2. Formula method (can solve all quadratic equations of one variable)
3. Factorization method (which can solve partial quadratic equation) (Factorization method is divided into "raising common factor method", "formula method" (divided into "square difference formula" and "complete square formula") and "cross multiplication".
4. Open method (you can solve all quadratic equations in one yuan) It is really not good to solve quadratic equations in one yuan (you can buy fx-500 calculator or Casio's 99 1 solution equation, but you need a general form).
5. Algebraic method (which can solve quadratic equation in one variable)
Introduction to algebraic methods
ax^2+bx+c=0
Divided by a at the same time, it becomes x 2+bx+c = 0.
Let: x = y-b/2
The equation becomes: (y 2+b 2/4-by)+(by+b 2/2)+c = 0.
Then it becomes: y 2+(b 2 * 3)/4+c = 0.
y= √[(b^2*3)/4+c]
How to choose the simplest scheme;
1, and see if we can directly root out the solution;
2. See if it can be solved by factorization (the common factor method is considered first, then the formula method is considered, and finally the cross multiplication is considered);
3. Solve by formula method;
4, unless the topic requires, finally consider the matching method (although the matching method can solve all quadratic equations with one variable, but the problem-solving steps are too troublesome).
First, the main points of knowledge:
One-dimensional quadratic equation and one-dimensional linear equation are both integral equations, which are a key content of junior high school mathematics and the basis of studying mathematics in the future, and should be paid attention to by students.
The general form of unary quadratic equation is: ax 2+bx+c = 0, (a ≠ 0), which is an integral equation with only one unknown and the highest degree of the unknown is 2.
The basic idea of solving quadratic equations with one variable is to simplify them into two quadratic equations with one variable. There are four solutions to the unary quadratic equation: 1, direct Kaiping method; 2. Matching method; 3. Formula method; 4. Factorial decomposition method.
Second, detailed methods and examples:
1, direct Kaiping method:
The direct Kaiping method is a method to solve a quadratic equation with a direct square root. The equation with the shape of (x-m)2=n (n≥0) is solved by direct Kaiping method, and its solution is x = m √ n.
Example 1. Solve the equation (1) (3x+1) 2 = 7 (2) 9x2-24x+16 =1.
Analysis: (1) This equation is obviously easy to do by direct flattening, (2) The left side of the equation is completely flat (3x-4) 2, and the right side =11>; 0, so this equation can also be solved by direct Kaiping method.
(1) solution: (3x+ 1) 2 = 7.
∴(3x+ 1)^2=7
∴ 3x+ 1 = √ 7 (be careful not to lose the solution)
∴x= ...
The solution of the original equation is x 1= ..., x2= ...
(2) Solution: 9X2-24x+16 =11.
∴(3x-4)^2= 1 1
∴3x-4= √ 1 1
∴x= ...
The solution of the original equation is x 1= ..., x2= ...
2. Matching method: use matching method to solve the equation AX 2+BX+C = 0 (A ≠ 0).
First, the fixed number c is moved to the right of the equation: ax 2+bx =-C.
Convert the quadratic term into1:x 2+(b/a) x =-c/a.
Add half the square of the first coefficient on both sides of the equation: x2+(b/a) x+0.5 (b/a) 2 =-c/a+0.5 (b/a) 2.
The left side of the equation becomes completely flat: [x+0.5 (b/a)] 2 =-c/a+0.5 (b/a) 2.
When b2-4ac≥0, x+= √ [-c/a+0.5 (b/a) 2]-0.5 (b/a).
∴x= ... (This is the root formula)
Example 2. Solving Equation 3x2-4x-2 = 0 by Matching Method
Solution: Move the constant term to the right of equation 3x 2-4x = 2.
Convert the quadratic term to 1: x 2-x =
Add half the square of the coefficient of the first order term on both sides of the equation: x 2-x+() 2 =+() 2.
Formula: (x-) 2 =
Direct square: x-=
∴x=
The solution of the original equation is x 1=, x2=.
3. Formula method: the quadratic equation of one variable is transformed into the general form of AX 2+BX+C, and then the values of various coefficients A, B and C are substituted into the formula for finding the root to get the root of the equation.
When b 2-4ac > 0, the root formula is x 1 = [-b+√ (b 2-4ac)]/2a, and x2 = [-b-√ (b 2-4ac)]/2a (two unequal real roots).
When b 2-4ac = 0, the root formula is x 1=x2=-b/2a (two equal real roots).
When b 2-4ac
Example 3. Solving Equation 2x 2-8x =-5 by Formula Method
Solution: Change the equation into a general form: 2x 2-8x+5 = 0.
∴a=2，b=-8，c=5
b^2-4ac=(-8)2-4×2×5=64-40=24>； 0
∴x= = =
The solution of the original equation is x 1=, x2=.
