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First, let's remember the basic formulas of powers and their properties.

Product of a number a occurs on itself n times, we can write this expression as a a … a=a n

1. a 0 = 1 (a ≠ 0)

3. a n a m = a n + m

4. (a n) m = a nm

5. a n b n = (ab) n

7. a n / a m = a n - m

Power or exponential equations– these are equations in which the variables are in powers (or exponents), and the base is a number.

Examples of exponential equations:

In this example, the number 6 is the base; it is always at the bottom, and the variable x degree or indicator.

Let us give more examples of exponential equations.
2 x *5=10
16 x - 4 x - 6=0

Now let's look at how exponential equations are solved?

Let's take a simple equation:

2 x = 2 3

This example can be solved even in your head. It can be seen that x=3. After all, in order for the left and right sides to be equal, you need to put the number 3 instead of x.
Now let’s see how to formalize this decision:

2 x = 2 3
x = 3

In order to solve such an equation, we removed identical grounds(that is, twos) and wrote down what was left, these are degrees. We got the answer we were looking for.

Now let's summarize our decision.

Algorithm for solving the exponential equation:
1. Need to check the same whether the equation has bases on the right and left. If the reasons are not the same, we are looking for options to solve this example.
2. After the bases become the same, equate degrees and solve the resulting new equation.

Now let's look at a few examples:

Let's start with something simple.

The bases on the left and right sides are equal to the number 2, which means we can discard the base and equate their degrees.

x+2=4 The simplest equation is obtained.
x=4 – 2
x=2
Answer: x=2

In the following example you can see that the bases are different: 3 and 9.

3 3x - 9 x+8 = 0

First, move the nine to the right side, we get:

Now you need to make the same bases. We know that 9=3 2. Let's use the power formula (a n) m = a nm.

3 3x = (3 2) x+8

We get 9 x+8 =(3 2) x+8 =3 2x+16

3 3x = 3 2x+16 Now it is clear that on the left and right sides the bases are the same and equal to three, which means we can discard them and equate the degrees.

3x=2x+16 we get the simplest equation
3x - 2x=16
x=16
Answer: x=16.

Let's look at the following example:

2 2x+4 - 10 4 x = 2 4

First of all, we look at the bases, bases two and four. And we need them to be the same. We transform the four using the formula (a n) m = a nm.

4 x = (2 2) x = 2 2x

And we also use one formula a n a m = a n + m:

2 2x+4 = 2 2x 2 4

Add to the equation:

2 2x 2 4 - 10 2 2x = 24

We gave an example for the same reasons. But other numbers 10 and 24 bother us. What to do with them? If you look closely you can see that on the left side we have 2 2x repeated, here is the answer - we can put 2 2x out of brackets:

2 2x (2 4 - 10) = 24

Let's calculate the expression in brackets:

2 4 — 10 = 16 — 10 = 6

We divide the entire equation by 6:

Let's imagine 4=2 2:

2 2x = 2 2 bases are the same, we discard them and equate the degrees.
2x = 2 is the simplest equation. Divide it by 2 and we get
x = 1
Answer: x = 1.

Let's solve the equation:

9 x – 12*3 x +27= 0

Let's convert:
9 x = (3 2) x = 3 2x

We get the equation:
3 2x - 12 3 x +27 = 0

Our bases are the same, equal to three. In this example, you can see that the first three has a degree twice (2x) than the second (just x). In this case, you can solve replacement method. We replace the number with the smallest degree:

Then 3 2x = (3 x) 2 = t 2

We replace all x powers in the equation with t:

t 2 - 12t+27 = 0
We get a quadratic equation. Solving through the discriminant, we get:
D=144-108=36
t 1 = 9
t2 = 3

Returning to the variable x.

Take t 1:
t 1 = 9 = 3 x

That is,

3 x = 9
3 x = 3 2
x 1 = 2

One root was found. We are looking for the second one from t 2:
t 2 = 3 = 3 x
3 x = 3 1
x 2 = 1
Answer: x 1 = 2; x 2 = 1.

