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Types of algebraic equations and ways to solve them
For students interested in mathematics, when solving algebraic equations of higher degrees, an effective method for quickly finding the roots, division with a remainder by the binomial x - or by ax + b, is Horner's scheme.
Consider Horner's scheme.
We denote the incomplete quotient when P (x) is divided by x - by
Q (x) = b 0 x n -1 + b 1 x n -2 +… + b n -1, and the remainder through b n.
Since Р (х) = Q (x) (х – ) + b n, then the equality
a 0 x n + a 1 x n -1 +… + a n = (b 0 x n -1 + b 1 x n -2 +… + b n -1) (x – ) + b n
Let's open the brackets on the right side and compare the coefficients at the same powers of x on the left and right. We obtain that a 0 = b 0 and for 1 k n the relations a k = b k - b k -1 hold. It follows that b 0 = a 0 and b k = a k + b k -1, 1 k n.
We write down the calculation of the coefficients of the polynomial Q (x) and the remainder b n in the form of a table:
a 0
a 1
a 2
a n-1
a n
b 0 = a 0
b 1 = a 1 + b 0
b 2 = a 2 + b 1
b n-1 = a n-1 + b n-2
b n = a n + b n-1
Example 1. Divide the polynomial 2x 4 - 7x 3 - 3x 2 + 5x - 1 by x + 1.
Solution. We use Horner's scheme.
When dividing 2x 4 - 7x 3 - 3x 2 + 5x - 1 by x + 1 we get 2x 3 - 9x 2 + 6x - 1
Answer: 2 x 3 - 9x 2 + 6x - 1
Example 2. Calculate P (3), where P (x) = 4x 5 - 7x 4 + 5x 3 - 2x + 1
Solution. Using Bezout's theorem and Horner's scheme, we get:
Answer: P (3) = 535
The exercise
Using Horner's scheme, divide the polynomial
4x 3 - x 5 + 132 - 8x 2 for x + 2;
2) Divide the polynomial
2x 2 - 3x 3 - x + x 5 + 1 on x + 1;
3) Find the value of the polynomial P 5 (x) = 2x 5 - 4x 4 - x 2 + 1 at x = 7.
1.1. Finding rational roots of equations with integer coefficients
A method for finding rational roots of an algebraic equation with integer coefficients is given by the following theorem.
Theorem: If an equation with integer coefficients has rational roots, then they are the quotient of dividing the divisor of the free term by the divisor of the leading coefficient.
Proof: a 0 x n + a 1 x n -1 + ... + a n = 0
Let x = p / q is a rational root, q, p are coprime.
Substituting the fraction p / q into the equation, and freeing ourselves from the denominator, we get
a 0 p n + a 1 p n -1 q + ... + a n -1 pq n -1 + a n q n = 0 (1)
We rewrite (1) in two ways:
a n q n = р (- а 0 р n -1 - а 1 р n -2 q - ... - а n -1 q n -1) (2)
a 0 p n = q (- а 1 р n -1 - ... - а n -1 рq n -2 - а n q n -1) (3)
From equality (2) it follows that a n q n is divisible by p, and since q n and p are coprime, then a n is divisible by p. Similarly, equality (3) implies that a 0 is divisible by q. The theorem is proved.
Example 1. Solve the equation 2x 3 - 7x 2 + 5x - 1 = 0.
Solution. The equation has no integral roots, we find the rational roots of the equation. Let p / q be an irreducible fraction as a root of the equation, then we find p among the divisors of the free term, i.e. among the numbers 1, and q among the positive divisors of the leading coefficient: 1; 2.
Those. rational roots of the equation must be sought among the numbers 1, 1/2, we denote P 3 (x) = 2x 3 - 7x 2 + 5x - 1, P 3 (1) 0, P 3 (–1) 0,
P 3 (1/2) = 2/8 - 7/4 + 5/2 - 1 = 0, 1/2 is the root of the equation.
2x 3 - 7x 2 + 5x - 1 = 2x 3 - x 2 - 6 x 2 + 3x + 2x - 1 = 0.
We get: x 2 (2x - 1) - 3x (2x - 1) + (2x - 1) = 0; (2x - 1) (x 2 - 3x + 1) = 0.
Equating the second factor to zero, and solving the equation, we get
Answer: 
,
Exercises
Solve equations:
6x 3 - 25x 2 + 3x + 4 = 0;
6x 4 - 7x 3 - 6x 2 + 2x + 1 = 0;
3x 4 - 8x 3 - 2x 2 + 7x - 1 = 0;
1.2. Return equations and solution methods
Definition. An equation with integer powers relative to the unknown is called returnable if its coefficients, equidistant from the ends of the left side, are equal to each other, i.e. equation of the form
a x n + bx n -1 + cx n -2 +… + cx 2 + bx + a = 0
Reverse equation of odd degree
a x 2 n +1 + bx 2 n + cx 2 n -1 +… + cx 2 + bx + a = 0
always has a root x = - 1. Therefore, it is equivalent to the union of the equation x + 1 = 0 and x 2 n + x 2 n -1 +… + x + = 0. The last equation is a recurrent equation of even degree. Thus, the solution of recurrent equations of any degree is reduced to solving the recurrent equation of an even degree.
How to solve it? Let the recurrent equation of even degree be given
a x 2 n + bx 2 n -1 +… + dx n +1 + ex n + dx n -1 +… + bx + a = 0
Note that x = 0 is not a root of the equation. Then we divide the equation by x n, we get
a x n + bx n -1 +… + dx + e + dx -1 +… + bx 1- n + ax -n = 0
Grouping in pairs the members of the left side
a( x n + x - n) + b (x n -1 + x - (n -1) +… + d (x + x -1) + e = 0
We make the substitution x + x -1 = y. After substitution of expressions x 2 + x -2 = y 2 - 2;
x 3 + x -3 = y 3 - 3y; x 4 + x -4 = y 4 - 4y + 2 into the equation we get the equation for at Ay n + By n -1 + Cy n -2 + ... + Ey + D = 0.
To solve this equation, you need to solve several quadratic equations of the form x + x -1 = y k, where k = 1, 2,… n. Thus, we get the roots of the original equation.
Example 1. Solve the equation x 7 + x 6 - 5x 5 - 13x 4 - 13x 3 - 5x 2 + 2x + 1 = 0.
