Solution of symmetric systems of equations. Symmetric equations. I. Motivation for learning activities of students
1.
The equations are called symmetric equations of the 3rd degree if they look like
ax 3 + bx 2 + bx + a = 0.
In order to successfully solve equations of this type, it is useful to know and be able to use the following simple properties of reciprocal equations:
a) Any reciprocal equation of odd degree always has a root equal to -1.
Indeed, if we group the terms on the left side as follows: a(x 3 + 1) + bx(x + 1) = 0, that is, it is possible to take out a common factor, i.e. (x + 1) (ax 2 + (b - a) x + a) \u003d 0, therefore,
x + 1 \u003d 0 or ax 2 + (b - a) x + a \u003d 0, the first equation and proves the statement of interest to us.
b) The reciprocal equation has no zero roots.
in) When dividing a polynomial of odd degree by (x + 1), the quotient is again a reciprocal polynomial, and this is proved by induction.
Example.
x 3 + 2x 2 + 2x + 1 = 0.
Solution.
The original equation necessarily has a root x \u003d -1, so we divide x 3 + 2x 2 + 2x + 1 by (x + 1) according to the Horner scheme:
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1 |
2 |
2 |
1 |
-1 |
1 |
2 – 1 = 1 | 2 – 1 = 1 | 1 – 1 = 0 |
x 3 + 2x 2 + 2x + 1 = (x + 1) (x 2 + x + 1) = 0.
The quadratic equation x 2 + x + 1 = 0 has no roots.
Answer: -1.
2.
The equations are called symmetric equations of the 4th degree if they look like
ax 4 + bx 3 + cx 2 + bx + a = 0.
Solution algorithm similar equations is:
a) Divide both sides of the original equation by x 2. This action will not lead to the loss of the root, because x \u003d 0 is not a solution to the given equation.
b) Using grouping, bring the equation to the form:
a(x 2 + 1/x 2) + b(x + 1/x) + c = 0.
in) Enter a new unknown: t = (x + 1/x).
Let's make transformations: t 2 = x 2 +2 + 1/x 2 . If we now express x 2 + 1/x 2, then t 2 - 2 = x 2 + 1/x 2.
G) Solve in new variables obtained quadratic equation:
at 2 + bt + c - 2a = 0.
e) Make a reverse substitution.
Example.
6x 4 - 5x 3 - 38x 2 - 5x + 6 = 0.
Solution.
6x 2 - 5x - 38 - 5 / x + 6 / x 2 \u003d 0.
6 (x 2 + 1 / x 2) - 5 (x + 1 / x) - 38 \u003d 0.
Enter t: substitution (x + 1/x) = t. Replacement: (x 2 + 1 / x 2) \u003d t 2 - 2, we have:
6t 2 – 5t – 50 = 0.
t = -5/2 or t = 10/3.
Let's go back to x. After the reverse substitution, we solve the two resulting equations:
1) x + 1/x = -5/2;
x 2 + 5/2 x +1 = 0;
x = -2 or x = -1/2.
2) x + 1/x = 10/3;
x 2 - 10/3 x + 1 = 0;
x = 3 or x = 1/3.
Answer: -2; -1/2; 1/3; 3.
Ways to solve some types of equations of higher degrees
1. Equations that look like (x + a) n + (x + b) n = c, are solved by substitution t = x + (a + b)/2. This method is called symmetrization method.
An example of such an equation would be an equation of the form (x + a) 4 + (x + b) 4 = c.
Example.
(x + 3) 4 + (x + 1) 4 = 272.
Solution.
We make the substitution mentioned above:
t \u003d x + (3 + 1) / 2 \u003d x + 2, after simplification: x \u003d t - 2.
(t - 2 + 3) 4 + (t - 2 + 1) 4 = 272.
(t + 1) 4 + (t - 1) 4 = 272.
Removing the brackets using formulas, we get:
t 4 + 4t 3 + 6t 2 + 4t + 1 + t 4 - 4t 3 + 6t 2 - 4t + 1 = 272.
2t 4 + 12t 2 - 270 = 0.
t 4 + 6t 2 - 135 = 0.
t 2 = 9 or t 2 = -15.
The second equation does not give roots, but from the first we have t = ±3.
After the reverse substitution, we get that x \u003d -5 or x \u003d 1.
Answer: -5; one.
To solve such equations, it often turns out to be effective and method of factorization of the left side of the equation.
2. Equations of the form (x + a)(x + b)(x + c)(x + d) = A, where a + d = c + b.
The technique for solving such equations is to partially open the brackets, and then introduce a new variable.
Example.
(x + 1)(x + 2)(x + 3)(x + 4) = 24.
Solution.
