Quadratic equations are studied in grade 8, so there is nothing complicated here. The ability to solve them is essential.

A quadratic equation is an equation of the form ax 2 + bx + c = 0, where the coefficients a , b and c are arbitrary numbers, and a ≠ 0.

Before studying specific solution methods, we note that all quadratic equations can be divided into three classes:

1. Have no roots;
2. They have exactly one root;
3. They have two different roots.

This is an important difference between quadratic and linear equations, where the root always exists and is unique. How to determine how many roots an equation has? There is a wonderful thing for this - discriminant.

Discriminant

Let it be given quadratic equation ax 2 + bx + c = 0. Then the discriminant is simply the number D = b 2 − 4ac .

This formula must be known by heart. Where it comes from is not important now. Another thing is important: by the sign of the discriminant, you can determine how many roots a quadratic equation has. Namely:

1. If D< 0, корней нет;
2. If D = 0, there is exactly one root;
3. If D > 0, there will be two roots.

Please note: the discriminant indicates the number of roots, and not at all their signs, as for some reason many people think. Take a look at the examples and you will understand everything yourself:

Task. How many roots do quadratic equations have:

1. x 2 - 8x + 12 = 0;
2. 5x2 + 3x + 7 = 0;
3. x 2 − 6x + 9 = 0.

We write the coefficients for the first equation and find the discriminant:
a = 1, b = −8, c = 12;
D = (−8) 2 − 4 1 12 = 64 − 48 = 16

So, the discriminant is positive, so the equation has two different roots. We analyze the second equation in the same way:
a = 5; b = 3; c = 7;
D \u003d 3 2 - 4 5 7 \u003d 9 - 140 \u003d -131.

The discriminant is negative, there are no roots. The last equation remains:
a = 1; b = -6; c = 9;
D = (−6) 2 − 4 1 9 = 36 − 36 = 0.

The discriminant is equal to zero - the root will be one.

Note that coefficients have been written out for each equation. Yes, it's long, yes, it's tedious - but you won't mix up the odds and don't make stupid mistakes. Choose for yourself: speed or quality.

By the way, if you “fill your hand”, after a while you will no longer need to write out all the coefficients. You will perform such operations in your head. Most people start doing this somewhere after 50-70 solved equations - in general, not so much.

The roots of a quadratic equation

Now let's move on to the solution. If the discriminant D > 0, the roots can be found using the formulas:

The basic formula for the roots of a quadratic equation

When D = 0, you can use any of these formulas - you get the same number, which will be the answer. Finally, if D< 0, корней нет — ничего считать не надо.

1. x 2 - 2x - 3 = 0;
2. 15 - 2x - x2 = 0;
3. x2 + 12x + 36 = 0.

First equation:
x 2 - 2x - 3 = 0 ⇒ a = 1; b = −2; c = -3;
D = (−2) 2 − 4 1 (−3) = 16.

D > 0 ⇒ the equation has two roots. Let's find them:

Second equation:
15 − 2x − x 2 = 0 ⇒ a = −1; b = −2; c = 15;
D = (−2) 2 − 4 (−1) 15 = 64.

D > 0 ⇒ the equation again has two roots. Let's find them

\[\begin(align) & ((x)_(1))=\frac(2+\sqrt(64))(2\cdot \left(-1 \right))=-5; \\ & ((x)_(2))=\frac(2-\sqrt(64))(2\cdot \left(-1 \right))=3. \\ \end(align)\]

Finally, the third equation:
x 2 + 12x + 36 = 0 ⇒ a = 1; b = 12; c = 36;
D = 12 2 − 4 1 36 = 0.

D = 0 ⇒ the equation has one root. Any formula can be used. For example, the first one:

As you can see from the examples, everything is very simple. If you know the formulas and be able to count, there will be no problems. Most often, errors occur when negative coefficients are substituted into the formula. Here, again, the technique described above will help: look at the formula literally, paint each step - and get rid of mistakes very soon.

Incomplete quadratic equations

It happens that the quadratic equation is somewhat different from what is given in the definition. For example:

1. x2 + 9x = 0;
2. x2 − 16 = 0.

It is easy to see that one of the terms is missing in these equations. Such quadratic equations are even easier to solve than standard ones: they do not even need to calculate the discriminant. So let's introduce a new concept:

The equation ax 2 + bx + c = 0 is called an incomplete quadratic equation if b = 0 or c = 0, i.e. the coefficient of the variable x or the free element is equal to zero.

Of course, a very difficult case is possible when both of these coefficients are equal to zero: b \u003d c \u003d 0. In this case, the equation takes the form ax 2 \u003d 0. Obviously, such an equation has a single root: x \u003d 0.

