To solve the indefinite equation, we should use the known conditions to find the ratio of each unknown, then use them to become X (or other) at the same time, calculate X (or other), and the others can be solved. For example, if X:Y is 3:2.
X:Y=3:2
Y=2/3X
X+Y= 10 means X+2/3X= 10, then X= 10/3/5, x = 6,2/3x = 4.
It's kind of like algebra, actually. Hmm. How interesting
Attached:
indeterminate equation
Indefinite equation is a branch of mathematical number theory, which has a long history and rich content. The so-called indefinite equation refers to an equation or system of equations whose solution ranges from integer, positive integer, rational number or algebraic integer, and the number of unknowns is usually more than the number of equations.
The ancient Greek mathematician Diophantine studied several such equations at the beginning of the third century, so the indefinite equation is also called Diophantine equation. From 65438 to 0969, Mo Deer systematically summarized the research results in this field.
In recent years, more important progress has been made in this field. But generally speaking, people don't know much about multivariate indefinite equations higher than quadratic. On the other hand, indefinite equations are closely related to other branches of mathematics, such as algebraic number theory, algebraic geometry and combinatorial mathematics. The problem of indefinite equation is often raised in finite group theory and optimization design, which makes indefinite equation, an ancient branch, still attract the attention of many mathematicians and become one of the important research topics in number theory.
Definition: An integral equation with only one unknown number and the highest degree of the unknown number is 2 is called a quadratic equation with one variable.
General form: ax +bx+c=0 squared (a, b, c are constants, a≠0.
Example: x2- 1 = 0
general solution
1。 Direct Kaiping method
2。 Method of completing a square
3。 Formula method
4。 Factorization method
Complicated.
Dealing with the integer solution problem of quadratic equation in one variable from another angle
Problem: A quadratic equation with one variable (in which at least one of A, B and C contains parameters), when the parameters are integers, the equation about X has an integer solution. The general idea of this kind of problem is: find the value of the parameter when △ = B2-4ac is a complete square number, and then substitute it into the root formula to make x satisfy an integer; Or by Vieta's theorem. This solution sometimes leads to very troublesome calculations, and sometimes even gets into trouble. This paper tries to talk about the solution to this kind of problem from another angle. Its thinking method is to transform it into an integer solution of an indefinite equation, which can reflect the essence of the problem and make it solve the problem faster, simpler and more accurate. The following will introduce several common concrete methods to deal with this kind of problem.
First, variable separation method-separate a variable from other variables, and then determine the integer solution.
Example 1 known equation x2+px-p+30 = 0. When p is an integer, two equations are positive integers, so find two equations.
Analysis: from x2+px-p+30 = 0, p (x- 1) =-x2-30, obviously x≠ 1, then
Since p is an integer and x is a positive integer, x = 2,32. At this time, P =-34, so when P =-34, the equation has two positive integer solutions, and x 1=2, x2=32.
When m is an integer, both roots of the equation about x are integers.
Analysis: from, m (x+ 1) =-x2+x- 1, obviously x≦- 1, then, because m and x are integers, x =-4, -2, 0, 2; Substitution gives the corresponding m=7,-1. Therefore, when m=7 or m =- 1, both roots of the equation are integers.
Example 3 Find all such positive integers A so that the quadratic equation about X has at least one integer root. (The Third "Zu Chongzhi Cup" Junior High School Mathematics Competition)
Analysis: a(x+2)2=2x+ 12, obviously x≦-2, then a=, because a is a positive integer, it is ≥ 1, so the solution is -4 ≤ x ≤ 2, and x ≤-2. The possible values of x are -4, -3,-1, 0, 1, 2; The replacement test results are a= 1, 3,6, 10. Therefore, when a= 1, 3,6, 10, the equation has at least one integer root.
Second, factorization method-through factorization, the equation is transformed into two indefinite equations or several indefinite equations to solve.
Example 4 It is known that the equation A2X2-(3A2-8A) x+2A2-13A+15 = 0 (where A is a non-negative integer) has at least one integer root, then a= (1998 National Junior High School Mathematics Competition).
Analysis: From the original equation to (AX-2A+3) (AX-A+5) = 0, then the original equation is transformed into two indefinite equations AX-2A+3 = 0 or AX-A+5 = 0. Obviously a≠0, then
① or ②
Because a is a non-negative integer, a = 1 3 is obtained from ①; Get a= 1, 5 from ②. Therefore, when a= 1, 3,5, the equation about x has at least one integer solution.
Example 5 When k is an integer, the quadratic equation x2-3kx+2k2-6 = 0 about x is an integer.
Analysis: (x-2k)(x-k)=6 is obtained from x2-3kx+2k2-6 = 0. Because x and k are integers, the original equation becomes
X-2k = 2 or X-2k = 3 or X-2k = 6 or X-2k = 1.
x-k= 3 x-k= 2 x-k= 1 x-k= 6
Moreover, x-2k and x-k have the same sign, so eight indefinite equations are obtained, and the solutions are k =- 1, 1,-5,5.
Third, the formula estimation method-the original equation is passed through the formula, so that one side becomes a flat road or the sum of several flat roads, and the other side is a simple algebraic formula or number, and then the range of some variables is narrowed by estimation and other methods, and then their integer solutions are determined.
Example 6 Let m be an integer, 4
Analysis: x2-2 (2m-3) x+(2m-3) 2 = 2m+1According to the original equation, that is, (x-2m+3) 2 = 2m+ 1, because 4.
Example 7 It is known that m is a positive integer. When m takes any value, the equation 3x2-4mx+3m2-35 = 0 about x has at least one integer solution.