Solution. x = - 1 is the root of the equation. Let's apply Horner's scheme.
Our equation will take the form:
(x + 1) (x 6 + x 5 - 6x 4 - 7x 3 - 6x 2 + x + 1) = 0
1) x + 1 = 0, x = -1;
2) x 6 + x 5 - 6x 4 - 7x 3 - 6x 2 + x + 1 = 0 | : x 3 0; x 3 + x 2 - 6x - 7 - 6 / x + 1 / x 2 + 1 / x 3 = 0.
By grouping, we get:.
We introduce a replacement: 
; 
; 
.
We get relatively at equation: y 3 - 3y + y 2 - 2 - 6y - 7 = 0;
for 3 + for 2 - 9y - 9 = 0; y 2 (y + 1) - 9 (y + 1) = 0; (y + 1) (y 2 - 9); y 1 = -1, y 2,3 = 3.
Solving equations 
, 
, 
,
we get the roots: 
, 
, 
, 
Answer: x 1 = -1, 
, 
Exercises
Solve equations.
2x 5 + 5x 4 - 13x 3 - 13x 2 + 5x + 2 = 0;
2x 4 + 3x 3 - 16x 2 + 3x + 2 = 0;
15x 5 + 34x 4 + 15x 3 - 15x 2 - 34x - 15 = 0.
1.3. Variable change method for solving equations
The variable replacement method is the most common method. The art of substituting a variable is to see which substitution is more rational and leads to success faster.
If the equation is given
F (f (x)) = 0, (1)
then by replacing the unknown y = f (x) it first reduces to the equation
F (y) = 0, (2)
and then after finding all the solutions of equation (2) at 1, at 2, ..., yn, ... is reduced to solving the set of equations f (x) = at 1, f (x) = at 2, ..., f (x) = at 2, ...
The main ways to implement the variable replacement method are:
using the basic property of a fraction;
allocation of a square of a binomial;
transition to a system of equations;
opening brackets in pairs;
opening brackets in pairs and dividing both sides of the equation;
lowering the degree of the equation;
double replacement.
1.3.1. Lowering the degree of an equation
Solve the equation (x 2 + x + 2) (x 2 + x + 3) = 6 (3)
Solution. We denote x 2 + x + 2 = y, then we will treat y (y + 1) = 6, solving the latter, we get y 1 = 2, y 2 = -3. This equation (3) is equivalent to a set of equations x 2 + x + 2 = 2
x 2 + x + 2 = -3
Solving the first, we get x 1 = 0, x 2 = -1. Solving the second, we get 
, 
Answer: x 1 = 0, x 2 = -1,
1.3.2. An equation of the fourth degree of the form (x + a) (x +b )(x + c )(x + d ) = m , where a + b = c + d, or a + c = b + d, or a + d = b + c.
Example. Solve the equation (x - 1) (x - 7) (x -4) (x + 2) = 40
Solution. - 1- 4 = - 7 + 2, - 5 = - 5, multiplying these pairs of brackets, we get the equation (x 2 - 5x - 14) (x 2 - 5x + 4) = 40
We introduce the replacement: x 2 - 5x - 14 = y, we get the equation y (y + 18) = 40, y 2 + 18y = 40, y 2 + 18y - 40 = 0.y 1 = -20, y 2 = 2. Returning to the original variable, we solve the set of equations:
X 2 - 5x - 14 = - 20 x 1 = 2; x 2 = 3
x 2 - 5x - 14 = 2 x 3.4 =
Answer: x 1 = 2; x 2 = 3 x 3.4 =
1.3.3. An equation of the form (x + a) (x + b) (x + c) (x + d) = Ex 2,
where ab = cd, or ac = bd, or ad = bc. Expand the parentheses in pairs and divide both sides by x 2 0.
Example. (x - 1) (x - 2) (x - 8) (x - 4) = 4x 2
Solution. The product of the numbers in the first and third, in the second and fourth brackets are equal, i.e. - 8 (- 1) = (- 2) (- 4). We multiply the indicated pairs of brackets and write the equation (x 2 - 9x + 8) (x 2 - 6x + 8) = 4x 2.
Since x = 0 is not a root of the equation, we divide both sides of the equation by x 2
0, we get: 
, replacement: 
, the original equation will take the form:t(t+3) =4,
t 2
+ 3
t=4,
t 2
+ 3
t – 4=0,
t 1
=1,
t 2
= - 4.
Let's go back to the original variable:
x 2 - 10x + 8 = 0
x 2 - 5x + 8 = 0
We solve the first equation, we get x 1,2
= 5
The second equation has no roots.
Answer: x 1.2 = 5
1.3.4. Equation of the fourth kind (ax 2 + b 1 x + c) (a x 2 + b 2 x + c) = A x 2
Equation (ax 2 + b 1 x + c)(a x 2 + b 2
x +
c) =
A x 2, where c
0, A
2
, which after replacing the unknown 
will be rewritten as a square and can be easily solved.
Example. (x 2 + x + 2) (x 2 + 2x + 2) = 2x 2
Solution. It is easy to see that x = 0 is not a root of this equation, dividing this equation by x 2
, we get the equation 
replacement 
, we obtain the equation (y + 1) (y + 2) = 2, solving it, we have the roots y 1 = 0; at 2 = - 3, therefore, the original equation is equivalent to a set of equations 
solving, we get x 1 = -1; x 2 = -2.
Answer: x 1 = -1; x 2 = -2
1.3.5. Equation of the form: a (cx 2 + p 1 x + q) 2 + b (cx 2 + p 2 x + q) 2 = Ax 2
The equation a(cx 2
+
p 1
x +
q)
2
+
b(cx 2
+
p 2
x +
q)
2
=
Ax 2, where a,
b,
c,
q,
A are such that q
0,
A
0,
c
0,
a
0,
b
0, does not have a root x = 0, therefore, dividing the equation by x 2
, we obtain the equivalent equation 
, which after replacement 
will be rewritten as a quadratic equation, which is easy to solve.