Let's consider other cases. Let b \u003d 0, then we get an incomplete quadratic equation of the form ax 2 + c \u003d 0. Let's slightly transform it:

Since the arithmetic square root exists only from a non-negative number, the last equality only makes sense when (−c / a ) ≥ 0. Conclusion:

1. If an incomplete quadratic equation of the form ax 2 + c = 0 satisfies the inequality (−c / a ) ≥ 0, there will be two roots. The formula is given above;
2. If (−c / a )< 0, корней нет.

As you can see, the discriminant was not required - there are no complex calculations at all in incomplete quadratic equations. In fact, it is not even necessary to remember the inequality (−c / a ) ≥ 0. It is enough to express the value of x 2 and see what is on the other side of the equal sign. If there is a positive number, there will be two roots. If negative, there will be no roots at all.

Now let's deal with equations of the form ax 2 + bx = 0, in which the free element is equal to zero. Everything is simple here: there will always be two roots. It is enough to factorize the polynomial:

Taking the common factor out of the bracket

The product is equal to zero when at least one of the factors is equal to zero. This is where the roots come from. In conclusion, we will analyze several of these equations:

Task. Solve quadratic equations:

1. x2 − 7x = 0;
2. 5x2 + 30 = 0;
3. 4x2 − 9 = 0.

x 2 − 7x = 0 ⇒ x (x − 7) = 0 ⇒ x 1 = 0; x2 = −(−7)/1 = 7.

5x2 + 30 = 0 ⇒ 5x2 = -30 ⇒ x2 = -6. There are no roots, because the square cannot be equal to a negative number.

4x 2 − 9 = 0 ⇒ 4x 2 = 9 ⇒ x 2 = 9/4 ⇒ x 1 = 3/2 = 1.5; x 2 \u003d -1.5.

After we have studied the concept of equalities, namely one of their types - numerical equalities, we can move on to another important view- equations. Within the framework of this material, we will explain what an equation is and its root, formulate the main definitions and give various examples equations and finding their roots.

Yandex.RTB R-A-339285-1

The concept of an equation

Usually the concept of an equation is studied at the very beginning. school course algebra. Then it is defined like this:

Definition 1

Equation called equality with an unknown number to be found.

It is customary to denote unknowns in small Latin letters, for example, t, r, m etc., but x, y, z are most often used. In other words, the equation determines the form of its recording, that is, equality will be an equation only when it is brought to a certain form - it must contain a letter, the value of which must be found.

Let us give some examples of the simplest equations. These can be equalities of the form x = 5, y = 6, etc., as well as those that include arithmetic operations, for example, x + 7 = 38, z − 4 = 2, 8 t = 4, 6:x=3.

After the concept of brackets is studied, the concept of equations with brackets appears. These include 7 (x − 1) = 19 , x + 6 (x + 6 (x − 8)) = 3, etc. The letter to be found may occur more than once, but several, as, for example, in the equation x + 2 + 4 x - 2 - x = 10 . Also, the unknowns can be located not only on the left, but also on the right, or in both parts at the same time, for example, x (8 + 1) - 7 = 8, 3 - 3 = z + 3 or 8 x - 9 = 2 (x + 17).

Further, after the students get acquainted with the concept of whole, real, rational, natural numbers, as well as logarithms, roots and powers, new equations appear that include all these objects. We have devoted a separate article to examples of such expressions.

In the program for the 7th grade, the concept of variables first appears. These are letters that can take on different values ​​(for more details, see the article on numeric, literal and expressions with variables). Based on this concept, we can redefine the equation:

Definition 2

The equation is an equality involving a variable whose value is to be calculated.

That is, for example, the expression x + 3 \u003d 6 x + 7 is an equation with a variable x, and 3 y − 1 + y \u003d 0 is an equation with a variable y.

In one equation, there can be not one variable, but two or more. They are called, respectively, equations with two, three variables, etc. Let's write the definition:

Definition 3

Equations with two (three, four or more) variables are called equations that include an appropriate number of unknowns.

For example, an equality of the form 3, 7 x + 0, 6 = 1 is an equation with one variable x, and x − z = 5 is an equation with two variables x and z. An example of an equation with three variables would be x 2 + (y − 6) 2 + (z + 0, 6) 2 = 26 .

Root of the equation

When we talk about an equation, it immediately becomes necessary to define the concept of its root. Let's try to explain what it means.

