Solving irrational equations. How to solve irrational equations. Examples

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

"Unspeakable in Words"

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

Based on the above, let's try to define an irrational equation. Such expressions contain so-called "unexpressible numbers", written using the square root sign. They can be all sorts of rather complex options, but in their simplest form they look like the one in the photo below.

When starting to solve irrational equations, first of all it is necessary to calculate the range of permissible values ​​of the variable.

Does the expression make sense?

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

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

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

Method of analysis

From the above, it becomes clear how to solve some types of irrational equations. Here a simple analysis can be an effective way.

Let us give a number of examples that will again clearly demonstrate this (pictured below).

In the first case, upon careful examination of the expression, it immediately turns out to be extremely clear that it cannot be true. Indeed, the left side of the equality should result in a positive number, which cannot possibly be equal to -1.

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

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

Unlimited zoom

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

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

Squaring

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

But let's continue looking at the examples and try to find the variables using the newly proposed method.

It is not at all difficult, using Vieta’s theorem, to find the desired values ​​of quantities after, as a result of certain operations, we have formed a quadratic equation. Here it turns out that among the roots there will be 2 and -19. However, when checking, substituting the resulting values ​​into the original expression, you can make sure that none of these roots are suitable. This is a common occurrence in irrational equations. This means that our dilemma again has no solutions, and the answer should indicate an empty set.

More complex examples

In some cases, it is necessary to square both sides of an expression not once, but several times. Let's look at examples where this is required. They can be seen below.

Having received the roots, do not forget to check them, because extra ones may appear. It should be explained why this is possible. When applying this method, the equation is somewhat rationalized. But by getting rid of roots we don’t like, which prevent us from performing arithmetic operations, we seem to expand the existing range of meanings, which is fraught (as one can understand) with consequences. Anticipating this, we carry out a check. In this case, there is a chance to make sure that only one of the roots is suitable: x = 0.

Systems

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

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

Irrational in mathematics

Each time, the need to create new types of numbers arose among humanity when it did not have enough “space” to solve some equations. Irrational numbers are no exception. As facts from history testify, the great sages first paid attention to this even before our era, in the 7th century. This was done by a mathematician from India known as Manava. He clearly understood that it was impossible to extract a root from some natural numbers. For example, these include 2; 17 or 61, as well as many others.

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

Sign of the Radical: Evolution

The root sign for expressing the numerical value of “deaf” numbers did not immediately begin to be used in solving irrational inequalities and equations. European, in particular Italian, mathematicians first began to think about the radical around the 13th century. At the same time, they came up with the idea of ​​using the Latin R for designation. But German mathematicians acted differently in their works. They liked the letter V better. In Germany, the designation V(2), V(3) soon spread, which was intended to express the square root of 2, 3, and so on. Later, the Dutch intervened and modified the sign of the radical. And Rene Descartes completed the evolution, bringing the square root sign to modern perfection.

Getting rid of the irrational

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

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

Lesson summary

"Methods for solving irrational equations"

11th grade physics and mathematics profile.

Zelenodolsk municipal district of the Republic of Tatarstan"

Valieva S.Z.

Lesson topic: Methods for solving irrational equations

The purpose of the lesson: 1.Study different ways to solve irrational equations.

1. 
   Develop the ability to generalize, correctly select methods for solving irrational equations.
2. 
   Develop independence, improve speech literacy

Lesson type: seminar.
Lesson plan:

1. 
   Organizing time
2. 
   Learning new material
3. 
   Consolidation
4. 
   Homework
5. 
   Lesson summary

During the classes
I. Organizing time: message of the topic of the lesson, the purpose of the lesson.

In the previous lesson, we looked at solving irrational equations containing square roots by squaring them. In this case, we obtain a corollary equation, which sometimes leads to the appearance of extraneous roots. And then a mandatory part of solving the equation is checking the roots. We also looked at solving equations using the definition of square roots. In this case, the check may not be performed. However, when solving equations, you should not always immediately start “blindly” applying algorithms for solving the equation. In the tasks of the Unified State Exam there are quite a lot of equations, when solving which it is necessary to choose a solution method that allows you to solve the equations easier and faster. Therefore, it is necessary to know other methods for solving irrational equations, which we will get acquainted with today. Previously, the class was divided into 8 creative groups, and they were given specific examples to reveal the essence of a particular method. We give them the floor.