+ 1)( x 2 - 14x + 15 = 0
x 2
– 7
x + 15 = 0 
Answer:
Ministry of General and Professional Education of the Russian Federation
Municipal educational institution
Gymnasium number 12
writing
on the topic: Equations and ways to solve them
Completed: student 10 "A" class
Krutko Evgeniy
Checked by: teacher of mathematics Iskhakova Gulsum Akramovna
Tyumen 2001
Plan................................................. .................................................. ................................ 1
Introduction ................................................. .................................................. ........................ 2
Main part................................................ .................................................. ............... 3
Conclusion................................................. .................................................. .................. 25
Application................................................. .................................................. ................ 26
List of used literature ............................................... ........................... 29
Plan.
Introduction.
Historical reference.
Equations. Algebraically equations.
a) Basic definitions.
b) Linear equation and a way to solve it.
c) Quadratic equations and methods for its solution.
d) Two-term equations, the way to solve them.
e) Cubic equations and ways to solve it.
f) Biquadratic equation and a way to solve it.
g) Equations of the fourth degree and methods of its solution.
g) Equations of high degrees and methods of solution.
h) Rational algebraic equation and its method
i) Irrational equations and ways to solve it.
j) Equations containing the unknown under the sign.
absolute value and a way to solve it.
Transcendental equations.
a) Exponential equations and a method for solving them.
b) Logarithmic equations and a way to solve them.
Introduction
Mathematical education received in a general education school is an essential component of general education and general culture of a modern person. Almost everything that surrounds a modern person is in one way or another connected with mathematics. And the latest achievements in physics, engineering and information technologies leave no doubt that the state of affairs will remain the same in the future. Therefore, the solution of many practical problems is reduced to solving various types of equations that must be learned to solve.
This work is an attempt to generalize and systematize the material studied on the above topic. I have arranged the material in order of difficulty, starting with the easiest. It included both the types of equations known to us from the school algebra course, and additional material. At the same time, I tried to show the types of equations that are not studied in the school course, but the knowledge of which may be needed when entering a higher educational institution. In my work, when solving equations, I did not limit myself only to a real solution, but also indicated a complex one, since I think that otherwise the equation is simply incomplete. After all, if there are no real roots in the equation, this does not mean that it has no solutions. Unfortunately, due to lack of time, I was not able to present all the material I have, but even with the material presented here, many questions may arise. I hope that my knowledge is enough to answer most of the questions. So, I proceed to present the material.
Mathematics ... brings out the order
symmetry and certainty,
and these are the most important types of beauty.
Aristotle.
Historical reference
In those distant times, when sages first began to think about equalities containing unknown quantities, there probably were no coins or wallets yet. But on the other hand, there were heaps, as well as pots, baskets, which perfectly suited the role of caches-storage, containing an unknown number of items. "A heap is being looked for, which together with two-thirds of it, half and one-seventh is 37 ...", the Egyptian scribe Ahmes taught in the II millennium BC. In the ancient mathematical problems of Mesopotamia, India, China, Greece, unknown values expressed the number of peacocks in the garden, the number of bulls in the herd, the totality of things taken into account when dividing property. Scribes, officials well trained in the science of counting, and priests initiated into secret knowledge were quite successful in coping with such tasks.
Sources that have come down to us testify that ancient scientists possessed some general methods of solving problems with unknown quantities. However, not a single papyrus or a single clay tablet contains a description of these techniques. The authors only occasionally supplied their numerical calculations with scanty comments such as: "Look!", "Do this!", "You found it right." In this sense, the exception is the "Arithmetic" of the Greek mathematician Diophantus of Alexandria (III century) - a collection of problems for the compilation of equations with a systematic presentation of their solutions.
However, the first widely known guide to solving problems was the work of a Baghdad scholar of the 9th century. Mohammed bin Musa al-Khwarizmi. The word "al-jabr" from the Arabic name of this treatise - "Kitab al-jerber wal-mukabala" ("The Book of Restoration and Opposition") - over time turned into the well-known word "algebra", and al-Khwarizmi's work itself served starting point in the formation of the science of solving equations.
equations. Algebraic equations
Basic definitions
In algebra, two types of equalities are considered - identities and equations.
Identity Is an equality that holds for all (valid) values of the letters included in it). To write an identity, a sign is also used along with a sign.
The equation Is an equality that holds only for some values of the letters included in it. The letters included in the equation, according to the condition of the problem, can be unequal: some can take all their permissible values (they are called parameters or coefficients equations and are usually denoted by the first letters of the Latin alphabet :,, ... - or by the same letters with indices:,, ... or,, ...); others whose values are to be found call unknown(they are usually denoted by the last letters of the Latin alphabet:,,, ... - or by the same letters with indices:,, ... or,, ...).
In general terms, the equation can be written as follows:
Depending on the number of unknowns, the equation is called an equation with one, two, etc. unknowns.
The value of the unknowns that turn the equation into identity, called solutions equations.
Solving an equation means finding a set of solutions to it, or proving that there are no solutions. Depending on the type of the equation, the set of solutions to the equation can be infinite, finite and empty.
If all solutions to the equation are solutions to the equation, then they say that the equation is a consequence of the equation, and write
Two equations
are called equivalent if each of them is a consequence of the other, and write
Thus, two equations are considered equivalent if the set of solutions to these equations coincide.
An equation is considered equivalent to two (or more) equations if the set of solutions to the equation coincides with the union of the sets of solutions to the equations,.
SOME EQUIVALENT EQUATIONS:
The equation is equivalent to the equation considered on the set of admissible values of the original equation.
Equivalent to two equations and.
An equation is equivalent to an equation.
The equation for odd n is equivalent to the equation, and for even n it is equivalent to two equations and.
Algebraic equation is called an equation of the form
where is a polynomial of degree n in one or several variables.
Algebraic Equation with One Unknown is called an equation that reduces to an equation of the form
where n is a non-negative integer; the coefficients of the polynomial,,, ...,, are called coefficients(or parameters) equations and are considered given; x is called unknown and is the desired one. The number n is called degree equations.
The values of the unknown x that make an algebraic equation an identity are called roots(less often decisions) of an algebraic equation.
There are several types of equations that can be solved using ready-made formulas. These are linear and quadratic equations, as well as equations of the form F (x), where F is one of the standard functions (power or exponential function, logarithm, sine, cosine, tangent or cotangent). Such equations are considered to be the simplest. There are also formulas for the cubic equation, but it is not considered to be the simplest one.