4. Factorial decomposition method: the quadratic trinomial on one side of the equation is decomposed into the product of two linear factors, so that the two linear factors are equal to zero respectively, and two linear equations are obtained. The roots obtained by solving these two linear equations are the two roots of the original equation. This method of solving a quadratic equation with one variable is called factorization.
Example 4. Solve the following equation by factorization:
( 1)(x+3)(x-6)=-8(2)2x^2+3x=0
(3) 6x 2+5x-50 = 0 (optional study) (4) x 2-4x+4 = 0 (optional study)
(1) Solution: (x+3)(x-6)=-8 Simplified sorting.
X 2-3x- 10 = 0 (the equation has a quadratic trinomial on the left and zero on the right).
(x-5)(x+2)=0 (factorization factor on the left side of the equation)
∴x-5=0 or x+2=0 (converted into two linear equations)
∴x 1=5,x2=-2 is the solution of the original equation.
(2) Solution: 2x 2+3x = 0
X(2x+3)=0 (factorize the left side of the equation by increasing the common factor)
∴x=0 or 2x+3=0 (converted into two linear equations)
∴x 1=0, x2=- is the solution of the original equation.
Note: Some students easily lose the solution of x=0 when doing this kind of problem. It should be remembered that there are two solutions to the quadratic equation of one variable.
(3) Solution: 6x2+5x-50=0
(2x-5)(3x+ 10)=0 (pay special attention to symbols when factorizing by cross multiplication).
2x-5 = 0 or 3x+ 10=0.
∴x 1=, x2=- is the solution of the original equation.
(4) Solution: x 2-4x+4 = 0 (∵ 4 can be decomposed into 2.2, ∴ this problem can be factorized).
(x-2)(x-2 )=0
∴x 1=2, x2=2 is the solution of the original equation.
Summary:
Usually, factorization is the most commonly used method to solve quadratic equations with one variable. When factorization is applied, the equation is written in a general form and the quadratic coefficient is turned into a positive number.
Direct leveling method is the most basic method.
Formula and collocation are the most important methods. Formula method is suitable for any quadratic equation with one variable (some people call it universal method). When using the formula method, the original equation must be transformed into a general form to determine the coefficient, and before using the formula, the value of the discriminant should be calculated to judge whether the equation has a solution.
Matching method is a tool to derive formulas. After mastering the formula method, we can directly use the formula method to solve the quadratic equation of one variable, so we generally don't need to use the matching method to solve the quadratic equation of one variable. However, collocation method is widely used in the study of other mathematical knowledge, and it is one of the three important mathematical methods required to be mastered in junior high school, so we must master it well. Three important mathematical methods: method of substitution, collocation method and undetermined coefficient method.
Example 5. Solve the following equations in an appropriate way. (optional research)
( 1)4(x+2)^2-9(x-3)^2=0(2)x^2+2x-3=0
(3)x2-2x =-(4)4x 2-4mx- 10x+m2+5m+6 = 0
Analysis: (1) First of all, we should observe whether the topic has characteristics, and don't blindly do multiplication first. After observation, it is found that the factor on the left side of the equation can be decomposed by the square difference formula and become the product of two linear factors.
(2) The left factor of the equation can be decomposed by cross multiplication.
(3) After it is transformed into a general form, it is solved by the formula method.
(4) Transform the equation into 4x 2-2 (2m+5) x+(m+2) (m+3) = 0, and then decompose it by cross factor.
(1) solution: 4 (x+2) 2-9 (x-3) 2 = 0.
[2(x+2)+3(x-3)][2(x+2)-3(x-3)]= 0
(5x-5)(-x+ 13)=0
5x-5=0 or -x+ 13=0.
∴x 1= 1,x2= 13
(2) Solution: x 2+2x-3 = 0
[x-(-3)](x- 1)=0
X-(-3)=0 or x- 1=0.
∴x 1=-3,x2= 1
(3) Solution: x 2-2 x =-
X 2-2 x+= 0 (general form first)
△=(-2 )^2-4×= 12-8 = 4 & gt； 0
∴x=
∴x 1=,x2=
(4) Solution: 4x 2-4mx- 10x+m 2+5m+6 = 0.
4x^2-2(2m+5)x+(m+2)(m+3)=0
[2x-(m+2)][2x-(m+3)]=0
2x-(m+2)=0 or 2x-(m+3)=0.
∴x 1=，x2=
Example 6. Find two roots of equation 3 (x+1) 2+5 (x+1) (x-4)+2 (x-4) 2 = 0. (optional research)
Analysis: If this equation is multiplied first, then multiplied, and then merged into a common form, it will be very complicated. Observing the topic carefully, we find that if x+ 1 and x-4 are regarded as a whole respectively, the left side of the equation can be decomposed by cross multiplication (in fact, method of substitution is used).
Solution: [3 (x+1)+2 (x-4)] [(x+1)+(x-4)] = 0.