From each group, 1 student explains to the children how to solve irrational equations. The whole class listens and takes notes on their story.

1 way. Introduction of a new variable.

Solve the equation: (2x + 3) 2 - 3

4x 2 + 12x + 9 - 3

4x 2 - 8x - 51 - 3

, t ≥0

x 2 – 2x – 6 = t 2;

4t 2 – 3t – 27 = 0

x 2 – 2x – 15 =0

x 2 – 2x – 6 =9;

Answer: -3; 5.

Method 2. DL research.

Solve the equation

ODZ:


x = 2. By checking we are convinced that x = 2 is the root of the equation.

3 way. Multiplying both sides of the equation by the conjugate factor.

+
(multiply both sides by -
)

x + 3 – x – 8 = 5(-)

2=4, hence x=1. By checking we are convinced that x = 1 is the root of this equation.

Solve the equation

Let = u,
=v.

We get the system:

Let's solve by substitution method. We get u = 2, v = 2. This means

we get x = 1.

Answer: x = 1.

5 way. Selecting a complete square.

Solve the equation

Let's expand the modules. Because -1≤сos0.5x≤1, then -4≤сos0.5x-3≤-2, which means . Likewise,

Then we get the equation

x = 4πn, nZ.

Answer: 4πn, nZ.

6 way. Evaluation method

Solve the equation

ODZ: x 3 - 2x 2 - 4x + 8 ≥ 0, by definition the right side is -x 3 + 2x 2 + 4x - 8 ≥ 0

we get
those. x 3 - 2x 2 - 4x + 8 = 0. Solving the equation by factoring, we get x = 2, x = -2

Method 7: Using the properties of monotonicity of functions.

Solve the equation. The functions are strictly increasing. The sum of increasing functions is increasing and this equation has at most one root. By selection we find x = 1.

8 way. Using vectors.

Solve the equation. ODZ: -1≤х≤3.

Let the vector
. The scalar product of vectors is the left side. Let's find the product of their lengths. This is the right side. Got
, i.e. vectors a and b are collinear. From here
. Let's square both sides. Solving the equation, we get x = 1 and x =
.

1. 
   Consolidation.(each student is given worksheets)

Find an idea for solving equations (1-10)

1.
(ODZ - )

2.
x = 2

3. x 2 – 3x +
(replacement)

4. (selecting a complete square)

5.
(Reducing an equation to a system by introducing a variable.)

6.
(multiplying by the conjugate expression)

7.
because
. Then this equation has no roots.

8. Because Each term is non-negative, we equate them to zero and solve the system.

9. 3

10. Find the root of the equation (or the product of the roots, if there are several) of the equation.

Written independent work followed by testing

solve equations numbered 11,13,17,19

12. (x + 6) 2 -

14.

1. 
   Which of these methods are used to solve other types of equations?
2. 
   Which of these methods did you like best and why?

1. 
   Homework: Solve the remaining equations.

1. 
   Algebra and the beginnings of mathematical analysis: textbook. for 11th grade general education institutions / S.M.Nikolsky, M.K.Potapov, N.N.Reshetnikov, A.V.Shevkin. M: Prsveshchenie, 2009

1. 
   Didactic materials on algebra and the beginnings of analysis for grade 11 / B.M. Ivlev, S.M. Sahakyan, S.I. Schwartzburd. – M.: Education, 2003.
2. 
   Mordkovich A. G. Algebra and the beginnings of analysis. 10 – 11 grades: Problem book for general education. institutions. – M.: Mnemosyne, 2000.
3. 
   Ershova A. P., Goloborodko V. V. Independent and test work on algebra and the beginnings of analysis for grades 10 – 11. – M.: Ilexa, 2004
4. 
   KIM Unified State Examination 2002 - 2010

As is already known (Chapter II, § 2), the equation

is called irrational if there is an irrational function of the unknowns.

