Irrational equations with roots. Solving an irrational equation

An irrational equation is any equation containing a function under the root sign. For example:

Such equations are always solved in 3 steps:

1. Seclude the root. In other words, if to the left of the equal sign, in addition to the root, there are other numbers or functions, all this must be moved to the right, changing the sign. In this case, only the radical should remain on the left - without any coefficients.
2. 2. Square both sides of the equation. At the same time, we remember that the range of values ​​of the root is all non-negative numbers. Therefore, the function on the right irrational equation must also be non-negative: g(x) ≥ 0.
3. The third step logically follows from the second: you need to perform a check. The fact is that in the second step we could have extra roots. And in order to cut them off, you need to substitute the resulting candidate numbers into the original equation and check whether the correct result is actually obtained numerical equality?

Solving an irrational equation

Let's deal with our ir rational equation given at the very beginning of the lesson. Here the root is already isolated: to the left of the equal sign there is nothing but the root. Square both sides:

2x 2 − 14x + 13 = (5 − x ) 2
2x 2 − 14x + 13 = 25 − 10x + x 2
x 2 − 4x − 12 = 0

We solve the resulting quadratic equation through the discriminant:

D = b 2 − 4ac = (−4) 2 − 4 1 (−12) = 16 + 48 = 64
x 1 = 6; x 2 = −2

All that remains is to substitute these numbers into the original equation, i.e. perform the check. But even here you can do the right thing to simplify the final decision.

How to simplify the solution

Let's think: why do we even perform a check at the end of solving an irrational equation? We want to make sure that when we substitute our roots, there will be a non-negative number to the right of the equals sign. After all, we already know for sure that there is a non-negative number on the left, because the arithmetic square root (which is why our equation is called irrational) by definition cannot be less than zero.

Therefore, all we need to check is that the function g (x) = 5 − x, which is to the right of the equal sign, is non-negative:

g(x) ≥ 0

We substitute our roots into this function and get:

g (x 1) = g (6) = 5 − 6 = −1< 0
g (x 2) = g (−2) = 5 − (−2) = 5 + 2 = 7 > 0

From the obtained values ​​it follows that the root x 1 = 6 does not suit us, since when substituting in right side from the original equation we get a negative number. But the root x 2 = −2 is quite suitable for us, because:

1. This root is the solution to the quadratic equation obtained by raising both sides irrational equation into a square.
2. The right side of the original irrational equation when substituting the root x 2 = −2 turns into positive number, i.e. range arithmetic root not broken.

That's the whole algorithm! As you can see, solving equations with radicals is not that difficult. The main thing is not to forget to check the received roots, otherwise there is a very high probability of receiving unnecessary answers.

The first part of the material in this article forms the idea of ​​irrational equations. After studying it, you will be able to easily distinguish irrational equations from equations of other types. The second part examines in detail the main methods for solving irrational equations and provides detailed solutions huge amount typical examples. If you master this information, you will almost certainly cope with almost any irrational equation from a school mathematics course. Good luck in gaining knowledge!

What are irrational equations?

Let's first clarify what irrational equations are. To do this, we will find the appropriate definitions in textbooks recommended by the Ministry of Education and Science of the Russian Federation.

A detailed conversation about irrational equations and their solution is conducted in algebra lessons and began analysis in high school. However, some authors introduce equations of this type earlier. For example, those who study using the textbooks of Mordkovich A.G. learn about irrational equations already in the 8th grade: the textbook states that

There are also examples of irrational equations, , , etc. Obviously, each of the above equations contains a variable x under the square root sign, which means that, according to the above definition, these equations are irrational. Here we immediately discuss one of the main methods for solving them -. But we will talk about solution methods a little lower, but for now we will give definitions of irrational equations from other textbooks.

In the textbooks of A. N. Kolmogorov and Yu. M. Kolyagin.

Definition

irrational are equations in which a variable is contained under the root sign.

Let's pay attention to the fundamental difference this definition from the previous one: it simply says the root, not the square root, that is, the degree of the root under which the variable is located is not specified. This means that the root can be not only square, but also third, fourth, etc. degrees. Thus, the last definition specifies a wider set of equations.

A natural question arises: why do we begin to use this broader definition of irrational equations in high school? Everything is understandable and simple: when we get acquainted with irrational equations in the 8th grade, we only know well the square root, not about any cubic roots, roots of the fourth or more high degrees we don't know yet. And in high school the concept of a root is generalized, we learn about , and when talking about irrational equations we are no longer limited to the square root, but we mean the root of an arbitrary degree.

For clarity, we will demonstrate several examples of irrational equations. - here the variable x is located under the cube root sign, so this equation is irrational. Another example: - here the variable x is under the sign of both the square root and the fourth root, that is, this is also an irrational equation. Here are a couple more examples of irrational equations more complex type: And .

The above definitions allow us to note that in the notation of any irrational equation there are signs of the roots. It is also clear that if there are no signs of the roots, then the equation is not irrational. However, not all equations containing root signs are irrational. Indeed, in an irrational equation there must be a variable under the root sign; if there is no variable under the root sign, then the equation is not irrational. As an illustration, we give examples of equations that contain roots, but are not irrational. Equations And are not irrational, since they do not contain variables under the root sign - there are numbers under the roots, but there are no variables under the root signs, therefore these equations are not irrational.

It is worth mentioning the number of variables that can participate in writing irrational equations. All the above irrational equations contain a single variable x, that is, they are equations with one variable. However, nothing prevents us from considering irrational equations with two, three, etc. variables. Let us give an example of an irrational equation with two variables and with three variables.

Note that in school you mainly have to work with irrational equations with one variable. Irrational equations with several variables are much less common. They can be found in the composition, as, for example, in the task “solve the system of equations "or, say, in the algebraic description of geometric objects, so a semicircle with a center at the origin, a radius of 3 units, lying in the upper half-plane, corresponds to the equation.

Some collections of problems for preparing for the Unified State Exam in the “irrational equations” section contain tasks in which the variable is not only under the root sign, but also under the sign of some other function, for example, modulus, logarithm, etc. Here's an example , taken from the book, but here - from the collection. In the first example, the variable x is under the logarithmic sign, and the logarithm is also under the root sign, that is, we have, so to speak, an irrational logarithmic (or logarithmic irrational) equation. In the second example, the variable is under the modulus sign, and the modulus is also under the root sign; with your permission, we will call it an irrational equation with a modulus.

Should equations of this type be considered irrational? Good question. It seems that there is a variable under the sign of the root, but it is confusing that it is not in its “pure form”, but under the sign of one or more functions. In other words, there seems to be no contradiction to how we defined the irrational equations above, but there is some degree of uncertainty due to the presence of other functions. From our point of view, one should not be fanatical about “calling a spade a spade.” In practice, it is enough to simply say “equation” without specifying what type it is. And all these additives are “irrational”, “logarithmic”, etc. serve mostly for convenience of presentation and grouping of material.

In light of the information in the last paragraph, the definition of irrational equations given in the textbook authored by A. G. Mordkovich for grade 11 is of interest

Definition

Irrational are called equations in which the variable is contained under the radical sign or under the sign of raising to a fractional power.

Here, in addition to equations with a variable under the sign of the root, equations with variables under the sign of raising to a fractional power are also considered irrational. For example, according to this definition, the equation is considered irrational. Why suddenly? We are already accustomed to roots in irrational equations, but here it is not a root, but a degree, and would you rather call this equation, for example, a power equation, rather than an irrational one? Everything is simple: it is determined through the roots, and on the variable x for a given equation (provided x 2 +2·x≥0) it can be rewritten using the root as , and the last equality is a familiar irrational equation with a variable under the root sign. And the methods for solving equations with variables in the base of fractional powers are absolutely the same as the methods for solving irrational equations (they will be discussed in the next paragraph). So it is convenient to call them irrational and consider them in this light. But let's be honest with ourselves: initially we have an equation in front of us, not , and the language is not very willing to call the original equation irrational due to the absence of a root in the notation. Get away from these controversial issues Regarding terminology, the same technique allows: to call the equation simply an equation without any specific clarifications.

The simplest irrational equations

It is worth mentioning about the so-called simplest irrational equations. Let’s say right away that this term does not appear in the main textbooks of algebra and elementary analysis, but is sometimes found in problem books and training manuals, as, for example, in. It should not be considered generally accepted, but it does not hurt to know what is usually understood by the simplest irrational equations. This is usually the name given to irrational equations of the form , where f(x) and g(x) are some . In this light, the simplest irrational equation can be called, for example, the equation or .

How can one explain the appearance of such a name as “the simplest irrational equations”? For example, the fact that solving irrational equations often requires their initial reduction to the form and further use of any standard methods solutions. Irrational equations in this form are called the simplest.