So, the main task in solving any equation is to reduce it to the simplest ones.
All the equations listed below also have their own graphical solution, which consists in representing the left and right sides of the equation as two identical functions of the unknown. Then a graph is built first of one function, and then another, and the point (s) of the intersection of the two graphs will give the solution (s) of the original equation. Examples of graphical solutions to all equations are given in the appendix.
Linear Equation
Linear equation is called an equation of the first degree.
where a and b are some real numbers.
A linear equation always has a single root, which is found as follows.
Adding a number to both sides of equation (1), we obtain the equation
equivalent to equation (1). Dividing both sides of equation (2) by the value, we obtain the root of equation (1):
Quadratic equation
Algebraic equation of the second degree.
, (3)
where,, are some real numbers, called quadratic... If, then the quadratic equation (3) is called given .
Quadratic roots are calculated by the formula
,
The expression is called discriminant quadratic equation.
Wherein:
if, then the equation has two different real roots;
if, then the equation has one real root of multiplicity 2;
if, then the equation has no real roots, but has two complex conjugate roots:
, 
,
Particular types of the quadratic equation (3) are:
1) The reduced quadratic equation (if), which is usually written in the form
.
The roots of the reduced quadratic equation are calculated by the formula
. (4)
This formula is called Vieta's formula - after the French mathematician of the late 16th century, who made a significant contribution to the formation of algebraic symbolism.
2) A quadratic equation with an even second coefficient, which is usually written as
(- integer).
It is convenient to calculate the roots of this quadratic equation by the formula
. (5)
Formulas (4) and (5) are particular forms of the formula for calculating the roots of a complete quadratic equation.
The roots of the reduced quadratic equation
related to its coefficients Vieta Formulas
,
.
If the given quadratic equation has real roots, Vieta's formulas make it possible to judge both the signs and the relative magnitude of the roots of the quadratic equation, namely:
if, then both roots are negative;
if, then both roots are positive;
if,, then the equation has roots of different signs, and the negative root is greater in absolute value than the positive one;
if,, the equation has roots of different signs, and the negative root is less than the positive root in absolute value.
Let's rewrite the quadratic equation
(6)
and show one more way how you can derive the roots of the quadratic equation (6) through its coefficients and a free term. If
then the roots of the quadratic equation are calculated by the formula
,
, .
which can be obtained as a result of the following transformations of the original equation, as well as taking into account formula (7).
,
Note that, therefore
,
.
,
but, from formula (7), therefore, finally
If we put that +, then
,
Note that, therefore
,
,
but, therefore finally
.
Two-term equations
Equations of the nth degree of the form
called two-term equation... With and replacement)
where is the arithmetic value of the root, equation (8) is reduced to the equation
The two-term equation for odd n has one real root. In the set of complex numbers, this equation has n roots (of which one is real and complex):
( 0, 1, 2, ..., ). (9)
The two-term equation for even n in the set of real numbers has two roots 
, and in the set of complex numbers there are n roots calculated by formula (9).
A two-term equation for even n has one real root, and in the set of complex numbers of roots calculated by the formula
( 0, 1, 2, ..., ). (10)
The two-term equation for even n has no real roots. In the set of complex numbers, the equation has roots calculated by formula (10).
Let us give a brief summary of the sets of roots of the two-term equation for some specific values of n.
The equation has two real roots.
.
The equation has two real roots and two complex roots.
The equation has no real roots. Complex roots:.
The equation has one real root and two complex roots
.
The equation has no real roots. Complex roots:
, 
.
Cubic Equations
If the mathematicians of Babylonia and Ancient India were able to solve quadratic equations, then cubic, i.e. equations of the form
turned out to be a tough nut to crack. At the end of the 15th century. Luca Pacioli, professor of mathematics at the universities of Rome and Milan, in his famous textbook "The Sum of Knowledge in Arithmetic, Geometry, Relations and Proportionality", put the problem of finding a general method for solving cubic equations on a par with the problem of squaring the circle. And yet, through the efforts of the Italian algebraists, such a method was soon found.
Let's start with simplification
If a cubic equation of general form
divided by, then the coefficient at becomes equal to 1. Therefore, in what follows, we will proceed from the equation
Just as the solution to the quadratic equation is based on the formula for the square of the sum, the solution to the cubic equation is based on the formula for the cube of the sum:
In order not to get confused in the coefficients, we replace here with and regroup the terms:
We see that by an appropriate choice, namely by taking, it is possible to achieve that the right-hand side of this formula will differ from the left-hand side of Eq. (11) only by the coefficient at and by the free term. Let us add equations (11) and (12) and give similar ones:
If we make a substitution here, we get a cubic equation for without a term with:
.
So, we have shown that in the cubic equation (11), using an appropriate substitution, one can get rid of the term containing the square of the unknown. Therefore, now we will solve an equation of the form
. (13)
Formula Cardano
Let's look at the sum cube formula again, but write it differently:
Compare this entry with equation (13) and try to establish a relationship between them. Even with a hint, it's not easy. We must pay tribute to the mathematicians of the Renaissance, who solved the cubic equation without knowing the letter symbols. Let's substitute in our formula:
Now it is already clear: in order to find the root of equation (13), it is enough to solve the system of equations
or 
and take as the amount and. By replacing, this system is reduced to a very simple form:
Then you can act in different ways, but all the "roads" will lead to the same quadratic equation. For example, according to Vieta's theorem, the sum of the roots of the reduced quadratic equation is equal to the coefficient at with a minus sign, and the product is equal to the free term. Hence it follows that and are the roots of the equation
.
Let's write out these roots:
The variables and are equal to the cubic roots of and, and the desired solution to the cubic equation (13) is the sum of these roots:
.
This formula is known as Cardano formula .
Trigonometric solution
, 
, 
. (14)
The roots,, of the "incomplete" cubic equation (14) are equal
, 
,
, 
,
.
Let the "incomplete" cubic equation (14) be valid.
a) If ("irreducible" case), then
,
,
.
(b) If, then
, ,
, 
.
(c) If, then
, 
,
, 
.
In all cases, the actual value of the cube root is taken.
Biquadratic equation
Algebraic equation of the fourth degree.