That is (5x-5)(2x-3)=0.
∴5(x- 1)(2x-3)=0
(x- 1)(2x-3)=0
∴x- 1=0 or 2x-3=0
∴ x 1 = 1, x2 = is the solution of the original equation.
Example 7. Solving quadratic equation with one variable by collocation method.
Solution: x 2+px+q = 0 can be transformed into
X 2+px =-q (constant term moved to the right of the equation)
X 2+px+() 2 =-q+() 2 (both sides of the equation plus the square of half the coefficient of the first term)
(x+)2= (formula)
When P 2-4q ≥0, ≥ 0 (P 2-4q must be discussed separately).
∴x=- =
∴x 1=，x2=
When p 2-4q
Note: this is called the letter coefficient equation, and there are no additional conditions for P and Q in the question. Therefore, in the process of solving problems, we should always pay attention to the requirements for letter values and discuss them in categories when necessary.
Exercise:
(1) Solve the following equation with an appropriate method:
1.6x^2-x-2=0 2。 (x+5)(x-5)=3
3.x^2-x=0 4。 x^2-4x+4=0
5.3x2+ 1=2x 6。 (2x+3)2+5(2x+3)-6=0
(2) Solve the following equation about x.
1.x^2-ax+-b2=0 2。 x^2-( + )ax+ a2=0
Practice reference answer:
( 1) 1.x 1 =- 1/2，x2 = 2/32。 x 1 = 2，x2 =-2。
3.x 1=0，x2 = 4 . x 1 = x2 = 2 5 . x 1 = x2 =
6. Solution: (Take 2x+3 as a whole and decompose the factors on the left side of the equation)
[(2x+3)+6][(2x+3)- 1]=0
That is, (2x+9)(2x+2)=0.
* 2x+9 = 0 or 2x+2=0
∴x 1=-,x2=- 1 is the solution of the original equation.
(2) 1. solution: x 2-ax+(+b) (-b) = 0 2, solution: x 2-(+) ax+a = 0.
[x-( +b)] [x-( -b)]=0 (x- a)(x-a)=0
∴x-( +b)=0 or x-( -b) =0 x- a=0 or x- a=0.
∴x 1= +b, x2= -b is ∴x 1= a, x2=a is.
The solution of the original equation. The solution of the original equation.
Test (answers below)
Multiple choice
1. The root of the equation x(x-5)=5(x-5) is ().
a、x=5 B、x=-5 C、x 1=x2=5 D、x 1=x2=-5
2. The value of polynomial a2+4a- 10 is equal to 1 1, so the value of a is ().
A, 3 or 7 B, -3 or 7 C, 3 or -7 D, -3 or -7
3. If the sum of quadratic coefficient, linear coefficient and constant term in unary quadratic equation AX 2+BX+C = 0 is equal to zero, then the equation must have a root ().
a、0 B、 1 C 、- 1 D、 1
4. The unary quadratic equation AX 2+BX+C = 0 has a root, if ().
A, b≠0 and c=0 B, b=0 and c≠0.
C and b=0 and c=0 D and c=0.
5. The two roots of the equation x 2-3x = 10 are ().
a 、-2，5 B、2 、-5 C、2，5 D、2
6. The solution of equation x 2-3x+3 = 0 is ().
A, B, C, D, have no real roots.
7. The solution of equation 2x 2-0. 15 = 0 is ().
a、x= B、x=-
c、x 1=0.27，x2=-0.27
8. After the left side of the equation x 2-x-4 = 0 is matched into a completely flat mode, the equation obtained is ().
a 、( x-)2= B 、( x- )2=-
C, (x- )2= D, none of the above answers are correct.
9. It is known that the unary quadratic equation x 2-2x-m = 0, and the equation after solving the formula of this equation by matching method is ().
a、(x- 1)^2=m2+ 1 B、(x- 1)^2=m- 1 C、(x- 1)^2= 1-m D、(x- 1)^2=m+ 1
Answer and analysis
Answer:1.c2.c3.b4.d5.a6.d7.d8.c9.d.
Analysis:
1. Analysis: (x-5) 2 = 0, then x 1=x2=5,
Note: Don't easily divide the two sides of the equation with an algebraic expression. Another quadratic equation with one variable has real roots, and it must be two.
2. analysis: according to the meaning of the question: a 2+4a-10 =1,the solution is a=3 or a=-7.
3. analysis: according to the meaning of the question: if there is a+b+c=0, the left side of the equation is a+b+c, only x= 1, ax 2+bx+c = a+b+c, that is to say, when x= 1, the equation holds, then there must be a root as X.
4. Analysis: the unary quadratic equation AX 2+BX+c=0. If a root is zero, AX 2+BX+C must have a factor X, and if and only if C = 0, there is a common factor X, so C = 0. In addition, it is easier to substitute x=0 to get C = 0!