Calculate: 1 + 4 = 2 + 3. Group the brackets in pairs:
((x + 1)(x + 4))((x + 2)(x + 3)) = 24,
(x 2 + 5x + 4) (x 2 + 5x + 6) = 24.
Making the change x 2 + 5x + 4 = t, we have the equation
t(t + 2) = 24, it is square:
t 2 + 2t - 24 = 0.
t = -6 or t = 4.
After performing the reverse substitution, we can easily find the roots of the original equation.
Answer: -5; 0.
3. Equations of the form (x + a) (x + b) (x + c) (x + d) \u003d Ax 2, where ad \u003d cb.
The solution method consists in partially opening the brackets, dividing both parts by x 2 and solving a set of quadratic equations.
Example.
(x + 12)(x + 2)(x + 3)(x + 8) = 4x 2.
Solution.
Multiplying the first two and the last two brackets on the left side, we get:
(x 2 + 14x + 24) (x 2 + 11x + 24) = 4x 2. Divide by x 2 ≠ 0.
(x + 14 + 24/x)(x + 11 + 24/x) = 4. Replacing (x + 24/x) = t, we arrive at the quadratic equation:
(t + 14)(t + 11) = 4;
t 2 + 25x + 150 = 0.
t=10 or t=15.
By making the reverse substitution x + 24 / x \u003d 10 or x + 24 / x \u003d 15, we find the roots.
Answer: (-15 ± √129)/2; -four; -6.
4. Solve the equation (3x + 5) 4 + (x + 6) 3 = 4x 2 + 1.
Solution.
This equation is immediately difficult to classify and choose a solution method. Therefore, we first transform using the difference of squares and the difference of cubes:
((3x + 5) 2 - 4x 2) + ((x + 6) 3 - 1) = 0. Then, after taking out the common factor, we come to a simple equation:
(x + 5) (x 2 + 18x + 48) = 0.
Answer: -5; -9±√33.
A task.
Compose a polynomial of the third degree, which has one root equal to 4, has a multiplicity of 2 and a root equal to -2.
Solution.
f(x)/((x - 4) 2 (x + 2)) = q(x) or f(x) = (x - 4) 2 (x + 2)q(x).
Multiplying the first two brackets, and bringing like terms, we get: f (x) \u003d (x 3 - 6x 2 + 32) q (x).
x 3 - 6x 2 + 32 is a polynomial of the third degree, therefore, q (x) is some number from R(i.e. valid). Let q(x) be one, then f(x) = x 3 - 6x 2 + 32.
Answer: f (x) \u003d x 3 - 6x 2 + 32.
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Home > SolutionRational equations and inequalities
I. Rational equations.
Linear equations.
Systems linear equations.
Return equations.
Vieta's formula for polynomials of higher degrees.
Systems of equations of the second degree.
Method for introducing new unknowns in solving equations and systems of equations.
Homogeneous equations.
Solution of symmetric systems of equations.
Equations and systems of equations with parameters.
Graphical method for solving systems of nonlinear equations.
Equations containing the modulus sign.
Basic methods for solving rational equations
II. Rational inequalities.
Properties of equivalent inequalities.
Algebraic inequalities.
interval method.
Fractional-rational inequalities.
Inequalities containing the unknown under the absolute value sign.
Inequalities with parameters.
Systems of rational inequalities.
Graphical solution of inequalities.
III. Verification test.
Rational Equations
view function
P(x) \u003d a 0 x n + a 1 x n - 1 + a 2 x n - 2 + ... + a n - 1 x + a n,
where n is a natural number, a 0 , a 1 ,…, a n are some real numbers, is called an entire rational function.
An equation of the form P(x) = 0, where P(x) is an entire rational function, is called an entire rational equation.
Type equation
P 1 (x) / Q 1 (x) + P 2 (x) / Q 2 (x) + ... + P m (x) / Q m (x) = 0,
where P 1 (x), P 2 (x), …, P m (x), Q 1 (x), Q 2 (x), …, Q m (x) are integers rational functions, is called a rational equation.
Solution rational equation P (x) / Q (x) = 0, where P (x) and Q (x) are polynomials (Q (x) 0), reduces to solving the equation P (x) = 0 and checking that the roots satisfy condition Q (x) 0.
Linear equations.
An equation of the form ax+b=0, where a and b are some constants, is called a linear equation.
If a0, then the linear equation has a single root: x = -b /a.
If a=0; b0, then the linear equation has no solutions.
If a=0; b=0, then, rewriting the original equation in the form ax = -b, it is easy to see that any x is a solution to a linear equation.
The straight line equation has the form: y = ax + b.
If the line passes through a point with coordinates X 0 and Y 0, then these coordinates satisfy the equation of the line, i.e. Y 0 = aX 0 + b.
Example 1.1. solve the equation
2x - 3 + 4(x - 1) = 5.