Basic methods for solving irrational equations

By definition of a root

One of the methods for solving irrational equations is based on. With its help, irrational equations of the simplest form are usually solved , where f(x) and g(x) are some rational expressions(we gave the definition of the simplest irrational equations in). Irrational equations of the form are solved in a similar way , but in which f(x) and/or g(x) are expressions other than rational. However, in many cases it is more convenient to solve such equations by other methods, which will be discussed in the following paragraphs.

For the convenience of presenting the material, we separate irrational equations with even root exponents, that is, the equations , 2·k=2, 4, 6, … , from equations with odd root exponents , 2 k+1=3, 5, 7, … Let’s immediately outline approaches to solving them:

The above approaches follow directly from And .

So, method for solving irrational equations by definition of a root is as follows:

By definition of a root, it is most convenient to solve the simplest irrational equations with numbers on the right sides, that is, equations of the form , where C is a certain number. When there is a number on the right side of the equation, then even if the root exponent is even, there is no need to go to the system: if C is a non-negative number, then, by definition, a root of even degree, and if C is a negative number, then we can immediately conclude that there are no roots of the equation, After all, by definition, a root of an even degree is a non-negative number, which means that the equation does not turn into a true numerical equality for any real values ​​of the variable x.

Let's move on to solving typical examples.

We will go from simple to complex. Let's start by solving the simplest irrational equation, on the left side of which there is a root of an even degree, and on the right side - a positive number, that is, by solving an equation of the form , where C is a positive number. Determining the root allows you to move from solving a given irrational equation to solving a simpler equation without roots С 2·k =f(x) .

The simplest irrational equations with zero on the right side are solved in a similar way by defining a root.

Let us dwell separately on irrational equations, on the left side of which there is a root of an even degree with a variable under its sign, and on the right side there is a negative number. Such equations have no solutions on the set of real numbers (about complex roots we will talk after meeting you complex numbers ). This is pretty obvious: an even root is by definition a non-negative number, which means it cannot be equal to a negative number.

The left sides of the irrational equations from the previous examples were roots of even powers, and the right sides were numbers. Now let's consider examples with variables on the right sides, that is, we will solve irrational equations of the form . To solve them, by determining the root, a transition is made to the system , which has the same set of solutions as the original equation.

It must be borne in mind that the system , to the solution of which the solution of the original irrational equation is reduced , it is advisable to solve not mechanically, but, if possible, rationally. It is clear that this is more of a question from the topic “ systems solution“, but still we list three frequently encountered situations with examples illustrating them:

1. For example, if its first equation g 2·k (x)=f(x) has no solutions, then there is no point in solving the inequality g(x)≥0, because from the absence of solutions to the equation one can conclude that there are no solutions to the system .

1. Similarly, if the inequality g(x)≥0 has no solutions, then it is not necessary to solve the equation g 2·k (x)=f(x), because even without this it is clear that in this case the system has no solutions.

1. Quite often, the inequality g(x)≥0 is not solved at all, but only checked which of the roots of the equation g 2·k (x)=f(x) satisfy it. The set of all those that satisfy the inequality is a solution to the system, which means it is also a solution to the original irrational equation equivalent to it.

Enough about equations with even exponents of roots. It's time to pay attention to irrational equations with roots of odd powers of the form . As we have already said, to solve them we move to the equivalent equation , which can be solved by any available methods.

To conclude this point, let us mention checking solutions. The method of solving irrational equations by determining the root guarantees the equivalence of transitions. This means that it is not necessary to check the solutions found. This point can be attributed to the advantages this method solving irrational equations, because in most other methods, verification is a mandatory stage of the solution, which allows you to cut off extraneous roots. But it should be remembered that checking by substituting the found solutions into the original equation is never superfluous: suddenly a computational error has crept in.

We also note that the issue of checking and filtering out extraneous roots is very important when solving irrational equations, so we will return to it in one of the next paragraphs of this article.

Method of raising both sides of an equation to the same power

Further presentation assumes that the reader has an idea of ​​equivalent equations and corollary equations.

The method of raising both sides of an equation to the same power is based on the following statement:

Statement

raising both sides of an equation to the same even number natural degree gives the corollary equation, and raising both sides of the equation to the same odd natural power gives an equivalent equation.

Proof

Let us prove it for equations with one variable. For equations with several variables, the principles of the proof are the same.

Let A(x)=B(x) be the original equation and x 0 be its root. Since x 0 is the root of this equation, then A(x 0)=B(x 0) – true numerical equality. We know this property of numerical equalities: term-by-term multiplication of true numerical equalities gives a true numerical equality. Let us multiply term by term 2·k, where k – natural number, correct numerical equalities A(x 0)=B(x 0), this will give us the correct numerical equality A 2·k (x 0)=B 2·k (x 0) . And the resulting equality means that x 0 is the root of the equation A 2·k (x)=B 2·k (x), which is obtained from the original equation by raising both sides to the same even natural power 2·k.

To justify the possibility of the existence of a root of the equation A 2·k (x)=B 2·k (x) , which is not the root of the original equation A(x)=B(x) , it is enough to give an example. Consider the irrational equation , and equation , which is obtained from the original by squaring both parts. It is easy to check that zero is the root of the equation , really, , that the same thing 4=4 is a true equality. But at the same time, zero is an extraneous root for the equation , since after substituting zero we obtain the equality , which is the same as 2=−2 , which is incorrect. This proves that an equation obtained from the original one by raising both sides to the same even power can have roots foreign to the original equation.

It has been proven that raising both sides of an equation to the same even natural power leads to a corollary equation.

It remains to prove that raising both sides of the equation to the same odd natural power gives an equivalent equation.

Let us show that each root of the equation is the root of an equation obtained from the original by raising both its parts to an odd power, and conversely, that each root of the equation obtained from the original by raising both its parts to an odd power is the root of the original equation.

Let us have the equation A(x)=B(x) . Let x 0 be its root. Then the numerical equality A(x 0)=B(x 0) is true. While studying the properties of true numerical equalities, we learned that true numerical equalities can be multiplied term by term. Multiplying term by term 2·k+1, where k is a natural number, the correct numerical equalities A(x 0)=B(x 0) we obtain the correct numerical equality A 2·k+1 (x 0)=B 2·k+1 ( x 0) , which means that x 0 is the root of the equation A 2·k+1 (x)=B 2·k+1 (x) . Now back. Let x 0 be the root of the equation A 2·k+1 (x)=B 2·k+1 (x) . This means that the numerical equality A 2·k+1 (x 0)=B 2·k+1 (x 0) is correct. Due to the existence of an odd root of any real number and its uniqueness, the equality will also be true. This, in turn, due to the identity , where a is any real number that follows from the properties of roots and powers, can be rewritten as A(x 0)=B(x 0) . This means that x 0 is the root of the equation A(x)=B(x) .

It has been proven that raising both sides of an irrational equation to an odd power gives an equivalent equation.

The proven statement replenishes the arsenal known to us, used to solve equations, with another transformation of equations - raising both sides of the equation to the same natural power. Raising both sides of an equation to the same odd power is a transformation leading to a corollary equation, and raising it to an even power is an equivalent transformation. The method of raising both sides of the equation to the same power is based on this transformation.

Raising both sides of an equation to the same natural power is mainly used to solve irrational equations, since in certain cases this transformation allows one to get rid of the signs of the roots. For example, raising both sides of the equation to the power of n gives the equation , which can later be transformed into the equation f(x)=g n (x) , which no longer contains a root on the left side. The above example illustrates the essence of the method of raising both sides of the equation to the same power: using an appropriate transformation, obtain a simpler equation that does not have radicals in its notation, and through its solution, obtain a solution to the original irrational equation.

Now we can proceed directly to the description of the method of raising both sides of the equation to the same natural power. Let's start with an algorithm for solving, using this method, the simplest irrational equations with even root exponents, that is, equations of the form , where k is a natural number, f(x) and g(x) are rational expressions. An algorithm for solving the simplest irrational equations with odd root exponents, that is, equations of the form , we'll give it a little later. Then let's go even further: let's extend the method of raising both sides of an equation to the same power to more complex irrational equations containing roots under the signs of the roots, several signs of the roots, etc.

method of raising both sides of the equation to the same even power:

From the above information it is clear that after the first step of the algorithm we will arrive at an equation whose roots contain all the roots of the original equation, but which may also have roots that are foreign to the original equation. Therefore, the algorithm contains a clause about filtering out extraneous roots.

Let's look at the application of the given algorithm for solving irrational equations using examples.

Let's start by solving a simple and fairly typical irrational equation, squaring both sides of which leads to a quadratic equation that has no roots.

Here is an example in which all the roots of the equation obtained from the original irrational equation by squaring both sides turn out to be extraneous to the original equation. Conclusion: it has no roots.

The next example is a little more complicated. Its solution, unlike the previous two, requires raising both parts not to the square, but to the sixth power, and this will no longer lead to a linear or quadratic equation, but to a cubic equation. Here a check will show us that all three of its roots will be the roots of the irrational equation given initially.