5. Analysis: The original equation becomes x 2-3x- 10 = 0,
Then (x-5)(x+2)=0.
X-5=0 or x+2=0.
x 1=5，x2=-2。
6. Analysis: δ = 9-4× 3 =-3
7. Analysis: 2x2=0. 15
x2=
x=
Pay attention to the simplification of roots, square directly and don't lose roots.
8. Analysis: multiply both sides by 3: x 2-3x- 12 = 0, and then according to the linear coefficient formula, x 2-3x+(-) 2 = 12+(-) 2,
The sort is: (x-)2=
The equation can be deformed by using the equality property. When X 2-BX is formulated, the term of the formula is the square of half the coefficient of the first term-B.
9. Analysis: x 2-2x = m, then x 2-2x+ 1 = m+ 1.
Then (x- 1) 2 = m+ 1.
Analysis of senior high school entrance examination
Comments on examination questions
The root of 1. (Gansu Province) equation is ()
(A) (B) (C) or (d) or
Comments: Since the quadratic equation of one variable has two roots, we exclude options A and B by exclusion, and then select the correct option from options C and D by verification ... This equation can also be solved by factorization, and the result can also be compared with the options. Options a and b only consider one hand and forget one yuan.
Quadratic equation has two roots, so it is wrong, and x =- 1 in option D can't make the left and right sides of the equation equal, so it is also wrong. The correct option is C.
In addition, students often use an algebraic expression to divide both sides of the equation at the same time, which makes the equation lose its roots. This kind of mistake should be avoided.
2. (Jilin Province) The root of the quadratic equation of one variable is _ _ _ _ _ _ _.
Comments: The train of thought can be solved by factorization or formula method according to the characteristics of the equation.
3. The root of (Liaoning Province) equation is ()
0(B)– 1(C)0，– 1(D)0， 1
Comments: Thinking: Because the equation is a quadratic equation with two real roots, the correct option can be selected through exclusion verification, while the two options A and B have only one root. D option a number is not the root of the equation. In addition, you can also use the method of directly finding the root of the equation.
4. (Henan Province) It is known that one root of the quadratic equation of X is–2, so k = _ _ _ _ _ _ _ _
Comment: k=4. Substitute x=-2 into the original equation, construct a quadratic equation about k, and then solve it.
5.(Xi 'an) Solve the equation (x-3)2=8 by direct Kaiping method, and the root of the equation is ().
(A)x=3+2 (B)x=3-2
x 1=3+2，x2=3-2
Comments: You can solve the equation directly, or you don't need to calculate it. If a quadratic equation with one variable has a solution, there must be two solutions and the square root of 8, and then you can choose the answer.
Extracurricular development
monadic quadratic equation
One-dimensional quadratic equation refers to an integral equation with an unknown number, and the highest degree of the unknown number is quadratic. The general form is ax 2+bx+c = 0, (a ≠ 0).
Around 2000 BC, a quadratic equation of one yuan appeared on the clay tablet of Babylon and its solution: find a number so that the sum of its reciprocal is equal to a given number, that is, find such a sum of x, so that,
x= 1，x+ =b，
x^2-bx+ 1=0，
They made () 2; Do it again and get the solutions:+and-. It can be seen that the Babylonians already knew the formula for finding the root of the quadratic equation of one variable. But they didn't accept negative numbers at that time, so they omitted negative roots.
Egyptian papyrus literature also involves the simplest quadratic equation, for example: ax 2 = b.
In the 4th and 5th centuries BC, China had mastered the formula for finding the root of a quadratic equation with one variable.
Diophantine (246-330) in Greece only takes the positive root of a quadratic equation, even if both of them are positive roots, he only takes one of them.
In 628 AD, a formula for finding the root of quadratic equation x 2+px+q = 0 was obtained from the Yarlung Zangbo River Correction System written by India.
In Algebra written by Al-Hualazimi of Arabia, the solutions of equations are discussed, and the first and second equations are solved, involving six different forms, so that A, B and C are positive numbers, such as AX 2 = Bx, AX 2 = C, AX2+C = Bx, AX2+Bx = C, AX. It is in line with Diophantine's practice to discuss quadratic equations in different forms. In addition to several special solutions of quadratic equation, Al-Hua Lazimi also gave the general solution of quadratic equation for the first time, admitting that the equation has two roots and irrational roots, but he didn't know the imaginary root. /kloc-in the 6th century, Italian mathematicians began to understand cubic equations with complex roots.
David (1540- 1603) not only knows that the unary equation always has a solution in the range of complex numbers, but also gives the relationship between roots and coefficients.
Chapter 9 Arithmetic China Pythagorean Theorem No.20, finding the positive root is equivalent to x 2+34x-7 1000 = 0. China mathematicians also applied interpolation in the study of equations.