Equation solution in 3 powers. Solution of equations of the third degree. Cardano and Vieta formulas for solving a cubic equation

Without the help of a script, you will have to perform rather complex calculations using the Cardano method, which includes at least 6 steps. The calculation begins by reducing the original equation to the form y³ + py + q = 0, etc.

The calculation of equations of the third degree is in demand in solving many fundamental and applied mathematical, physical, statistical, research and engineering problems.

Equation of the third degree online

The cubic equation has the form:

$$ x^3 + a \cdot x^2 + b \cdot x +c =0 $$

where a, b, c are numerical coefficients at x.

x is a variable whose value, which turns the cubic polynomial into an identity, will be the root of the cubic equation.

In order to solve a cubic equation online, you need to set the coefficients of the equation one by one.

A cubic equation can have three real roots, or one (or two for the degenerate case) and two complex conjugate roots.

An equation has three real roots if $$R^2< Q^3$$

$$ R $$ is found by the following formula:

$$ Q $$ can be found using the formula:

If $$ R^2< Q^3 $$ , то уравнение имеет три действительных корня:

If $$ R^2 >= Q^3 $$ , then the equation has one real root (or two, for degenerate cases) and two complex conjugates:

Function y = x³ and its graph

Let's make a table of values of the function y = x 3: We see that for x > 0 and y > 0 (the cube of a positive number is positive), and for x< 0 и y < 0 (куб отрицательного числа отрицателен). Следовательно, график расположится на coordinate plane in I and III quarters. Let's replace the value of the argument x with the opposite value -x , then the function will take opposite meaning; since if y = x 3 then

This means that each point (x; y) of the graph corresponds to a point (–x; –y) of the same graph, located symmetrically with respect to the origin.

Thus, the origin of coordinates is the center of symmetry of the graph.

The graph of the function y \u003d x 3 is shown in drawing 81. This line is called a cubic parabola.

In the first quarter, the cubic parabola (for x > 0) "steeply" rises (the value of y "rapidly" increases with increasing x, see table), for small values of x, the line "closely" approaches the abscissa axis (for "small" values x value of y is “very small”, see table). The left side of the cubic parabola (in the third quadrant) is symmetrical to the right side with respect to the origin.

A neatly drawn graph can serve as a means of approximate raising numbers to a cube. So, for example, setting x = 1.6, we find from the graph y ≈ 4.1.

Special tables have been compiled for the approximate calculation of cubes.

Such a table is also available in the manual by V. M. Bradis “Four-digit mathematical tables”.

This table contains approximate values of cubes of numbers from 1 to 10, rounded to 4 significant figures.

The structure of the table of cubes and the rules for using it are the same as the tables of squares. However, when the number increases (or decreases) by 10, 100, etc. Once its cube increases (or decreases) by 1000, etc. times. So, when using the table of cubes, you need to keep in mind the following comma wrapping rule:

If you move a comma to several digits in a number, then in the cube of this number you need to move the comma in the same direction to a triple number of digits.

Let's explain with examples:

1) Calculate 2.2353. According to the table we find: 2.233 ≈ 11.09; add correction 8 to the last digit to the last digit: 2.2353 ≈ 11.17.

2) Calculate (-179.8) 3 . Since (–a) 3 = –a 3 , we find (179.8) 3 .

According to the table, we find 1.798 3 ≈ 5.813, moving the comma, we get 179.8 3 ≈.

Hence, (–179.8) 3 ≈ –.

Approximate formulas. If in identity

(1 ± α)³ ≈ 1 ± 3α ± 3α² ± α³

the number α is small compared to unity, then discarding the terms with α² and α³, we obtain approximate formulas:

Using these formulas, it is easy to find approximate cubes of numbers close to one, for example:

1.02³ ≈ 1 + 3 * 0.02 = 1.06; exact cube: 1.061208;

1.03³ ≈ 1 + 3 * 0.03 = 1.09; exact cube: 1.092727;

0.98³ ≈ 1 - 3 * 0.02 = 0.94; exact cube: 0.941192;

0.97³ ≈ 1 - 3 * 0.03 = 0.91; exact cube: 0.912673.

Raising numbers into a cube on a counting ruler. For raising numbers to a cube, there is a scale of cubes on the body of the ruler K. The scale of cubes consists of three parts: left, middle and right (see Fig. 82); each of these parts is a basic D scale, but scaled down by a factor of three.

The value of the number raised to the cube is marked with a sight on the main scale D, and the result is read on the scale of cubes K.

For example, 2³ = 8 (see Fig. 39).

A few examples of cubed numbers are shown in the following table. For comparison, the values of the cubes of the same numbers, calculated from four-digit tables, are given.