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In general, an equation of degree greater than 4 cannot be solved in radicals. But sometimes we can still find the roots of a polynomial on the left in an equation of the highest degree if we represent it as a product of polynomials to a degree of no more than 4. Solving such equations is based on factoring a polynomial, so we advise you to review this topic before studying this article.

Most often you have to deal with equations of higher degrees with integer coefficients. In these cases, we can try to find rational roots and then factor the polynomial so that we can then transform it into a lower degree equation that is easy to solve. In this material we will look at just such examples.

Higher degree equations with integer coefficients

All equations of the form a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 = 0 , we can produce an equation of the same degree by multiplying both sides by a n n - 1 and making a variable change of the form y = a n x:

a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 = 0 a n n · x n + a n - 1 · a n n - 1 · x n - 1 + … + a 1 · (a n) n - 1 · x + a 0 · (a n) n - 1 = 0 y = a n x ⇒ y n + b n - 1 y n - 1 + … + b 1 y + b 0 = 0

The resulting coefficients will also be integer. Thus, we will need to solve the reduced equation of the nth degree with integer coefficients, having the form x n + a n x n - 1 + ... + a 1 x + a 0 = 0.

We calculate the integer roots of the equation. If the equation has integer roots, you need to look for them among the divisors of the free term a 0 . Let's write them down and substitute them into the original equality one by one, checking the result. Once we have obtained the identity and found one of the roots of the equation, we can write it in the form x - x 1 · P n - 1 (x) = 0. Here x 1 is the root of the equation, and P n - 1 (x) is the quotient of x n + a n x n - 1 + ... + a 1 x + a 0 divided by x - x 1 .

We substitute the remaining divisors written out into P n - 1 (x) = 0, starting with x 1, since the roots can be repeated. After obtaining the identity, the root x 2 is considered found, and the equation can be written in the form (x - x 1) (x - x 2) · P n - 2 (x) = 0. Here P n - 2 (x) will be the quotient of dividing P n - 1 (x) by x - x 2.

We continue to sort through the divisors. Let's find all the whole roots and denote their number as m. After this, the original equation can be represented as x - x 1 x - x 2 · … · x - x m · P n - m (x) = 0. Here P n - m (x) is a polynomial of n - m degree. For calculation it is convenient to use Horner's scheme.

If our original equation has integer coefficients, we cannot ultimately obtain fractional roots.

We ended up with the equation P n - m (x) = 0, the roots of which can be found in any convenient way. They can be irrational or complex.

Let us show with a specific example how this solution scheme is used.

Example 1

Condition: find the solution to the equation x 4 + x 3 + 2 x 2 - x - 3 = 0.

Solution

Let's start by finding whole roots.

We have a free term equal to minus three. It has divisors equal to 1, - 1, 3 and - 3. Let's substitute them into the original equation and see which of them give the resulting identities.

With x equal to one, we get 1 4 + 1 3 + 2 · 1 2 - 1 - 3 = 0, which means that one will be the root of this equation.

Now let's divide the polynomial x 4 + x 3 + 2 x 2 - x - 3 by (x - 1) in a column:

So x 4 + x 3 + 2 x 2 - x - 3 = x - 1 x 3 + 2 x 2 + 4 x + 3.

1 3 + 2 1 2 + 4 1 + 3 = 10 ≠ 0 (- 1) 3 + 2 (- 1) 2 + 4 - 1 + 3 = 0

We got an identity, which means we found another root of the equation, equal to - 1.

Divide the polynomial x 3 + 2 x 2 + 4 x + 3 by (x + 1) in a column:

We get that

x 4 + x 3 + 2 x 2 - x - 3 = (x - 1) (x 3 + 2 x 2 + 4 x + 3) = = (x - 1) (x + 1) (x 2 + x + 3)

We substitute the next divisor into the equality x 2 + x + 3 = 0, starting from - 1:

1 2 + (- 1) + 3 = 3 ≠ 0 3 2 + 3 + 3 = 15 ≠ 0 (- 3) 2 + (- 3) + 3 = 9 ≠ 0

The resulting equalities will be incorrect, which means that the equation no longer has integer roots.