Analysis: From the original equation, 9x2- 12mx+9m2 = 105, then (3x-2m) 2+5m2 = 105, because (3x-2m) 2 ≥ 0 and 5m2≤ 105.
Examples 8 a, b and c, which integers satisfy the inequality A2+B2+C2+3?
Analysis: If there are integers A, B, C, A2+B2+C2+3.
Fourth, comprehensive analysis method-sometimes we use the above methods at the same time to get integer solutions.
Example 9 Given an equation, what is the integer A? The equation about X has at least one integer solution.
Analysis: Based on the formula of the original equation, let x+a=p and ax=t, then the above formula is 3p2-7p-9t=0 ①, and p2≥4t ②.
It is easy to know from the original equation that if a=0, the equation has an integer solution of 0, and if a≠0, it is obtained from ①.
Because A and X are integers, P is a positive integer, T is an integer, and 3p+2 and 9 are coprime, so P can be divisible by 9, that is, p=9, 18, 27 ... and it is solved from ① and ② with 4t-p2 =, so P = 9, t=20, and substituted into ①. Therefore, when a = 0,4,5, the equation has an integer solution.
As we have seen above, the integer solution problem of quadratic equation with one variable is essentially a case of integer solution problem of indefinite equation. Of course, some special cases can be easily solved by conventional thinking, but it is natural and concise to transform the general situation into an integer solution of an indefinite equation. This is my opinion.
monadic quadratic equation
Definition: An integral equation with only one unknown number and the highest degree of the unknown number is 2 is called a quadratic equation with one variable.
General form: ax +bx+c=0 squared (a, b, c are constants, a≠0.
For example: x2- 1 = 0
general solution
1。 Direct Kaiping method
2。 Method of completing a square
3。 Formula method
4。 Factorization method
1. Definition of quadratic equation in one variable
The quadratic equation with one variable has three characteristics: (1) contains only one unknown; (2) The maximum number of unknowns is 2; (3) It is an integral equation. To judge whether an equation is an unary quadratic equation, first look at whether it is an integral equation, and if it is, then sort it out. If it can be sorted out in the form of (a≠0), then this equation is a quadratic equation.
2. The general form of quadratic equation with one variable
We call (a≠0) the general form of a quadratic equation with one variable, paying special attention to the fact that the quadratic coefficient must not be 0, and b and c can be any real numbers, including 0, that is, the quadratic equation with one variable can have no linear terms and constant terms. (a≠0)，(a≠0)，(a≦)。
3. Solution of quadratic equation in one variable
There are four solutions to the quadratic equation of one variable: (1) direct Kaiping method; (2) factorization; (3) Matching method; (4) Formula method. We should choose the method flexibly according to the characteristics of the equation, in which the formula method is a general method and can solve any quadratic equation with one variable.
4. Discriminating formula for roots of quadratic equation with one variable.
The discriminant of the root of a quadratic equation with one variable is.
△& gt； The 0 equation has two unequal real roots.
Delta = 0 equation has two equal real roots.
△& lt； Equation 0 has no real root.
The above can be pushed from left to right and vice versa.
5. The relationship between the roots and coefficients of a quadratic equation.
If the two roots of the unary quadratic equation (a≠0) are, then.
6. Steps to solve application problems
(1) Analyze the meaning of the problem and find out the equivalent relationship between the unknown in the problem and the conditions given in the problem;
(2) Set unknowns, using algebraic expressions of set unknowns to represent other unknowns;
(3) Find out the equation relation and use it to list the equations;
(4) Solving the equation to find the value of the unknown quantity in the problem;
(5) Check whether the required answer meets the meaning of the question and make an answer.
Problem solving thinking
1. Change your mind
Transforming thinking is the most common way of thinking in junior high school mathematics.
With the idea of transformation, unknown problems can be transformed into known problems, and complex problems can be transformed into simple problems. In this chapter, the solution of a quadratic equation is transformed into a square root problem, and the quadratic equation is transformed into a linear equation through factorization.
2. Ideas from special to general
From the special to the general is the universal law of our understanding of the world. Through the study of special phenomena, we can draw general conclusions, such as using direct Kaiping method to solve special problems, using formula matching method to explore the relationship between roots and coefficients of quadratic equations in one variable.
3. The idea of classified discussion
The discriminant of the roots of a quadratic equation in one variable embodies the idea of classified discussion.
Elaboration of classic examples.
1. For the definition of a univariate quadratic equation, we should fully consider the three characteristics of the definition, and don't ignore that the quadratic term coefficient is not 0.
2. When solving a quadratic equation with one variable, according to the characteristics of the equation, choose the solution method flexibly, first consider whether the direct Kaiping method and factorization method can be used, and then consider the formula method.
3. There are both positive and negative discriminant of the root of the unary quadratic equation (a≠0), which can be used to solve the equation and determine the root of the equation (1); (2) Determine the range of roots according to the properties of parameter coefficients; (3) Solve the problem of proof related to roots.
4. The roots and coefficients of a quadratic equation with one variable have many applications: (1) Knowing one root of the equation, finding another root and parameter coefficients without solving the equation; (2) Knowing the equation, finding the value of algebraic expression with two symmetrical expressions and related unknown coefficients; (3) Given two equations, find the root of an unary quadratic equation.
Vieta's Theorem: In the unary quadratic equation AX 2+BX+C (A is not 0).
Let two roots be X 1 and X2.
Then x1+x2 =-b/a.
X 1*X2=c/a
See:/view/375208.htm.
See:/view/397767.html.