± (b ^ 2-4ac) ^ (1 / 2)] / (2a), (b ^ 2-4ac ≥ 0) to get the root of the equation
Solving equation 2x2-8x = - 5 by formula method
The equation is reduced to a general form: 2x2-8x + 5 = 0
∴a=2,b=-8,c=5
b^2-4ac=(-8)2-4×2×5=64-40=24>0
∴x=[(-b±(b^2-4ac)^(1/2)]/(2a)
The solution of the original equation is X1 =, X2 =
4. Factorization: transform the equation into zero on one side, decompose the quadratic trinomial on the other side into the product of two quadratic factors, let the two quadratic factors equal to zero respectively, and get two linear equations with one variable. The roots obtained by solving the two linear equations with one variable are the two roots of the original equation. This method of solving quadratic equations with one variable is called factorization

A problem of solving quadratic equation of one variable
We can take X & sup2; - 1 as a whole, and then let X & sup2; - 1 = y.①, then the original equation can be changed into Y & sup2; - 5Y = 4 = 0, the solution is y = 1, y2 = 4, when y = 1, X & sup2; - 5 (X & sup2; - 1 = 1, we can take X & sup2; - 1 as a whole, and then let X & sup2; - 1 = 1 as a whole, and then let X & sup2; - 1 = 1 = 1 = X-1 = 1 = 1 as a whole, and then let X & sup2; - 1 = 1 = 1 = 1, and then let X & sup2; (X & sup2; - 1 = 4, 1 = 4, \\\\\\\\\\\\\\\theroot is 5
(1) In the process of getting equation (1) from the original equation, we use the________ The method achieves the purpose of understanding the equation and embodies the mathematical thought of transformation;
(2) Please use the above knowledge to solve the fourth power of equation x - X & sup2; - 6 = 0

(1) Exchange yuan
(2) To solve x ^ 4 - X & sup2; - 6 = 0, we can change y = x & sup2; ≥ 0
Then y & sup2; - y-6 = 0, that is, (Y-3) (y + 2) = 0, y = 3 or y = - 2 (rounding off)
So y = x & sup2; = 3: x = √ 3, or x = - √ 3

Application of quadratic equation of one variable
In July 2006, an electrical factory produced 450 sets of electrical appliances. Later, due to market demand, the factory increased its production. It is known that the total number of electrical appliances in the third quarter is 1638 sets. If the monthly increase rate is the same, the monthly increase rate in the quarter can be calculated
In the middle of a 55 meter long and 45 meter wide rectangular green space, build two mutually vertical paths with the same width. The area that can be used for greening is 2000 m2. What is the width of the path

Let the growth rate be X
450+450*（1+x）+450*（1+x）^2=1638
x=20%