The remaining roots will be the roots of the expression x 2 + x + 3.

D = 1 2 - 4 1 3 = - 11< 0

It follows from this that this quadratic trinomial has no real roots, but there are complex conjugate ones: x = - 1 2 ± i 11 2.

Let us clarify that instead of dividing into a column, Horner’s scheme can be used. This is done like this: after we have determined the first root of the equation, we fill out the table.

In the table of coefficients we can immediately see the coefficients of the quotient of the division of polynomials, which means x 4 + x 3 + 2 x 2 - x - 3 = x - 1 x 3 + 2 x 2 + 4 x + 3.

After finding the next root, which is - 1, we get the following:

Answer: x = - 1, x = 1, x = - 1 2 ± i 11 2.

Example 2

Condition: solve the equation x 4 - x 3 - 5 x 2 + 12 = 0.

Solution

The free term has divisors 1, - 1, 2, - 2, 3, - 3, 4, - 4, 6, - 6, 12, - 12.

Let's check them in order:

1 4 - 1 3 - 5 1 2 + 12 = 7 ≠ 0 (- 1) 4 - (- 1) 3 - 5 (- 1) 2 + 12 = 9 ≠ 0 2 4 2 3 - 5 2 2 + 12 = 0

This means x = 2 will be the root of the equation. Divide x 4 - x 3 - 5 x 2 + 12 by x - 2 using Horner's scheme:

As a result, we get x - 2 (x 3 + x 2 - 3 x - 6) = 0.

2 3 + 2 2 - 3 2 - 6 = 0

This means that 2 will again be the root. Divide x 3 + x 2 - 3 x - 6 = 0 by x - 2:

As a result, we get (x - 2) 2 · (x 2 + 3 x + 3) = 0.

Checking the remaining divisors does not make sense, since the equality x 2 + 3 x + 3 = 0 is faster and more convenient to solve using the discriminant.

Let's solve the quadratic equation:

x 2 + 3 x + 3 = 0 D = 3 2 - 4 1 3 = - 3< 0

We obtain a complex conjugate pair of roots: x = - 3 2 ± i 3 2 .

Answer: x = - 3 2 ± i 3 2 .

Example 3

Condition: Find the real roots for the equation x 4 + 1 2 x 3 - 5 2 x - 3 = 0.

Solution

x 4 + 1 2 x 3 - 5 2 x - 3 = 0 2 x 4 + x 3 - 5 x - 6 = 0

We multiply 2 3 on both sides of the equation:

2 x 4 + x 3 - 5 x - 6 = 0 2 4 x 4 + 2 3 x 3 - 20 2 x - 48 = 0

Replace variables y = 2 x:

2 4 x 4 + 2 3 x 3 - 20 2 x - 48 = 0 y 4 + y 3 - 20 y - 48 = 0

As a result, we got a standard equation of the 4th degree, which can be solved according to the standard scheme. Let's check the divisors, divide and ultimately get that it has 2 real roots y = - 2, y = 3 and two complex ones. We will not present the entire solution here. Due to the substitution, the real roots of this equation will be x = y 2 = - 2 2 = - 1 and x = y 2 = 3 2.

Answer: x 1 = - 1 , x 2 = 3 2

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Let's consider solving equations with one variable of degree higher than the second.

The degree of the equation P(x) = 0 is the degree of the polynomial P(x), i.e. the greatest of the powers of its terms with a coefficient not equal to zero.

So, for example, the equation (x 3 – 1) 2 + x 5 = x 6 – 2 has the fifth degree, because after the operations of opening the brackets and bringing similar ones, we obtain the equivalent equation x 5 – 2x 3 + 3 = 0 of the fifth degree.

Let us recall the rules that will be needed to solve equations of degree higher than two.

