Equations with x to the power of 3. Various methods for solving equations of the third degree. Solution of the reciprocal cubic equation

We get the answer: \

Equation of the third degree online

Consider two examples of cubic equations that the equation calculator can easily solve with a detailed solution:

An example of a simple cubic equation

The first example will be simple:

49*x^3 - x = 0

After you click "Solve Equation!", you will get an answer with a detailed explanation:

Given the equation:

transform

Let's take out common factor x out of brackets

we get the equation

This is an equation of the form

using the discriminant.

The roots of the quadratic equation:

(0)^2 — 4 * (49) * (-1) = 196

Because D>

We get the final answer for -x + 49*x^3 = 0:

The second simple example of a cubic equation would be:

8 = (1/2 + 3*x)^3

We get a detailed solution:

Given the equation:

transform:

Let's take the common factor out of brackets

/ 2\ -9*(-1 + 2*x)*\7 + 12*x + 12*x / ——————————= 0 8

cubic equation

the right side of the equation is equal to zero, then the equation will have a solution if at least one of the factors on the left side of the equation is equal to zero.

Let's get ur-tion

we solve the resulting equations:

Transferring free terms (without x)

from the left to the right, we get:

Divide both parts of the equation by -9/4

We get the answer: x1 = 1/2

This is an equation of the form

The quadratic equation can be solved

using the discriminant.

The roots of the quadratic equation:

___ \/ D — b x2 = ——— 2*a ___ -b — \/ D x3 = ———- 2*a

where D = b^2 - 4*a*c is the discriminant.

(12)^2 — 4 * (12) * (7) = -192

Because D< 0, то уравнение

has no real roots

but there are complex roots.

x2 = (-b + sqrt(D)) / (2*a) x3 = (-b - sqrt(D)) / (2*a)

Then, the final answer is:

1 I*\/ 3 x2 = — — + ——- 2 3 ___ 1 I*\/ 3 x3 = — — — ——- 2 3

An example of a complex cubic equation

The third example would be more complex − reciprocal cubic equation online.

5*x^3 -8*x^2 - 8*x + 5 = 0

To solve such a reciprocal cubic equation, then enter this equation into the calculator:

Given the equation:

2 3 5 - 8*x - 8*x + 5*x = 0

transform

3 2 5*x + 5 - 8*x + 8 - 8*x - 8 = 0

3 3 2 2 5*x - 5*(-1) - 8*x - -8*(-1) - 8*x - 8 = 0 / 3 3\ / 2 2\ 5*\x - (-1 ) / - 8*\x - (-1) / - 8*(x + 1) = 0 / 2 2\ 5*(x + 1)*\x - x + (-1) / + -8*( x + 1)*(x - 1) - 8*(x + 1) = 0

Let's take the common factor 1 + x out of brackets

/ / 2 2\ \ (x + 1)*\5*\x - x + (-1) / - 8*(x - 1) - 8/ = 0

/ 2\ (1 + x)*\5 - 13*x + 5*x / = 0

we get the equation

This is an equation of the form

The quadratic equation can be solved

using the discriminant.

The roots of the quadratic equation:

___ \/ D — b x2 = ——— 2*a ___ -b — \/ D x3 = ———- 2*a

where D = b^2 - 4*a*c is the discriminant.

(-13)^2 — 4 * (5) * (5) = 69

Because D > 0, then the equation has two roots.

x2 = (-b + sqrt(D)) / (2*a) x3 = (-b - sqrt(D)) / (2*a)

We get the final answer for 5 - 8*x - 8*x^2 + 5*x^3 = 0:

13 \/ 69 x2 = — + —— 10 10 ____ 13 \/ 69 x3 = — — —— 10 10

Tags: the equation

The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Equations have been used by man since ancient times and since then their use has only increased. The cubic equation is a third order equation and has next view:

\ where \ A number \ is called the root of a cubic equation if, when it is substituted, the equation becomes a true equality.

Also read our article "Solve the equation online grade 9 solver"

This kind of equation always has 3 roots. Roots can be either real or complex. If the initial data allows you to choose one of the roots of the cubic equation \ then you can divide the cubic polynomial by \ and solve the resulting quadratic equation.

Suppose we are given an equation of the form:

Let's do the grouping:

After analyzing the equation, it is clear that \ is the root of the equation

Find the roots of the resulting square trinomial \

We get the answer: \

Where can I solve a 3rd degree equation online with a solver?

You can solve the equation on our website pocketteacher.ru. Free online solver will allow you to solve an online equation of any complexity in seconds. All you have to do is just enter your data into the solver.

How to solve a third degree equation

You can also watch the video instruction and learn how to solve the equation on our website. And if you still have questions, then you can ask them in our Vkontakte group: pocketteacher. Join our group, we are always happy to help you.