Solution of cubic equations.

Any cubic equation with real coefficients has at least one real root, the other two are either also real or are complex conjugates.

Let's start the review with the simplest cases - binomial and returnable equations. Then we turn to finding rational roots (if any). We finish with an example of finding the roots of a cubic equation using Cardano's formula for the general case.

Page navigation.

Solution of a two-term cubic equation.

The two-term cubic equation has the form.

This equation is reduced to the form by dividing by a non-zero factor 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.

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 its discriminant is negative.

Solution of the reciprocal cubic equation.

The reciprocal cubic equation has the form, where A and B are coefficients.

Obviously, x \u003d -1 is the root of such an equation, and the roots of the resulting square trinomial are easily found through the discriminant.

Solve a cubic equation.

This equation is reciprocal. Let's group:

Obviously, x = -1 is the root of the equation.

We find the roots of a square trinomial:

Solution of cubic equations with rational roots.

Let's start with the simplest case, when x=0 is the root of a cubic equation.

In this case, the free term D is equal to zero, that is, the equation has the form.

If we take x out of brackets, then the square trinomial remains in brackets, the roots of which are easy to find either through the discriminant or using the Vieta theorem.

Find the real roots of the equation.

x=0 is the root of the equation. Find the roots of a square trinomial.

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

If the coefficients of a cubic equation are integers, then the equation can have rational roots.

For , multiply both sides of the equation by and change the variables y = Ax:

We arrived at the above cubic equation. It can have integer roots that are divisors of the free term. So we write out all the divisors and start substituting them into the resulting equation until an identical equality is obtained. The divisor for which the identity is obtained is the root of the equation. Therefore, the root of the original equation is.

Find the roots of a cubic equation.

Let's transform the equation to the given one: multiply by both parts and change the variable y = 2x .

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

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

So y = -1 is the root. It matches him.

It remains to find the roots of the square trinomial.

Obviously, that is, its multiple root is x=3.

This algorithm can be used to solve reciprocal equations. Since -1 is the root of any recurrent cubic equation, we can divide the left side of the original equation by x + 1 and find the roots of the resulting square trinomial.

In the case when the cubic equation does not have rational roots, other methods of solving are used, for example, specific methods of factoring a polynomial into factors.

Solution of cubic equations by Cardano's formula.

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

For a cubic equation, the values are found. Next, we find and

We substitute the obtained p and q into the Cardano formula:

The values of cube roots should be taken such that their product is equal. As a result, we find the roots of the original equation by the formula.

We solve the previous example using the Cardano formula.

How to solve cubic equations

Cubic equations have the form ax 3 + bx 2 + cx + d= 0. A method for solving such equations has been known for several centuries (it was discovered in the 16th century by Italian mathematicians). Solving some cubic equations is quite difficult, but with the right approach (and a good level of theoretical knowledge) you can solve even the most complex cubic equations.

Steps Edit

Method 1 of 3:

Solving with a formula for solving a quadratic equation Edit

Method 2 of 3:

Finding entire solutions by factoring Edit

Cubic Equations

where $a\ne 0,\ b,\ c,\ d$ are some numbers.

A cubic equation always has at least one root $x_1$ .

Hence, it always holds: $ax^3+bx^2+cx+d=a(x-x_1)(x^2+mx+n)$ , where $m, n$ are some numbers.

for any number $a$ have a single root

The solution to the equation $x^3=-8$ is $x=\sqrt=-2$ .

$>$ Cubic equations of the form $ax^3+bx^2+cx+d=0$ in some cases can be solved by factoring the left side.

Solve the equation $5x^3-x^2-20x+4=0$ .

Group the terms on the left side and factorize it: \[(5x^3-20x)-(x^2-4)=0 \quad \Leftrightarrow \quad 5x(x^2-4)-(x^2- 4)=0 \quad \Leftrightarrow \quad (x^2-4)(5x-1)=0\]

Then the roots of this equation are $x_1=-2, x_2=2, x_3=\frac15$ .

In some problems, abbreviated multiplication formulas may be useful:

$>$ Cubic equations of the form $ax^3+bx^2+cx+d=0$ , in which it is not possible to factorize the left side, can be solved in another way: choose a rational root, if there is one.

The following assertions can be used for this:

$\blacktriangleright$ If the sum is $a+b+c+d=0$ , then the root of the equation is the number $1$ .

$\blacktriangleright$ If $b+d=a+c$ , then the root of the equation is the number $-1$ .