Statements about the roots of a polynomial and its divisors:

1. A polynomial of nth degree has a number of roots not exceeding n, and roots of multiplicity m occur exactly m times.

2. A polynomial of odd degree has at least one real root.

3. If α is the root of P(x), then P n (x) = (x – α) · Q n – 1 (x), where Q n – 1 (x) is a polynomial of degree (n – 1).

4.

5. The reduced polynomial with integer coefficients cannot have fractional rational roots.

6. For a third degree polynomial

P 3 (x) = ax 3 + bx 2 + cx + d one of two things is possible: either it is decomposed into the product of three binomials

Р 3 (x) = а(х – α)(х – β)(х – γ), or decomposes into the product of a binomial and a square trinomial Р 3 (x) = а(х – α)(х 2 + βх + γ ).

7. Any polynomial of the fourth degree can be expanded into the product of two square trinomials.

8. A polynomial f(x) is divisible by a polynomial g(x) without a remainder if there is a polynomial q(x) such that f(x) = g(x) · q(x). To divide polynomials, the “corner division” rule is used.

9. For the polynomial P(x) to be divisible by a binomial (x – c), it is necessary and sufficient that the number c be the root of P(x) (Corollary of Bezout’s theorem).

10. Vieta's theorem: If x 1, x 2, ..., x n are real roots of the polynomial

P(x) = a 0 x n + a 1 x n - 1 + ... + a n, then the following equalities hold:

x 1 + x 2 + … + x n = -a 1 /a 0,

x 1 x 2 + x 1 x 3 + … + x n – 1 x n = a 2 /a 0,

x 1 x 2 x 3 + … + x n – 2 x n – 1 x n = -a 3 / a 0,

x 1 · x 2 · x 3 · x n = (-1) n a n / a 0 .

Solving Examples

Example 1.

Find the remainder of division P(x) = x 3 + 2/3 x 2 – 1/9 by (x – 1/3).

Solution.

By corollary to Bezout’s theorem: “The remainder of a polynomial divided by a binomial (x – c) is equal to the value of the polynomial of c.” Let's find P(1/3) = 0. Therefore, the remainder is 0 and the number 1/3 is the root of the polynomial.

Answer: R = 0.

Example 2.

Divide with a “corner” 2x 3 + 3x 2 – 2x + 3 by (x + 2). Find the remainder and incomplete quotient.

Solution:

2x 3 + 3x 2 – 2x + 3| x + 2

2x 3 + 4 x 2 2x 2 – x

X 2 – 2 x

Answer: R = 3; quotient: 2x 2 – x.

Basic methods for solving higher degree equations

1. Introduction of a new variable

The method of introducing a new variable is already familiar from the example of biquadratic equations. It consists in the fact that to solve the equation f(x) = 0, a new variable (substitution) t = x n or t = g(x) is introduced and f(x) is expressed through t, obtaining a new equation r(t). Then solving the equation r(t), the roots are found:

(t 1, t 2, …, t n). After this, a set of n equations q(x) = t 1 , q(x) = t 2 , … , q(x) = t n is obtained, from which the roots of the original equation are found.

Example 1.

(x 2 + x + 1) 2 – 3x 2 – 3x – 1 = 0.

Solution:

(x 2 + x + 1) 2 – 3(x 2 + x) – 1 = 0.

(x 2 + x + 1) 2 – 3(x 2 + x + 1) + 3 – 1 = 0.

Substitution (x 2 + x + 1) = t.

t 2 – 3t + 2 = 0.

t 1 = 2, t 2 = 1. Reverse substitution:

x 2 + x + 1 = 2 or x 2 + x + 1 = 1;

x 2 + x - 1 = 0 or x 2 + x = 0;

Answer: From the first equation: x 1, 2 = (-1 ± √5)/2, from the second: 0 and -1.

2. Factorization by grouping and abbreviated multiplication formulas

The basis of this method is also not new and consists in grouping terms in such a way that each group contains a common factor. To do this, sometimes it is necessary to use some artificial techniques.

Example 1.

x 4 – 3x 2 + 4x – 3 = 0.

Solution.