,
where a, b, c are some real numbers, called biquadratic equation... The substitution reduces the equation to a quadratic equation 
with the subsequent solution of two two-term equations and (and are the roots of the corresponding quadratic equation).
If and, then the biquadratic equation has four real roots:
, 
.
If,), then the biquadratic equation has two real roots and imaginary conjugate roots:
.
If and, then the biquadratic equation has four purely imaginary pairwise conjugate roots:
, .
Equations of the fourth degree
The method for solving equations of the fourth degree was found in the 16th century. Ludovico Ferrari, pupil of Gerolamo Cardano. It is called that - method Ferrari .
As in the solution of cubic and quadratic equations, in the equation of the fourth degree
you can get rid of the member by substitution. Therefore, we will assume that the coefficient at the cube of the unknown is zero:
Ferrari's idea was to represent the equation in the form, where the left side is the square of the expression, and the right side is the square of the linear equation from, the coefficients of which depend on. After that, it remains to solve two quadratic equations: and. Of course, such a representation is only possible with a special choice of the parameter. It is convenient to take it in the form, then the equation will be rewritten as follows:
. (15)
The right side of this equation is the quadratic trinomial of. It will be a perfect square when its discriminant is equal to zero, i.e.
, or
This equation is called resolvent(ie "permissive"). It is relatively cubic, and Cardano's formula allows you to find some of its roots. For, the right-hand side of Eq. (15) takes the form
,
and the equation itself is reduced to two square ones:
.
Their roots give all solutions to the original equation.
For example, let's solve the equation
Here it will be more convenient to use not ready-made formulas, but the very idea of the solution. We rewrite the equation as
and add an expression to both sides so that a complete square is formed on the left side:
Now let us equate the discriminant of the right-hand side of the equation to zero:
or, after simplification,
One of the roots of the resulting equation can be guessed by going through the divisors of the free term:. After substituting this value, we obtain the equation
where 
... The roots of the resulting quadratic equations are 
and 
... Of course, in the general case, complex roots can also be obtained.
Descartes-Euler solution
substitution is reduced to "incomplete" form
. (16)
The roots,,, of the "incomplete" fourth-degree equation (16) are equal to one of the expressions
in which combinations of signs are chosen so that the condition
moreover, and are the roots of the cubic equation
.
High degree equations
Solvability in radicals
The formula for the roots of the quadratic equation has been known since time immemorial, and in the 16th century. Italian algebraists have solved equations of the third and fourth degrees in radicals. Thus, it was found that the roots of any equation not higher than the fourth degree are expressed through the coefficients of the equation by a formula that uses only four arithmetic operations (addition, subtraction, multiplication, division) and the extraction of the roots of a degree not exceeding the degree of the equation. Moreover, all equations of a given degree () can be "served" by one general formula. Substituting the coefficients of the equation into it, we obtain all roots - both real and complex.
After that, the question naturally arose: are there similar general formulas for solving equations of the fifth degree and higher? The answer to it was found by the Norwegian mathematician Niels Henrik Abel at the beginning of the 19th century. Earlier this result was indicated, but insufficiently substantiated by the Italian Paolo Ruffini. The Abel-Ruffini theorem reads like this:
The general equation of power at is unsolvable in radicals.
Thus, there is no general formula applicable to all equations of a given degree. However, this does not mean that it is impossible to solve in radicals certain particular types of equations of high degrees. Abel himself found such a solution for a wide class of equations of arbitrarily high degree - the so-called Abelian equations. The Abel-Ruffini theorem does not even exclude the fact that the roots of each specific algebraic equation can be written in terms of its coefficients using the signs of arithmetic operations and radicals, in particular, that any algebraic number, i.e. the root of an equation of the form
with integer coefficients, can be expressed in radicals in terms of rational numbers. In fact, such an expression does not always exist. This follows from the theorem of solvability of algebraic equations, constructed by the outstanding French mathematician Évariste Galois in his "Memoir on conditions for the solvability of equations in radicals" (1832; published in 1846).
We emphasize that in applied problems we are only interested in the approximate values of the roots of the equation. Therefore, its solvability in radicals usually does not play a role here. There are special computational methods that make it possible to find the roots of any equation with any predetermined accuracy, no less than that given by calculations using ready-made formulas.
Equations being solved
They want high-degree equations in the general case insoluble in radicals, and Cardano's and Ferrari's formulas for equations of the third and fourth degrees do not work at school, in algebra textbooks, at university entrance exams, sometimes problems are encountered where it is required to solve equations higher than the second degree. Usually they are specially selected so that the roots of the equations can be found using some elementary techniques.
One of these techniques is based on the theorem on rational roots of a polynomial:
If an irreducible fraction is a root of a polynomial with integer coefficients, then its numerator is the divisor of the free term, and the denominator is the divisor of the leading coefficient.
To prove it, it is enough to substitute in the equation and multiply the equation by. We get
All the terms on the left side, except for the last one, are divisible by, therefore, it is divisible by, and since and are mutually prime numbers, it is a divisor. The proof for is similar.
Using this theorem, one can find all rational roots of an equation with integer coefficients by testing a finite number of "candidates". For example, for the equation
whose leading coefficient is 1, the "candidates" will be the divisors of the number –2. There are only four of them: 1, -1, 2 and –2. The check shows that only one of these numbers is a root:.
If one root is found, you can lower the degree of the equation. According to Bezout's theorem,
the remainder of dividing a polynomial by a binomial is equal, i.e.
The theorem immediately implies that
If is a root of a polynomial, then the polynomial is divisible by, i.e., where is a polynomial of degree 1 less than.
Continuing our example, we take from the polynomial
factor 
... To find the quotient, you can do the division "corner":
But there is also an easier way. It will become clear from the example:
Now it remains to solve the quadratic equation 
... Its roots:
.
Undefined coefficient method
If a polynomial with integer coefficients has no rational roots, you can try to decompose it into factors of lesser degree with integer coefficients. Consider, for example, the equation
We represent the left side as a product of two square trinomials with unknown (undefined) coefficients:
Let's open the brackets on the right side and give similar ones:
Now, equating the coefficients at the same degrees in both parts, we obtain the system of equations
An attempt to solve this system in general form would bring us back to solving the original equation. But whole roots, if they exist, are not difficult to find by selection. Without loss of generality, we can assume that, then the last equation shows that only two options need to be considered:, and. Substituting these pairs of values into the rest of the equations, we make sure that the first of them gives the required expansion:. This solution is called method of undefined coefficients .