How to solve 3rd degree equations

The use of equations is widespread in our lives.

How to solve 3rd degree equations

They are used in many calculations, construction of structures and even sports. Equations have been used by man since ancient times and since then their use has only increased. The cubic equation is a third-order equation and has the following form:

\ where \ A number \ is called the root of a cubic equation if, when it is substituted, the equation becomes a true equality.

Also read our article "Solve the equation online grade 9 solver"

This kind of equation always has 3 roots. Roots can be either real or complex. If the initial data allows you to choose one of the roots of the cubic equation \ then you can divide the cubic polynomial by \ and solve the resulting quadratic equation.

Suppose we are given an equation of the form:

Let's do the grouping:

After analyzing the equation, it is clear that \ is the root of the equation

Find the roots of the resulting square trinomial \

We get the answer: \

Where can I solve a 3rd degree equation online with a solver?

You can solve the equation on our website pocketteacher.ru. Free online solver will allow you to solve an online equation of any complexity in seconds. All you have to do is just enter your data into the solver. You can also watch the video instruction and learn how to solve the equation on our website. And if you still have questions, then you can ask them in our Vkontakte group: pocketteacher. Join our group, we are always happy to help you.

Abbreviated multiplication formulas

Abbreviated multiplication formulas.

- Studying the formulas of abbreviated multiplication: the square of the sum and the square of the difference of two expressions; difference of squares of two expressions; the cube of the sum and the cube of the difference of two expressions; sums and differences of cubes of two expressions.

- Application of abbreviated multiplication formulas when solving examples.

To simplify expressions, factorize polynomials, and reduce polynomials to a standard form, abbreviated multiplication formulas are used. Abbreviated multiplication formulas you need to know by heart.

Let a, b R. Then:

1. The square of the sum of two expressions is the square of the first expression plus twice the product of the first expression and the second plus the square of the second expression.

(a + b)2 = a2 + 2ab + b2

2. The square of the difference of two expressions is the square of the first expression minus twice the product of the first expression and the second plus the square of the second expression.

(a - b)2 = a2 - 2ab + b2

5. Equations of the third and fourth degree

Difference of squares two expressions is equal to the product of the difference of these expressions and their sum.

a2 - b2 = (a-b) (a+b)

4. sum cube of two expressions is equal to the cube of the first expression plus three times the square of the first expression times the second plus three times the product of the first expression times the square of the second plus the cube of the second expression.

(a + b)3 = a3 + 3a2b + 3ab2 + b3

5. difference cube of two expressions is equal to the cube of the first expression minus three times the product of the square of the first expression and the second plus three times the product of the first expression and the square of the second minus the cube of the second expression.

(a - b)3 = a3 - 3a2b + 3ab2 - b3

6. Sum of cubes two expressions is equal to the product of the sum of the first and second expressions by the incomplete square of the difference of these expressions.

a3 + b3 = (a + b) (a2 - ab + b2)

7. Difference of cubes of two expressions is equal to the product of the difference of the first and second expressions by the incomplete square of the sum of these expressions.

Application of abbreviated multiplication formulas when solving examples.

Example 1

Calculate

a) Using the formula for the square of the sum of two expressions, we have

(40+1)2 = 402 + 2 40 1 + 12 = 1600 + 80 + 1 = 1681

b) Using the formula for the squared difference of two expressions, we obtain

982 \u003d (100 - 2) 2 \u003d 1002 - 2 100 2 + 22 \u003d 10000 - 400 + 4 \u003d 9604

Example 2

Calculate

Using the formula for the difference of the squares of two expressions, we obtain

Example 3

Simplify Expression

(x - y)2 + (x + y)2

We use the formulas for the square of the sum and the square of the difference of two expressions

(x - y)2 + (x + y)2 = x2 - 2xy + y2 + x2 + 2xy + y2 = 2x2 + 2y2

Abbreviated multiplication formulas in one table:

(a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 - 2ab + b2
a2 - b2 = (a - b) (a+b)
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a - b)3 = a3 - 3a2b + 3ab2 - b3
a3 + b3 = (a + b) (a2 - ab + b2)
a3 - b3 = (a - b) (a2 + ab + b2)

Algebraic equation of the fourth degree.

1. Reduction of the equation to the canonical form.

Let's make a change of variable according to the formula:

We get the equation:

Let's expand the brackets:

We get the equation:

Equation reduced to canonical form:

2.

"Solution of Equations of Higher Degrees". 9th grade

Equation solution

Method number 1.
Solving by Decomposing into Two Quadratic Equations

Consider the case when q not equal to zero.