Solution. Let's expand the brackets one by one, give like terms and find x: 2x - 3 + 4x - 4 = 5, 2x + 4x = 5 + 4 + 3,
Example 1.2. solve the equation
2x - 3 + 2(x - 1) = 4(x - 1) - 7.
Solution. 2x + 2x - 4x = 3 +2 - 4 - 7, 0x = - 6.
Answer: .
Example 1.3. Solve the equation.
2x + 3 - 6(x - 1) = 4(x - 1) + 5.
Solution. 2x - 6x + 3 + 6 = 4 - 4x + 5,
– 4x + 9 = 9 – 4x,
4x + 4x = 9 - 9,
Answer: Any number.
Systems of linear equations.
Type equation
a 1 x 1 + a 2 x 2 + … + a n x n = b,
where a 1 , b 1 , … ,a n , b are some constants, is called a linear equation with n unknowns x 1 , x 2 , …, x n .
A system of equations is called linear if all the equations in the system are linear. If the system consists of n unknowns, then the following three cases are possible:
the system has no solutions;
the system has exactly one solution;
The system has infinitely many solutions.
Example 2.4. solve the system of equations
Solution. It is possible to solve a system of linear equations by the substitution method, which consists in the fact that any equation of the system expresses one unknown in terms of other unknowns, and then substitutes the value of this unknown in the rest of the equations.
From the first equation we express: x = (8 - 3y) / 2. We substitute this expression into the second equation and get a system of equations
X \u003d (8 - 3y) / 2, 3 (8 - 3y) / 2 + 2y \u003d 7. From the second equation we get y \u003d 2. Taking this into account, from the first equation x \u003d 1. Answer: (1; 2). Example 2.5. Solve a system of equations
Solution. The system has no solutions, since two equations of the system cannot be satisfied simultaneously (from the first equation x + y = 3, and from the second x + y = 3.5).
Answer: There are no solutions.
Example 2.6. solve the system of equations
Solution. The system has infinitely many solutions, since the second equation is obtained from the first by multiplying by 2 (i.e., in fact, there is only one equation with two unknowns).
Answer: Infinitely many solutions.
Example 2.7. solve the system of equations
x + y - z = 2,
2x – y + 4z = 1,
Solution. When solving systems of linear equations, it is convenient to use the Gauss method, which consists in transforming the system to a triangular form.
We multiply the first equation of the system by - 2 and, adding the result obtained with the second equation, we get - 3y + 6z \u003d - 3. This equation can be rewritten as y - 2z \u003d 1. Adding the first equation with the third one, we get 7y \u003d 7, or y = 1.
Thus, the system acquired a triangular form
x + y - z = 2,
Substituting y = 1 into the second equation, we find z = 0. Substituting y =1 and z = 0 into the first equation, we find x = 1. Answer: (1; 1; 0). Example 2.8. for what values of the parameter a the system of equations
2x + ay = a + 2,
(a + 1)x + 2ay = 2a + 4
has infinitely many solutions? Solution. From the first equation we express x:
x = - (a / 2)y + a / 2 +1.
Substituting this expression into the second equation, we get
(a + 1)(– (a / 2)y + a / 2 +1) + 2ay = 2a + 4.
(a + 1)(a + 2 – ay) + 4ay = 4a + 8,
4ay – a(a + 1)y = 4(a + 2) – (a + 1)(a + 2),
ya(4 – a – 1) = (a + 2)(4 – a – 1),
ya(3 – a) = (a + 2)(3 – a).
Analyzing the last equation, we note that for a = 3 it has the form 0y = 0, i.e. it is satisfied for any values of y. Answer: 3.
Quadratic equations and equations reducing to them.
An equation of the form ax 2 + bx + c = 0, where a, b and c are some numbers (a0);
x is a variable, called a quadratic equation.
The formula for solving a quadratic equation.
First, we divide both sides of the equation ax 2 + bx + c = 0 by a - this will not change its roots. To solve the resulting equation
x 2 + (b / a)x + (c / a) = 0
select a full square on the left side
x 2 + (b / a) + (c / a) = (x 2 + 2(b / 2a)x + (b / 2a) 2) - (b / 2a) 2 + (c / a) =
= (x + (b / 2a)) 2 - (b 2) / (4a 2) + (c / a) = (x + (b / 2a)) 2 - ((b 2 - 4ac) / (4a 2 )).
For brevity, we denote the expression (b 2 - 4ac) by D. Then the resulting identity takes the form
Three cases are possible:
if the number D is positive (D > 0), then in this case it is possible to take the square root of D and write D as D = (D) 2 . Then
D / (4a 2) = (D) 2 / (2a) 2 = (D / 2a) 2 , therefore the identity takes the form
x 2 + (b / a)x + (c / a) = (x + (b / 2a)) 2 - (D / 2a) 2 .