And here we will go even further. To get rid of the root, you will have to raise both sides of the irrational equation to the fourth power, which in turn will lead to an equation of the fourth power. Checking will show that only one of the four potential roots will be the desired root of the irrational equation, and the rest will be extraneous.

Three recent examples are an illustration of the following statement: if, when raising both sides of an irrational equation to the same even power, an equation is obtained that has roots, then their subsequent verification can show that

• or they are all extraneous roots for the original equation, and it has no roots,
• or there are no extraneous roots among them at all, and they are all roots of the original equation,
• or only some of them are outsiders.

The time has come to move on to solving the simplest irrational equations with an odd root exponent, that is, equations of the form . Let's write down the corresponding algorithm.

Algorithm for solving irrational equations method of raising both sides of an equation to the same odd power:

• Both sides of the irrational equation are raised to the same odd power 2·k+1.
• The resulting equation is solved. Its solution is the solution to the original equation.

Please note: the above algorithm, in contrast to the algorithm for solving the simplest irrational equations with an even root exponent, does not contain a clause regarding the elimination of extraneous roots. We showed above that raising both sides of the equation to an odd power is an equivalent transformation of the equation, which means that such a transformation does not lead to the appearance of extraneous roots, so there is no need to filter them out.

Thus, solving irrational equations by raising both sides to the same odd power can be carried out without eliminating outsiders. At the same time, do not forget that when raising to an even power, verification is required.

Knowing this fact allows us to legally do not filter out extraneous roots when solving an irrational equation . Especially in in this case the check involves “unpleasant” calculations. There will be no extraneous roots anyway, since it is raised to an odd power, namely to a cube, which is an equivalent transformation. It is clear that the check can be performed, but more for self-control, in order to further verify the correctness of the solution found.

Let's sum up the intermediate results. At this point, we, firstly, expanded the already known arsenal of solving various equations with another transformation, which consists in raising both sides of the equation to the same power. When raised to an even power, this transformation may be unequal, and when using it, it is necessary to check to filter out extraneous roots. When raised to an odd power, the specified transformation is equivalent, and it is not necessary to filter out extraneous roots. And secondly, we learned to use this transformation to solve the simplest irrational equations of the form , where n is the root exponent, f(x) and g(x) are rational expressions.

Now it's time to look at raising both sides of the equation to the same power from a general perspective. This will allow us to extend the method of solving irrational equations based on it from the simplest irrational equations to irrational equations of a more complex type. Let's do this.

In fact, when solving equations by raising both sides of the equation to the same power, we use the already known general approach: the original equation, through some transformations, is transformed into a simpler equation, it is transformed into an even simpler one, and so on, up to an equation that we are able to solve. It is clear that if in a chain of such transformations we resort to raising both sides of the equation to the same power, then we can say that we are following the same method of raising both sides of the equation to the same power. All that remains is to figure out exactly what transformations and in what sequence need to be carried out to solve irrational equations by raising both sides of the equation to the same power.

Here is a general approach to solving irrational equations by raising both sides of the equation to the same power:

• First, we need to move from the original irrational equation to a more simple equation, which can usually be achieved by cyclically performing the following three actions:
  • Isolation of the radical (or similar techniques, for example, isolation of the product of radicals, isolation of a fraction whose numerator and/or denominator is a root, which allows, upon subsequent raising of both sides of the equation to a power, to get rid of the root).
  • Simplifying the form of the equation.
• Secondly, you need to solve the resulting equation.
• Finally, if during the solution there were transitions to corollary equations (in particular, if both sides of the equation were raised to an even power), then extraneous roots need to be eliminated.

Let's put the acquired knowledge into practice.

Let's solve an example in which the solitude of the radical brings the irrational equation to its simplest form, after which all that remains is to square both sides, solve the resulting equation and weed out extraneous roots using a check.

The following irrational equation can be solved by isolating the fraction with a radical in the denominator, which can be eliminated by subsequent squaring of both sides of the equation. And then everything is simple: the result is solved fractional rational equation and a check is made to ensure that no extraneous roots are included in the response.

Irrational equations that contain two roots are quite typical. They are usually successfully solved by raising both sides of the equation to the same power. If the roots have same degree, and besides them there are no other terms, then to get rid of radicals it is enough to isolate the radical and perform exponentiation once, as in the following example.

And here is an example in which there are also two roots, besides them there are also no terms, but the degrees of the roots are different. In this case, after isolating the radical, it is advisable to raise both sides of the equation to a power that eliminates both radicals at once. Such a degree serves, for example, as indicators of roots. In our case, the degrees of the roots are 2 and 3, LCM(2, 3) = 6, therefore, we will raise both sides to the sixth power. Note that we can also act along the standard path, but in this case we will have to resort to raising both parts to a power twice: first to the second, then to the third. We will show both solutions.

In more complex cases, when solving irrational equations by raising both sides of the equation to the same power, you have to resort to raising the power twice, less often - three times, and even less often - larger number once. The first irrational equation, illustrating what has been said, contains two radicals and one more term.

Solving the following irrational equation also requires two successive exponentiations. If you do not forget to isolate radicals, then two exponentiations are enough to get rid of the three radicals present in its notation.

The method of raising both sides of an irrational equation to the same power allows one to cope with irrational equations in which under the root there is another root. Here is the solution to a typical example.

Finally, before moving on to the analysis of the following methods for solving irrational equations, it is necessary to note the fact that raising both sides of an irrational equation to the same power can, as a result of further transformations, give an equation that has an infinite number of solutions. An equation that has infinitely many roots is obtained, for example, by squaring both sides of the irrational equation and subsequent simplification of the form of the resulting equation. However, for obvious reasons, we are not able to perform a substitution check. In such cases, you have to either resort to other verification methods, which we will talk about, or abandon the method of raising both sides of the equation to the same power in favor of another solution method, for example, in favor of a method that assumes.

We examined solutions to the most typical irrational equations by raising both sides of the equation to the same power. The general approach studied makes it possible to cope with other irrational equations, if this solution method is suitable for them at all.

Method for introducing a new variable

When is it still quite easy to see the possibility of introducing a new variable? When the equation contains “inverted” fractions and (with your permission, we will call them mutually inverse by analogy with ). How would we solve a rational equation with fractions like these? We would take one of these fractions as a new variable t, while the other fraction would be expressed through the new variable as 1/t. In irrational equations, introducing a new variable in this way is not entirely practical, since in order to further get rid of the roots, most likely, you will have to introduce another variable. It is better to immediately accept the root of the fraction as a new variable. Well, then transform the original equation using one of the equalities And , which will allow you to move to an equation with a new variable. Let's look at an example.

Do not forget about already known replacement options. For example, the expression x+1/x and x 2 +1/x 2 may appear in the recording of an irrational equation, which makes one think about the possibility of introducing a new variable x+1/x=t. This thought does not arise by chance, because we already did this when we decided reciprocal equations. This method of introducing a new variable, like other methods already known to us, should be kept in mind when solving irrational equations, as well as equations of other types.

We move on to more complex irrational equations, in which it is suitable for introducing a new variable expression harder to see. And let's start with equations in which the radical expressions are the same, but, unlike the case discussed above, the larger exponent of one root is not completely divided by the smaller exponent of the other root. Let's figure out how to choose the right expression to introduce a new variable in such cases.

When the radical expressions are the same, and the larger exponent of one root k 1 is not completely divided by the smaller exponent of the other root k 2 , the root of the degree LCM (k 1 , k 2) can be taken as a new variable, where LCM is . For example, in an irrational equation the roots are equal to 2 and 3, three is not a multiple of two, LCM(3, 2)=6, so a new variable can be introduced as . Further, the definition of the root, as well as the properties of the roots, allows you to transform the original equation in order to explicitly select the expression and then replace it with a new variable. We present the complete and detailed solution this equation.

Using similar principles, a new variable is introduced in cases where the expressions under the roots differ in degrees. For example, if in an irrational equation the variable is contained only under the roots, and the roots themselves have the form and , then you should calculate the least common multiple of the roots LCM(3, 4) = 12 and take . Moreover, according to the properties of roots and powers, the roots should be transformed as And accordingly, which will allow you to introduce a new variable.

You can act in a similar way in irrational equations, in which under the roots with different indicators there are mutually inverse fractions and . That is, it is advisable to take a root with an indicator equal to the LCM of the root indicators as a new variable. Well, then move on to the equation with a new variable, which allows us to make equalities And , definition of a root, as well as properties of roots and powers. Let's look at an example.

Now let's talk about equations in which the possibility of introducing a new variable can only be suspected, and which, if successful, opens only after quite serious transformations. For example, an irrational equation is only reduced to the form after a series of not so obvious transformations, which opens the way to the replacement . Let's give a solution to this example.