Let's imagine - 3x 2 = -2x 2 – x 2 and group:

(x 4 – 2x 2) – (x 2 – 4x + 3) = 0.

(x 4 – 2x 2 +1 – 1) – (x 2 – 4x + 3 + 1 – 1) = 0.

(x 2 – 1) 2 – 1 – (x – 2) 2 + 1 = 0.

(x 2 – 1) 2 – (x – 2) 2 = 0.

(x 2 – 1 – x + 2)(x 2 – 1 + x - 2) = 0.

(x 2 – x + 1)(x 2 + x – 3) = 0.

x 2 – x + 1 = 0 or x 2 + x – 3 = 0.

Answer: There are no roots in the first equation, from the second: x 1, 2 = (-1 ± √13)/2.

3. Factorization by the method of undetermined coefficients

The essence of the method is that the original polynomial is factorized with unknown coefficients. Using the property that polynomials are equal if their coefficients are equal at the same powers, the unknown expansion coefficients are found.

Example 1.

x 3 + 4x 2 + 5x + 2 = 0.

Solution.

A polynomial of degree 3 can be expanded into the product of linear and quadratic factors.

x 3 + 4x 2 + 5x + 2 = (x – a)(x 2 + bx + c),

x 3 + 4x 2 + 5x + 2 = x 3 +bx 2 + cx – ax 2 – abx – ac,

x 3 + 4x 2 + 5x + 2 = x 3 + (b – a)x 2 + (cx – ab)x – ac.

Having solved the system:

(b – a = 4,
(c – ab = 5,
(-ac = 2,

(a = -1,
(b = 3,
(c = 2, i.e.

x 3 + 4x 2 + 5x + 2 = (x + 1)(x 2 + 3x + 2).

The roots of the equation (x + 1)(x 2 + 3x + 2) = 0 are easy to find.

Answer: -1; -2.

4. Method of selecting a root using the highest and free coefficient

The method is based on the application of theorems:

1) Every integer root of a polynomial with integer coefficients is a divisor of the free term.

2) In order for the irreducible fraction p/q (p is an integer, q is a natural number) to be the root of an equation with integer coefficients, it is necessary that the number p be an integer divisor of the free term a 0, and q be a natural divisor of the leading coefficient.

Example 1.

6x 3 + 7x 2 – 9x + 2 = 0.

Solution:

6: q = 1, 2, 3, 6.

Therefore, p/q = ±1, ±2, ±1/2, ±1/3, ±2/3, ±1/6.

Having found one root, for example – 2, we will find other roots using corner division, the method of indefinite coefficients or Horner’s scheme.

Answer: -2; 1/2; 1/3.

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Class: 9

Basic goals:

1. Reinforce the concept of an entire rational equation of the th degree.
2. Formulate the basic methods for solving equations of higher degrees (n > 3).
3. Teach basic methods for solving higher-order equations.
4. Learn to use the type of equation to determine the most effective way to solve it.

Forms, methods and pedagogical techniques used by the teacher in the classroom:

• Lecture-seminar teaching system (lectures - explanation of new material, seminars - problem solving).
• Information and communication technologies (frontal survey, oral work with the class).
• Differentiated learning, group and individual forms.
• Using a research method in teaching aimed at developing the mathematical apparatus and thinking abilities of each individual student.
• Printed material – an individual brief summary of the lesson (basic concepts, formulas, statements, lecture material condensed in the form of diagrams or tables).

Lesson plan:

1. Organizing time.
   The purpose of the stage: to include students in educational activities, to determine the content of the lesson.
2. Updating students' knowledge.
   The purpose of the stage: to update students’ knowledge on previously studied related topics
3. Studying a new topic (lecture). Goal of the stage: to formulate the basic methods for solving equations of higher degrees (n > 3)
4. Summarizing.
   The purpose of the stage: to once again highlight the key points in the material studied in the lesson.
5. Homework.
   The purpose of the stage: to formulate homework for students.

Lesson summary

1. Organizational moment.

Formulation of the lesson topic: “Equations of higher powers. Methods for solving them.”