Example 1

We are given an equation that includes one variable. If we substitute instead unknown letter number, then the equation will become a numerical equality - true or false. So, if in the equation a + 1 \u003d 5 we replace the letter with the number 2, then the equality will become incorrect, and if 4, then we will get the correct equality 4 + 1 \u003d 5.

We are more interested in precisely those values ​​with which the variable will turn into true equality. They are called roots or solutions. Let's write down the definition.

Definition 4

The root of the equation name the value of the variable that turns the given equation into a true equality.

The root can also be called a decision, or vice versa - both of these concepts mean the same thing.

Example 2

Let's take an example to clarify this definition. Above we gave the equation a + 1 = 5 . According to the definition, the root in this case will be 4, because when substituting for a letter, it gives the correct numerical equality, and two will not be a solution, since it corresponds to an incorrect equality 2 + 1 \u003d 5.

How many roots can one equation have? Does every equation have a root? Let's answer these questions.

Equations that do not have a single root also exist. An example would be 0 x = 5 . We can plug in infinitely many different numbers into it, but none of them will turn it into a true equality, since multiplying by 0 always gives 0 .

There are also equations that have multiple roots. They can have both finite and infinitely many roots.

Example 3

So, in the equation x - 2 \u003d 4 there is only one root - six, in x 2 \u003d 9 two roots - three and minus three, in x (x - 1) (x - 2) \u003d 0 three roots - zero, one and two, in the equation x=x there are infinitely many roots.

Now we will explain how to write the roots of the equation correctly. If there are none, then we write like this: "the equation has no roots." It is also possible in this case to indicate the sign of the empty set ∅ . If there are roots, then we write them separated by commas or indicate them as elements of the set, enclosing them in curly brackets. So, if any equation has three roots - 2, 1 and 5, then we write - 2, 1, 5 or (- 2, 1, 5) .

It is allowed to write the roots in the form of the simplest equalities. So, if the unknown in the equation is denoted by the letter y, and the roots are 2 and 7, then we write y \u003d 2 and y \u003d 7. Sometimes subscripts are added to letters, for example, x 1 \u003d 3, x 2 \u003d 5. Thus we indicate the numbers of the roots. If the equation has infinitely many solutions, then we write the answer as a numerical interval or use generally accepted notation: the set of natural numbers is denoted N, integers - Z, real numbers - R. Let's say, if we need to write that any integer will be the solution to the equation, then we write that x ∈ Z, and if any real number is from one to nine, then y ∈ 1, 9.

When an equation has two, three or more roots, then, as a rule, they are not talking about roots, but about solutions to the equation. We formulate the definition of a solution to an equation with several variables.

Definition 5

The solution of an equation with two, three or more variables is two, three or more values ​​of the variables that turn this equation into a true numerical equality.

Let us explain the definition with examples.

Example 4

Let's say we have an expression x + y = 7 , which is an equation with two variables. Substitute one for the first, and two for the second. We get an incorrect equality, which means that this pair of values ​​\u200b\u200bwill not be a solution to this equation. If we take a pair of 3 and 4, then the equality becomes true, which means we have found a solution.

Such equations may also have no roots or have an infinite number of them. If we need to write down two, three, four or more values, then we write them separated by commas in parentheses. That is, in the example above, the answer will look like (3 , 4) .

In practice, most often one has to deal with equations containing one variable. We will consider the algorithm for solving them in detail in an article devoted to solving equations.

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

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. Quite often, the root sign is found in equations, and many mistakenly believe that such equations are difficult to solve. For such equations in mathematics there is a special term, which is called equations with a root - irrational equations.

The main difference in solving equations with a root from other equations, for example, square, logarithmic, linear, is that they do not have a standard solution algorithm. Therefore, in order to solve an irrational equation, it is necessary to analyze the initial data and choose a more suitable solution.

In most cases, to solve this kind of equations, the method of raising both parts of the equation to the same power is used.

Let's say the following equation is given:

\[\sqrt((5x-16))=x-2\]

We square both sides of the equation:

\[\sqrt((5x-16)))^2 =(x-2)^2\], whence successively we obtain:

Having received a quadratic equation, we find its roots:

Answer: \

If we substitute these values ​​into the equation, we will get the correct equality, which indicates the correctness of the data obtained.

Where can I solve an equation with roots with an online 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.

