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How to solve 4th degree equations. Equation of the fourth degree. Solving biquadratic equations of the fourth degree

2. Equation If an equality includes a letter, then the equality is called an equation.
The equation can be true for some values ​​of this letter
and incorrect for its other meanings.

For example, the equation x + 6 = 7
true for x = 1
and false for x = 2.

3. Equivalent equations The linear equation is ax + by + c = 0.
For example: 5x – 4y + 6 = 0.
Let's express y:
⇒ 4y = 5x + 6 ⇒ y =

5x+6
4

⇒ y = 1.25x + 1.5.
The resulting equation, equivalent to the first one, has the form
y = kx + m,
where: x - independent variable (argument);
y - dependent variable (function);
k and m are coefficients (parameters).

4 Equivalent equations

The two equations are called equivalent (equivalent), if the sets of all their solutions coincide or both of them have no solutions and denote .

5/Equation of the first degree.

The first degree equation can be reduced to the form:

ax+b = 0,

Where x– variable, a And b– some numbers, and a ≠ 0.

From here it is easy to derive the value x:

b
x = – -
a

This is the meaning x is the root of the equation.

Equations of the first degree have one root.

Equation of the second degree.

The second degree equation can be reduced to the form:

ax 2 + bx + c = 0,

Where x– variable, a, b, c– some numbers, and a ≠ 0.

The number of roots of the second degree equation depends on the discriminant:

If D > 0, then the equation has two roots;

If D = 0, then the equation has one root;

If D< 0, то уравнение корней не имеет.

An equation of the second degree can have no more than two roots.

(about what a discriminant is and how to find the roots of an equation, see the sections “Formula for the roots of a quadratic equation. Discriminant” and “Another way to solve a quadratic equation”).

Equation of the third degree.

The third degree equation can be reduced to the form:

ax 3 + bx 2 + cx + d = 0,

Where x– variable, a, b, c, d– some numbers, and a ≠ 0.

An equation of the third degree can have no more than three roots.

Equation of the fourth degree.

The fourth degree equation can be reduced to the form:

ax 4 + bx 3 + cx 2 + dx + e = 0,

Where x– variable, a, b, c, d, e– some numbers, and a ≠ 0.

An equation of the third degree can have no more than four roots.

Summary:

1) equation of the fifth, sixth, etc. degrees can be easily derived independently by following the above diagram;


2) equation n- degree can have no more n roots.

6/An equation with one variable is an equality containing only one variable. The root (or solution) of an equation is the value of the variable at which the equation turns into a true numerical equality.

1. 8/-11/Systems linear equations: basic concepts System of linear equations.

Inconsistent and indefinite systems of linear equations. Set of linear equations. Consistent and inconsistent set of linear equations.

System of linear equations is a union of n linear equations, each of which contains k variables. It is written like this:

Many, when encountering higher algebra for the first time, mistakenly believe that the number of equations must necessarily coincide with the number of variables. In school algebra this usually happens, but for higher algebra this is generally not true.

Solving a system of equations is a sequence of numbers ( k 1 , k 2 , ..., k n), which is the solution to each equation of the system, i.e. when substituting into this equation instead of variables x 1 , x 2 , ..., x n gives the correct numerical equality.

Accordingly, solving a system of equations means finding the set of all its solutions or proving that this set is empty. Since the number of equations and the number of unknowns may not coincide, three cases are possible:

1. The system is inconsistent, i.e. the set of all solutions is empty. A rather rare case that is easily detected no matter what method is used to solve the system.

2. The system is consistent and defined, i.e. has exactly one solution. The classic version, well known since school.

3. The system is consistent and undefined, i.e. has infinitely many solutions. This is the toughest option. It is not enough to indicate that “the system has an infinite set of solutions” - it is necessary to describe how this set is structured.

Variable x i called permitted, if it is included in only one equation of the system, and with a coefficient of 1. In other words, in the remaining equations the coefficient of the variable x i must be equal to zero.

If we select one allowed variable in each equation, we obtain a set of allowed variables for the entire system of equations. The system itself, written in this form, will also be called resolved. Generally speaking, one and the same original system can be reduced to different permitted ones, but for now we are not concerned about this. Here are examples of permitted systems:

Both systems are variable-resolved x 1 , x 3 and x 4 . However, with the same success it can be argued that the second system is permitted relatively x 1 , x 3 and x 5 . It is enough to rewrite the very last equation in the form x 5 = x 4 .

Now let's consider a more general case. May we have everything k variables, of which r are allowed. Then two cases are possible:

1. Number of allowed variables r equal to the total number of variables k: r = k. We obtain the system from k equations in which r = k allowed variables. Such a system is joint and definite, because x 1 = b 1 , x 2 = b 2 , ..., x k = b k;

2. Number of allowed variables r less than total number of variables k: r < k. The rest ( k − r) variables are called free - they can take any values, from which allowed variables can be easily calculated.