True identity:

We got the equation:

Let's choose the parameter z so that the right side of this equation is a perfect square with respect to y . For this it is necessary and sufficient that the discriminant from the coefficients of the trinomial with respect to y , standing on the right, vanished: z to plus infinity, the value of the polynomial on the left side of the equation also tends to plus infinity, that is, it becomes positive at some positive z=M , and since continuous on the interval the function takes on the interval (0;M) any intermediate, including zero, value, then there is a positive root of this cubic equation. Such a positive root is either the first root in the program for solving a cubic equation, where the argument is under the cosine sign f/3 , as Cos(F/3)0 at 0F3/2*Pi , if the cubic equation has three different real roots, or the only real root of this cubic equation.

If any of the real roots of the cubic equation takes on a zero value, then the biquadratic equation is solved

Method number 2.
Descartes-Euler solution.

After bringing the algebraic equation of the fourth degree to the canonical form, the program finds three roots of the cubic equation

If this cubic equation has three real positive roots, then the fourth degree equation has four real roots.

If this cubic equation has three real roots, one positive and two negative, then the quartic equation has two pairs of complex conjugate roots.

If this cubic equation has one positive real root and two complex conjugate roots, then the quartic equation has two real and two complex conjugate roots. Javascript program "Solution of the equation of the fourth degree Ax4+Bx3+Cx2+Dx+E=0" Program "Solution of the equation of the fourth degree Ax4+Bx3+Cx2+Dx+E=0".Program code "Solution of the equation of the fourth degree Ax4+Bx3+ Cx2+Dx+E=0»Derivation of the roots of a cubic equation.To the main page.

A cubic equation is an algebraic equation of the third degree. General view of the cubic equation: ax3 + bx2 + cx + d = 0, a ≠ 0

Replacing x in this equation with a new unknown y associated with x by the equality x = y - (b / 3a), the cubic equation can be reduced to a simpler (canonical) form: y3 + pu + q = 0, where p = - b2 + c , q = 2b – bc + d

3a2 a 27a3 3a2 a the solution of this equation can be obtained using the Cardano formula.

1.1 History of cubic equations

The term "cubic equation" was introduced by R. Descartes (1619) and W. Outred (1631).

The first attempts to find solutions to problems that reduce to cubic equations were made by ancient mathematicians (for example, the problems of doubling a cube and trisecting an angle).

Mathematicians of the Middle Ages of the East created a fairly developed theory (in geometric form) of cubic equations; it is most detailed in the treatise on the proofs of problems in algebra and almukabala "Omar Khaya" (circa 1070), where the question of finding positive roots of 14 types of cubic equations containing in both parts only terms with positive coefficients is considered.

In Europe, for the first time in trigonometric form, the solution of one case of a cubic equation was given by Viet (1953).

The first solution in radicals of one of the types of cubic equations was found by S. Ferro (circa 1515), but it was not published. The discovery was independently repeated by Tartaglia (1535), indicating a rule for solving two other types of cubic equations. These discoveries were published in 1545 by G. Cardano, who mentioned the authorship of N. Tartaglia.

At the end of the XV century. Professor of Mathematics at the Universities of Rome and Milan Luca Pacioli in his famous textbook "The sum of knowledge in arithmetic, geometry, relations and proportionality" the problem of finding general method for solving cubic equations, he put it on a par with the problem of squaring a circle. And yet, through the efforts of Italian algebraists, such a method was soon found.

Let's start with simplification

If a general cubic equation ax3 + bx2 + cx + d = 0, where a ≠ 0, is divided by a, then the coefficient at x3 becomes equal to 1. Therefore, in what follows we will proceed from the equation x3 + Px2 + Qx + R = 0. ( one)

Just as the solution of a quadratic equation is based on the formula of the square of the sum, the solution of the cubic equation is based on the formula of the cube of the sum:

(a + b)3 = a3 + 3a2b + 3ab2 + b3.

In order not to get confused in the coefficients, here we replace a by x and rearrange the terms:

(x + b)3 = x3 + 3bx2 + 3b2x + b3. (2)

We see that in a proper way b, namely by taking b = P/3, it is possible to ensure that the right side of this formula will differ from the left side of the equation x3 + Px2 + Qx + R = 0 only by the coefficient at x and the free term. We add the equation x3 + Px2 + Qx + R = 0 and (x + b)3 = x3 + 3bx2 + 3b2x + b3 and give similar ones:

(x + b)3 + (Q - 3b2)x + R - b3 = 0.

If we make the change here y = x + b, we get a cubic equation for y without a term with y2: y3 + py + q = 0.

So, we have shown that in the cubic equation x3 + Px2 + Qx + R = 0, using an appropriate substitution, you can get rid of the term containing the square of the unknown. Therefore, now we will solve an equation of the form x3 + px + q = 0. (3)

1.2 History of the Cardano formula

The Cardano formula is named after J. Cardano, who first published it in 1545.