If the equation has the form, where and are polynomials, then the replacement reduces its solution to the solution of two equations of lower degrees: and.
Return equations
A reversible algebraic equation is an equation of even degree of the form
in which the coefficients, equally spaced from the ends, are equal:, etc. Such an equation is reduced to an equation of half the degree by dividing by and subsequent replacement.
Consider, for example, the equation
Dividing it by (which is legal, since it is not a root), we get
.
notice, that
.
Therefore, the quantity satisfies the quadratic equation
,
solving which can be found from the equation 
.
When solving recurrent equations of higher degrees, they usually use the fact that the expression for any can be represented as a polynomial of degree from.
Rational algebraic equations
Rational an algebraic equation is an equation of the form
The set of admissible values of the rational algebraic equation (17)
is given by the condition, i.e.,, ..., where,, ..., are the roots of the polynomial.
The method for solving equation (17) is as follows. Solving the equation
whose roots are denoted by
.
Compare the sets of roots of the polynomials and. If no root of the polynomial is a root of the polynomial, then all the roots of the polynomial are the roots of equation (17). If any root of a polynomial is a root of a polynomial, then it is necessary to compare from the multiplicity: if the multiplicity of the root of the polynomial is greater than the multiplicity of the root of the polynomial, then this root is a root (17) with multiplicity equal to the difference of the multiplicities of the roots of the dividend and the divisor; otherwise, the root of the polynomial is not a root of the rational equation (17).
EXAMPLE Find the real roots of the equation
where 
, .
The polynomial has two real roots (both simple):
The polynomial has one simple root. Therefore, the equation has one real root.
Solving the same equation in the set of complex numbers, we find that the equation has, in addition to the indicated real root, two complex conjugate roots:
Irrational equations
An equation containing an unknown (or a rational algebraic expression of the unknown) under the radical sign is called irrational equation... In elementary mathematics, solutions to irrational equations are found in the set of real numbers.
Any irrational equation with the help of elementary algebraic operations (multiplication, division, raising to an integer power of both sides of the equation) can be reduced to a rational algebraic equation. It should be borne in mind that the resulting rational algebraic equation may not be equivalent to the original irrational equation, namely, it may contain "extra" roots that will not be the roots of the original irrational equation. Therefore, having found the roots of the resulting rational algebraic equation, it is necessary to check whether all the roots of the rational equation will be the roots of the irrational equation.
In the general case, it is difficult to indicate any universal method for solving any irrational equation, since it is desirable that, as a result of transformations of the original irrational equation, one obtains not just some rational algebraic equation, among the roots of which there will be the roots of this irrational equation, but a rational algebraic equation formed from polynomials of the least degree possible. The desire to obtain that rational algebraic equation formed from polynomials of the least degree possible is quite natural, since finding all the roots of a rational algebraic equation in itself can be a rather difficult task, which we can completely solve only in a very limited number of cases.
Here are some standard, most frequently used methods for solving irrational algebraic equations.
1) One of the simplest methods for solving irrational equations is the method of freeing from radicals by sequentially raising both sides of the equation to the corresponding natural power. It should be borne in mind that when raising both sides of the equation to an odd power, the resulting equation is equivalent to the original one, and when raising both sides of the equation to an even power, the resulting equation will, generally speaking, be nonequivalent to the original equation. This is easy to verify by raising both sides of the equation
to any even power. As a result of this operation, the equation is obtained
the set of solutions of which is the union of the sets of solutions:
AND 
.
However, despite this drawback, it is the procedure for raising both sides of the equation to a certain (often even) power that is the most common procedure for reducing an irrational equation to a rational equation.
where,, are some polynomials.
By virtue of the definition of the operation of extracting a root in the set of real numbers, the admissible values of the unknown are determined by the conditions
Squaring both sides of equation (18), we obtain the equation
After re-squaring, the equation turns into an algebraic equation
Since both sides of equation (18) were squared, it may turn out that not all roots of equation (19) will be solutions of the original equation, it is necessary to check the roots.
2) Another example of solving irrational equations is the method of introducing new unknowns, for which either a simpler irrational equation or a rational equation is obtained.
PRI me R 2. Solve the irrational equation
.
The set of valid values for this equation:
Putting, after substitution, we obtain the equation
or its equivalent equation
which can be viewed as a quadratic equation with respect to. Solving this equation, we get
Consequently, the set of solutions to the original irrational equation is the union of the sets of solutions of the following two equations:
, 
.
Raising both sides of each of these equations to a cube, we get two rational algebraic equations:
, 
.
Solving these equations, we find that this irrational equation has a single root.
In conclusion, we note that when solving irrational equations, one should not start solving an equation by raising both sides of the equations to a natural power, trying to reduce the solution of an irrational equation to solving a rational algebraic equation. First, you need to see if you can make some identical transformation of the equation, which can significantly simplify its solution.
. (20)
The set of valid values for this equation:. Let's make the following transformations of this equation:
.
,
at, the equation of solutions will not have;
at, the equation can be written in the form
.
For, this equation has no solutions, since for any, belonging to the set of admissible values of the equation, the expression on the left side of the equation is positive.
For, the equation has a solution
.
Taking into account that the set of admissible solutions of the equation is determined by the condition, we finally obtain:
When solving the irrational equation (20), it will be
.
For all other values, the equation has no solutions, that is, the set of its solutions is an empty set.
Equations containing the unknown under the absolute value sign
Equations containing the unknown under the absolute value sign can be reduced to equations not containing the absolute value sign using the definition of the modulus. So, for example, the solution to the equation
(21)
is reduced to solving two equations with additional conditions.
1) If, then equation (21) is reduced to the form
. (22)
Solutions to this equation:,. The condition is satisfied by the second root of the quadratic equation (22), and the number 3 is the root of equation (21).
2) If, equation (21) is reduced to the form
.
The roots of this equation will be the numbers 
and 
... First root 
does not satisfy the condition and therefore is not a solution to this equation (21).
Thus, the solutions of equation (21) will be the numbers 3 and.