2. Updating students' knowledge.

Theoretical survey - conversation. Repetition of some previously studied information from the theory. Students formulate basic definitions and formulate the necessary theorems. Give examples to demonstrate the level of previously acquired knowledge.

• The concept of an equation with one variable.
• The concept of the root of an equation, the solution of an equation.
• The concept of a linear equation with one variable, the concept of a quadratic equation with one variable.
• The concept of equivalence of equations, equations-consequences (the concept of extraneous roots), transition not by consequence (the case of loss of roots).
• The concept of a whole rational expression with one variable.
• The concept of a whole rational equation n th degree. Standard form of a whole rational equation. Reduced whole rational equation.
• Transition to a set of equations of lower degrees by factoring the original equation.
• The concept of a polynomial n th degree from x. Bezout's theorem. Corollaries from Bezout's theorem. Root theorems ( Z-roots and Q-roots) of an entire rational equation with integer coefficients (reduced and unreduced, respectively).
• Horner's scheme.

3. Studying a new topic.

We will consider the whole rational equation n-th power of standard form with one unknown variable x:Pn(x)= 0, where P n (x) = a n x n + a n-1 x n-1 + a 1 x + a 0– polynomial n th degree from x, a n ≠ 0. If a n = 1 then such an equation is called a reduced integer rational equation n th degree. Let us consider such equations for various values n and list the main methods for solving them.

n= 1 – linear equation.

n= 2 – quadratic equation. Discriminant formula. Formula for calculating roots. Vieta's theorem. Selecting a complete square.

n= 3 – cubic equation.

Grouping method.

Example: x 3 – 4x 2 – x+ 4 = 0 (x – 4)(x 2– 1) = 0 x 1 = 4 , x 2 = 1,x 3 = -1.

Reciprocal cubic equation of the form ax 3 + bx 2 + bx + a= 0. We solve by combining terms with the same coefficients.

Example: x 3 – 5x 2 – 5x + 1 = 0 (x + 1)(x 2 – 6x + 1) = 0 x 1 = -1, x 2 = 3 + 2, x 3 = 3 – 2.

Selection of Z-roots based on the theorem. Horner's scheme. When applying this method, it is necessary to emphasize that the search in this case is finite, and we select the roots using a certain algorithm in accordance with the theorem Z-roots of the given whole rational equation with integer coefficients.

Example: x 3 – 9x 2 + 23x– 15 = 0. The equation is given. Let us write down the divisors of the free term ( + 1; + 3; + 5; + 15). Let's apply Horner's scheme:

	x 3	x 2	x 1	x 0	conclusion
	1	-9	23	-15
1	1	1 x 1 – 9 = -8	1 x (-8) + 23 = 15	1 x 15 – 15 = 0	1 – root
	x 2	x 1	x 0

We get ( x – 1)(x 2 – 8x + 15) = 0 x 1 = 1, x 2 = 3, x 3 = 5.

Equation with integer coefficients. Selection of Q-roots based on the theorem. Horner's scheme. When applying this method, it is necessary to emphasize that the search in this case is finite and we select the roots using a certain algorithm in accordance with the theorem about Q-roots of an unreduced integer rational equation with integer coefficients.

Example: 9 x 3 + 27x 2 – x– 3 = 0. The equation is unreduced. Let us write down the divisors of the free term ( + 1; + 3). Let us write down the divisors of the coefficient at the highest power of the unknown. ( + 1; + 3; + 9) Consequently, we will look for roots among the values ​​( + 1; + ; + ; + 3). Let's apply Horner's scheme:

	x 3	x 2	x 1	x 0	conclusion
	9	27	-1	-3
1	9	1 x 9 + 27 = 36	1 x 36 – 1 = 35	1 x 35 – 3 = 32 ≠ 0	1 – not a root
-1	9	-1 x 9 + 27 = 18	-1 x 18 – 1 = -19	-1 x (-19) – 3 = 16 ≠ 0	-1 – not a root
	9	x 9 + 27 = 30	x 30 – 1 = 9	x 9 – 3 = 0	root
	x 2	x 1	x 0

We get ( x – )(9x 2 + 30x + 9) = 0 x 1 = , x 2 = - , x 3 = -3.