Finally, let's add a little exoticism. Sometimes an irrational equation can be solved by introducing more than one variable. This approach to solving equations is proposed in the textbook. There to solve the irrational equation it is proposed to enter two variables . The textbook provides a short solution, let's restore the details.

Solving irrational equations using the factorization method

In addition to the method of introducing a new variable, other methods are used to solve irrational equations. general methods, in particular, the factorization method. The article at the link indicated in the previous sentence discusses in detail when the factorization method is used, what its essence is and what it is based on. Here we are more interested not in the method itself, but in its use in solving irrational equations. Therefore, we will present the material as follows: we will briefly recall the main provisions of the method, after which we will analyze in detail the solutions to characteristic irrational equations using the method of factorization.

The factorization method is used to solve equations, the left sides of which contain a certain product, and the right sides contain zeros, that is, to solve equations of the form f 1 (x) f 2 (x) f n (x)=0, where f 1, f 2, …, f n are some functions. The essence of the method is to replace the equation f 1 (x) f 2 (x) f n (x)=0 on the variable x for the original equation.

First part last sentence about the transition to the totality follows from the known primary school fact: the product of several numbers is equal to zero if and only if at least one of the numbers is equal to zero. The presence of the second part about ODZ is explained by the fact that the transition from the equation f 1 (x) f 2 (x) f n (x)=0 to a set of equations f 1 (x)=0, f 2 (x)=0, …, f n (x)=0 may be unequal and lead to the appearance of extraneous roots, which in this case can be eliminated by taking into account ODZ. It is worth noting that screening out extraneous roots, if convenient, can be carried out not only through ODZ, but also in other ways, for example, by checking by substituting the found roots into the original equation.

So, to solve the equation f 1 (x) f 2 (x) f n (x)=0 using the method of factorization, including irrational, it is necessary

• Go to set of equations f 1 (x)=0, f 2 (x)=0, …, f n (x)=0,
• Solve the composed set,
• If the set of solutions does not have, then conclude that the original equation has no roots. If there are roots, then weed out extraneous roots.

Let's move on to the practical part.

The left-hand sides of typical irrational equations, which are solved by factorization, are the products of several algebraic expressions, usually linear binomials and quadratic trinomials, and several roots with algebraic expressions underneath them. There are zeros on the right sides. Such equations are ideal for gaining initial skills in solving them. We will begin by solving a similar equation. In doing so, we will try to achieve two goals:

• consider all steps of the factorization method algorithm when solving an irrational equation,
• recall the three main ways of sifting out extraneous roots (by ODZ, by ODZ conditions, and by directly substituting solutions into the original equation).

The following irrational equation is typical in the sense that when solving it using the method of factorization, it is convenient to filter out extraneous roots according to the conditions of the ODZ, and not according to the ODZ in the form of a numerical set, since it is difficult to obtain the ODZ in the form of a numerical factor. The difficulty is that one of the conditions defining DL is irrational inequality . This approach to sifting out extraneous roots makes it possible to do without solving it; moreover, sometimes in the school course mathematicians are not taught at all about solving irrational inequalities.

It’s good when the equation has a product on the left side and a zero on the right. In this case, you can immediately go to the set of equations, solve it, find and discard roots extraneous to the original equation, which will give the desired solution. But more often the equations have a different form. If at the same time there is an opportunity to transform them into a form suitable for applying the factorization method, then why not try to carry out the appropriate transformations. For example, to obtain the product on the left side of the following irrational equation, it is enough to resort to the difference of squares.

There is another class of equations that are usually solved by factorization. It includes equations, both sides of which are products that have the same factor in the form of an expression with a variable. This is, for example, the irrational equation . You can go by dividing both sides of the equation by the same factor, but you must not forget to separately check the values ​​that make these expressions vanish, otherwise you may lose solutions, because dividing both sides of the equation by the same expression may be an unequal transformation. It is more reliable to use the factorization method; this makes it possible to guarantee that the roots will not be lost during further correct solution. It is clear that to do this, you first need to get the product on the left side of the equation, and zero on the right side. It's easy: just move the expression from the right side to the left, changing its sign, and take out common multiplier out of brackets. We'll show you complete solution a similar, but slightly more complex irrational equation.

It is useful to start solving any equation (as, indeed, solving many other problems) by finding the ODZ, especially if the ODZ is easy to find. Let us give some of the most obvious arguments in favor of this.

So, having received the task of solving an equation, you should not rush into transformations and calculations without looking back, maybe just look at the ODZ? This is clearly demonstrated by the following irrational equation.

Functional graphic method

Graphical method

Using the properties of increasing and decreasing functions

As we have already noted, graphic method solving irrational equations is inconvenient in cases where the expressions on the left and right sides of the equation are quite complex in the sense that it is not easy to construct the corresponding function graphs. But quite often, instead of graphs, you can refer to the properties of functions. There is a method for solving equations that uses the monotonicity of functions corresponding to parts of the equation. In particular, this method allows you to solve irrational equations. It is based on the following statement:

Statement

if on a set X the function f is defined and strictly monotonic (increases or decreases), then the equation f(x)=C, where C is a certain number, either has one single root or has no roots on the specified set.

The following statement boils down to it:

Statement

if the functions f and g are defined on a set X and one of them increases and the other decreases, then the equation f(x)=g(x) either has one single root or has no roots on the set X.

These statements are usually used to solve equations when it is possible to somehow determine one root of the equation, and it is possible to prove the increase and decrease of the corresponding functions.

As for finding the root of the equation, in typical cases it is obvious or easy to guess. Usually, the root of an irrational equation is some number from the ODZ, when substituting it into the original equation under the roots, we obtain numbers whose roots can be easily extracted.

As for the proof of increasing-decreasing functions, it is usually carried out based on the properties of basic elementary functions and known properties of increasing and decreasing functions(such as the root of an increasing function is an increasing function), or in more complex cases the derivative is used for proof.

Let's look at these points when solving irrational equations.

Let's start with solving a typical irrational equation: the increase of the function corresponding to one of its parts is proved, the decrease of the function corresponding to the other part of the equation is proved, and a root is selected from the ODZ of the variable for the equation, which in this case will be unique.

The following irrational equation also has to be solved using the functional-graphical method. The root of the equation is easy to find, as in the previous example, but here the increase of one function and the decrease of another function have to be proven using the derivative.

Let us summarize the issue of using the properties of increasing and decreasing functions when solving equations:

• if the root of the equation is visible, then you can try to examine the functions corresponding to the left and right sides of the equation for increasing and decreasing. Perhaps this will allow us to prove the uniqueness of the found root.
• if it is clear that one of the functions f and g is decreasing and the other is increasing, then you should try to find the only possible root of the equation by any in an accessible way. If we can find this root, then the equation will be solved.

Evaluation method

Finally, we come to the last of the three main varieties of the functional-graphical method for solving equations, which is based on the use of boundedness of functions. Let's agree to call this type of functional-graphic method the assessment method.

The estimation method is usually used to solve equations of the form f(x)=C, where f(x) is some expression with variable x (and f is the corresponding function), C is some number, or the form g(x)=h(x) , where g(x) and h(x) are some expressions with variable x (and g and h are the corresponding functions). Note that the equation g(x)=h(x) can always be reduced to an equivalent equation of the form f(x)=C (in particular, by transferring the expression h(x) from the right side to the left side opposite sign), that is, we can limit ourselves to considering the estimation method only for equations of the form f(x)=C. However, sometimes it is quite convenient to work with equations of the form g(x)=h(x) , so we will not refuse to consider them.

Solving equations using the estimation method is carried out in two stages. The first stage is estimating the values ​​of the function f (or the corresponding expression f(x), which is essentially the same thing), if the equation f(x)=C is solved, or estimating the values ​​of the functions g and h (or the corresponding expressions f(x ) and g(x) ), if the equation g(x)=h(x) is solved. The second stage is the use of the obtained estimates to further search for the roots of the equation or justify their absence. Let's clarify these points.