The author of this formula is Niccolò Tartaglia. He created this solution in 1535 specifically for participation in a mathematical competition, in which, of course, he won. Tartaglia, reporting the formula (in poetic form) to Cardano, presented only that part of the solution of the cubic equation in which the root has one (real) value.

Cardano's results in this formula refer to the consideration of the so-called irreducible case, in which the equation has three values ​​( real values, in those days there were no imaginary or even negative numbers, although there were attempts in this direction). However, contrary to the fact that Cardano indicated in his publication the authorship of Tartaglia, the formula is called by the name of Cardano.

1. 3 Cardano Formula

Now let's look at the sum cube formula again, but write it differently:

(a + b)3 = a3 + b3 + 3ab(a + b).

Compare this entry with the equation x3 + px + q = 0 and try to establish a connection between them. Substitute in our formula x = a + b: x3 = a3 + b3 + 3abx, or x3 - 3abx - (a3 + b3) = 0

Now it is already clear: in order to find the root of the equation x3 + px + q = 0, it is enough to solve the system of equations a3 + b3 = - q, a3 + b3 = - q, or

3аb \u003d - p, a3b3 \u003d - p 3,

3 and take as x the sum of a and b. By changing u = a3, v = b3 this system is reduced to a completely plain sight: and + v = - q, and v = - p 3.

Then you can act in different ways, but all the "roads" will lead to the same quadratic equation. For example, according to the Vieta theorem, the sum of the roots of the given quadratic equation is equal to the coefficient at x with a minus sign, and the product is the free term. This implies that and and v are the roots of the equation t2 + qt – (p/3)3 = 0.

Let's write out these roots: t1,2 = - q ± q 2 + p 3.

Variables a and b are equal to the cube roots from t1 and t2, and the desired solution of the cubic equation x3 + px + q = 0 is the sum of these roots: x = 3 - q + q 2 + p 3+ 3 - q - q 2 + p 3 .

This formula is known as the Cardano formula.

Solving Equations

Before looking at the Cardano formula in the work, let us explain how to find its other roots, if any, from one root of the cubic equation x3 + px + q = 0.

Let it be known that our equation has a root h. Then it left side can be decomposed into linear and square factors. This is done very simply. We substitute the expression of the free term through the root q \u003d - h3 - ph into the equation and use the formula for the difference of cubes:

0 \u003d x3 - h3 + px - ph \u003d (x - h) (x2 + hx + h2) + p (x - h) \u003d (x - h) (x2 + hx + h2 + p).

Now you can solve the quadratic equation x2 + hx + h2 + p = 0 and find the rest of the roots of this cubic equation.

So, we are fully armed and, it would seem, we can cope with any cubic equation. Let's try our hand.

1. Let's start with the equation x3 + 6x - 2 = 0

We substitute p = 6 and q = -2 into the Cardano formula and after simple reductions we get the answer: x = 3√4 - 3√2. Well, the formula is quite nice. Only the prospect of taking the factor x - (3√4 - 3√2) from the left side of the equation and solving the remaining quadratic equation with "terrible" coefficients to calculate other roots is not very inspiring. However, looking at the equation more closely, we can calm down: the function on the left side is strictly increasing and therefore can vanish only once. This means that the number found is the only real root of the equation.

y y \u003d x3 + 6x - 2

3√4 – 3√2 x

Rice. 1 The graph of the function y \u003d x3 + 6x - 2 crosses the x-axis at one point - 3√4 - 3√2.

2. The next example is the equation x3 + 3x - 4 = 0.

Cardano's formula gives x = 3 2 + √5 + 3 2 - √5.

As in the previous example, we see that this root is unique. But you don't have to be super insightful to look at the equation and guess its root: x = 1. We have to admit that the formula gave out the usual unit in such a bizarre form. By the way, it is not possible to simplify this cumbersome, but not devoid of elegance, expression by algebraic transformations - the cubic irrationality in it cannot be eliminated.

3. Well, now let's take an equation that obviously has three real roots. It is easy to compose it - just multiply three brackets of the form x - b. You just need to take care that the sum of the roots is equal to zero, because, according to the general Vieta theorem, it differs from the coefficient at x2 only in sign. The simplest set of such roots is 0, 1, and -1.

Let's apply the Cardano formula to the equation x (x - 1) (x + 1) = 0, or x3 - x = 0.

Assuming p = -1 and q = 0 in it, we get x = 3 √ - 1/27 + 3 - √ - 1/27.

y y \u003d x (x - 1) (x + 1)

Rice. 2 The equation x (x - 1) (x + 1) \u003d 0 has three real roots: -1, 0 and 1. Accordingly, the graph of the function y \u003d x (x - 1) (x + 1) intersects the x-axis at three points.