According to the formula for the difference of squares, we derive from here:
x 2 + (b / a)x + (c / a) = (x + (b / 2a) – (D / 2a))(x + (b / 2a) + (D / 2a)) =
= (x - ((-b + D) / 2a)) (x - ((- b - D) / 2a)).
Theorem: If the identity holds
ax 2 + bx + c \u003d a (x - x 1) (x - x 2),
then the quadratic equation ax 2 + bx + c \u003d 0 for X 1 X 2 has two roots X 1 and X 2, and for X 1 \u003d X 2 - only one root X 1.
By virtue of this theorem, it follows from the identity derived above that the equation
x 2 + (b / a)x + (c / a) = 0,
and thus the equation ax 2 + bx + c = 0 has two roots:
X 1 \u003d (-b + D) / 2a; X 2 \u003d (-b - D) / 2a.
Thus x 2 + (b / a)x + (c / a) = (x - x1)(x - x2).
Usually these roots are written in one formula:
where b 2 - 4ac \u003d D.
if the number D is equal to zero (D = 0), then the identity
x 2 + (b / a)x + (c / a) = (x + (b / 2a)) 2 - (D / (4a 2))
takes the form x 2 + (b / a)x + (c / a) = (x + (b / 2a)) 2 .
It follows that for D = 0, the equation ax 2 + bx + c = 0 has one root of multiplicity 2: X 1 = - b / 2a
3) If the number D is negative (D< 0), то – D >0, and therefore the expression
x 2 + (b / a)x + (c / a) = (x + (b / 2a)) 2 - (D / (4a 2))
is the sum of two terms, one of which is non-negative and the other positive. Such a sum cannot be equal to zero, so the equation
x 2 + (b / a)x + (c / a) = 0
has no real roots. Neither does the equation ax 2 + bx + c = 0.
Thus, to solve the quadratic equation, one should calculate the discriminant
D \u003d b 2 - 4ac.
If D = 0, then the quadratic equation has a unique solution:
If D > 0, then the quadratic equation has two roots:
X 1 \u003d (-b + D) / (2a); X 2 \u003d (-b - D) / (2a).
If D< 0, то квадратное уравнение не имеет корней.
If one of the coefficients b or c zero, then the quadratic equation can be solved without calculating the discriminant:
b = 0; c 0; c/a<0; X1,2 = (-c / a)
b 0; c = 0; X1 = 0, X2= -b / a.
The roots of a general quadratic equation ax 2 + bx + c = 0 are found by the formula
A quadratic equation in which the coefficient at x 2 is equal to 1 is called reduced. Usually the given quadratic equation is denoted as follows:
x 2 + px + q = 0.
Vieta's theorem.
We have derived the identity
x 2 + (b / a)x + (c / a) \u003d (x - x1) (x - x2),
where X 1 and X 2 are the roots of the quadratic equation ax 2 + bx + c =0. Let us expand the brackets on the right side of this identity.
x 2 + (b / a)x + (c / a) \u003d x 2 - x 1 x - x 2 x + x 1 x 2 \u003d x 2 - (x 1 + x 2) x + x 1 x 2.
It follows that X 1 + X 2 = - b / a and X 1 X 2 = c / a. We have proved the following theorem, first established by the French mathematician F. Viet (1540 - 1603):
Theorem 1 (Vieta). The sum of the roots of the quadratic equation is equal to the coefficient at X, taken with the opposite sign and divided by the coefficient at X 2; the product of the roots of this equation is equal to the free term divided by the coefficient at X 2 .
Theorem 2 (reverse). If the equalities
X 1 + X 2 \u003d - b / a and X 1 X 2 \u003d c / a,
then the numbers X 1 and X 2 are the roots of the quadratic equation ax 2 + bx + c = 0.
Comment. Formulas X 1 + X 2 \u003d - b / a and X 1 X 2 \u003d c / a remain true even in the case when the equation ax 2 + bx + c \u003d 0 has one root X 1 of multiplicity 2, if we put in the indicated formulas X 2 = X 1 . Therefore, it is generally accepted that for D = 0, the equation ax 2 + bx + c = 0 has two roots that coincide with each other.
When solving problems related to the Vieta theorem, it is useful to use the relations
(1 / X 1) + (1 / X 2) \u003d (X 1 + X 2) / X 1 X 2;
X 1 2 + X 2 2 \u003d (X 1 + X 2) 2 - 2 X 1 X 2;
X 1 / X 2 + X 2 / X 1 \u003d (X 1 2 + X 2 2) / X 1 X 2 \u003d ((X 1 + X 2) 2 - 2X 1 X 2) / X 1 X 2;
X 1 3 + X 2 3 = (X 1 + X 2)(X 1 2 - X 1 X 2 + X 2 2) =
\u003d (X 1 + X 2) ((X 1 + X 2) 2 - 3X 1 X 2).