$\blacktriangleright$ Let $a,b,c,d$ be $>>$ numbers. Then if the equation has a rational root $\large >$ , then the following will be true for it:

$d$ is evenly divisible by $p$ ; $a$ is evenly divisible by $q$ .

1. The equation $7x^3+3x^2-x-9=0$ has the sum of the coefficients equal to $7+3-1-9=0$ , so $x=1$ is the root ( not necessarily the only one) of this equation.

2. The equation $4.5x^3-3x^2-0.5x+7=0$ has: $4.5-0.5=-3+7$ , so $x= -1$ is the root of this equation.

3. The equation $2x^3+5x^2+3x-3=0$ has integer coefficients, so you can choose the root: divisors of the free term $-3$ : $\pm 1, \pm 3 $ ; leading coefficient divisors $2$ : $\pm1, \pm2$ . So the possible combinations of rational roots are: \[\pm 1, \ \pm\dfrac12, \ \pm 3, \ \pm \dfrac32\]

Substituting each number in turn into the equation, we make sure that $x=\frac12$ is the root (because after substituting this number into the equation, it turns into a true equality):

Note that if the equation has coefficients - rational numbers, then by multiplying the equation by their common denominator, one can obtain an equivalent equation with integer coefficients. For example, the equation $\frac12x^3+\frac16x+2=0$ after multiplication by $6$ reduces to an equation with integer coefficients: $3x^3+x+12=0$ .

Find the root of the equation $(2x + 1)^3 = 27$ . If the equation has more than one root, write down the larger one in your answer.

The original equation $(2x + 1)^3 = 3^3$ is of standard form, it is equivalent to the equation $2x + 1 = 3$ , from which we conclude that $x = 1$ is suitable according to the ODZ.

Find the root of the equation $(2x + 1)^3 = -27$ . If the equation has more than one root, write down the larger one in your answer.

ODZ: $x$ - arbitrary. Let's decide on ODZ:

The original equation $(2x + 1)^3 = (-3)^3$ is of standard form, it is equivalent to the equation $2x + 1 = -3$ , whence we conclude that $x = -2$ – suitable for ODZ.

Find the root of the equation $(3x + 2)^3 = -64$ . If the equation has more than one root, write down the larger one in your answer.

ODZ: $x$ - arbitrary. Let's decide on ODZ:

The original equation $(3x + 2)^3 = (-4)^3$ is of standard form, it is equivalent to the equation $3x + 2 = -4$ , from which we conclude that $x = -2$ – suitable for ODZ.

Find the root of the equation $(7x + 11)^3 = 64$ . If the equation has more than one root, write down the larger one in your answer.

ODZ: $x$ - arbitrary. Let's decide on ODZ:

The original equation $(7x + 11)^3 = 4^3$ is of standard form, it is equivalent to the equation $7x + 11 = 4$ , from which we conclude that $x = -1$ is suitable according to the ODZ.

Find the root of the equation $(-x - 11)^3 = 216$ . If the equation has more than one root, write down the larger one in your answer.

ODZ: $x$ - arbitrary. Let's decide on ODZ:

The original equation $(-x - 11)^3 = 6^3$ is of standard form, it is equivalent to the equation $-x - 11 = 6$ , from which we conclude that $x = -17$ is suitable for ODZ.

Solve the equation $8x^3-36x^2+54x-27=0$ .

Note that the left side is the difference cube: \[(2x)^3-3\cdot (2x)^2\cdot 3+3\cdot (2x)\cdot3^2-3^3=0\quad\ Leftrightarrow\quad (2x-3)^3=0\quad\Leftrightarrow\quad x=\frac32.\]

Find the larger root of the equation $8x^3+12x^2+6x+1=0$ .

Note that the left side is the cube of the sum: \[(2x)^3+3\cdot (2x)^2\cdot 1+3\cdot (2x)\cdot1^2+1^3=0\quad\ Leftrightarrow\quad (2x+1)^3=0\quad\Leftrightarrow\quad x=-\frac12.\]

In the Unified State Examination, cubic equations are found both in the profile and in the basic level. This means that every student needs to be able to correctly solve such tasks. Some may say that the number of points in the exam for solving equations of the third degree is small and it is inappropriate to spend time on them. It is difficult to agree with this. Firstly, in the exam, every score is extremely important, and secondly, the equations of the third degree are not so complicated if they are given due attention during preparation. In order for the student to be able to quickly and, most importantly, correctly complete such tasks, it is worth using our educational resource.

Shkolkovo is a unique platform that allows graduates from Moscow and other regions with any level of mathematical knowledge to learn how to solve cubic equations and effectively prepare for passing the exam. First of all, we recommend that you start by reviewing or studying the theoretical material on the topic. "Shkolkovo" presents to the attention of students from Moscow and other cities who are preparing for the exam, in fact, the author's manual, which clearly and easily presents the material on the topic "Cubic Equations".