A negative number appeared under the square root sign. This also happens when solving quadratic equations. But the quadratic equation in this case does not have real roots, while the cubic one has three of them!

A closer analysis shows that we did not fall into this trap by chance. The equation x3 + px + q = 0 has three real roots if and only if the expression Δ = (q/2)2 + (p/3)3 under square root in the Cardano formula is negative. If Δ > 0, then there is one real root (Fig. 3b), and if Δ = 0, then there are two of them (one of them is double), except for the case p = q = 0, when all three roots merge.

y Δ 0 y \u003d -px - q y \u003d x3

0 x 0 x y \u003d -px - q y \u003d x3 a) b)

Rice. 3 The cubic equation x3 + px + q = 0 can be represented as x3 = -px - q. This shows that the roots of the equation will correspond to the abscissas of the intersection points of the two graphs: y \u003d x3 and y \u003d -px - q. If Δ 0 is one.

1.4 Vieta's theorem

Vieta's theorem. If an integer rational equation degree n, reduced to standard form, has n different real roots x1, x2,. xn, then they satisfy the equalities: x1 + x2 + + xn = - a1, a0 x1x2 + x1x3 + + xn-1xn = a2 a0 x1 x2 xn = (-1)nаn.

For the roots of the equation of the third degree a0x3 + a1x2 + a2x + a3 = 0, where a0 ≠ 0, the equalities x1 + x2 + x3 = - a1, a0 x1x2 + x1x3 + x2x3 = a2, a0 x1x2x3 = - a3 are valid.

1. 5 Bezout's theorem. Horner's scheme

The solution of equations is closely related to the factorization of polynomials. Therefore, when solving equations, everything that is connected with the selection of linear factors in the polynomial, i.e., with the division of the polynomial A (x) by the binomial x - α, is important. The basis of much knowledge about the division of the polynomial A (x) by the binomial x - α is a theorem belonging to the French mathematician Etienne Bezout (1730-1783) and bearing his name.

Bezout's theorem. The remainder of the division of the polynomial A (x) by the binomial x - α is equal to A (α) (i.e., the value of the polynomial A (x) at x = α).

Find the remainder after dividing the polynomial A(x) = x4 - 6x3 + 8 by x + 2.

Decision. According to the Bezout theorem, the remainder of the division by x + 2 is A (-2) \u003d (-2) 4 - 6 (-2) 3 + 8 \u003d 72.

A convenient way to find the values ​​of a polynomial for a given value of the variable x was introduced by the English mathematician Williams George Horner (1786-1837). This method was later called Horner's scheme. It consists in filling in some table of two lines. For example, to calculate A (-2) in the previous example, in the top line of the table we list the coefficients of this polynomial, written in the standard form x4 - 6x3 + 8 = x4 + (-6) x3 + 0 x2 + 0 x + 8.

We duplicate the coefficient at the highest degree in the bottom line, and before it we write the value of the variable x = -2, at which the value of the polynomial is calculated. This results in the following table:

Empty cells of the table are filled according to the following rule: the rightmost number of the bottom row is multiplied by -2 and added to the number above the empty cell. According to this rule, the first empty cell contains the number (-2) 1 + (-6) = -8, the second cell contains the number (-2) (-8) + 0 = 16, the third cell contains the number (- 2) 16 + 0 = - 32, in the last cell - the number (-2) (-32) + 8 = 72. The table completely filled out according to Horner's scheme looks like this:

2 1 -8 16 -32 72

The number in the last cell is the remainder of dividing the polynomial by x + 2, A(-2) = 72.

In fact, from the resulting table, filled in according to Horner's scheme, one can write down not only the remainder, but also the incomplete quotient

Q(x) \u003d x3 - 8x2 + 16x - 32, since the number on the second line (not counting from the last one) is the coefficients of the polynomial Q (x) - the incomplete quotient of division by x + 2.

Solve the equation x3 - 2x2 - 5x + 6 = 0

We write out all the divisors of the free term of the equation: ± 1, ± 2, ± 3, ± 6.

x=1, x=-2, x=3

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

2. CONCLUSION

I will formulate the main conclusions about the work done.

In the process of work, I got acquainted with the history of the development of the problem of solving an equation of the third degree. The theoretical significance of the results obtained lies in the fact that it deliberately takes the place of the Cardano formula in solving some equations of the third degree. I made sure that the formula for solving the equation of the third degree exists, but because of its cumbersomeness it is not popular and not very reliable, since it does not always reach the final result.

In the future, we can consider such questions: how to find out in advance what roots an equation of the third degree has; is it possible to solve a cubic equation graphically, if so, how; how to estimate approximately the roots of a cubic equation?