So, in the above systems the variables x 2 , x 5 , x 6 (for the first system) and x 2 , x 5 (for the second) are free. The case when there are free variables is better formulated as a theorem:

Please note: this is very important point! Depending on how you write the resulting system, the same variable can be either allowed or free. Most higher mathematics tutors recommend writing out variables in lexicographic order, i.e. ascending index. However, you are under no obligation to follow this advice.

Theorem. If the system is from n equation variables x 1 , x 2 , ..., x r- permitted, and x r + 1 , x r + 2 , ..., x k- free, then:

1. If you set the values ​​of free variables ( x r + 1 = t r + 1 , x r + 2 = t r + 2 , ..., x k = tk), and then find the values x 1 , x 2 , ..., x r, we get one of the solutions.

2. If in two solutions the values ​​of free variables coincide, then the values ​​of allowed variables also coincide, i.e. solutions are equal.

What is the meaning of this theorem? To obtain all solutions to a resolved system of equations, it is enough to isolate the free variables. Then, assigning different values ​​to the free variables, we will obtain ready-made solutions. That's all - in this way you can get all the solutions of the system. There are no other solutions.

Conclusion: the resolved system of equations is always consistent. If the number of equations in a resolved system is equal to the number of variables, the system will be definite; if less, it will be indefinite.

Several equations form Set of equations

2. 12,13/ Linear inequality./ Strict and non-strict inequalities What is inequality? Any equation is taken, the "=" ("equals") sign is replaced with another sign ( > ;≥ ; < ; ≤ ; ≠ ) and an inequality is obtained.) The equation can be anything: linear, quadratic, fractional, exponential, trigonometric, logarithmic, etc. and so on. Accordingly, our inequalities will be linear, quadratic, etc.

What you need to know about inequality icons? Inequalities with icon more (> ), or less (< ) are called strict. With icons more or equal (≥ ), less or equal (≤ ) are called not strict. Icon not equal (≠ ) stands apart, but you also have to solve examples with this icon all the time. And we will decide.)

The icon itself does not have much influence on the solution process. But at the end of the decision, when choosing the final answer, the meaning of the icon appears in full force! This is what we will see below in examples. There are some jokes there...

Inequalities, like equalities, exist faithful and unfaithful. Everything is simple here, no tricks. Let's say 5 > 2 is a true inequality. 5 < 2 - incorrect.

Linear, quadratic, fractional, exponential, trigonometric and other inequalities are solved in different ways. Each type has its own method, its own special technique. But! All of these special techniques can be used only to someone standard view inequalities. Those. inequality of any kind must first prepare to use your method.

3. 14,16/Basic properties of inequalities/. Actions with two inequalities.

1) If

2) Property of transitivity. If

3) If you add the same number to both sides of a true inequality, you get a true inequality, i.e. If

4) If we transfer any term from one part of a true inequality to another, changing its sign to the opposite, then we get a true inequality, i.e. If

5) If both sides of a true inequality are multiplied by the same positive number, you get a true inequality. For example, if

6) If both sides of a true inequality are multiplied by the same negative number and change the inequality sign to the opposite, the result is a true inequality. For example, if

7) Similar to rules 5) and 6), the rules for dividing by the same number apply. If

Descartes-Euler solution

Having made the substitution, we obtain an equation in the following form (it is called “incomplete”):

y 4 + py 2 + qy + r = 0 .

Roots y 1 , y 2 , y 3 , y 4 of such an equation are equal to one of the following expressions:

in which combinations of characters are selected in such a way that the following relationship is satisfied:

,

and z 1 , z 2 and z 3 are roots cubic equation

Ferrari's solution

Main article: Ferrari method

Let's represent the fourth degree equation in the form:

Ax 4 + Bx 3 + Cx 2 + Dx + E = 0,

Its solution can be found from the following expressions:

if β = 0, solving u 4 + α u 2 + γ = 0 and, making the substitution , let's find the roots: . , (any sign square root will do), (three complex roots, one of which will do) Two ± s must have the same sign, ± t - are independent. In order to find all the roots, you need to find x for signed combinations ± s ,± t = +,+ for +,− for −,+ for −,−. Double roots will appear twice, triple roots three times, and quaternary roots four times. The order of the roots depends on which cube root U selected.

see also

  • Easily solved types of 4th degree equations: Biquadratic equation, reciprocal equation of the fourth degree

Literature

  • Korn G., Korn T. (1974) Handbook of Mathematics.

Links

  • Ferrari's decision

Wikimedia Foundation.

2010.

Goals:

  1. Systematize and generalize knowledge and skills on the topic: Solutions of equations of the third and fourth degree.
  2. Deepen your knowledge by completing a number of tasks, some of which are unfamiliar either in type or method of solution.
  3. Forming an interest in mathematics through the study of new chapters of mathematics, nurturing a graphic culture through the construction of graphs of equations.