In addition to presenting the basic definitions and formulas, you will be able to get acquainted with examples on the topic and learn how to solve them. It should be noted that our experts have selected very interesting options. In order for you to learn how to confidently solve exam problems, you need practice. Therefore, we recommend that you then go to the "Catalogue" section and proceed to independent work with equations of the third degree.

X to the third power

The function y is equal to x cubed

Properties of the function y equals x cubed

The function y equals x cubed has the following properties:

2. The function y equals x cubed increases along the entire number line;

3. The domain of the function y = x 3 - the entire number line;

4. The set of values of the function of the function y = x 3 is the entire number line.

The graph of the function y is equal to x cubed

The graph of the function y \u003d x 3 is called a cubic parabola:

You can build a graph of the function y \u003d x 3 yourself right now using the graph builder. Select in it the type of function "Power: y \u003d k * x n + b", specify the value of "n" equal to three and click the "Build Graph" button.

The function y = x 3 is a special case of a power function.

These are the properties and the graph of the function y is equal to x cubed.

cubic equation

Solution of a cubic equation using the Vieta formula. Created at the request of the user.

The canonical form of the cubic equation:

We will solve the cubic equation using the Vieta formula.

Vieta formula - a way to solve a cubic equation of the form

The calculator is below, and the description of the Vieta formula is below it

cubic equation

By the way, for some reason, other sites use the Cardano formula to solve cubic equations, but I agree with Wikipedia that the Vieta formula is more convenient for practical use. So why is Cardano's formula everywhere - it is not clear, except that people are too lazy to implement Hyperbolic functions and Inverse hyperbolic functions. Well, I was not lazy.

So, Vieta's formula (from Wikipedia)

Note that in the Vieta representation, a is the second coefficient, and the coefficient before x3 is always considered equal to 1. The calculator allows you to enter a as a coefficient in front of x3, but immediately divides the equation by it to get 1

If S > 0, then we calculate:

If S< 0, то заменяем trigonometric functions hyperbolic. There are two possible cases depending on the sign of Q

(pair of complex roots)

If S = 0, then the equation is degenerate and has less than 3 different solutions (second root of multiplicity 2):

According to these formulas, the calculator works. It seems to solve correctly, although I did not check the solutions with the imaginary part. If so, write.

Roots and degrees

Degree

The degree is an expression of the form: , where:

Let us define the concept of degree, the exponent of which is - natural number(i.e. integer and positive).

By definition: .
To square a number is to multiply it by itself:
To cube a number is to multiply it by itself three times: .

To raise a number to a natural power is to multiply the number by itself times:

Degree with integer exponent

If the exponent is a positive integer:

Raising to the zero power:

If the exponent is a negative integer:

Note: the expression is not defined, in the case of n ≤ 0. If n > 0, then

Degree with rational exponent

Degree properties

Root

The equation has two solutions: x=2 and x=-2. These are numbers whose square is 4.

Consider the equation. Let's draw a graph of the function and see that this equation also has two solutions, one positive, the other negative.

But in this case the solutions are not integers. Moreover, they are not rational. In order to write down these irrational decisions, we introduce a special square root symbol.

An arithmetic square root is a non-negative number whose square is, a ≥ 0. For a< 0 - выражение не определено, т.к. нет такого действительного числа, квадрат которого равен отрицательному числу.

The root of the square

For instance, . And the solutions of the equation, respectively, and

cube root

The cube root of a number is the number whose cube is. The cube root is defined for everyone. It can be extracted from any number: .

nth root

The th root of a number is the number whose th power is equal to.

Then if a< 0 корень n-ой степени из a не определен.
Or if a ≥ 0, then the non-negative root of the equation is called the arithmetic root of the nth degree of a and is denoted

Then the equation has a single root for any.

Roots and degrees

An expression of the form is called a degree.

Here - the base of the degree, - the exponent.

Degree with a natural indicator

The easiest way is to determine the degree with a natural (that is, a positive integer) exponent.

The expressions "square" and "cube" have long been familiar to us.

To square a number is to multiply it by itself.

To cube a number is to multiply it by itself three times.

To raise a number to a power is to multiply it by itself once:

Degree with integer exponent

The exponent can be not only natural (that is, a positive integer), but also equal to zero, as well as a negative integer.

This is true for. The expression is not defined.

We also define what a degree with a negative integer exponent is.

Of course, all this is true for, since you cannot divide by zero.

Note that when raised to a minus first power, the fraction is flipped.

