(1) Key points of knowledge:
First, the quadratic equation definition:
The integral equation of 1 contains only one unknown, and the highest order of the unknown is quadratic, which is called the unary quadratic equation.
2. General form: A x 2+B x+C = 0 (A ≠ 0)
Where a is called quadratic coefficient, b is called linear coefficient and c is called constant term;
Solutions of quadratic and quadratic equations:
1, direct Kaiping method: the shape is A x 2 = b or a (x+c) 2 = b. If a and b have the same sign and a≠0, the equation can be solved directly by Kaiping method. If the signs of A and B are different, it can be directly judged that the equation has no solution.
The direct Kaiping method is based on the roots of numbers, and is suitable for A×2 = b or A (x+c) 2 = b type equations.
2. Matching method: the quadratic equation of one variable is arranged in the form that the right side contains only constant terms and the left side is completely square. If the right side is a non-negative constant, we can further find its root by direct Kaiping method. This method of solving a quadratic equation with one variable is called collocation method.
Matching method itself is a method, the basis of formula method and a basic algebraic method. Based on formula and direct Kaiping method, it is suitable for quadratic equations with arbitrary characteristics, but the process is complicated.
3. Root formula method: The method of solving a quadratic equation with a root formula is called formula method. The root formula of a quadratic equation AX 2+BX+C = 0 (A ≠ 0) is X = (B2-4AC ≥ 0). Formula method is the direct result of applying collocation method to the general form of quadratic equation with the same applicability as collocation method.
4. Factorization: Factorization is a method to solve a quadratic equation with one variable. It is a method to decompose the non-zero edge of a quadratic equation of one variable into the product of two linear factors, and solve the equation according to the fact that at least one of the two factors is equal to zero.
The main points of the four schemes are different, but the basic idea is the same, all of which are reduction. The four solutions have their own advantages, and which one is simpler depends on the different situations of specific equations. However, their solutions are interrelated. Generally speaking, it is convenient to decompose factors quickly and apply factorization method. Otherwise, the root formula method is suitable; Matching method is the basis of deducing formula and an important mathematical method. We should master it skillfully and use it flexibly. Although the process of solving problems is complicated, it is generally not necessary to use matching method. However, for some special equations or even the coefficients of the first term, this formula can be easily used. The key to a simple solution is to observe the characteristics of the equation and choose the best solution.
(2) Typical examples:
Example 1. When k is what value, the equation (k-2) x k-2-k x+6 = 0 is a quadratic equation. Solve this equation.
Solution: from the meaning of the question: solution: ∴ k =-2
When k =-2, the original equation is: -4x2+2x+6 = 0, that is, 2x2-x-3 = 0.
The factorization is: (2x-3) (x+1) = 0 ∴ x1= 3/2x2 =-1.
Explanation: correctly understand the concept of quadratic equation with one variable: the maximum unknown degree is 2, and the coefficient of quadratic term is not 0; Is the key to solve this problem. It is relatively simple to solve this equation by quadratic trinomial decomposition of cross multiplication.
Example 2. Fill in the blanks:
1, the root of equation (x-3) 2 = 8 is. (x = 3)
2. If the value of algebraic expression (2x+ 1) 2 is 9, then the value of x is. (1 or -2).
3. The root of the unary quadratic equation X2+4x- 12 = 0 is. (X 1 = 2，X2 =-6)
4. The root of the equation X4-5x2+6 = 0 is. (x =，x =)
5. If x = 1 is a root of the equation x 2+kx+k-5 = 0, then the value of k is equal to 2.
6. If one root of the equation x 2+ax+2 = 0 is+1, then the value of a is. (-2).
7. It is known that α and β are two roots of equation X2+ 1998X+ 1 = 0, then (1+2000α+α 2) (1+2000β+β 2) =;
Solution: ∫ α and β are the two roots of the equation X 2+ 1998X+ 1 = 0.
∴α2+ 1998α+ 1=0,β2+ 1998β+ 1=0 ,α? β= 1
The original formula = (1+1998α+α 2+2α) (1+kloc-0/998β+β 2+2β) = 2α× 2β = 4α? β=4
Example 3, multiple-choice questions:
1. Solve the equation (x-3) 2 = 8 by direct Kaiping method, and the root of the equation is: (c).
a、x=3+2 B、x=3-2
c、x 1=3+2，x2=3-2
2. Among the following equations, the equation with real number solution is: (d)
a 、=0 B 、+ 1=0 C、x 2+ 1=0 D、x 2- 1=0
3. The root of equation = 0 is: (d)
a、x 1= 1，x ^ 2 =-2 B、x=2 C、x =- 1 D、x=-2
4. It is known that only one of the two real roots of the equation X 2-(2A+ 1) X+A2+A = 0 is greater than 5, so the value range of a is: (d).
a、a & gt4 B、4 & lta & lt5 C、a & gt5 D、4 & lta ≤5
Solution: Factorization gives: x 1 = a, x 2 = a+1:Only one root is greater than 5.