Example 3.9. Solve the equation 2x 2 + 5x - 1 = 0.
Solution. D = 25 – 42(– 1) = 33 >0;
X 1 \u003d (- 5 + 33) / 4; X 2 \u003d (- 5 -33) / 4.
Answer: X 1 \u003d (- 5 + 33) / 4; X 2 \u003d (- 5 -33) / 4.
Example 3.10. Solve the equation x 3 - 5x 2 + 6x = 0
Solution. Let's factorize the left side of the equation x(x 2 - 5x + 6) = 0,
hence x \u003d 0 or x 2 - 5x + 6 \u003d 0.
Solving the quadratic equation, we get X 1 \u003d 2, X 2 \u003d 3.
Answer: 0; 2; 3.
Example 3.11.
x 3 - 3x + 2 = 0. Solution. Let's rewrite the equation, writing -3x \u003d - x - 2x, x 3 - x - 2x + 2 \u003d 0, and now we group x (x 2 - 1) - 2 (x - 1) \u003d 0, (x - 1) (x( x + 1) - 2) = 0,x - 1 = 0, x 1 = 1,x 2 + x - 2 = 0, x 2 = - 2, x 3 = 1. Answer: x 1 = x 3 = 1 , x 2 = - 2. Example 3.12. Solve Equation7
(x - 1)(x - 3)(x - 4)
(2x – 7)(x + 2)(x – 6)Solution. Let's find the range of admissible values x:X + 2 0; x – 6 0; 2x – 7 0 or x – 2; x 6; x 3.5. We bring the equation to the form (7x - 14) (x 2 - 7x + 12) \u003d (14 - 4x) (x 2 - 4x - 12), open the brackets. 7x 3 - 49x 2 + 84x - 14x 2 + 98x – 168 + 4x 3 – 16x 2 – 48x – 14x 2 + 56x + 168 = 0.11x 3 – 93x 2 + 190x = 0.x(11x 2 – 93x + 190) = 0.x 1 = 011x2 – 93x + 190 = 0.93(8649 - 8360) 93 17 x 2.3 = = ,
Those. x 1 = 5; x 2 = 38 / 11.
The found values satisfy the ODZ.
Answer: x 1 = 0; x 2 \u003d 5; x 3 \u003d 38 / 11.
Example 3.13. Solve the equation x 6 - 5x 3 + 4 = 0
Solution. Denote y = x 3 , then the original equation takes the form
y 2 - 5y + 4 = 0, solving which we get Y 1 = 1; Y2=4.
Thus, the original equation is equivalent to the set
equations: x 3 \u003d 1 or x 3 \u003d 4, i.e. X 1 \u003d 1 or X 2 \u003d 3 4
Answer: 1; 3 4.
Example 3.14. Solve the equation (x 3 - 27) / (x - 3) = 27
Solution. We decompose the numerator into factors (according to the formula for the difference of cubes):
ReportSupervisor: Kulabukhov Sergey Yurievich, Candidate of Physical and Mathematical Sciences, teacher additional education MOU DOD DTDiM, Rostov-on-Don.
Lesson Objectives:
- educational: learning to solve systems of equations containing homogeneous equation, symmetric systems of equations;
- developing: development of thinking, attention, memory, ability to highlight the main thing;
- educational: development of communication skills.
Lesson type: lesson learning new material.
Used learning technologies:
- work in groups;
- design method.
Equipment: computer, multimedia projector.
A week before the lesson, students receive topics for creative assignments (according to options).
I option. Symmetric systems of equations. Solutions.
II option. Systems containing a homogeneous equation. Solutions.
Each student, using an additional educational literature, must find the appropriate educational material, select a system of equations and solve it.
One student from each option creates multimedia presentations on the topic of the creative task. The teacher provides guidance to students as needed.
I. Motivation for learning activities of students
Introductory speech of the teacher
In the previous lesson, we considered the solution of systems of equations by the method of replacing unknowns. general rule there is no choice of new variables. However, two types of systems of equations can be distinguished when there is a reasonable choice of variables:
- symmetric systems of equations;
- systems of equations, one of which is homogeneous.
II. Learning new material
Students of the second option report on their homework.
1. Slideshow of a multimedia presentation "Systems containing a homogeneous equation" (presentation 1).
2. Work in pairs of students sitting at the same desk: a student of the second option explains to a neighbor in the desk the solution to a system containing a homogeneous equation.
Report of students of the 1st option.
1. Slideshow of the multimedia presentation "Symmetric systems of equations" (presentation 2).
Students write in their notebooks:
2. Work in pairs of students sitting at the same desk: a student of option I explains to a neighbor in the desk the solution of a symmetric system of equations.