Note that the coefficients of the equation containing the unknown under the sign of the absolute value can be selected in such a way that the solutions of the equation are all values of the unknown belonging to a certain interval of the numerical axis. For example, let's solve the equation
. (23)
Consider the numerical axis Ox and mark on it the points 0 and 3 (the zeros of the functions under the sign of the absolute value). These points will split the number axis into three intervals (Fig. 1):
1) For, equation (23) is reduced to the form
In the interval, the last equation has no solutions.
Similarly, for, equation (23) is reduced to the form
and has no solutions in between.
2) For, equation (23) is reduced to the form
,
that is, it turns into identity. Therefore, any value is a solution to equation (23).
Transcendental Equations
An equation that is not reducible to an algebraic equation using algebraic transformations is called transcendental equation ).
Simple transcendental equations are exponential, logarithmic, and trigonometric equations.
Exponential Equations
The exponential equation is called an equation in which the unknown is included only in exponents for some constant bases.
The simplest exponential equation, the solution of which reduces to solving an algebraic equation, is an equation of the form
where and are some positive numbers. The exponential equation (24) is equivalent to the algebraic equation
.
In the simplest case, when the exponential equation (24) has a solution
The set of solutions to an exponential equation of the form
where is some polynomial, is found as follows.
A new variable is introduced, and equation (25) is solved as algebraic with respect to the unknown. After that, the solution of the original equation (25) is reduced to the solution of the simplest exponential equations of the form (24).
EXAMPLE 1. Solve the equation
Writing the equation in the form
and introducing a new variable, we get a cubic equation for the variable:
It is easy to verify that this cubic equation has a single rational root and two irrational roots: and.
Thus, the solution of the original equation is reduced to the solution of the simplest exponential equations:
The last of the listed equations has no solutions. The set of solutions to the first and second equations:
Some simple performance levels:
1) Equation of the form
.
2) An equation of the form
by replacement reduces to the quadratic equation
.
3) Equation of the form
by replacement reduces to the quadratic equation
.
Logarithmic Equations
Logarithmic an equation is an equation in which the unknown enters in the form of an argument of a logarithmic function.
The simplest logarithmic equation is an equation of the form
, (26)
where is some positive number other than one, is any real number. Logarithmic equation (26) is equivalent to the algebraic equation
In the simplest case, when the logarithmic equation (26) has a solution
The set of solutions to a logarithmic equation of the form 
, where is some polynomial of the specified unknown, is found as follows.
A new variable is introduced, and equation (25) is solved as an algebraic equation for. After that, the simplest logarithmic equations of the form (25) are solved.
EXAMPLE 1. Solve the equation
With respect to the unknown, this equation is quadratic:
.
The roots of this equation are:,.
Solving logarithmic equations
we obtain solutions of the logarithmic equation (27):,.
In some cases, in order to reduce the solution of the logarithmic equation to the sequential solution of the algebraic and simplest logarithmic equations, it is necessary to first make suitable transformations of the logarithms included in the equation. Such transformations can be the transformation of the sum of the logarithms of two quantities into the logarithm of the product of these quantities, the transition from the logarithm with one base to the logarithm with another base, etc.
PRI me R 2. Solve the equation
In order to reduce the solution of this equation to the sequential solution of the algebraic and the simplest logarithmic equations, it is necessary first of all to bring all logarithms to one base (here, for example, to base 2). For this we use the formula
,
by virtue of which 
... Substituting into equation (28) instead of an equal value, we obtain the equation
Replacement 
this equation is reduced to a quadratic equation for the unknown:
.
The roots of this quadratic equation are:,. We solve the equations and 
:
,
PRI me R 3. Solve the equation
Converting the difference between the logarithms of two quantities into the logarithm of the quotient of these quantities:
we reduce this equation to the simplest logarithmic equation
.
Conclusion
Mathematics, like any other science, does not stand still, along with the development of society, people's views also change, new thoughts and ideas arise. And the XX century was no exception in this sense. The advent of computers made their own adjustments to the methods of solving equations and greatly facilitated them. But a computer may not always be at hand (exam, test), therefore, knowledge of at least the most important methods of solving equations must be known. The use of equations in everyday life is rare. They have found their application in many sectors of the economy and in almost all the latest technologies.
In this work, not all were presented, methods of solving equations and not even all of their types, but only the most basic ones. I hope that my essay can serve as a good reference material when solving certain equations. In conclusion, I would like to note that when writing this essay, I did not set myself the goal of showing all types of equations, but expounded only the material I have.
List of used literature
Chapters ed. M. D. Aksenova. Encyclopedia for children. Volume 11. Mathematics. - M .: Avanta +, 1998 .-- 688 p.
Tsypkin A.G. Ed. S. A. Stepanova. Handbook of Mathematics for High School. - Moscow: Nauka, 1980. - 400 p.
G. Korn and T. Korn. The Mathematics Handbook for Beginners and Engineers. - Moscow: Nauka, 1970. - 720 p.
) Under admissible the numerical values of the letters are understood for which all operations performed on the letters included in the equality are feasible. For example, the valid values of the letters included in the equality
there will be the following; for 
; for, for
) If a and b have different signs, then.
) A case similar to the one discussed above.
) Under algebraic transformations equations
Understand the following transformations:
1) adding the same algebraic expression to both sides of the equation;
2) multiplying both sides of the equation by the same algebraic expression;
3) raising both sides of the equation to a rational power.
In algebra, two types of equalities are considered - identities and equations.
Identity is an equality that holds for all (admissible) values of the letters included in it.
An equation is an equality that holds only for some values of the letters included in it.
The letters included in the equation can be unequal: some can take all their admissible values, which are called the coefficients (sometimes parameters) of the equation, others whose values you want to find are called unknowns of this equation (as a rule, they are denoted by the last letters of the Latin alphabet x , y, z, u, v, w, or the same letters with indices.
Equations are:
Quadratic equations
Rational equations
Equations containing a variable under the modulus sign
Irrational equations
Exponential Equations
Logarithmic Equations
Systems of equations:
Systems of rational equations
Systems of nonlinear equations
Symmetric systems
Mixed systems
Extraneous roots that have arisen in the process of transformations can be identified by verification. Of course, if all the transformations led us to a chain of equivalent equations, then verification is unnecessary. However, this is not always possible to achieve, it is easier to ensure that each equation in the chain is a consequence of the previous one, i.e. so that no root loss occurs. In this case, verification is part of the solution. It should be noted that it is often easier to make a check than to justify that it is not necessary. In addition, validation is a means of verifying the correctness of the calculations performed. Sometimes it is useful to do this: at each stage of solving the equation, determine the intervals in which the roots of the equation can be. All roots that do not belong to these gaps are extraneous and must be discarded. However, the rest of the roots still need to be checked by substitution in the original equation.