How are function values ​​evaluated? This issue is discussed in detail in. Here we will limit ourselves to listing the estimation methods that are most often used when solving irrational equations using the estimation method. Here is the list of evaluation methods:

• Evaluation based on the definition of a root with an even exponent. Since, by definition, a root with an even exponent is a non-negative number, then for any x from the ODZ for the expression , where n is a natural number, p(x) is some expression, the inequality is true, and if and only if p(x)= 0 .
• Estimation based on the following property of roots: for any non-negative numbers a and b, a , ≥ ), the inequality (≤ , > , ≥ ) is satisfied. If for any x from the OD the inequality p(x) is satisfied for the expression , ≥ ), where c is some non-negative number, then for any x from the ODZ the inequality (≤ , > , ≥ ) is true.
• An estimate based on the fact that the power of any number with an even exponent is a non-negative number. For any x from the ODZ, for the expression p 2·n (x) the inequality p 2·n (x)≥0 is true, and p 2·n (x)=0 if and only if p(x)=0.
• Estimating the values ​​of a quadratic trinomial. To estimate, you can use the ordinate of the vertex of the parabola, and with a negative discriminant - zero.
  • If a>0, then a x 2 +b x+c≥y 0, where y 0 is the ordinate of the vertex of the parabola, and if a<0 , то a·x 2 +b·x+c≤y 0 .
  • If a>0 and discriminant D<0 , то a·x 2 +b·x+c>0 , and if a<0 и D<0 , то a·x 2 +b·x+c<0 .
• Estimation based on properties of numerical inequalities.
• Estimation through the largest and smallest value of a function found using the derivative. If A is the smallest value of a function p on a set X, then the inequality p(x)≥A is true on X. If B is the largest value of a function p on a set X, then the inequality p(x)≤B holds on X.

Let's say we have completed the first stage, that is, we have estimated the values ​​of the functions. A logical question arises about how to further use the obtained estimates to solve the equation. And then you need to refer to one of the following statements:

The provisions of the second block of statements follow from the properties of addition and multiplication of true numerical inequalities of the same meaning.

The first block of positions becomes clear if you imagine the relative position of the graph of the function f and the straight line y=C, and the positions of the remaining blocks - if you imagine the relative position of the graphs of the functions g and h.

Let's look at the first block of statements. When the graph of a function f is below or not above the line y=A, which in turn is below the line y=C, then it is clear that it does not intersect with the line y=C, which implies the absence of roots of the equation f(x)=C. When the graph of a function f is higher or not lower than the straight line y=B, which in turn is higher than the straight line y=C, then it is clear that it does not intersect with the straight line y=C, this implies the absence of roots of the equation f(x)=C. When the graph of a function f is below or above the line y=C, then it is clear that it does not intersect with this line; this also implies the absence of roots of the equation f(x)=C.

Now let's justify the third block of statements. Let on the set X the values ​​of the function g be less than or not greater than the number A, and the values ​​of the function h be greater than or not less than the number B. This means that all points on the graph of function g are below or not above the line y=A, and points on the graph of function h are above or not below the line y=B. It is clear that on the set X for A

Let's move on to the fourth block of statements. Here, in the first case, one graph is located below this line, the other is located above this line. In the second case, one graph is not above this line, the other is above this line. In the third case, one graph is below this line, the other is not below this line. It is clear that in all cases the graphs do not have common points, which means that the equation g(x) = h(x) has no solutions.

In the latter situation, the graph of one function is not higher than the straight line y=C, and the graph of the other function is not lower than this straight line. It is clear that graphs can have common points only on this line. This explains the transition from the equation g(x)=h(x) to the system.

You can move on to practice. Let us consider solutions to characteristic irrational equations using the estimation method.

First, it’s worth understanding the issue of the accuracy of estimating the values ​​of expressions. To make it clear where this question comes from, look at three estimates of the root values: first , second, third, and tell me which one to prefer? Well, we’ll discard the first one, since it is mostly far-fetched, but the second and third estimates are quite workable, and depending on the situation, both the first of them, relatively rough, and the second can be used. Let's look at this issue from a practical perspective.

To prove that an equation has no solutions, rough estimates are sufficient. The main advantage of rough estimates over more precise estimates is their relative ease of obtaining. Rough estimates are practically obvious and do not require additional research, since they are based on well-known facts, such as: the square root is a non-negative number, the modulus is a non-negative number, the square of a number is a non-negative number, the sum of positive reciprocals is not less than two, the values ​​of a quadratic trinomial with a negative leading term and a negative discriminant are negative, etc. So, to solve the following irrational equation by the estimation method, a rough estimate of the root on the one hand and the quadratic trinomial on the other hand is sufficient.

It is usually easier to obtain rough estimates of the values ​​of functions or expressions than accurate ones. But quite often, rough estimates do not allow us to draw conclusions about the roots of the equations being solved, while more accurate estimates make this possible. Let's solve a typical irrational equation.

Let's start with solving a simple but very characteristic irrational equation: the estimate of the values ​​of its left side follows from the estimates of its constituent roots, and from the resulting estimate the conclusion follows that there are no roots of the equation.

The situation is more interesting when the expression corresponding to the left side of the irrational equation f(x)=C is the sum or product of several expressions and its values ​​are estimated as f(x)≤C or f(x)≥C. In such cases, the statements written above prescribe the transition from the original irrational equation to an equivalent system of equations. Let us present the solution to a characteristic irrational equation.

Let's consolidate the skills of transition using the estimation method from the irrational equation f(x) = C with a sum or product on the left side to an equivalent system of equations. To do this, we will solve a relatively complex irrational equation, the left side of which is the sum of two irrational expressions, one of which is the product of two expressions. The solution principle is the same: we obtain an estimate that allows us to move from the original equation to an equivalent system.

Let's move on to irrational equations of the form g(x)=h(x) .

The previous examples were quite simple in terms of evaluating the values ​​of expressions and functions. It's time to work on the evaluation aspect in more detail. For obvious reasons, we will focus on estimation methods that have to be resorted to most often when solving irrational equations using the estimation method. Let's start with estimation methods that do not require finding the derivative. So, in order to solve the following irrational equation, you will have to use almost all known means: from the property of powers with an even exponent and the property of monotonicity of the root extraction function to estimates based on the properties of numerical equalities.

The methods for obtaining estimates that we used in all previous examples do not completely cover the issue of estimating values. In other words, it is not always possible to evaluate the values ​​of functions and expressions with their help. In particular, the considered methods are not good when the range of permissible values ​​of the variable x for the irrational equation being solved is different from the set of all real numbers R. As an example, we give an estimate of the root in two cases: when the ODZ is a set R and when the ODZ is a segment from 3 to 5. Based on the evaluation methods we used above, we can get the estimate . For the case when the ODZ is a set R, this estimate is very good. But for the case when the ODZ is a segment, the recorded estimate already turns out to be relatively rough, and it is possible to estimate the root more accurately, namely as . But it’s not just the DL that limits the possibilities of obtaining estimates using the methods discussed above. Often these methods do not provide the ability to estimate function values ​​due to the type of function being estimated. For example, the estimation methods we are talking about allow us to estimate the values ​​of the roots and , as well as their sum: , , whence and further . But these estimation methods no longer allow us to estimate the difference between the indicated roots. In such situations, one has to resort to studying the function, finding its largest and smallest values, through which to evaluate the values ​​of the function. Sometimes it is convenient to combine different methods of obtaining estimates. Let us show the solution to a characteristic irrational equation.

Concluding the conversation about solving irrational equations using the functional-graphical method and the estimation method in particular, let us remember one promise given at the end of the paragraph dedicated to. Remember, we solved the irrational equation in a rather exotic way through the introduction of two new variables (which still needed to be thought of), and they promised to show its solution using a more standard method. This method in this case is the assessment method. So let's fulfill our promise.

Solving irrational equations through ODZ

Very often part of the process of solving equations is. The reasons that force you to look for DL ​​can be different: it is necessary to carry out transformations of the equation, and they, as is known, are carried out on DL, the chosen solution method involves finding DL, checking using DL, etc. And in certain cases, ODZ acts not only as an auxiliary or control tool, but also allows one to obtain a solution to the equation. Here we mean two situations: when the ODZ is an empty set and when the ODZ is a finite set of numbers.

It is clear that if the ODZ of an equation, in particular an irrational one, is an empty set, then the equation has no solutions. So the ODZ of variable x for the following irrational equation is an empty set, which means that the equation has no solutions.

When the ODZ of a variable for an equation is a finite set of numbers, then by sequentially checking by substituting these numbers, one can obtain a solution to the equation. For example, consider an irrational equation for which the ODZ consists of two numbers, and substitution shows that only one of them is the root of the equation, from which it is concluded that this root is the only solution to the equation.

Solving irrational equations of the form “fraction equals zero”

Irrational equations reducing to numerical equalities

Go to modules

If in the notation of an irrational equation under the sign of a root of an even degree there is a degree of some expression with an exponent equal to the exponent of the root, then you can go to the modulus. This transformation takes place due to one of the formulas, where 2·m is an even number, a is any real number. It is worth noting that this transformation is an equivalent transformation of the equation. Indeed, with such a transformation, the root is replaced by an identically equal module, while the ODZ does not change.

Let us consider a characteristic irrational equation, which can be solved by passing to the modulus.

Is it always worth switching to modules when possible? In the vast majority of cases, such a transition is justified. The exception is those cases when it is obvious that alternative methods for solving an irrational equation require relatively less labor. Let's take an irrational equation that can be solved through the transition to modules and some other methods, for example, by squaring both sides of the equation or by determining the root, and see which solution will be the simplest and most compact.