For ease of calculation when selecting Q -roots It can be convenient to make a change of variable, go to the given equation and select Z -roots.

• If the dummy term is 1

.

• If you can use a replacement of the form y = kx

.

Cardano formula. There is a universal method for solving cubic equations - this is the Cardano formula. This formula is associated with the names of Italian mathematicians Gerolamo Cardano (1501–1576), Nicolo Tartaglia (1500–1557), and Scipione del Ferro (1465–1526). This formula is beyond the scope of our course.

n= 4 – equation of the fourth degree.

Grouping method.

Example: x 4 + 2x 3 + 5x 2 + 4x – 12 = 0 (x 4 + 2x 3) + (5x 2 + 10x) – (6x + 12) = 0 (x + 2)(x 3 + 5x – 6) = 0 (x + 2)(x– 1)(x 2 + x + 6) = 0 x 1 = -2, x 2 = 1.

Variable replacement method.

• Biquadratic equation of the form ax 4 + bx 2 + s = 0 .

Example: x 4 + 5x 2 – 36 = 0. Replacement y = x 2. From here y 1 = 4, y 2 = -9. That's why x 1,2 = + 2 .

• Reciprocal equation of the fourth degree of the form ax 4 + bx 3+c x 2 + bx + a = 0.

We solve by combining terms with the same coefficients by replacing the form

• ax 4 + bx 3 + cx 2 – bx + a = 0.

• Generalized recurrent equation of the fourth degree of the form ax 4 + bx 3 + cx 2 + kbx + k 2 a = 0.

• General replacement. Some standard replacements.

Example 3 . General view replacement(follows from the type of specific equation).

n = 3.

Equation with integer coefficients. Selection of Q-roots n = 3.

General formula. There is a universal method for solving fourth degree equations. This formula is associated with the name of Ludovico Ferrari (1522–1565). This formula is beyond the scope of our course.

n > 5 – equations of the fifth and higher degrees.

Equation with integer coefficients. Selection of Z-roots based on the theorem. Horner's scheme. The algorithm is similar to that discussed above for n = 3.

Equation with integer coefficients. Selection of Q-roots based on the theorem. Horner's scheme. The algorithm is similar to that discussed above for n = 3.

Symmetric equations. Any reciprocal equation of odd degree has a root x= -1 and after factoring it into factors we find that one factor has the form ( x+ 1), and the second factor is a reciprocal equation of even degree (its degree is one less than the degree of the original equation). Any reciprocal equation of even degree together with a root of the form x = φ also contains the root of the species. Using these statements, we solve the problem by lowering the degree of the equation under study.

Variable replacement method. Use of homogeneity.

There is no general formula for solving entire equations of the fifth degree (this was shown by the Italian mathematician Paolo Ruffini (1765–1822) and the Norwegian mathematician Niels Henrik Abel (1802–1829)) and higher degrees (this was shown by the French mathematician Evariste Galois (1811–1832) )).

• Let us recall once again that in practice it is possible to use combinations the methods listed above. It is convenient to pass to a set of equations of lower degrees by factoring the original equation.
• Outside the scope of our discussion today are those widely used in practice. graphical methods solving equations and approximate solution methods equations of higher degrees.
• There are situations when the equation does not have R-roots.
Then the solution comes down to showing that the equation has no roots. To prove this, we analyze the behavior of the functions under consideration on monotonicity intervals. Example: equation x 8 – x 3 + 1 = 0 has no roots.
• Using the property of monotonicity of functions
. There are situations when using various properties of functions allows you to simplify the task.
Example 1: Equation x 5 + 3x– 4 = 0 has one root x= 1. Due to the property of monotonicity of the analyzed functions, there are no other roots.
Example 2: Equation x 4 + (x– 1) 4 = 97 has roots x 1 = -2 and x 2 = 3. Having analyzed the behavior of the corresponding functions on intervals of monotonicity, we conclude that there are no other roots.