Introduction

1. Theoretical part

1.1 Basic concepts and definitions

1.3 Cardano formula

2. Problem solving

Conclusion

Introduction

Equations. It can be said for sure that there is not a single person who would not be familiar with them. From an early age, children begin to solve “problems with X”. Further more. True, for many, acquaintance with equations ends with school affairs. The famous German mathematician Courant wrote: “For more than two thousand years, the possession of some, not too superficial, knowledge in the field of mathematics was a necessary integral part in the intellectual inventory of each educated person". And among this knowledge was the ability to solve equations.

Already in ancient times, people realized how important it is to learn how to solve algebraic equations of the form

a0xn + a1xn ​​- 1 + ... + an = 0

after all, very many and very diverse questions of practice and natural science are reduced to them (of course, here we can immediately assume that a0 ¹ 0, since otherwise the degree of the equation is actually not n, but less). Many, of course, came up with the tempting idea to find formulas for any power of n that would express the roots of the equation in terms of its coefficients, i.e., would solve the equation in radicals. However, the "gloomy Middle Ages" turned out to be as gloomy as possible in relation to the problem under discussion - for seven whole centuries no one found the required formulas! Only in the 16th century, Italian mathematicians managed to move further - to find formulas for n \u003d 3 and 4. The history of their discoveries and even the authorship of the formulas found are rather obscure to this day, and we will not here clarify the complex relationship between Ferro, Cardano, Tartaglia and Ferrari, but let us state the mathematical essence of the matter better.

The purpose of the work is to explore various methods for solving equations of the third degree.

To achieve this goal, it is necessary to perform a number of tasks:

-Analysis scientific literature;

-Analysis of school textbooks;

-Selection of examples for solution;

-Solution of equations by various methods.

The work consists of two parts. The first deals with various methods for solving equations. The second part is devoted to solving equations in various ways.

1. Theoretical part

1 Basic concepts and definitions

A cubic equation is an equation of the third degree of the form:

The number x that turns the equation into an identity is called the root or solution of the equation. It is also the root of a polynomial of the third degree, which is on the left side of the canonical notation.

Over the field of complex numbers, according to the fundamental theorem of algebra, a cubic equation always has 3 roots (taking into account the multiplicity).

Since every real polynomial of odd degree has at least one real root, all possible cases of the composition of the roots of a cubic equation are limited to the three described below. These cases are easily distinguished using the discriminant

So there are only three possible cases:

If a? > 0, then the equation has three different real roots.

If a?< 0, то уравнение имеет один вещественный и пару комплексно сопряжённых корней.

If a? = 0, then at least two roots coincide. This can be when the equation has a double real root and another real root different from them; or, all three roots coincide, forming a root of multiplicity 3. The resultant of the cubic equation and its second derivative helps to separate these two cases: the polynomial has a root of multiplicity 3 if and only if the indicated resultant is also equal to zero.

The roots of a cubic equation are related to the coefficients as follows:

1.2 Methods for solving cubic equations

The most common method for solving cubic equations is the enumeration method.

First, by enumeration, we find one of the roots of the equation. The point is that cubic equations always have at least one real root, and the integer root of a cubic equation with integer coefficients is a divisor of the free term d. The coefficients of these equations are usually chosen so that the desired root lies among small integers, such as: 0, ± 1, ± 2, ± 3. Therefore, we will look for the root among these numbers and check it by substituting it into the equation. The success rate with this approach is very high. Let's assume this root.

The second stage of the solution is the division of the polynomial by the binomial x - x1. According to Bezout's theorem, this division without a remainder is possible, and as a result we get a polynomial of the second degree, which must be equated to zero. By solving the resulting quadratic equation, we will find (or not) the remaining two roots.

Solution of a two-term cubic equation

The two-term cubic equation has the form (2)

This equation is reduced to the form by dividing by a non-zero coefficient A. Next, the formula for the abbreviated multiplication of the sum of cubes is applied:

From the first bracket we find, and the square trinomial has only complex roots.

Recurrent cubic equations

The reciprocal cubic equation has the form and B-coefficients.

Let's group:

Obviously, x=-1 is the root of such an equation, and the roots of the resulting square trinomial are easily found through the discriminant.

1.3 Cardano formula

In the general case, the roots of a cubic equation are found using the Cardano formula.

For the cubic equation (1), the values ​​are found using the substitution: x= (2), and the equation is reduced to the form:

an incomplete cubic equation in which there will be no term containing the second degree.

We assume that the equation has complex numbers as coefficients. This equation will always have complex roots.

Let's denote one of these roots: . We introduce an auxiliary unknown u and consider the polynomial f(u)=.