Municipal educational institution

"Kudinskaya secondary school No. 2"

Ways to solve irrational equations

Completed by: Egorova Olga,

Supervisor:

Teacher

mathematics,

higher qualification

Introduction....……………………………………………………………………………………… 3

Section 1. Methods for solving irrational equations…………………………………6

1.1 Solving the irrational equations of part C……….….….……………………21

Section 2. Individual tasks…………………………………………….....………...24

Answers………………………………………………………………………………………….25

Bibliography…….…………………………………………………………………….26

Introduction

Mathematics education received in general education school, is the most important component general education and general culture modern man. Almost everything that surrounds a modern person is all connected in one way or another with mathematics. And the latest advances in physics, engineering and information technology leave no doubt that in the future the state of affairs will remain the same. Therefore, the decision of many practical tasks comes down to a decision various kinds equations to learn how to solve. One of these types are irrational equations.

Irrational equations

An equation containing an unknown (or a rational algebraic expression from an unknown) under the radical sign is called irrational equation. In elementary mathematics, solutions to irrational equations are sought in the set of real numbers.

Any irrational equation with the help of elementary algebraic operations (multiplication, division, raising both parts of the equation to an integer power) can be reduced to a rational algebraic equation. It should be borne in mind that the resulting rational algebraic equation may not be equivalent to the original irrational equation, namely, it may contain "extra" roots that will not be the roots of the original irrational equation. Therefore, finding the roots of the obtained rational algebraic equation, it is necessary to check whether all the roots of a rational equation are the roots of an irrational equation.

In general, it is difficult to specify any universal method solution of any irrational equation, since it is desirable that as a result of transformations of the original irrational equation, not just some kind of rational algebraic equation is obtained, among the roots of which there will be the roots of this irrational equation, but a rational algebraic equation formed from polynomials of the least possible degree. The desire to obtain that rational algebraic equation formed from polynomials of the smallest possible degree is quite natural, since finding all the roots of a rational algebraic equation can in itself be a rather difficult task, which we can completely solve only in a very limited number of cases.

Types of irrational equations

Solving irrational equations even degree always calls more problems than the solution of irrational equations of odd degree. When solving irrational equations of odd degree, the ODZ does not change. Therefore, below we will consider irrational equations, the degree of which is even. There are two kinds of irrational equations:

2..

Let's consider the first of them.

odz equation: f(x)≥ 0. In ODZ, the left side of the equation is always non-negative, so a solution can only exist when g(x)≥ 0. In this case, both sides of the equation are non-negative, and exponentiation 2 n gives an equivalent equation. We get that

Let us pay attention to the fact that while ODZ is performed automatically, and you can not write it, but the conditiong(x) ≥ 0 must be checked.

Note: This is a very important condition of equivalence. Firstly, it frees the student from the need to investigate, and after finding solutions, check the condition f(x) ≥ 0 - the non-negativity of the root expression. Secondly, it focuses on checking the conditiong(x) ≥ 0 are the nonnegativity of the right side. After all, after squaring, the equation is solved i.e., two equations are solved at once (but on different intervals of the numerical axis!):

1. - where g(x)≥ 0 and

2. - where g(x) ≤ 0.

Meanwhile, many, according to the school habit of finding ODZ, do exactly the opposite when solving such equations:

a) check, after finding solutions, the condition f(x) ≥ 0 (which is automatically satisfied), make arithmetic errors and get an incorrect result;

b) ignore the conditiong(x) ≥ 0 - and again the answer may be wrong.

Note: The equivalence condition is especially useful when solving trigonometric equations, in which finding the ODZ is associated with solving trigonometric inequalities, which is much more difficult than solving trigonometric equations. Check in trigonometric equations even conditions g(x)≥ 0 is not always easy to do.

Consider the second kind of irrational equations.

. Let the equation . His ODZ:

In the ODZ, both sides are non-negative, and squaring gives the equivalent equation f(x) =g(x). Therefore, in the ODZ or

With this method of solution, it is enough to check the non-negativity of one of the functions - you can choose a simpler one.

Section 1. Methods for solving irrational equations

1 method. Liberation from radicals by successively raising both sides of the equation to the corresponding natural power

The most commonly used method for solving irrational equations is the method of freeing from radicals by successively raising both parts of the equation to the corresponding natural degree. In this case, it should be borne in mind that when both parts of the equation are raised to an odd power, the resulting equation is equivalent to the original one, and when both parts of the equation are raised to an even power, the resulting equation will, generally speaking, be nonequivalent to the original equation. This can be easily verified by raising both sides of the equation to any even power. This operation results in the equation , whose set of solutions is the union of sets of solutions: https://pandia.ru/text/78/021/images/image013_50.gif" width="95" height="21 src=">. However, despite this drawback , it is the procedure for raising both parts of the equation to some (often even) power that is the most common procedure for reducing an irrational equation to a rational equation.

