related notion
1. An equation with an unknown number is called an equation, or an equation with an unknown number is an equation.
3. The value of the unknown quantity that holds the equation is called the solution of the equation or the root of the equation.
3. Solving the equation is the process of finding the values of all the unknowns in the equation.
An equation must be an equation, and an equation is not necessarily an equation. An equation without unknowns is not an equation.
5. Verification: Generally, after solving the equation, verification is needed. Verification is to substitute the unknown value into the original equation to see if both sides of the equation are equal. If they are equal, then the value obtained is the solution of the equation.
[6] Note: Write the word "Xie", align the equal sign and tick.
⒎ The equation depends on the relationship between the parts of the equation and the relationship between addition, subtraction, multiplication and division (addend+addend = sum, and- one of addend = another addend, difference+subtraction = minuend, minuend-difference = subtraction, factor × factor = product, product ÷ a factor =
Solution process
way
1. estimation method: an introductory method when you are just learning to solve equations. Estimate the solution of the equation directly, and then substitute it into the original equation for verification.
3. Apply the properties of the equation to solve the equation.
13. Merge similar terms: transform the equation into a monomial.
4. Move terms: move terms containing unknown to the left and constant items to the right.
For example: 3+x= 18
Solution: x = 18-3
x = 15
⒌ Parenthesis deletion: Use the rule of parenthesis deletion to delete the parentheses in the equation.
4x+2(79-x)= 192 Solution: 4x+ 158-2x= 192.
4x-2x+ 158= 192
2x+ 158= 192
2x= 192- 158
x= 17
6. Formula method: There are some equations, and the general forms of solutions have been studied and become fixed formulas, which can be used directly. Generally solvable multivariate higher-order equations have formulas to follow.
7. Function image method: Use the solution of the equation to solve the geometric meaning of the intersection of two or more related function images.
The equation is positive thinking.
step
(1) If there is a denominator, go to the denominator first.
(2) If there are brackets, remove them.
(3) If it is necessary to move items, move items.
(4) merging similar projects
5] coefficient to 1 to get an unknown value.
[6] Write "Xie" at the beginning.
For example:
3+x= 18
Solution:
x = 18-3
x = 15
——————————
4x+2(79-x)= 192
Solution:
4x+ 158-2x= 192
4x-2x+ 158= 192
2x+ 158= 192
2x= 192- 158
2x=34
x= 17
——————————
πr=6.28 (only two decimal places)
To solve this problem, we must first know how much π equals, π = 3. 14 1592 ... only 3. 14.
Solution: 3. 14r=6.28
r=6.28/3. 14=2
But x is not necessarily placed on the left side of the equation, or there are two x in an equation, so it needs to be solved by simple calculation methods in mathematics. Some formulas have an X on the right, which can be simply calculated in a different position.
Equation classification
monadic quadratic equation
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. Factorization method.
1. Direct Kaiping method: Direct Kaiping method is a method of solving quadratic equation with direct square root.
The equation in the form of (x-m) 2 = n (n ≥ 0) is solved by direct Kaiping method, and its solution is x = √ n+m.
Example 1. Solve the equation (1) (x-2) 2 = 929x2-24x+16 =11.
Analysis: (1) This equation is obviously easy to do by direct Kaiping method, (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: (x-2) 2 = 9 ∴ x-2 = 9 ∴ x-2 = 3 ∴ x1= 3+2x2 =-3+2 ∴ x/kloc-0.
(2) Solution: 9x^2；; -24x+ 16 = 1 1∴(3x-4)2 = 1 1∴3x-4 =∴ 1 1 ∴.
2. Matching method: use matching method to solve the equation AX 2+BX+C = 0 (A ≠ 0).
First, move the constant c 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: x 2+b/ax+(b/2a) 2 =-c/a+(b/2a) 2.
The left side of the equation becomes completely flat: (x+b/2a) 2 =-c/a-b/2a) 2;
When b 2-4ac ≥ 0, X+B/2A = √-C/A-B/2A-2;
∴x=﹛﹣b[√﹙b^2； -4ac]-/2a (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 into 1: x 2-4/3-x = 2/3.
Both sides of the equation, plus the square of half the coefficient of the first term: x 2-4/3-x+(4/6) 2 =? +(4/6)^2
Formula: (X-4/6) 2 =+(4/6) 2.
Direct square: x-4/6 = √ [? +(4/6)^2 ]
∴x= 4/6 √[？ +(4/6)^2 ]
∴ The solution of the original equation is x? =4/6﹢√﹙ 10/6﹚,x？ =4/6﹣√﹙ 10/6﹚ .
3. Formula method: convert the quadratic equation of one variable into a general form, and then calculate the value of the discriminant △ = b 2-4ac. When b 2-4ac ≥ 0, substitute the values of the coefficients A, B and C into the root-finding formula X = [-b √ (b 2-4ac)]/(2a).
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 = 5b2-4ac = (-8) 2-4× 2× 5 = 64-40 = 24 > 0 ∴ X = [(-b] =，x？ = .
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) 2x2+3x = 0 (3) 6x2+5x-50 = 0 (optional) (4) x2-2 (+) x+4 = 0 (optional).
(1) solution: (x+3)(x-6)=-8, which is simplified to get x 2-3x- 10 = 0 (the left side of the equation is quadratic trinomial and the right side is zero) (x-5)(x+2)=0 (the left side of the equation is factorization)
⑵ Solution: 2x 2+3x = 0x (2x+3) = 0 (factorizing the left side of the equation by raising the common factor) ∴x=0 or 2x+3=0 (transforming 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 the symbols when factorizing cross multiplication) ∴2x-5=0 or 3x 3x+ 10 = 0 ∴ x65438++.