The exponent can be not only an integer, but also a fractional, that is, a rational number. In the article "Number Sets" we talked about what rational numbers are. These are numbers that can be written as a fraction, where is an integer, is a natural number.

Here we need a new concept - the root -degree. Roots and degrees are two related topics. Let's start with the already familiar arithmetic square root.

Arithmetic square root

The equation has two solutions: and.

These are numbers whose square is equal.

How to solve the equation?

If we draw a graph of a function, we will see that this equation also has two solutions, one of which is positive and the other is negative.

But these solutions are not integers. Moreover, they are not rational. In order to write down these solutions, we introduce a special square root symbol.

The arithmetic square root of a number is a non-negative number whose square is equal to.

Remember this definition.

The arithmetic square root is denoted.

1) The square root can only be taken from non-negative numbers

2) The expression is always non-negative. For instance, .

We list the properties of the arithmetic square root:

Remember that the expression is not equal. It's easy to check:

Got a different answer.

cube root

Similarly, the cube root of is a number that, when raised to the third power, gives a number.

For example, since;

Note that the third root can be taken from both positive and negative numbers.

Now we can define the root of the -th power for any integer.

root degree

The root of a number is a number that, when raised to the -th power, produces a number.

Note that the root of the third, fifth, ninth - in a word, any odd degree - can be extracted from both positive and negative numbers.

The square root, as well as the root of the fourth, tenth, in general, of any even degree can only be extracted from non-negative numbers.

So, - such a number that. It turns out that roots can be written as powers with a rational exponent. It's comfortable.

Immediately agree that the base of the degree is greater.

The expression is equal by definition.

In this case, the condition that is greater is also satisfied.

Remember the rules for actions with degrees:

When multiplying exponents, the exponents add up

When dividing the degree by the degree, the indicators are subtracted

When raising a power to a power, the exponents are multiplied

Let us show how these formulas are applied in USE assignments mathematics:

They brought everything under a common root, factored it out, reduced the fraction and took the root.

Here we have written the roots as powers and used the power action formulas.

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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 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.

This kind of equation always has 3 roots. Roots can be either real or complex. If the initial data allow you to choose one of the roots of the cubic equation \ then you can divide the cubic polynomial into \ [(x - x1) \] 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 can be seen 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 https: // site. 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 have any questions, you can ask them in our Vkontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.

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 the 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

\ where \ A number \ is called a 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?

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 a 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"

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: \

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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 , because 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 the cast algebraic equation 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 quartic 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".Code of the program "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.

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 Researchers and Engineers, 2012.

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First, let's recall the basic formulas of degrees and their properties.

Product of a number a happens 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 \u003d 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 measure.

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

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

Let's take a simple equation:

2 x = 2 3

Such an example can be solved even in the mind. 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 this decision should be made:

2 x = 2 3
x = 3

To solve this equation, we removed same grounds(that is, deuces) and wrote down what was left, these are degrees. We got the answer we were looking for.

Now let's summarize our solution.

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

Now let's solve some examples:

Let's start simple.

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

x+2=4 The simplest equation has turned out.
x=4 - 2
x=2
Answer: x=2

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

3 3x - 9 x + 8 = 0

To begin with, we transfer 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 \u003d (3 2) x + 8

We get 9 x + 8 \u003d (3 2) x + 8 \u003d 3 2 x + 16

3 3x \u003d 3 2x + 16 now it is clear that the bases on the left and right sides are the same and equal to three, which means we can discard them and equate the degrees.

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

Let's look at the following example:

2 2x + 4 - 10 4 x \u003d 2 4

First of all, we look at the bases, the bases are different two and four. And we need to be the same. We transform the quadruple according to 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 interfere with us. What to do with them? If you look closely, you can see that on the left side we repeat 2 2x, 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 whole equation by 6:

Imagine 4=2 2:

2 2x \u003d 2 2 bases are the same, discard them and equate the degrees.
2x \u003d 2 turned out to be the simplest equation. We divide it by 2, 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, it is clear that the first triple has a degree twice (2x) than the second (just x). In this case, you can decide substitution method. The number with the smallest degree is replaced by:

Then 3 2x \u003d (3 x) 2 \u003d t 2

We replace all degrees with x's in the equation with t:

t 2 - 12t + 27 \u003d 0
We get a quadratic equation. We solve through the discriminant, we get:
D=144-108=36
t1 = 9
t2 = 3

Back to Variable x.

We take t 1:
t 1 \u003d 9 \u003d 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 \u003d 3 \u003d 3 x
3 x = 3 1
x 2 = 1
Answer: x 1 \u003d 2; x 2 = 1.

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