Solve the equation:

Where are some polynomials. By virtue of the definition of the operation of extracting the root in the set of real numbers, the admissible values ​​of the unknown https://pandia.ru/text/78/021/images/image017_32.gif" width="123 height=21" height="21">..gif " width="243" height="28 src=">.

Since both parts of the 1st equation were squared, it may turn out that not all roots of the 2nd equation will be solutions to the original equation, it is necessary to check the roots.

Solve the equation:

https://pandia.ru/text/78/021/images/image021_21.gif" width="137" height="25">

Raising both sides of the equation into a cube, we get

Given that https://pandia.ru/text/78/021/images/image024_19.gif" width="195" height="27">(The last equation may have roots that, generally speaking, are not roots of the equation ).

We raise both sides of this equation to a cube: . We rewrite the equation in the form x3 - x2 = 0 ↔ x1 = 0, x2 = 1. By checking, we establish that x1 = 0 is an extraneous root of the equation (-2 ≠ 1), and x2 = 1 satisfies the original equation.

Answer: x = 1.

2 method. Replacing an adjacent system of conditions

When solving irrational equations containing even-order radicals, extraneous roots may appear in the answers, which are not always easy to identify. To make it easier to identify and discard extraneous roots, in the course of solving irrational equations it is immediately replaced by an adjacent system of conditions. Additional inequalities in the system actually take into account the ODZ of the equation being solved. You can find the ODZ separately and take it into account later, but it is preferable to use mixed systems of conditions: there is less danger of forgetting something, not taking it into account in the process of solving the equation. Therefore, in some cases it is more rational to use the method of transition to mixed systems.

Solve the equation:

Answer: https://pandia.ru/text/78/021/images/image029_13.gif" width="109 height=27" height="27">

This equation is equivalent to the system

Answer: the equation has no solutions.

3 method. Using the properties of the nth root

When solving irrational equations, the properties of the root of the nth degree are used. arithmetic root n- th degrees from among A call a non-negative number, n- i whose degree is equal to A. If n- even( 2n), then a ≥ 0, otherwise the root does not exist. If n- odd( 2 n+1), then a is any and = - ..gif" width="45" height="19"> Then:

2.

3.

4.

5.

Applying any of these formulas, formally (without taking into account the indicated restrictions), it should be borne in mind that the ODZ of the left and right parts of each of them can be different. For example, the expression is defined with f ≥ 0 And g ≥ 0, and the expression is as in f ≥ 0 And g ≥ 0, as well as f ≤ 0 And g ≤ 0.

For each of the formulas 1-5 (without taking into account the indicated restrictions), the ODZ of its right part may be wider than the ODZ of the left. It follows that transformations of the equation with the formal use of formulas 1-5 "from left to right" (as they are written) lead to an equation that is a consequence of the original one. In this case, extraneous roots of the original equation may appear, so verification is a mandatory step in solving the original equation.

Transformations of equations with the formal use of formulas 1-5 "from right to left" are unacceptable, since it is possible to judge the ODZ of the original equation, and, consequently, the loss of roots.

https://pandia.ru/text/78/021/images/image041_8.gif" width="247" height="61 src=">,

which is a consequence of the original. The solution of this equation is reduced to solving the set of equations .

From the first equation of this set we find https://pandia.ru/text/78/021/images/image044_7.gif" width="89" height="27"> from where we find . Thus, the roots of this equation can only be numbers ( -1) and (-2) Verification shows that both found roots satisfy this equation.

Answer: -1,-2.

Solve the equation: .

Solution: based on the identities, replace the first term with . Note that as the sum of two non-negative numbers on the left side. “Remove” the module and, after bringing like terms, solve the equation. Since , we get the equation . Since and , then https://pandia.ru/text/78/021/images/image055_6.gif" width="89" height="27 src=">.gif" width="39" height="19 src= ">.gif" width="145" height="21 src=">

Answer: x = 4.25.

4 method. Introduction of new variables

Another example of solving irrational equations is the way in which new variables are introduced, with respect to which either a simpler irrational equation or a rational equation is obtained.

The solution of irrational equations by replacing the equation with its consequence (with subsequent checking of the roots) can be carried out as follows:

1. Find the ODZ of the original equation.

2. Go from the equation to its corollary.

3. Find the roots of the resulting equation.

4. Check if the found roots are the roots of the original equation.

The check is as follows:

A) the belonging of each found root of the ODZ to the original equation is checked. Those roots that do not belong to the ODZ are extraneous for the original equation.

B) for each root included in the ODZ of the original equation, it is checked whether the left and right parts of each of the equations that arise in the process of solving the original equation and are raised to an even power have the same signs. Those roots for which parts of any equation raised to an even power have different signs are extraneous for the original equation.