Every algebraic equation always has at least one solution, real or complex.
In analytic geometry, one equation with two unknowns is interpreted as a curve on a plane, the coordinates of all points of which satisfy the given Equation. One Equation with three unknowns is interpreted using a surface in three-dimensional space. With this interpretation, the solution of the system Equation coincides with the problem of finding the intersection points of lines, surfaces, etc. Equations with a large number of unknowns are interpreted using manifolds in n-dimensional spaces.
Welcome!
Equations of mathematical physics are partial differential equations, as well as some related equations of other types (integral, integro-differential, etc.), which leads to mathematical analysis of physical phenomena. The theory of the Equations of Mathematical Physics is characterized by the formulation of problems in such a form as is necessary in the study of a physical phenomenon. The circle of Equations of mathematical physics with the expansion of the field of application of mathematical analysis is also steadily expanding. When systematizing the results obtained, it becomes necessary to include in the theory of equations of mathematical physics equations and problems of a more general form than those that appear in the analysis of specific phenomena; however, even for such equations and problems, it is characteristic that their properties admit more or less visual physical interpretation.
Chemical equations - images of chemical reactions by means of chemical signs, chemical formulas, numbers and mathematical signs. The possibility of such a description of chemical reactions was pointed out in 1789 by A. Lavoisier, based on the conservation of mass by the law; however, chemical equations received universal application only in the first half of the 19th century.
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equation inequality mathematics
The concept of "equation" refers to the most important general mathematical concepts.
There are different interpretations of the concept of "equation".
AND I. Vilenkin et al. Gives a logical - mathematical definition of the equation. Let a set of algebraic operations be fixed on the set M, x is a variable on M; then an equation on the set M with respect to x is called a predicate of the form, where and are terms with respect to given operations, the notation of which includes a symbol. Similarly, an equation in two or more variables is defined.
Accepted in logic, the terms "term" and "predicate" correspond to such terms of school mathematics as "expression" and "sentence with a variable". Therefore, the following definition can be considered the closest to the given formal definition: "A sentence with a variable, which has the form of equality between two expressions with this variable, is called an equation." This definition is given in the textbook "Algebra and the beginning of analysis" by AN Kolmogorov and others. Equality with a variable is called an equation. The value of a variable at which equality with a variable turns into a true numerical equality is called the root of the equation.
Often, especially at the beginning of a systematic course in algebra, the concept of an equation is introduced by separating it from the algebraic method of solving problems. For example, in the textbook by Sh.A. Alimov et al. The concept of an equation is introduced on the basis of a word problem. The transition to the concept of an equation is carried out on the basis of the analysis of some formal features of the record, expressing the content of this problem in algebraic form: "An equality containing an unknown number, indicated by a letter, is called an equation." The indicated way of introducing the concept of an equation corresponds to another component of the concept of an equation - the applied one.
Another approach to the concept of an equation is obtained by compiling the domain of the equation and the set of its roots. For example, in the textbook by DK Fadeev "Letter equality, which does not necessarily turn into a true numerical equality with admissible sets of letters, is called an equation."
You can also find a third version of the definition, the role of which is manifested in the study of the graphical method for solving equations: "An equation is the equality of two functions."
Among all the types of equations studied in the course of mathematics, V.I. Mishin emphasizes a relatively limited number of basic types. these include: a linear equation with one unknown, a system of two linear equations with two unknowns, quadratic equations, the simplest irrational and transcendental.
Yu.M. Kolyagin and others are classified according to the form of functions representing the right and left sides of the equations:
The equation is called:
algebraic if and are algebraic functions;
transcendental if at least one of the functions is transcendental;
rational algebraic (or simply rational), if algebraic functions and rational;
irrational algebraic (or simply irrational), if at least one of the algebraic functions is irrational;
a rational whole if the function and the whole are rational;
fractional rational if at least one of the rational functions and fractional rational.
The equation, where is a polynomial of the standard form, is called linear (first degree), square (second degree), cubic (third degree) and, in general, degree, if the polynomial has, respectively, first, second, third and, in general, degree.
Several types of equations are studied at school. These include: linear equations with one unknown, quadratic equations, irrational and transcendental equations, rational equations. These types of equations are studied with great care, for them the execution of the solution algorithm is indicated and brought to automatism, the form in which the answer should be written is indicated.
Types of equations and methods of solution:
1) Linear equation
An equation with one variable is an equality that contains only one variable.
The root (or solution) of an equation is the value of a variable at which the equation turns into a true numerical equality.
Finding all the roots of an equation, or proving that they do not exist, means solving the equation.
Example 1: Solve an equation.
2) Quadratic equation
A quadratic equation is an equation of the form, where the coefficients a, b and c are any real numbers, and a? 0.
The roots of a quadratic equation are those values of a variable at which the quadratic equation turns into a true numerical equality.
Solving a quadratic equation means finding all its roots or establishing that there are no roots.
Example 2: Solve an equation
This equation can be solved either through Vieta's Theorem or through the discriminant.
Answer: x 1 = -1, x 2 = -2.
3) Rational equations
rational equations - equations of the form
where and are polynomials, but are equations of the form, where and are rational.
Example 3: Solve an equation
4) Irrational equations
Irrational equations are equations in which the variable is contained under the root sign or under the sign of the operation of raising to a fractional power.
Example 4: Solve an equation
Let's square both sides:
5) Exponential and logarithmic equations
When solving exponential equations, two main methods are used: a) transition from equation to equation; b) introduction of new variables. Sometimes you have to use artificial techniques.
Logarithmic equations are solved by three methods, that is, the transition from equation to equation - the consequence; the method of introducing new variables of the logarithm, that is, the transition from equation to equation.
And also in many cases, when solving a logarithmic equation, you have to use the properties of the logarithm of the product, quotient, degree, root.