In the solved example, the solution to determine the root looks preferable: it is shorter and simpler than both the solution through the transition to the module, and the solution by squaring both sides of the equation. Could we have known this before solving the equation using all three methods? Let's face it, it wasn't obvious. So when you look at several solution methods and it’s not immediately clear which one to prefer, you should try to get a solution with any of them. If this works out, then good. If the chosen method does not lead to results or the solution turns out to be very difficult, then you should try another method.

At the end of this point, let's return to the irrational equation. In the previous paragraph, we already solved it and saw that an attempt to solve it through isolating the radical and squaring both sides of the equation led to the numerical equality 0=0 and the impossibility of drawing a conclusion about the roots. And the solution to determining the root involved solving an irrational inequality, which in itself is quite difficult. A good method for solving this irrational equation is to go to moduli. Let's give a detailed solution.

Transformation of irrational equations

The solution of irrational equations is almost never complete without transforming them. By the time we study irrational equations, we are already familiar with equivalent transformations of equations. When solving irrational equations, they are used in the same way as when solving previously studied types of equations. You saw examples of such transformations of irrational equations in the previous paragraphs, and, you see, they were perceived quite naturally, since they are familiar to us. Above, we also learned about a new transformation for us - raising both sides of the equation to the same power, which is typical for irrational equations; in the general case, it is not equivalent. It’s worth talking about all these transformations in detail in order to know all the subtle points that arise during their implementation and avoid mistakes.

We will analyze transformations of irrational equations in the following sequence:

1. Replacing expressions with identically equal expressions that do not change the ODZ.
2. Adding the same number to both sides of an equation or subtracting the same number from both sides of an equation.
3. Adding the same expression, without changing the property value, to both sides of the equation, or subtracting the same expression, without changing the property value, from both sides of the equation.
4. Transferring terms from one side of the equation to another with the opposite sign.
5. Multiplying and dividing both sides of an equation by the same number other than zero.
6. Multiplying and dividing both sides of an equation by the same expression, which does not change the range of permissible values ​​of the variable and does not turn to zero on it.
7. Raising both sides of an equation to the same power.

So, the range of questions is outlined. Let's start understanding them with examples.

The first transformation that interests us is the replacement of expressions in the equation with identically equal expressions. We know that it is equivalent if the VA for the equation obtained as a result of the transformation is the same as the VA for the original equation. From this it is clear that there are two main reasons for the occurrence of errors when carrying out this transformation: the first is a change in the OD that occurs as a result of the transformation, the second is the replacement of an expression with an expression that is not identically equal to it. Let us examine these aspects in detail and in order, considering examples of typical transformations of this type.

First, let's go over the typical transformations of equations, which consist in replacing an expression with an identically equal expression, which are always equivalent. Here is the relevant list.

• Rearranging terms and factors. This transformation can be carried out on both the left and right sides of the irrational equation. It can be used, for example, to group and then reduce similar terms in order to simplify the form of the equation. Rearranging terms or factors is obviously an equivalent transformation of the equation. This is understandable: the original expression and the expression with the terms or factors rearranged are identically equal (if, of course, the rearrangement is carried out correctly), and it is obvious that such a transformation does not change the ODZ. Let's give an example. On the left side of the irrational equation in the product x·3·x, you can swap the first and second factors x and 3, which will subsequently allow you to represent the polynomial under the root sign in a standard form. And on the right side of the equation in the sum 4+x+5, you can swap the terms 4 and x, which in the future will allow you to add the numbers 4 and 5. After these rearrangements, the irrational equation will take the form , the resulting equation is equivalent to the original one.
• Expanding parentheses. The equivalence of this transformation of equations is obvious: the expressions before and after opening the brackets are identically equal and have the same range of permissible values. For example, let's take the irrational equation . His solution requires opening the parentheses. Opening the brackets on the left side of the equation, as well as on the right side of the equation, we arrive at an equivalent equation.
• Grouping of terms and/or factors. This transformation of an equation essentially represents the replacement of any expression that is part of the equation with an identically equal expression with grouped terms or factors. Obviously, this does not change the ODZ. This means that the indicated transformation of the equation is equivalent. For illustration, let's take an irrational equation. Rearranging the terms (we talked about it two paragraphs above) and grouping the terms allows us to move on to an equivalent equation. The purpose of such a grouping of terms is clearly visible - to carry out the following equivalent transformation, which will allow the introduction of a new variable.
• Bracketing out the common factor. It is clear that the expressions before putting the common factor out of brackets and after putting the common factor out of brackets are identically equal. It is also clear that putting the common factor out of brackets does not change the VA. Therefore, taking the common factor out of brackets in an expression that is part of an equation is an equivalent transformation of the equation. This transformation is used, for example, to represent the left side of an equation as a product in order to solve it by factorization. Here's a concrete example. Let's consider the irrational equation. The left side of this equation can be represented as a product; to do this, you need to take the common factor out of brackets. As a result of this transformation, the irrational equation will be obtained , equivalent to the original one, which can be solved by factorization.
• Replacing numeric expressions with their values. It is clear that if the equation contains a certain numerical expression, and we replace this numerical expression with its value (correctly calculated), then such a replacement will be equivalent. Indeed, in essence, an expression is replaced by an identically equal expression, and at the same time the ODZ of the equation does not change. Thus, replacing in the irrational equation the sum of two numbers −3 and 1 and the value of this sum, which is equal to −2, we obtain an equivalent irrational equation. Similarly, one can carry out an equivalent transformation of the irrational equation , performing operations with numbers under the root sign (1+2=3 and ), this transformation will lead us to the equivalent equation .
• Performing operations with monomials and polynomials found in the notation of an irrational equation. It is clear that the correct implementation of these actions will lead to an equivalent equation. Indeed, in this case the expression will be replaced by an identically equal expression and the OD will not change. For example, in the irrational equation you can add the monomials x 2 and 3 x 2 and go to the equivalent equation . Another example: subtracting polynomials on the left side of an irrational equation is an equivalent transformation that leads to an equivalent equation .

We continue to consider transformations of equations, which consist in replacing expressions with identically equal expressions. Such transformations may also be unequal, since they can change the ODZ. In particular, there may be an expansion of ODZ. This can occur when reducing similar terms, when reducing fractions, when replacing a product with several zero factors or a fraction with a numerator equal to zero by zero, and most often when using formulas corresponding to the properties of roots. By the way, careless use of the properties of roots can also lead to a narrowing of the ODZ. And if transformations that expand the ODZ are acceptable when solving equations (they can cause the appearance of extraneous roots, which are eliminated in a certain way), then transformations that narrow the ODZ must be abandoned, since they can cause the loss of roots. Let's dwell on these points.

The first irrational equation is . Its solution begins by transforming the equation to the form based on one of the properties of degrees. This transformation is equivalent, since the expression is replaced by an identically equal expression, and the ODZ does not change. But the next transition to the equation, carried out on the basis of the definition of the root, may already be an unequal transformation of the equation, since with such a transformation the ODZ is expanded. Let us show the complete solution to this equation.

The second irrational equation, well suited to illustrate that transformations of irrational equations using the properties of roots and the definition of a root can be unequal, is of the form . It's good if you don't allow yourself to start the solution like this

Or so

Let's start with the first case. The first transformation is the transition from the original irrational equation to the equation consists of replacing the expression x+3 with the expression . These expressions are identically equal. But with such a replacement, the ODZ narrows from the set (−∞, −3)∪[−1, +∞) to the set [−1, +∞) . And we agreed to abandon reforms that narrow the ODZ, since they can lead to the loss of roots.

What's wrong in the second case? Expansion of ODZ during the last transition from to the number −3? Not only that. Of great concern is the first transition from the original irrational equation to the equation . The essence of this transition is the replacement of the expression x+3 with the expression . But these expressions are not identically equal: for x+3<0 значения этих выражений не совпадают. Действительно, согласно свойству квадратного корня из квадрата , from which it follows that .

So how then to solve this irrational equation ? Here it is best to immediately introduce a new variable , in this case (x+3)·(x+1)=t 2. Let's give a detailed solution.

Let us summarize the first of the transformations of equations being analyzed - replacing an expression that is part of an equation with an expression identical to it. Each time when it is carried out, it is necessary to fulfill two conditions: first, that the expression is replaced by an identically equal expression, and second, that in this case a narrowing of the ODZ does not occur. If such a replacement does not change the ODZ, then the result of the transformation will be an equivalent equation. If during such a replacement the ODZ expands, then extraneous roots may appear, and care must be taken to filter them out.