When solving irrational equations and systems that include irrational equations in the field of real numbers, those and only those systems of real values ​​in which the values ​​of the radical expressions of all roots of even degree are non-negative are considered acceptable systems of values ​​of unknowns; by the values ​​of roots of even degree we mean their arithmetic values, and by the values ​​of roots of odd degree we mean the real values ​​of these roots. Let's consider algebraic methods for solving irrational equations.

1. Freeing an irrational equation from radicals by raising both of its parts to the same power. When solving an irrational equation in this way, as a rule, one radical is isolated sequentially (i.e., the selected radical is left in one part, and all other terms of the equation are transferred to another part) and then both parts of the equation are raised to a power, the exponent of which is equal to indicator of the isolated radical. The most complex radical is usually isolated each time. This continues until they are completely free of radicals. As a result, an algebraic equation is obtained, which is a consequence of the given irrational one. Then the resulting algebraic equation is solved.

In some cases (see example 4 below), in order to quickly get rid of radicals, it is advisable to separate not one, but two radicals at once.

When solving irrational equations in this way, the domain of definition of the equation can expand, since for some systems of unknown values

Some radicals included in a given equation may not make sense in the field of real numbers, but these systems of unknown values ​​may be valid for the resulting algebraic equation. Expanding the domain of definition of an equation, as is known, can lead to the appearance of extraneous solutions that will not belong to the domain of definition of a given equation (see example 2 below).

In addition, raising both sides of the equation to an even power can also lead to the appearance of extraneous solutions that belong to the domain of definition of the given equation. The appearance of these extraneous solutions will not be caused by an expansion of the domain of definition of this equation, but by reasons of a different nature (see example 3 below).

Therefore, having found solutions to an algebraic equation obtained from a given irrational equation, it is necessary, by substituting each of them into the given equation, to check which of them satisfy it and which are foreign to it.

Examples. 1. Solve the equation

Solution. Let's select the radical, i.e., leave it on the left side of the equation, and move the radical to the right side. We will have: or after simplifications: Reducing by 2 and again separating the radical, we will have:

By squaring both sides of this equation, we get:

The solutions to this equation are Substitution into the given equation we make sure that each of these solutions satisfies it.

2. Solve the equation

Solution. Moving V to the right side of the equation we have:

We square both sides of this equation:

By squaring both sides of the resulting equation, we get: or after simplifications:

Hence the solutions to this equation are:

The second of these solutions satisfies the given equation, and the first is foreign to it.

The appearance of an extraneous solution is caused by an expansion of the domain of definition of the equation. Indeed, the number 0 is not included in the domain of definition of the given equation, but it is included in the domain of definition of the equation. A value cannot be a solution to a given equation because it does not belong to its domain.

3. Solve the equation

Solution. By squaring both sides of the equation, we have:

The solutions to this equation are: The first of these solutions satisfies the given equation, and the second is foreign to it.

The appearance of an extraneous solution is not caused by an expansion of the domain of definition of a given equation, but by the fact that the equation is not equivalent to the original one, but only

deducible from it. It is a consequence not only of the given equation, but also of the equation

The solution satisfies the equation. The solution to this equation is extraneous.

4. Solve the equation

Solution. Let's move the radicals into one part

By squaring both sides of this equation, we get:

or after simplifications:

The test shows that it satisfies the given equation.

2. Reducing an irrational equation to a mixed rational system by introducing new unknowns.

A set of one or more equations of the form

and one or more inequalities of the form

is called a mixed system if the requirement is to establish which systems of values ​​of unknowns satisfy simultaneously all these equations and inequalities. A system of values ​​of unknowns that satisfies all the equations and inequalities of a mixed system is called a solution of the mixed system. To solve a mixed system means to establish whether it has solutions or not, and if it does, then find all of them.

Theorem. Any irrational equation

(click to view scan)

Since in equation (1) for any admissible system of values ​​of unknowns, the radical of even degree denotes the arithmetic value of the root, and of odd degree the only real value of the root, then auxiliary unknowns can only take real values ​​and, in addition,

Let us add the inequalities to system (2). We obtain a mixed rational system

(see scan)

Let us now prove that the solution to the irrational equation (1) reduces to the solution to the mixed rational system (3).