4. Summing up.

Summary: Now we have mastered the basic methods for solving various equations of higher degrees (for n > 3). Our task is to learn how to effectively use the algorithms listed above. Depending on the type of equation, we will have to learn to determine which method of solution in a given case is the most effective, as well as correctly apply the chosen method.

5. Homework.

: paragraph 7, pp. 164–174, nos. 33–36, 39–44, 46.47.

: №№ 9.1–9.4, 9.6–9.8, 9.12, 9.14–9.16, 9.24–9.27.

Possible topics for reports or abstracts on this topic:

• Cardano formula
• Graphical method for solving equations. Examples of solutions.
• Methods for approximate solution of equations.

Analysis of student learning and interest in the topic:

Experience shows that students’ interest is primarily aroused by the possibility of selecting Z-roots and Q-roots of equations using a fairly simple algorithm using Horner’s scheme. Students are also interested in various standard types of substitution of variables, which can significantly simplify the type of problem. Graphical solution methods are usually of particular interest. In this case, you can additionally analyze problems using a graphical method for solving equations; discuss the general form of the graph for a polynomial of degree 3, 4, 5; analyze how the number of roots of equations of 3, 4, 5 degrees is related to the appearance of the corresponding graph. Below is a list of books where you can find additional information on this topic.

Bibliography:

1. Vilenkin N.Ya. and others. “Algebra. Textbook for 9th grade students with in-depth study of mathematics” - M., Prosveshchenie, 2007 - 367 p.
2. Vilenkin N.Ya., Shibasov L.P., Shibasova Z.F.“Behind the pages of a mathematics textbook. Arithmetic. Algebra. 10-11 grades” – M., Education, 2008 – 192 p.
3. Vygodsky M.Ya.“Handbook of Mathematics” – M., AST, 2010 – 1055 p.
4. Galitsky M.L.“Collection of problems in algebra. Textbook for grades 8-9 with in-depth study of mathematics” - M., Prosveshchenie, 2008 - 301 p.
5. Zvavich L.I. and others. “Algebra and the beginnings of analysis. 8–11 grades A manual for schools and classes with advanced study of mathematics” - M., Drofa, 1999 - 352 p.
6. Zvavich L.I., Averyanov D.I., Pigarev B.P., Trushanina T.N.“Mathematics assignments for preparing for the written exam in 9th grade” - M., Prosveshchenie, 2007 - 112 p.
7. Ivanov A.A., Ivanov A.P.“Thematic tests for systematizing knowledge in mathematics” part 1 – M., Fizmatkniga, 2006 – 176 p.
8. Ivanov A.A., Ivanov A.P.“Thematic tests for systematizing knowledge in mathematics” part 2 – M., Fizmatkniga, 2006 – 176 p.
9. Ivanov A.P.“Tests and tests in mathematics. Tutorial". – M., Fizmatkniga, 2008 – 304 p.
10. Leibson K.L.“Collection of practical tasks in mathematics. Part 2–9 grades” – M., MTSNM, 2009 – 184 p.
11. Makarychev Yu.N., Mindyuk N.G."Algebra. Additional chapters for the 9th grade school textbook. A textbook for students in schools and classes with in-depth study of mathematics.” – M., Education, 2006 – 224 p.
12. Mordkovich A.G."Algebra. In-depth study. 8th grade. Textbook” – M., Mnemosyne, 2006 – 296 p.
13. Savin A.P.“Encyclopedic Dictionary of a Young Mathematician” - M., Pedagogy, 1985 - 352 p.
14. Survillo G.S., Simonov A.S.“Didactic materials on algebra for grade 9 with in-depth study of mathematics” - M., Prosveshchenie, 2006 - 95 p.
15. Chulkov P.V.“Equations and inequalities in the school mathematics course. Lectures 1–4” – M., September 1st, 2006 – 88 p.
16. Chulkov P.V.“Equations and inequalities in the school mathematics course. Lectures 5–8” – M., September 1, 2009 – 84 p.