C) only those roots that belong to the ODZ of the original equation and for which both parts of each of the equations that arise in the process of solving the original equation and raised to an even power have the same signs are checked by direct substitution into the original equation.

Such a solution method with the indicated verification method allows avoiding cumbersome calculations in the case of direct substitution of each of the found roots of the last equation into the original one.

Solve the irrational equation:

.

The set of admissible values ​​of this equation:

Setting , after substitution we obtain the equation

or its equivalent equation

which can be viewed as a quadratic equation for . Solving this equation, we get

.

Therefore, the solution set of the original irrational equation is the union of the solution sets of the following two equations:

, .

Cube both sides of each of these equations, and we get two rational algebraic equations:

, .

Solving these equations, we find that this irrational equation has a single root x = 2 (no verification is required, since all transformations are equivalent).

Answer: x = 2.

Solve the irrational equation:

Denote 2x2 + 5x - 2 = t. Then the original equation will take the form . By squaring both parts of the resulting equation and bringing like terms, we obtain the equation , which is a consequence of the previous one. From it we find t=16.

Returning to the unknown x, we get the equation 2x2 + 5x - 2 = 16, which is a consequence of the original one. By checking, we make sure that its roots x1 \u003d 2 and x2 \u003d - 9/2 are the roots of the original equation.

Answer: x1 = 2, x2 = -9/2.

5 method. Identity Equation Transformation

When solving irrational equations, one should not start solving an equation by raising both parts of the equations to a natural power, trying to reduce the solution of an irrational equation to solving a rational algebraic equation. First, it is necessary to see if it is possible to make some identical transformation of the equation, which can significantly simplify its solution.

Solve the equation:

The set of valid values ​​for this equation: https://pandia.ru/text/78/021/images/image074_1.gif" width="292" height="45"> Divide this equation by .

.

We get:

For a = 0, the equation will have no solutions; for , the equation can be written as

for this equation has no solutions, since for any X, belonging to the set of admissible values ​​of the equation, the expression on the left side of the equation is positive;

when the equation has a solution

Taking into account that the set of admissible solutions of the equation is determined by the condition , we finally obtain:

When solving this irrational equation, https://pandia.ru/text/78/021/images/image084_2.gif" width="60" height="19"> the solution to the equation will be . For all other values X the equation has no solutions.

EXAMPLE 10:

Solve the irrational equation: https://pandia.ru/text/78/021/images/image086_2.gif" width="381" height="51">

The solution of the quadratic equation of the system gives two roots: x1 \u003d 1 and x2 \u003d 4. The first of the obtained roots does not satisfy the inequality of the system, therefore x \u003d 4.

Notes.

1) Carrying out identical transformations allows us to do without verification.

2) The inequality x - 3 ≥0 refers to identical transformations, and not to the domain of the equation.

3) There is a decreasing function on the left side of the equation, and an increasing function on the right side of this equation. Graphs of decreasing and increasing functions at the intersection of their domains of definition can have no more than one common point. Obviously, in our case, x = 4 is the abscissa of the intersection point of the graphs.

Answer: x = 4.

6 method. Using the domain of definition of functions when solving equations

This method is most effective when solving equations that include functions https://pandia.ru/text/78/021/images/image088_2.gif" width="36" height="21 src="> and find its area definitions (f)..gif" width="53" height="21"> .gif" width="88" height="21 src=">, then you need to check whether the equation is true at the ends of the interval, moreover, if a< 0, а b >0, then it is necessary to check on the intervals (a;0) And . The smallest integer in E(y) is 3.

Answer: x = 3.

8 method. Application of the derivative in solving irrational equations

Most often, when solving equations using the derivative method, the estimation method is used.

EXAMPLE 15:

Solve the equation: (1)

Solution: Since https://pandia.ru/text/78/021/images/image122_1.gif" width="371" height="29">, or (2). Consider the function ..gif" width="400" height="23 src=">.gif" width="215" height="49"> at all and therefore increasing. Therefore, the equation is equivalent to an equation that has a root that is the root of the original equation.

Answer:

EXAMPLE 16:

Solve the irrational equation:

The domain of definition of the function is a segment. Find the largest and smallest value the values ​​of this function on the interval . To do this, we find the derivative of the function f(x): https://pandia.ru/text/78/021/images/image136_1.gif" width="37 height=19" height="19">. Let's find the values ​​of the function f(x) at the ends of the segment and at the point : So, But and, therefore, equality is possible only under the condition https://pandia.ru/text/78/021/images/image136_1.gif" width="37" height="19 src=" > Verification shows that the number 3 is the root of this equation.