III. Consolidation of the studied material
Work in groups (in a group of 4 students unite students sitting at adjacent desks).
Each of the 6 groups performs the following task.
Determine the type of system and solve it:
Students in groups analyze systems, determine their type, then, in the course of frontal work, discuss solutions to systems.
a) system
symmetric, we introduce new variables x+y=u, xy=v
b) system
contains a homogeneous equation.
A pair of numbers (0;0) is not a solution to the system.
IV. Control of students' knowledge
Independent work on options.
Solve the system of equations:
Students hand in their notebooks to the teacher for review.
V. Homework
1. Performed by all students.
Solve the system of equations:
2. Perform "strong" students.
Solve the system of equations:
VI. Lesson summary
Questions:
What types of systems of equations did you learn in class?
What method of solving systems of equations is used to solve them?
Reporting grades received by students during the lesson.
Introduction The problem with my project is that for a successful passing the exam need to be able to decide various systems equations, but in the know high school they have not been given enough time to delve deeper into the matter. The purpose of the work: to prepare for the successful delivery of the exam. Tasks of the work: Expand your knowledge in the field of mathematics related to the concept of "symmetry". Improve your mathematical culture, using the concept of "symmetry" when solving systems of equations, called symmetric, as well as other problems of mathematics.
The concept of symmetry. Symmetry - (ancient Greek συμμετρία), in a broad sense - immutability under any transformations. So, for example, the spherical symmetry of a body means that the appearance of the body will not change if it is rotated in space at arbitrary angles. Bilateral symmetry means that right and left look the same with respect to some plane.
Problem solving using symmetry. Problem 1 Two people take turns putting identical coins on a round table, and the coins should not cover each other. The one who cannot make a move loses. Who wins at the right game? (In other words, which player has a winning strategy?)
Methods for solving symmetric systems. Symmetric systems can be solved by the change of variables, which are the main symmetric polynomials. A symmetric system of two equations with two unknowns x and y is solved by substituting u = x + y, v = xy.
Example No. 2 3 x 2y - 2xy + 3xy 2 \u003d 78, 2x - 3xy + 2y + 8 \u003d 0 Using the basic symmetric polynomials, the system can be written in the following form 3uv - 2v \u003d 78, 2u - 3v \u003d -8. Expressing u = from the second equation and substituting it into the first equation, we obtain 9v2– 28v – 156 = 0. The roots of this equation v 1 = 6 and v 2 = - allow us to find the corresponding values u1 = 5, u2= - from the expression u = .
Let us now solve the following set of systems Let us now solve the following set of systems x + y = 5, and x + y = - , xy = 6 xy = - . x \u003d 5 - y, and y \u003d -x -, xy \u003d 6 xy \u003d -. x \u003d 5 - y, and y \u003d -x -, y (5 - y) \u003d 6 x (-x -) \u003d -. x \u003d 5 - y, and y \u003d -x -, y 1 \u003d 3, y 2 \u003d 2 x 1 \u003d, x 2 \u003d - x 1 \u003d 2, x 2 \u003d 3, and x 1 \u003d, x 2 \u003d - y 1= 3, y 2 =2 y 1 = -, y 2= Answer: (2; 3), (3; 2), (; -), (- ;).
Theorems used in solving symmetric systems. Theorem 1. (on symmetric polynomials) Any symmetric polynomial in two variables can be represented as a function of two basic symmetric polynomials In other words, for any symmetric polynomial f (x, y) there exists a function of two variables φ (u, v) such that
Theorem 2. (on symmetric polynomials) Theorem 2. (on symmetric polynomials) Any symmetric polynomial in three variables can be represented as a function of three basic symmetric polynomials: In other words, for any symmetric polynomial f (x, y) there exists such function of three variables θ (u, v, w), which
More complex symmetrical systems - systems containing the module: | x – y | + y2 = 3, | x – 1 | + | y-1 | = 2. Consider this system separately for x< 1 и при х ≥ 1.
Если х < 1, то:
а) при у < х система принимает вид
х – у + у 2 = 3,
- х + 1 – у + 1 = 2,
или
х – у + у 2 = 3,
х + у = 0,
откуда находим х 1 = 1, у 1 = - 1, х 2 = - 3, у2 = 3. Эти пары чисел не принадлежат к рассматриваемой области;
b) for x ≤ y< 1 система принимает вид
б) при х ≤ у < 1 система принимает вид
- х + у + у 2 = 3,
- х + 1 – у + 1 = 2,
или
- х + у + у 2 = 3,
х + у = 0,
откуда находим х 1 = 3, у 1 = - 3; х 2 = - 1, у 2 = 1.