Let's move on to the second transformation of the list - adding the same number to both sides of the equation and subtracting the same number from both sides of the equation. This is an equivalent transformation of the equation. We usually resort to it when there are identical numbers on the left and right sides of the equation; subtracting these numbers from both sides of the equation allows us to get rid of them in the future. For example, on both the left and right sides of the irrational equation there is a term 3. Subtracting a triple from both sides of the equation results in an equation that, after performing manipulations with numbers, takes the form and further simplified to . According to the result, the transformation in question has something in common with the transfer of a term from one part of the equation to another with the opposite sign, but more on this transformation a little later. There are other examples of this transformation being used. For example, in an irrational equation, adding the number 3 to both sides is necessary to organize a perfect square on the left side of the equation and further transform the equation into form in order to introduce a new variable.

A generalization of the transformation just discussed is adding to or subtracting the same expression from both sides of the equation. This transformation of equations is equivalent when the ODZ does not change. This transformation is carried out mainly in order to subsequently get rid of identical terms that are simultaneously on the left and right sides of the equation. Let's give an example. Let us assume that we have an irrational equation. It is obvious that there is a term on both the left and right sides of the equation. It is reasonable to subtract this expression from both sides of the equation: . In our case, such a transition does not change the ODZ, so the transformation performed is equivalent. And this is done in order to further move on to a simpler irrational equation.

The next transformation of equations, which we will touch on in this paragraph, is the transfer of terms from one part of the equation to another with the opposite sign. This transformation of the equation is always equivalent. The scope of its application is quite wide. With its help, you can, for example, isolate the radical or collect similar terms in one part of the equation, so that you can then reduce them and thereby simplify the form of the equation. Let's give an example. To solve an irrational equation you can move the terms −1 to the right side, changing their sign, this will give an equivalent equation , which can be solved further, for example, by squaring both sides of the equation.

We move further along the path of considering transformations of equations to multiply or divide both sides of the equation by the same number, different from zero. This transformation is an equivalent transformation of the equation. Multiplying both sides of an equation by the same number is used primarily to move from fractions to whole numbers. For example, so that in the irrational equation to get rid of fractions, you should multiply both parts by 8, which gives an equivalent equation , which is further reduced to the form . The division of both sides of the equation is carried out mainly for the purpose of reducing the numerical coefficients. For example, both sides of the irrational equation It is advisable to divide by numerical coefficients 18 and 12, that is, by 6, such division gives the equivalent equation , from which we can later move on to the equation , which has smaller, but also integer coefficients.

The next transformation of an equation is to multiply and divide both sides of the equation by the same expression. This transformation is equivalent when the expression by which the multiplication or division is performed does not change the range of permissible values ​​of the variable and does not turn to zero on it. Typically, multiplying both sides by the same expression is similar for purposes to multiplying both sides of an equation by the same number. Most often, this transformation is resorted to in order to get rid of fractions by further transformations. Let's show this with an example.

We will not ignore irrational equations, to solve which we have to resort to dividing both sides of the equation by the same expression. We noted a little higher that such a division is an equivalent transformation if it does not affect the ODZ and this expression on the ODZ does not vanish. But sometimes division has to be carried out by an expression that vanishes in the ODZ. This is quite possible to do if at the same time you separately check the zeros of this expression to see if there are any roots of the equation being solved among them, otherwise these roots may be lost during such a division.

The last transformation of irrational equations that we will touch on in this paragraph is to raise both sides of the equation to the same power. This transformation can be called typical for irrational equations, since it is practically not used when solving equations of other types. We have already mentioned this transformation in the current article, when we examined . There are also many examples of this transformation. We will not repeat ourselves here, but just recall that in the general case this transformation is not equivalent. It can lead to the appearance of extraneous roots. Therefore, if during the solution process we turned to this transformation, then the found roots must be checked for the presence of extraneous roots among them.

About losing roots

What can cause loss of roots when solving an equation? The main reason for the loss of roots is the transformation of the equation, which narrows the OD. To understand this point, let's look at an example.

Let's take the irrational equation , which we have already solved within the current article. We began solving it with a warning against carrying out the following transformations of the equation

The very first transformation is the transition from the equation to the equation – narrows the ODZ. Indeed, the ODZ for the original equation is (−∞, −3)∪[−1, +∞) , and for the resulting equation it is [−1, +∞) . This entails the exclusion of the interval (−∞, −3) from consideration and, as a consequence, the loss of all roots of the equation from this interval. In our case, when carrying out this transformation, all the roots of the equation will be lost, of which there are two and .

So, if the transformation of an equation leads to a narrowing of the OD, then all the roots of the equation located in the part to which the narrowing occurred will be lost. That is why we call not to resort to reforms that narrow the DZ. However, there is one caveat.

This clause applies to transformations in which the ODZ is narrowed by one or more numbers. The most typical transformation, in which several individual numbers drop out of the ODZ, is the division of both sides of the equation by the same expression. It is clear that when carrying out such a transformation, only the roots that are among this finite set of numbers that drop out when narrowing the ODZ can be lost. Therefore, if you separately check all the numbers in this set to see if there are roots of the equation being solved among them, for example, by substitution, and include the found roots in the answer, then you can then carry out the intended transformation without fear of losing roots. Let us illustrate this with an example.

Let's consider the irrational equation, which has also already been solved in the previous paragraph. To solve this equation by introducing a new variable, it is useful to first divide both sides of the equation by 1+x. With this division, the number −1 drops out of the ODZ. Substituting this value into the original equation gives the incorrect numerical equality (), which means that −1 is not the root of the equation. After such a check, you can safely carry out the intended division without fear of losing the root.

In conclusion of this point, we note that most often, when solving irrational equations, the division of both sides of the equation by the same expression, as well as transformations based on the properties of the roots, leads to a narrowing of the ODZ. So you need to be very careful when carrying out such transformations and avoid losing roots.

About extraneous roots and methods of screening them out

The solution of the overwhelming number of equations is carried out through transformation of equations. Certain transformations can lead to corollary equations, and among the solutions to the corollary equation there may be roots that are foreign to the original equation. Extraneous roots are not roots of the original equation, therefore, they should not appear in the answer. In other words, they must be weeded out.

So, if in the chain of transformations of the equation being solved there is at least one corollary equation, then you need to take care of detecting and filtering out extraneous roots.

Methods for detecting and screening out foreign roots depend on the reasons causing their potential appearance. And there are two reasons for the possible appearance of extraneous roots when solving irrational equations: the first is the expansion of the ODZ as a result of transforming the equation, the second is the raising of both sides of the equation to an even power. Let's look at the corresponding methods.

Let's start with methods for sifting out extraneous roots, when the reason for their possible appearance is only the expansion of the ODZ. In this case, screening out extraneous roots is carried out in one of the following three ways:

• According to ODZ. To do this, the ODZ of the variable for the original equation is found and the belonging of the found roots is checked. Those roots that belong to the ODZ are roots of the original equation, and those that do not belong to the ODZ are extraneous roots for the original equation.
• Through the conditions of ODZ. The conditions that determine the ODZ of the variable for the original equation are written down, and the found roots are substituted into them one by one. Those roots that satisfy all conditions are roots, and those that do not satisfy at least one condition are extraneous roots for the original equation.
• Through substitution into the original equation (or into any equivalent equation). The found roots are substituted in turn into the original equation, those of them, upon substitution of which the equation turns into a correct numerical equality, are roots, and those of them, upon substitution of which an expression that does not make sense is obtained, are extraneous roots for the original equation.

When solving the following irrational equation, let's filter out extraneous roots using each of the indicated methods in order to get a general idea of ​​each of them.

It is clear that we will not identify and weed out extraneous roots every time using all known methods. To weed out extraneous roots, we will choose the most appropriate method in each specific case. For example, in the following example, it is most convenient to filter out extraneous roots through the conditions of the ODZ, since under these conditions it is difficult to find the ODZ in the form of a numerical set.

Now let's talk about sifting out extraneous roots, when solving an irrational equation is carried out by raising both sides of the equation to an even power. Here, sifting through ODZ or through ODZ conditions will no longer help, since it will not allow us to weed out extraneous roots that arise for another reason - due to raising both sides of the equation to the same even power. Why do extraneous roots appear when both sides of an equation are raised to the same even power? The appearance of extraneous roots in this case follows from the fact that raising both parts of an incorrect numerical equality to the same even power can give a correct numerical equality. For example, the incorrect numerical equality 3=−3 after squaring both sides becomes the correct numerical equality 3 2 =(−3) 2, which is the same as 9=9.

We have figured out the reasons for the appearance of extraneous roots when raising both sides of the equation to the same power. It remains to indicate how extraneous roots are eliminated in this case. Screening is mainly carried out by substituting the found potential roots into the original equation or into any equation equivalent to it. Let's demonstrate this with an example.

But it is worth keeping in mind one more method that allows you to weed out extraneous roots in cases when both sides of an irrational equation with a solitary radical are raised to the same even power. When solving irrational equations , where 2·k is an even number, by raising both sides of the equations to the same power, weeding out extraneous roots can be done through the condition g(x)≥0 (that is, actually solving an irrational equation by determining the root). This method often comes to the rescue when filtering out extraneous roots through substitution turns out to involve complex calculations. The following example is a good illustration of this.