Answer: x = 3.

9 method. Functional

In exams, they sometimes offer to solve equations that can be written in the form , where is a certain function.

For example, some equations: 1) 2) . Indeed, in the first case , in the second case . Therefore, solve irrational equations using the following statement: if a function is strictly increasing on the set X and for any , then the equations, etc., are equivalent on the set X .

Solve the irrational equation: https://pandia.ru/text/78/021/images/image145_1.gif" width="103" height="25"> strictly increasing on the set R, and https://pandia.ru/text/78/021/images/image153_1.gif" width="45" height="24 src=">..gif" width="104" height="24 src=" > which has a unique root Therefore, the equivalent equation (1) also has a unique root

Answer: x = 3.

EXAMPLE 18:

Solve the irrational equation: (1)

By definition square root we get that if equation (1) has roots, then they belong to the set https://pandia.ru/text/78/021/images/image159_0.gif" width="163" height="47">. (2)

Consider the function https://pandia.ru/text/78/021/images/image147_1.gif" width="35" height="21"> strictly increasing on this set for any ..gif" width="100" height ="41"> which has a single root Therefore, and equivalent to it on the set X equation (1) has a single root

Answer: https://pandia.ru/text/78/021/images/image165_0.gif" width="145" height="27 src=">

Solution: This equation is equivalent to a mixed system

When studying algebra, students are faced with equations of many kinds. Among those that are the simplest, one can name linear ones containing one unknown. If a variable in a mathematical expression is raised to a certain power, then the equation is called quadratic, cubic, biquadratic, and so on. These expressions may contain rational numbers. But there are also irrational equations. They differ from others by the presence of a function where the unknown is under the sign of the radical (that is, purely externally, the variable here can be seen written under the square root). Solving irrational equations has its own characteristics. When calculating the value of a variable to obtain the correct answer, they must be taken into account.

"Unspeakable in words"

It is no secret that ancient mathematicians operated mainly with rational numbers. These include, as you know, integers, expressed through ordinary and decimal periodic fractions, representatives of this community. However, scientists of the Middle and Near East, as well as India, developing trigonometry, astronomy and algebra, also learned to solve irrational equations. For example, the Greeks knew such quantities, but, putting them into a verbal form, they used the concept of “alogos”, which meant “inexpressible”. Somewhat later, Europeans, imitating them, called such numbers "deaf". They differ from all the others in that they can only be represented in the form of an infinite non-periodic fraction, the final numerical expression of which is simply impossible to obtain. Therefore, more often such representatives of the realm of numbers are written in the form of numbers and signs as some expression that is under the root of the second or greater degree.

Based on the foregoing, we will try to define the irrational equation. Such expressions contain the so-called "inexpressible numbers", written using the square root sign. They can be all sorts of rather complex options, but in their simplest form they look like the photo below.

Transgressing to the solution of irrational equations, first of all it is necessary to calculate the range of admissible values ​​of the variable.

Does the expression make sense?

The need to check the obtained values ​​follows from the properties. As is known, such an expression is acceptable and has any meaning only under certain conditions. In cases of an even root, all radical expressions must be positive or equal to zero. If this condition is not satisfied, then the presented mathematical notation cannot be considered meaningful.

Let's give a specific example of how to solve irrational equations (pictured below).

In this case, it is obvious that these conditions cannot be satisfied for any values ​​taken by the desired value, since it turns out that 11 ≤ x ≤ 4. This means that only Ø can be a solution.

Analysis method

From the above, it becomes clear how to solve some types of irrational equations. A simple analysis can be effective here.

We give a number of examples that again clearly demonstrate this (in the photo below).

In the first case, upon careful consideration of the expression, it immediately becomes extremely clear that it cannot be true. Indeed, after all, a positive number should be obtained on the left side of the equality, which cannot be equal to -1 in any way.

In the second case, the sum of two positive expressions can be considered equal to zero only when x - 3 = 0 and x + 3 = 0 at the same time. Again, this is impossible. And so, in the answer, you should write Ø again.

The third example is very similar to the previous one. Indeed, here the conditions of the ODZ require that the following absurd inequality be satisfied: 5 ≤ x ≤ 2. And such an equation in a similar way cannot have sound solutions.

Unlimited Zoom

The nature of the irrational can be most clearly and fully explained and known only through an endless series of numbers. decimal fraction. And a specific, striking example of the members of this family is pi. Not without reason, it is assumed that this mathematical constant has been known since ancient times, being used in calculating the circumference and area of ​​a circle. But among Europeans, it was first put into practice by the Englishman William Jones and the Swiss Leonard Euler.