∴a+ 1 & gt； 5 and a ≤ 5, then 4
5. The roots of equation x 3-5x2+6x = 0 are: (b)
a，0，-2，-3 B，0，2，3 C，0， 1，-6 D，0，- 1，-6
6. When solving the equation x 2+px+q = 0 by collocation method, the equation can be transformed into: (a)
a 、( x+ ) 2= B 、( x+ ) 2=
c 、( x- ) 2= D 、( x- ) 2=
Note: Matching method is an important mathematical method and the basis of formula method. We are willing to solve problems flexibly to improve our ability to solve problems.
Steps of the matching method: (1) Move the constant term to the right side of the equation, and arrange it on the left side in descending order, and convert the quadratic term coefficient into1;
(2) Add the square of half the coefficient of the first term to both sides of the equation, and the left side becomes completely flat;
(3) If the right side of the equation is non-negative, solve it by direct Kaiping method; If it is negative, it is pointed out that the original equation has no real root;
Example 4. Solve the equation in different ways: 2x2+7x-4 = 0.
Scheme 1: (allocation method) (1) Transfer term: 2x2+7x = 4 (2) Multiply both sides by 2:
Divide the two sides by 2:x2+7/2x = 2(2x)2+7×(2x)-8 = 0.
Formula: x2 ++ factorization: (2x- 1) (2x+8) = 0.
∴ (x+= 2 x = 1 or 2 x =-8.
Square root: x+∴ x 1 = x 2 =-4.
Solution 2: (formula method) ∫A = 2 b = 7 c =-4 Solution 3: (factorization method)
B2-4ac = 72-4×2×(-4)= 8 1 & gt； 0 factorization: (2x- 1) (x+4) = 0.
∴x= ∴ x 1= x 2=-4
Explanation: In order to cultivate students' ability to observe, analyze, find and solve problems, it is necessary to train students to solve problems in specified ways. It is necessary not only to cultivate students' good quality of mathematical thinking methods, but also to make students understand and master the significance of a certain thinking method, which often exceeds the application of knowledge itself.
Example 5: Solve the following equation by the optimal method: (1) (2x-1) 2 = 6 (2) 9x2-8 = 0 (3) 4x2 = 3x.
(4)x2+6x- 1 1 = 0(5)3 x2-5x+2 = 0(6) 1997 x2+2000 x+3 = 0
Solution: Equation (1) is a completely flat way with unknowns. Equation (2) lacks a linear term, so it is very simple to use the direct Kaiping method. Equation (3) lacks a constant term, the sum of the coefficients in Equation (5) is 0, and the parity terms in Equation (6) are equal, so factorization (cross multiplication) is needed. The quadratic coefficient of Equation (4) is 1, and the linear coefficient is even, so collocation method or formula method is more suitable.
Explanation: It is very skillful to solve the quadratic equation of one variable. We should not only be proficient in the four conventional methods, but also be good at analyzing the characteristics of the purpose of the problem, and adopt some unconventional and simple methods to solve the problem flexibly and improve the problem-solving ability.
Example 6: The quadratic equation of x (m+1) x2+3x+m2-3m-4 = 0 has a root of zero. How to find the value of m?
Solution: from the definition of the root of the equation: substitute x = 0 into the equation to get: m 2-3 m-4 = 0.
∴m 1=4 square meters =- 1
However, when m =- 1 and the quadratic coefficient m+ 1 = 0, the equation given at this time is not a quadratic equation and should be discarded.
So take m = 4.
Example 7: Given that the sum of two roots of AX2+BX+C = 0 (a ≠ 0) is S 1, the sum of squares of two roots is S 2, and the cubic sum of two roots is S 3, what is the value of AS3+BS2+CS 1?
Solution: Let m and n be the two roots of the known equation, then there are:
am2+bm+c=0 ( 1)
an2+bn+c=0 (2)
a(m3+n3)+b(m2+N2)+c(m+n)= 0 from( 1)×m+(2)×n
That is as 3+b S2+c s 1 = 0.
Explanation: the root of an equation is an unknown value that means that the values on the left and right sides of the equation are equal. If we can understand the definition of root thoroughly, use it well and pour it skillfully, it will bring great convenience to solve some problems. We call it the definition of the root of the equation. Using the definition of root, we can find the value of parameter, the value of algebra and the range of algebra.
Understand? Hope to use+++++++++++++++++++++++++++++++++++++.