Эти пары чисел не принадлежат к рассматриваемой области;
в) при у ≥ 1 (тогда у >x) the system takes the form - x + y + y 2 \u003d 3, - x + 1 + y - 1 \u003d 2, or - x + y + y 2 \u003d 3, x - y \u003d - 2, from where we find x 1 \u003d - 3, y 1 \u003d - 1, x 2 \u003d - 1, y 2 \u003d 1. The second pair of numbers belongs to the area under consideration, that is, it is a solution to this system.
If x ≥ 1, then: If x ≥ 1, then: a) x > y and y< 1 система принимает вид
х – у + у 2 = 3,
х – 1 – у = 1 = 2,
или
х – у + у 2= 3,
х – у = 2,
откуда находим х 1 = 1, у 1 = - 1, х 2 = 4, у 2 = 2. Первая пара чисел принадлежит рассматриваемой области, т. Е. является решением данной системы;
б) при х >y and y ≥ 1 the system takes the form x - y + y 2 = 3, x - 1 + y - 1 = 2, or x - y + y 2 = 3, x + y = 4, from which we find x = 1, y = 3. This pair of numbers does not belong to the area under consideration;
c) for x ≤ y (then y ≥ 1), the system takes the form c) for x ≤ y (then y ≥ 1), the system takes the form - x + y + y 2 = 3, x - 1 + y - 1 = 2, or - x + y + y 2 = 3, x + y = 4, from where we find x 1 = 5 + √8, y 1 = - 1 - √8; x 2 = 5 - √8, y 2 = - 1 + √8. These pairs of numbers do not belong to the area under consideration. Thus, x 1 \u003d - 1, y 1 \u003d 1; x 2 \u003d 1, y 2 \u003d - 1. Answer: (- 1; 1); (eleven).
Conclusion Mathematics develops human thinking, teaches through logic to find different solutions. So, having learned to solve symmetrical systems, I realized that they can be used not only to perform concrete examples, but I'm for solving various kinds of problems. I think that the project can benefit not only me. For those who also want to get acquainted with this topic, my work will be a good helper.
List of used literature: Bashmakov M.I., "Algebra and the beginnings of analysis", 2nd edition, Moscow, "Prosveshchenie", 1992, 350 pages. Rudchenko P.A., Yaremchuk F.P., "Algebra and elementary functions ", directory; third edition, revised and enlarged; Kyiv, Naukova, Dumka, 1987, 648 pages. Sharygin I. F., “Mathematics for high school students”, Moscow, Drofa publishing house, 1995, 490 pages. Internet resources: http://www.college. en/
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1.
The equations are called symmetric equations of the 3rd degree if they look like
ax 3 + bx 2 + bx + a = 0.
In order to successfully solve equations of this type, it is useful to know and be able to use the following simple properties of reciprocal equations:
a) Any reciprocal equation of odd degree always has a root equal to -1.
Indeed, if we group the terms on the left side as follows: a(x 3 + 1) + bx(x + 1) = 0, that is, it is possible to take out a common factor, i.e. (x + 1) (ax 2 + (b - a) x + a) \u003d 0, therefore,
x + 1 \u003d 0 or ax 2 + (b - a) x + a \u003d 0, the first equation and proves the statement of interest to us.
b) The reciprocal equation has no zero roots.
in) When dividing a polynomial of odd degree by (x + 1), the quotient is again a reciprocal polynomial, and this is proved by induction.
Example.
x 3 + 2x 2 + 2x + 1 = 0.
Solution.
The original equation necessarily has a root x \u003d -1, so we divide x 3 + 2x 2 + 2x + 1 by (x + 1) according to the Horner scheme:
. |
1 |
2 |
2 |
1 |
-1 |
1 |
2 – 1 = 1 | 2 – 1 = 1 | 1 – 1 = 0 |
x 3 + 2x 2 + 2x + 1 = (x + 1) (x 2 + x + 1) = 0.
The quadratic equation x 2 + x + 1 = 0 has no roots.
Answer: -1.
2.
The equations are called symmetric equations of the 4th degree if they look like
ax 4 + bx 3 + cx 2 + bx + a = 0.
Solution algorithm similar equations is:
a) Divide both sides of the original equation by x 2. This action will not lead to the loss of the root, because x \u003d 0 is not a solution to the given equation.
b) Using grouping, bring the equation to the form:
a(x 2 + 1/x 2) + b(x + 1/x) + c = 0.
in) Enter a new unknown: t = (x + 1/x).
Let's make transformations: t 2 = x 2 +2 + 1/x 2 . If we now express x 2 + 1/x 2, then t 2 - 2 = x 2 + 1/x 2.
G) Solve the resulting quadratic equation in new variables:
at 2 + bt + c - 2a = 0.
e) Make a reverse substitution.
Example.
6x 4 - 5x 3 - 38x 2 - 5x + 6 = 0.
Solution.
6x 2 - 5x - 38 - 5 / x + 6 / x 2 \u003d 0.