Indeed, if there is a solution to equation (1), then

there is a solution to the mixed system (3).

On the contrary, if the system of real numbers is a solution to the mixed system (3), then

Moreover, since that is the arithmetic value of the root of the power of

Likewise, a real number is the only real value of the root of the power of i.e.

From relations (4), (5) and (6) it follows that

and, therefore, the number system is a solution to equation (1).

From the above it follows that to solve equation (1) it is enough to find all solutions of the mixed system (3). The systems of unknown values ​​included in the found solutions of system (3) will be solutions to equation (1), and they exhaust all solutions to equation (1). If it turns out that the mixed system (3) is inconsistent, then equation (1) has no solutions. In the case considered, the irrational equation includes

only simple radicals were included. If the left side of an irrational equation contains radicals, the radical expressions of which in turn contain radicals, but the operation of extracting the root is performed a finite number of times, then by successively introducing auxiliary unknowns, the solution of such an equation is also reduced to the solution of a mixed rational system.

Examples. 1. Solve the equation:

Solution. Assuming that

compose a mixed rational system

Substituting instead into the second equation we obtain a system equivalent to system (7):

From the second equation of system (8) we subtract the third equation by parts, which gives an equation with integer coefficients:

Direct verification shows that divisor 2 of the free term satisfies the equation, i.e. equation (9) has a solution. Therefore, equation (9) can be written as follows:

and therefore

The solutions to equation (10) are and Therefore, equation (9) in the field of real numbers has only one solution. This solution satisfies the inequality

Substituting the value into the equations we find the values ​​namely:

Thus, the mixed rational system (7) has a unique solution. It follows that the given irrational equation also has a unique solution in the field of real numbers

2. Solve the equation

Solution. Putting

let's create a mixed rational system

Solving the first equation relatively and substituting the found value into the third equation, we obtain a mixed system equivalent to system (11):

Substituting the values ​​from the second and fourth equations into the third equation (12), we obtain a system equivalent to system (12):

By squaring both sides of the third equation of system (13), we obtain a system that is a consequence of system (13):

From the last three equations of this system we obtain: or after simplifications:

Obviously, what can be a solution to a given equation, since no system of values ​​can satisfy the third equation of the system satisfies the given equation. Consequently, the given irrational equation has a unique solution in the field of real numbers

Sometimes, when solving an irrational equation, it is advisable to combine the method of introducing new unknowns with the method of raising both sides of the equation to a power.

Example. Solve the equation

Solution. Assuming that we have:

We replace equation (15) with a mixed system

Separating the radical in the second equation of system (16) and squaring both sides of the equation, we obtain: or after simplifications:

Hence, both of these solutions satisfy the equation and inequality. Substituting the values ​​into the first equation of system (16), we obtain the following two equations:

Therefore, mixed system (16) has four solutions:

and, therefore, equation (15) also has four solutions, namely:

Artificial techniques. In the practice of solving irrational equations, individual, so-called artificial techniques are sometimes successfully used. Let's look at some of them with examples.

a) Solve the equation

Solution. Let's multiply both sides of the equation by the factor conjugate to its left side. Will have:

or after transformations:

Adding equations (17) and (18) by parts, we obtain:

Both solutions satisfy the given equation, which can be easily verified by substituting them into the equation,

b) Solve the equation

Solution. Let's take the identity

and write it like this:

Equality (20) is satisfied for any values, in particular, for values ​​satisfying equation (19). Therefore, if we replace its second factor on the left side of identity (20), which is the left side of equation (19), with the expression, we obtain the equation

which will be satisfied by all solutions of equation (19).

Equation (21) is thus a consequence of equation (19), and, therefore, solutions to equation (19) should be sought among the solutions to equation (21). We write Equation (21) as follows:

This shows that equation (21) splits into two equations:

From the above it follows that solutions to equation (19) must be sought among solutions to equation (22) and solutions to equation (23). The solution to equation (22) is This solution also satisfies the given equation (19). To find other solutions to equation (19), we add the parts of equations (19) and (23). We get the equation

which will be satisfied by all solutions of equation (19), different from the solution

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 radical 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 radical 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).