Lesson type: combined.

Equipment: graphic projector.

Visibility: table "Viete's Theorem".

During the classes

1. Oral counting

a) What is the remainder of the division of the polynomial p n (x) = a n x n + a n-1 x n-1 + ... + a 1 x 1 + a 0 by the binomial x-a?

b) How many roots can a cubic equation have?

c) How do we solve equations of the third and fourth degrees?

d) If b even number in a quadratic equation, what is equal to D and x 1; x 2

2. Independent work(in groups)

Write an equation if the roots are known (answers to tasks are coded) “Vieta’s Theorem” is used

1 group

Roots: x 1 = 1; x 2 = -2; x 3 = -3; x 4 = 6

Make up an equation:

B=1 -2-3+6=2; b=-2

c=-2-3+6+6-12-18= -23; c= -23

d=6-12+36-18=12; d= -12

e=1(-2)(-3)6=36

x 4 -2 x 3 - 23x 2 - 12 x + 36 = 0(this equation is then solved by group 2 on the board)

Solution . We look for whole roots among the divisors of the number 36.

р = ±1;±2;±3;±4;±6…

p 4 (1)=1-2-23-12+36=0 The number 1 satisfies the equation, therefore =1 is the root of the equation. According to Horner's scheme

p 3 (x) = x 3 - x 2 -24x -36

p 3 (-2) = -8 -4 +48 -36 = 0, x 2 = -2

p 2 (x) = x 2 -3x -18=0

x 3 =-3, x 4 =6

Answer: 1;-2;-3;6 sum of roots 2 (P)

2nd group

Roots: x 1 = -1; x 2 = x 3 =2; x 4 =5

Make up an equation:

B=-1+2+2+5-8; b= -8

c=2(-1)+4+10-2-5+10=15; c=15

D=-4-10+20-10= -4; d=4

e=2(-1)2*5=-20;e=-20

8+15+4x-20=0 (group 3 solves this equation on the board)

р = ±1;±2;±4;±5;±10;±20.

p 4 (1)=1-8+15+4-20=-8

р 4 (-1)=1+8+15-4-20=0

p 3 (x) = x 3 -9x 2 +24x -20

p 3 (2) = 8 -36+48 -20=0

p 2 (x) = x 2 -7x +10 = 0 x 1 =2; x 2 =5

Answer: -1;2;2;5 sum of roots 8(P)

3 group

Roots: x 1 = -1; x 2 =1; x 3 = -2; x 4 =3

Make up an equation:

В=-1+1-2+3=1;В=-1

с=-1+2-3-2+3-6=-7;с=-7

D=2+6-3-6=-1; d=1

e=-1*1*(-2)*3=6

x 4 - x 3- 7x 2 + x + 6 = 0(group 4 solves this equation later on the board)

Solution. We look for whole roots among the divisors of the number 6.

р = ±1;±2;±3;±6

p 4 (1)=1-1-7+1+6=0

p 3 (x) = x 3 - 7x -6

р 3 (-1) = -1+7-6=0

p 2 (x) = x 2 - x -6 = 0; x 1 = -2; x 2 =3

Answer: -1;1;-2;3 Sum of roots 1(O)

4 group

Roots: x 1 = -2; x 2 = -2; x 3 = -3; x 4 = -3

Make up an equation:

B=-2-2-3+3=-4; b=4

c=4+6-6+6-6-9=-5; с=-5

D=-12+12+18+18=36; d=-36

e=-2*(-2)*(-3)*3=-36;e=-36

x 4 +4x 3 – 5x 2 – 36x -36 = 0(this equation is then solved by group 5 on the board)

Solution. We look for whole roots among the divisors of the number -36

р = ±1;±2;±3…

p(1)= 1 + 4-5-36-36 = -72

p 4 (-2) = 16 -32 -20 + 72 -36 = 0

p 3 (x) = x 3 +2x 2 -9x-18 = 0

p 3 (-2) = -8 + 8 + 18-18 = 0

p 2 (x) = x 2 -9 = 0; x=±3

Answer: -2; -2; -3; 3 Sum of roots-4 (F)

5 group

Roots: x 1 = -1; x 2 = -2; x 3 = -3; x 4 = -4

Write an equation

x 4+ 10x 3 + 35x 2 + 50x + 24 = 0(this equation is then solved by group 6 on the board)

Solution . We look for whole roots among the divisors of the number 24.