6 (x 2 + 1 / x 2) - 5 (x + 1 / x) - 38 \u003d 0.
Enter t: substitution (x + 1/x) = t. Replacement: (x 2 + 1 / x 2) \u003d t 2 - 2, we have:
6t 2 – 5t – 50 = 0.
t = -5/2 or t = 10/3.
Let's go back to x. After the reverse substitution, we solve the two resulting equations:
1) x + 1/x = -5/2;
x 2 + 5/2 x +1 = 0;
x = -2 or x = -1/2.
2) x + 1/x = 10/3;
x 2 - 10/3 x + 1 = 0;
x = 3 or x = 1/3.
Answer: -2; -1/2; 1/3; 3.
Ways to solve some types of equations of higher degrees
1. Equations that look like (x + a) n + (x + b) n = c, are solved by substitution t = x + (a + b)/2. This method is called symmetrization method.
An example of such an equation would be an equation of the form (x + a) 4 + (x + b) 4 = c.
Example.
(x + 3) 4 + (x + 1) 4 = 272.
Solution.
We make the substitution mentioned above:
t \u003d x + (3 + 1) / 2 \u003d x + 2, after simplification: x \u003d t - 2.
(t - 2 + 3) 4 + (t - 2 + 1) 4 = 272.
(t + 1) 4 + (t - 1) 4 = 272.
Removing the brackets using formulas, we get:
t 4 + 4t 3 + 6t 2 + 4t + 1 + t 4 - 4t 3 + 6t 2 - 4t + 1 = 272.
2t 4 + 12t 2 - 270 = 0.
t 4 + 6t 2 - 135 = 0.
t 2 = 9 or t 2 = -15.
The second equation does not give roots, but from the first we have t = ±3.
After the reverse substitution, we get that x \u003d -5 or x \u003d 1.
Answer: -5; one.
To solve such equations, it often turns out to be effective and method of factorization of the left side of the equation.
2. Equations of the form (x + a)(x + b)(x + c)(x + d) = A, where a + d = c + b.
The technique for solving such equations is to partially open the brackets, and then introduce a new variable.
Example.
(x + 1)(x + 2)(x + 3)(x + 4) = 24.
Solution.
Calculate: 1 + 4 = 2 + 3. Group the brackets in pairs:
((x + 1)(x + 4))((x + 2)(x + 3)) = 24,
(x 2 + 5x + 4) (x 2 + 5x + 6) = 24.
Making the change x 2 + 5x + 4 = t, we have the equation
t(t + 2) = 24, it is square:
t 2 + 2t - 24 = 0.
t = -6 or t = 4.
After performing the reverse substitution, we can easily find the roots of the original equation.
Answer: -5; 0.
3. Equations of the form (x + a) (x + b) (x + c) (x + d) \u003d Ax 2, where ad \u003d cb.
The solution method consists in partially opening the brackets, dividing both parts by x 2 and solving a set of quadratic equations.
Example.
(x + 12)(x + 2)(x + 3)(x + 8) = 4x 2.
Solution.
Multiplying the first two and the last two brackets on the left side, we get:
(x 2 + 14x + 24) (x 2 + 11x + 24) = 4x 2. Divide by x 2 ≠ 0.
(x + 14 + 24/x)(x + 11 + 24/x) = 4. Replacing (x + 24/x) = t, we arrive at the quadratic equation:
(t + 14)(t + 11) = 4;
t 2 + 25x + 150 = 0.
t=10 or t=15.
By making the reverse substitution x + 24 / x \u003d 10 or x + 24 / x \u003d 15, we find the roots.
Answer: (-15 ± √129)/2; -four; -6.
4. Solve the equation (3x + 5) 4 + (x + 6) 3 = 4x 2 + 1.
Solution.
This equation is immediately difficult to classify and choose a solution method. Therefore, we first transform using the difference of squares and the difference of cubes:
((3x + 5) 2 - 4x 2) + ((x + 6) 3 - 1) = 0. Then, after taking out the common factor, we come to a simple equation:
(x + 5) (x 2 + 18x + 48) = 0.
Answer: -5; -9±√33.
A task.
Compose a polynomial of the third degree, which has one root equal to 4, has a multiplicity of 2 and a root equal to -2.
Solution.
f(x)/((x - 4) 2 (x + 2)) = q(x) or f(x) = (x - 4) 2 (x + 2)q(x).
Multiplying the first two brackets, and bringing like terms, we get: f (x) \u003d (x 3 - 6x 2 + 32) q (x).
x 3 - 6x 2 + 32 is a polynomial of the third degree, therefore, q (x) is some number from R(i.e. valid). Let q(x) be one, then f(x) = x 3 - 6x 2 + 32.
Answer: f (x) \u003d x 3 - 6x 2 + 32.
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