(4) solution: x2-2(+)x+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 original equation.
Summary: Generally speaking, 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: element method, collocation method and undetermined coefficient method.
one variable cubic equation
The formula for finding the root of the univariate cubic equation cannot be deduced by ordinary deductive thinking. The standard univariate cubic equation of AX 3+BX 2+CX+D = 0 can only be formalized as a special type of X 3+PX+Q = 0 by using a matching method similar to the formula for finding the root of the univariate quadratic equation.
The solution of the solution formula of the univariate cubic equation can only be obtained by inductive thinking, that is, the form of the root formula of the univariate cubic equation is summarized according to the form of the root formula of the univariate quadratic equation and the special higher order equation. The formula for finding the root of a univariate cubic equation in the form of x 3+px+q = 0 should be X = A (1/3)+B (1/3), which is the sum of two exponents. Summarized the form of the root formula of the univariate cubic equation, and the next step is to find the content of the square, that is, to represent a and b by P and Q, as follows:
(1) cube the two sides of x = a (1/3)+b (1/3) at the same time.
⑵x^3=(a+b)+3(ab)^( 1/3)(a^( 1/3)+b^( 1/3))
(3) Because x = a (1/3)+b (1/3), (2) can be changed to
X 3 = (a+b)+3 (ab) (1/3) x, transpositions are available.
(4) x 3-3 (ab) (1/3) x-(a+b) = 0. Comparing the univariate cubic equation with the special type x 3+px+q = 0,
5]-3 (AB) (1/3) = P, -(A+B) = Q, simplified.
⑹A+B=-q,AB=-(p/3)^3
(7) The roots of the univariate cubic equation are formulated into the formula of the univariate quadratic equation, because A and B can be regarded as the two roots of the univariate quadratic equation, (6) Vieta's theorem of the two roots of the univariate quadratic equation exists in the form of ay 2+by+c = 0, that is,
⑻y 1+y2=-(b/a),y 1*y2=c/a
Comparing [6] and [8], we can make A=y 1, B=y2, q=b/a,-(p/3) 3 = c/a.
⑽ Because the formula for finding the root of unary quadratic equation of AY 2+BY+C = 0 is
y 1=-(b+(b^2-4ac)^( 1/2))/(2a)
y2=-(b-(b^2-4ac)^( 1/2))/(2a)
Can become
⑾y 1=-(b/2a)-((b/2a)^2-(c/a))^( 1/2)
y2=-(b/2a)+((b/2a)^2-(c/a))^( 1/2)
Substitute A=y 1, B=y2, q=b/a and -(p/3) 3 = c/a in ⑾.
⑿a=-(q/2)-((q/2)^2+(p/3)^3)^( 1/2)
b=-(q/2)+((q/2)^2+(p/3)^3)^( 1/2)
[13] substitute a and b into X = A (1/3)+B (1/3).
⒁x=(-(q/2)-((q/2)^2+(p/3)^3)^( 1/2))^( 1/3)+(-(q/2)+((q/2)^2+(p/3)^3)^( 1/2))^( 1/3)
[14] formula is only the real root solution of a cubic equation with one variable. According to the cubic equation of Vieta's theorem, there should be three roots. But according to the cubic equation of Vieta's theorem, only one root is needed, and the other two roots are easy to find.
X y is the y power of x, which is very complicated to say. The solution of the univariate cubic equation discovered by Tattaglia is the general form of the univariate cubic equation.
x3+sx2+tx+u=0
If we do an abscissa translation y=x+s/3, then we can eliminate the quadratic term of the equation. So we only need to consider the cubic equation in the form of x3 = px+q.
Suppose that the solution X of the equation can be written in the form of x=a-b, where A and B are undetermined parameters.
Substitute it into the equation and we have it.
a3-3a2b+3ab2-b3=p(a-b)+q
Complete sorting
a3-b3 =(a-b)(p+3ab)+q
According to the theory of quadratic equation, A and B can be properly selected, so that when x=a-b,
3ab+p=0. So the above formula becomes
a3-b3=q
Multiply both sides by 27a3, and you get
27a6-27a3b3=27qa3
P=-3ab indicates that
27a6 + p3 = 27qa3
This is a quadratic equation about a3, so A can be solved. Then b and root x can be solved.
quartic equation of one unknown
Ferrari found that the solution of the quartic equation of one yuan is the same as that of the cubic equation, and the quartic equation can be eliminated by one coordinate translation.
Cubic term in general form. So as long as we consider the quartic equation of the following form:
x4=px2+qx+r
The key is to match the two sides of the equation into a complete square by using parameters. Consider a parameter
A: We do.
(x2+a)2 = (p+2a)x2+qx+r+a2
The right side of the equation is completely flat if and only if its discriminant is 0, that is,
q2 = 4(p+2a)(r+a2)
This is a cubic equation about A. By using the solution of the cubic equation in one variable above, we can
Solve the parameter A, so that both sides of the original equation are completely flat, and after finding the root, it is a formula about X.
One-dimensional quadratic equation, so we can find the root x of the original equation.
Finally, there is no general algebraic solution to the higher-order equation with a unary degree of five or more (that is, the coefficient is subjected to finite four operations and power and root operations), which is called Abel theorem.