р = ±1;±2;±3

p 4 (-1) = 1 -10 + 35 -50 + 24 = 0

p 3 (x) = x- 3 + 9x 2 + 26x+ 24 = 0

p 3 (-2) = -8 + 36-52 + 24 = O

p 2 (x) = x 2 + 7x+ 12 = 0

Answer: -1;-2;-3;-4 sum-10 (I)

6 group

Roots: x 1 = 1; x 2 = 1; x 3 = -3; x 4 = 8

Write an equation

B=1+1-3+8=7;b=-7

c=1 -3+8-3+8-24= -13

D=-3-24+8-24= -43; d=43

x 4 - 7x 3- 13x 2 + 43x - 24 = 0 (this equation is then solved by group 1 on the board)

Solution . We look for whole roots among the divisors of the number -24.

p 4 (1)=1-7-13+43-24=0

p 3 (1)=1-6-19+24=0

p 2 (x)= x 2 -5x - 24 = 0

x 3 =-3, x 4 =8

Answer: 1;1;-3;8 sum 7 (L)

3. Solving equations with a parameter

1. Solve the equation x 3 + 3x 2 + mx - 15 = 0; if one of the roots is equal to (-1)

Write the answer in ascending order

R=P 3 (-1)=-1+3-m-15=0

x 3 + 3x 2 -13x - 15 = 0; -1+3+13-15=0

By condition x 1 = - 1; D=1+15=16

P 2 (x) = x 2 +2x-15 = 0

x 2 = -1-4 = -5;

x 3 = -1 + 4 = 3;

Answer: - 1; -5; 3

In ascending order: -5;-1;3. (b N S)

2. Find all the roots of the polynomial x 3 - 3x 2 + ax - 2a + 6, if the remainders from its division into binomials x-1 and x +2 are equal.

Solution: R=P 3 (1) = P 3 (-2)

P 3 (1) = 1-3 + a- 2a + 6 = 4-a

P 3 (-2) = -8-12-2a-2a + 6 = -14-4a

x 3 -Zx 2 -6x + 12 + 6 = x 3 -Zx 2 -6x + 18

x 2 (x-3)-6(x-3) = 0

(x-3)(x 2 -6) = 0

3) a=0, x 2 -0*x 2 +0 = 0; x 2 =0; x 4 =0

a=0; x=0; x=1

a>0; x=1; x=a ± √a

2. Write an equation

1 group. Roots: -4; -2; 1; 7;

2nd group. Roots: -3; -2; 1; 2;

3 group. Roots: -1; 2; 6; 10;

4 group. Roots: -3; 2; 2; 5;

5 group. Roots: -5; -2; 2; 4;

6 group. Roots: -8; -2; 6; 7.

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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 powers.

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 you can see that in the left and right side 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 transform:
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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Soon after Cardano published a method for solving cubic equations, his students and followers found ways to reduce the general equation of the fourth degree to a cubic equation. Let us present the simplest method, which belongs to L. Ferrari.

When presenting the method, you will need to use the following elementary lemma.

Lemma. In order to quadratic trinomial was the square of a linear binomial, it is necessary and sufficient that its discriminant be equal to zero.

Proof. Necessity. Let . Then Sufficiency. Let Then

The idea of ​​the presented method is to present the left side of the equation as the difference of two squares. Then it can be decomposed into two factors of the second degree, and the solution of the equation will lead to the solution of two quadratic equations. To achieve the goal, let's represent the left side in the form:

Here y is an auxiliary unknown, which must be selected so that the expression in square brackets turns out to be the square of a linear binomial. By virtue of the lemma, for this it is necessary and sufficient to satisfy the condition

This condition is an equation of the third degree with respect to y. After opening the parentheses, it is converted to the form

Let be one of the roots of this equation. Then the condition will be satisfied, so it holds

for some k and I. The original equation takes the form

Equating each of the factors to zero, we will find the four roots of the original equation.

Let's make one more remark. Let be the roots of the first factor, and let be the roots of the second. Then, adding these equalities, we get that

Thus, we have obtained an expression for the root of the auxiliary cubic equation in terms of the roots of the original equation of the fourth degree.

Example. Solve the equation. According to the method outlined above, we transform the left side:

Now let's put . After formations we get the equation

It is easy to see that one of the roots of this equation is the number . Substituting it into the transformed left side of the original equation, we get:

Equating the factors to zero, we get

As for equations higher than the fourth degree, some classes of equations of a relatively particular form were known, admitting algebraic solutions in radicals, i.e. in the form of the results of arithmetic operations and the action of extracting the root. However, attempts to provide a solution general equations fifth degree and higher were unsuccessful, until, finally, at the beginning of the 19th century. Ruffini and Abel did not prove that a solution of this kind for general equations above the fourth degree is impossible. Finally, in 1830, the brilliant French mathematician E. Galois managed to find the necessary and sufficient conditions(which are quite difficult to verify) for the solvability of a specific equation in radicals. At the same time, Galois created and used the theory of permutation groups, which was new for his time.


Heading: Tolstoy L.N.
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