Let's denote the roots of this polynomial through? and?, according to the Viette theorem (see p. 8):

Substitute in equation (3), expression (4), we obtain:

From the other side of (5): (7)

It follows from here, i.e. from formulas (6), (7), that the numbers are the roots of the equation:

From the last equation:

The other two roots are found by the formula:

1.4 Vieta trigonometric formula

This formula finds solutions to the reduced cubic equation, that is, an equation of the form

Obviously, any cubic equation can be reduced to an equation of the form (4) by simply dividing it by the coefficient a. So, the algorithm for applying this formula:

Calculate

2. Calculate

3. a) If, then compute

And our equation has 3 roots (real):

b) If, then replace trigonometric functions hyperbolic.

Calculate

Then the only root (real):

Imaginary roots:

C) If, then the equation has less than three different solutions:

2. Problem solving

Example 1. Find the real roots of a cubic equation

We apply the formula for abbreviated multiplication of the difference of cubes:

From the first bracket we find that the square trinomial in the second bracket has no real roots, since the discriminant is negative.

Example 2. Solve the equation

This equation is reciprocal. Let's group:

is the root of the equation. Finding the roots of a square trinomial

Example 3. Find the roots of a cubic equation

Let's transform the equation to the reduced one: multiply by both parts and make a change of variable.

The free term is 36. Let's write down all its divisors:

We substitute them in turn into equality until we get the identity:

Thus, is the root. It matches

Divide by using Horner's scheme.

Polynomial coefficients2-11129-0.52-11+2*(-0.5)=-1212-12*(-0.5)=189+18*(-0.5)=0

We get

Let's find the roots of the square trinomial:

Obviously, that is, its multiple root is.

Example 4. Find the real roots of the equation

is the root of the equation. Find the roots of a square trinomial.

Since the discriminant is less than zero, the trinomial has no real roots.

Example 5. Find the roots of the cubic equation 2.

Hence,

We substitute into the Cardano formula:

takes three values. Let's write them down.

When we have

When we have

When we have

Let's break these values ​​into pairs, which in the product give

The first pair of values ​​and

The second pair of values ​​and

The third pair of values ​​and

Back to the Cardano formula

Thus,

Conclusion

cubic trinomial equation

As a result of the execution term paper various methods for solving equations of the third degree were investigated, such as the enumeration method, Carano's formula, Vieta's formula, methods for solving reciprocal, two-term equations.

List of sources used

1)Bronstein I.N., Semendyaev K.A. "Handbook of mathematics for engineers and students of technical universities", M., 1986.

2)Kolmogorov A.N. Algebra and the beginnings of analysis. Study guide for 9th grade high school, 1977.

)Omelchenko V.P. Mathematics: tutorial/ V.P. Omelchenko, E.V. Kurbatova. - Rostov n / a.: Phoenix, 2005.- 380s.

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Learn how to solve cubic equations. The case when one root is known is considered. Methods for finding integer and rational roots. Application of the Cardano and Vieta formulas to solve any cubic equation.

Content

Here we consider the solution of cubic equations of the form
(1) .
Further, we assume that these are real numbers.

(2) ,
then dividing it by , we obtain an equation of the form (1) with coefficients
.

Equation (1) has three roots: , and . One of the roots is always real. We denote the real root as . The roots and can be either real or complex conjugate. Real roots can be multiple. For example, if , then and are double roots (or roots of multiplicity 2), and is a simple root.

If only one root is known

Let us know one root of the cubic equation (1). Let's denote the known root as . Then dividing equation (1) by , we obtain a quadratic equation. Solving the quadratic equation, we find two more roots and .

For the proof, we use the fact that the cubic polynomial can be represented as:
.
Then, dividing (1) by , we obtain a quadratic equation.

Examples of division of polynomials are presented on the page
“Division and multiplication of a polynomial by a polynomial by a corner and a column”.
The solution of quadratic equations is considered on the page
"The roots of a quadratic equation".

If one of the roots is

If the original equation is:
(2) ,
and its coefficients , , , are integers, then you can try to find an integer root. If this equation has an integer root, then it is a divisor of the coefficient. The method of searching for integer roots is that we find all the divisors of a number and check if equation (2) holds for them. If equation (2) is satisfied, then we have found its root. Let's denote it as . Next, we divide equation (2) by . We get a quadratic equation. Solving it, we find two more roots.

Examples of defining integer roots are given on the page
Examples of factorization of polynomials > > > .

Finding Rational Roots

If in equation (2) , , , are integers, and , and there are no integer roots, then you can try to find rational roots, that is, roots of the form , where and are integers.

To do this, we multiply equation (2) by and make the substitution:
;
(3) .
Next, we look for integer roots of equation (3) among the divisors of the free term.

If we have found an integer root of equation (3), then, returning to the variable , we obtain a rational root of equation (2):
.