Literature

1. Mordkovich A. G. Algebra. 8th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemosyne, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
2. Mordkovich A. G. Algebra and the beginnings of mathematical analysis. 11th grade. In 2 hours. Part 1. Textbook for students of general education institutions (profile level) / A. G. Mordkovich, P. V. Semenov. - 2nd ed., erased. - M.: Mnemosyne, 2008. - 287 p.: ill. ISBN 978-5-346-01027-2.
3. Algebra and the beginning of analysis: Proc. for 10-11 grades. general education institutions / A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn and others; Ed. A. N. Kolmogorov. - 14th ed. - M.: Education, 2004. - 384 pp.: ill. - ISBN 5-09-013651-3.
4. Algebra and the beginning of mathematical analysis. 10th grade: textbook. for general education institutions: basic and profile. levels / [Yu. M. Kolyagin, M. V. Tkacheva, N. E. Fedorova, M. I. Shabunin]; edited by A. B. Zhizhchenko. - 3rd ed. - M.: Education, 2010.- 368 p.: ill.-ISBN 978-5-09-022771-1.
5. Mathematics. Increased level of Unified State Exam-2012 (C1, C3). Thematic tests. Equations, inequalities, systems / edited by F. F. Lysenko, S. Yu. Kulabukhov. - Rostov-on-Don: Legion-M, 2011. - 112 pp. - (Preparing for the Unified State Exam) ISBN 978-5-91724-094-7
6. Graduate of 2004. Mathematics. Collection of problems for preparing for the Unified State Exam. Part 1. I. V. Boykov, L. D. Romanova.

Solving irrational equations.

In this article we will talk about solutions simplest irrational equations.

Irrational equation is an equation that contains an unknown under the root sign.

Let's look at two types irrational equations, which are very similar at first glance, but in essence are very different from each other.

(1)

(2)

In the first equation we see that the unknown is under the sign of the root of the third degree. We can take the odd root of a negative number, so in this equation there are no restrictions on either the expression under the root sign or the expression on the right side of the equation. We can raise both sides of the equation to the third power to get rid of the root. We get an equivalent equation:

When raising the right and left sides of the equation to an odd power, we can not be afraid of getting extraneous roots.

Example 1. Let's solve the equation

Let's raise both sides of the equation to the third power. We get an equivalent equation:

Let's move all the terms to one side and put x out of brackets:

Equating each factor to zero, we get:

Answer: (0;1;2)

Let's look closely at the second equation: . On the left side of the equation is the square root, which takes only non-negative values. Therefore, for the equation to have solutions, the right-hand side must also be non-negative. Therefore, the condition is imposed on the right side of the equation:

Title="g(x)>=0"> - это !} condition for the existence of roots.

To solve an equation of this type, you need to square both sides of the equation:

(3)

Squaring can lead to the appearance of extraneous roots, so we need the equations:

Title="f(x)>=0"> (4)!}

However, inequality (4) follows from condition (3): if the right side of the equality contains the square of some expression, and the square of any expression can only take non-negative values, therefore the left side must also be non-negative. Therefore, condition (4) automatically follows from condition (3) and our equation is equivalent to the system:

Title="delim(lbrace)(matrix(2)(1)((f(x)=g^2((x))) (g(x)>=0) ))( )">!}

Example 2. Let's solve the equation:

.

Let's move on to an equivalent system:

Title="delim(lbrace)(matrix(2)(1)((2x^2-7x+5=((1-x))^2) (1-x>=0) ))( )">!}

Let's solve the first equation of the system and check which roots satisfy the inequality.

Inequality title="1-x>=0">удовлетворяет только корень !}

Answer: x=1

Attention! If in the process of solving we square both sides of the equation, then we must remember that extraneous roots may appear. Therefore, you either need to move on to an equivalent system, or at the end of the solution, DO A CHECK: find the roots and substitute them into the original equation.

Example 3. Let's solve the equation:

To solve this equation, we also need to square both sides. Let's not bother with the ODZ and the condition for the existence of roots in this equation, but simply do a check at the end of the solution.

Let's square both sides of the equation:

Let's move the term containing the root to the left, and all other terms to the right:

Let's square both sides of the equation again:

According to Vieta's theme:

Let's do a check. To do this, we substitute the found roots into the original equation. Obviously, at , the right-hand side of the original equation is negative, and the left-hand side is positive.

At we obtain the correct equality.

Equations in which a variable is contained under the root sign are called irrational.

Methods for solving irrational equations are usually based on the possibility of replacing (with the help of some transformations) an irrational equation with a rational equation that is either equivalent to the original irrational equation or is a consequence of it. Most often, both sides of the equation are raised to the same power. This produces an equation that is a consequence of the original one.

When solving irrational equations, the following must be taken into account:

1) if the root exponent is an even number, then the radical expression must be non-negative; in this case, the value of the root is also non-negative (definition of a root with an even exponent);

2) if the root exponent is an odd number, then the radical expression can be any real number; in this case, the sign of the root coincides with the sign of the radical expression.

Example 1. Solve the equation

Let's square both sides of the equation.
x 2 - 3 = 1;
Let's move -3 from the left side of the equation to the right and perform a reduction of similar terms.
x 2 = 4;
The resulting incomplete quadratic equation has two roots -2 and 2.

Let's check the obtained roots by substituting the values ​​of the variable x into the original equation.
Examination.
When x 1 = -2 - true:
When x 2 = -2- true.
It follows that the original irrational equation has two roots -2 and 2.

Example 2. Solve the equation .

This equation can be solved using the same method as in the first example, but we will do it differently.

Let's find the ODZ of this equation. From the definition of the square root it follows that in this equation two conditions must be simultaneously satisfied:

ODZ of this level: x.

Answer: no roots.

Example 3. Solve the equation =+ 2.

Finding the ODZ in this equation is a rather difficult task. Let's square both sides of the equation:
x 3 + 4x - 1 - 8= x 3 - 1 + 4+ 4x;
=0;
x 1 =1; x 2 =0.
After checking, we establish that x 2 =0 is an extra root.
Answer: x 1 =1.

Example 4. Solve the equation x =.

In this example, the ODZ is easy to find. ODZ of this equation: x[-1;).

Let's square both sides of this equation, and as a result we get the equation x 2 = x + 1. The roots of this equation are:

It is difficult to verify the roots found. But, despite the fact that both roots belong to the ODZ, it is impossible to assert that both roots are roots of the original equation. This will result in an error. In this case, the irrational equation is equivalent to a combination of two inequalities and one equation:

x+10 And x0 And x 2 = x + 1, from which it follows that the negative root for the irrational equation is extraneous and must be discarded.

Example 5. Solve equation += 7.

Let's square both sides of the equation and perform the reduction of similar terms, transfer the terms from one side of the equation to the other and multiply both sides by 0.5. As a result, we get the equation
= 12, (*) which is a consequence of the original one. Let's square both sides of the equation again. We obtain the equation (x + 5)(20 - x) = 144, which is a consequence of the original one. The resulting equation is reduced to the form x 2 - 15x + 44 =0.

This equation (also a consequence of the original one) has roots x 1 = 4, x 2 = 11. Both roots, as verification shows, satisfy the original equation.

Rep. x 1 = 4, x 2 = 11.

Comment. When squaring equations, students often multiply radical expressions in equations like (*), i.e., instead of equation = 12, they write the equation = 12. This does not lead to errors, since the equations are consequences of the equations. It should, however, be borne in mind that in the general case, such multiplication of radical expressions gives unequal equations.

In the examples discussed above, one could first move one of the radicals to the right side of the equation. Then there will be one radical left on the left side of the equation, and after squaring both sides of the equation, a rational function will be obtained on the left side of the equation. This technique (isolation of the radical) is quite often used when solving irrational equations.

Example 6. Solve equation-= 3.

Isolating the first radical, we obtain the equation
=+ 3, equivalent to the original one.

By squaring both sides of this equation, we get the equation

x 2 + 5x + 2 = x 2 - 3x + 3 + 6, equivalent to the equation

4x - 5 = 3(*). This equation is a consequence of the original equation. By squaring both sides of the equation, we arrive at the equation
16x 2 - 40x + 25 = 9(x 2 - 3x + 3), or

7x 2 - 13x - 2 = 0.

This equation is a consequence of equation (*) (and therefore the original equation) and has roots. The first root x 1 = 2 satisfies the original equation, but the second root x 2 = does not.

Answer: x = 2.

Note that if we immediately, without isolating one of the radicals, squared both sides of the original equation, we would have to perform rather cumbersome transformations.

When solving irrational equations, in addition to the isolation of radicals, other methods are used. Let's consider an example of using the method of replacing the unknown (method of introducing an auxiliary variable).