This constant arises as follows. If we compare the most different circumferences, then the ratio of their lengths and diameters is necessarily equal to the same number. This is pi. If expressed in terms of common fraction, then approximately we get 22/7. This was first done by the great Archimedes, whose portrait is shown in the figure above. That is why a similar number got his name. But this is not an explicit, but an approximate value of perhaps the most amazing of numbers. The brilliant scientist found the desired value with an accuracy of 0.02, but, in fact, this constant has no real value, but is expressed as 3.1415926535 ... It is an endless series of numbers, indefinitely approaching a certain mythical value.

Squaring

But back to irrational equations. To find the unknown, in this case very often resort to simple method: square both sides of the existing equality. This method usually gives good results. But one should take into account the insidiousness of irrational values. All the roots obtained as a result of this must be checked, because they may not be suitable.

But let's continue the consideration of examples and try to find the variables in the newly proposed way.

It is quite easy, using the Vieta theorem, to find the desired values ​​​​of the quantities after we have formed a quadratic equation as a result of certain operations. Here it turns out that among the roots there will be 2 and -19. However, when checking, substituting the obtained values ​​into the original expression, you can make sure that none of these roots is suitable. This is a common occurrence in irrational equations. This means that our dilemma again has no solutions, and the empty set should be indicated in the answer.

More complicated examples

In some cases, it is required to square both sides of the expression not once, but several times. Consider examples where the above is required. They can be seen below.

Having received the roots, do not forget to check them, because extra ones may arise. It should be explained why this is possible. When applying such a method, a rationalization of the equation occurs in some way. But getting rid of the roots that are objectionable to us, which prevent us from performing arithmetic operations, we, as it were, expand the existing range of values, which is fraught (as you can understand) with consequences. Anticipating this, we make a check. In this case, there is a chance to make sure that only one of the roots fits: x = 0.

Systems

What to do in cases when it is required to solve systems of irrational equations, and we have not one, but two whole unknowns? Here we proceed in the same way as in ordinary cases, but taking into account the above properties of the data mathematical expressions. And in each new task, of course, you should apply a creative approach. But, again, it is better to consider everything on specific example below. Here it is not only required to find the variables x and y, but also to indicate their sum in the answer. So, there is a system containing irrational quantities (see photo below).

As you can see, such a task is not supernaturally difficult. You just need to be smart and guess that the left side of the first equation is the square of the sum. Similar tasks are found in the exam.

Irrational in mathematics

Each time, the need to create new types of numbers arose for humanity when it lacked the “space” to solve some equations. Irrational numbers are no exception. As facts from history testify, for the first time the great sages drew attention to this even before our era, in the 7th century. This was done by a mathematician from India, known as Manava. He clearly understood that it is impossible to extract a root from some natural numbers. For example, these include 2; 17 or 61, as well as many others.

One of the Pythagoreans, a thinker named Hippasus, came to the same conclusion, trying to make calculations with the numerical expressions of the sides of the pentagram. By discovering mathematical elements that cannot be expressed numerically and do not have properties ordinary numbers, he so angered his colleagues that he was thrown overboard into the sea. The fact is that other Pythagoreans considered his reasoning a rebellion against the laws of the universe.

Radical Sign: Evolution

The root sign for expressing the numerical value of "deaf" numbers began to be used in solving irrational inequalities and equations far from immediately. For the first time, European, in particular Italian, mathematicians began to think about the radical around the 13th century. At the same time, they came up with the idea to use the Latin R for designation. But German mathematicians acted differently in their works. They liked the letter V more. In Germany, the designation V (2), V (3) soon spread, which was intended to express the square root of 2, 3, and so on. Later, the Dutch intervened and changed the sign of the radical. And Rene Descartes completed the evolution, bringing the square root sign to modern perfection.

Getting rid of the irrational

Irrational equations and inequalities may include a variable not only under the square root sign. It can be of any degree. The most common way to get rid of it is to raise both sides of the equation to the appropriate power. This is the main action that helps with operations with the irrational. The actions in even cases are not particularly different from those that have already been analyzed by us earlier. Here, the conditions for the non-negativity of the radical expression should be taken into account, and also, at the end of the solution, it is necessary to screen out extraneous values ​​of the variables in the way that was shown in the examples already considered.

Of the additional transformations that help find the correct answer, multiplication of the expression by the conjugate is often used, and it is also often necessary to introduce a new variable, which makes the solution easier. In some cases, to find the value of the unknowns, it is advisable to use graphs.