Cardano and Vieta formulas for solving a cubic equation

If we do not know a single root, and there are no integer roots, then we can find the roots of a cubic equation using Cardano's formulas.

Consider the cubic equation:
(1) .
Let's make a substitution:
.
After that, the equation is reduced to an incomplete or reduced form:
(4) ,
where
(5) ; .

References:
I.N. Bronstein, K.A. Semendyaev, Handbook of Mathematics for Engineers and Students of Higher Educational Institutions, Lan, 2009.
G. Korn, Handbook of Mathematics for scientists and engineers, 2012.

The equation of the third degree with complex coefficients has the form:

We subject (1) to simplification - we make the term with the square of the unknown equal to zero, for which we set and find .

Thus, by substituting in (1), we obtain an incomplete cubic equation:

To find the roots of equation (2), we set , where u and v are two new auxiliary unknowns. (2) we write in the form:

opening the brackets and rearranging the terms, we get:

We require that or . This requirement is always feasible, because together with the condition it means that u and v are the roots of a quadratic equation.

Then equation (2) will be reduced to the equations:

Hence, according to the Vieta formulas, are the roots of the quadratic equation:

So, the incomplete equation (2) is solved in radicals:

(3) is the Cardan formula.

The Cardan formula consists of the sum of two cubic radicals. Each of them has three meanings. Combining values u and v, we get nine sums u + v, but among them there are only three roots of equation (2). These will be the amounts u + v, which u and v related by the ratio:

Denote by , any pair of values ​​satisfying (4), and by - one of the primitive roots of the third degree of unity. For example: .

Then , . Let's find . Since and , then

Where

Where .

Thus, we obtain all values ​​of the roots of the incomplete cubic equation (2):

Considering that , , we have: (5)

Example. Determine the roots of the equation using the Cardan formula:

Denote - the expression standing under the sign of the square radical in the Cardan formula.

Offer If , then equation (2) has three different roots.

Let us show that , , , where is the primitive third root of 1.

Let be , , . Raising both parts of the equation into a cube, we get: , i.e. the quadratic equation has two equal roots: , which is impossible, because the discriminant of this quadratic equation is . Then from formulas (5) , because at . If , then , i.e.

Which is impossible.

Similarly, it turns out that .

If for and , then

Since , then . Hence .

Whence one of the values: . Corresponding value :

Turning to formulas (5) we get:

Offer: For ( and ), equation (2) has two equal roots: , and in this case the roots (2) can be found without resorting to extracting the roots of the second and third degrees, namely: , (6)

Example: Solve the equation: .

EQUATIONS OF THE THIRD DEGREE WITH REAL COEFFICIENTS.

Let (7) be an incomplete third-degree cubic equation with real coefficients and .

Theorem: If , then equation (7) has one real and two imaginary conjugate roots;

if , then the roots of equation (7) are real and at least one of them is a multiple;

if , then all roots (7) are real and distinct.

one. . Since , then all three roots of equation (7) must be different.

Let's consider the expression.

Since , then - real number. Therefore, one of the values ​​must be valid. Let , then . Based on (5), equation (7) has only one real root: , and the other two roots will be conjugate purely complex numbers:

2. . For , , the equation has two equal roots. Since (7) is an equation with real coefficients, then for , , all three roots of the equation are real, and two of them are equal.

For , , equation (7) has three roots equal to zero: .

3. (irreducible case). Since , then , where . Then . Find the module and argument radical expression:

Assuming we get:

Work complex number to the conjugate is equal to the square of the modulus:

Those. , but . Means . Then

Then the roots (7) have the form:

So, in the case equation (7) has three real roots.

The disadvantage of Cardan's formula is that it often presents rational roots in an irrational way.

Example. Obviously a real root.

(one real and two conjugate imaginary roots)

According to Cardan's formula:- irrational numbers

With approximate calculations , . As a result of this shortcoming, the rational roots of a cubic equation with rational coefficients are not determined by the Cardan formula.

FOURTH POWER EQUATIONS.

Let (1) -

Equation of the fourth degree with complex coefficients. The earliest way to solve (1) belongs to Ferrari, a student of Cardan.

Let us choose an auxiliary unknown so that the right side of (2) turns into a full square. Which is possible provided that , where , , . If , comparing the coefficients at : , , , whence . Conversely, if , then .

Substituting expressions A, B, C into equality, we find that .

(3) - cubic resolution.

Let be some root of equation (3). Substituting into (2) on the right side we get a full square:

These two quadratic equations give all four roots of equation (1). So, the solution of an equation of the fourth degree is reduced to the solution of one equation of the third degree and two equations of the second degree, and is also solved in radicals. When finding the roots of an equation of type (1) by the Ferrari method, all transformations are carried out sequentially, without remembering the cubic resolution.

Example.

- (terms of degree no more than two), leaving