Methods for solving irrational equations. Irrational equations. Comprehensive guide

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

Methods for solving irrational equations, as a rule, are 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 its consequence. Most often, both sides of the equation are raised to the same power. In this case, an equation is obtained, which is a consequence of the original one.

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

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

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

Example 1 solve the equation

Let's square both sides of the equation.
x 2 - 3 \u003d 1;
We transfer -3 from the left side of the equation to the right side and perform the reduction of similar terms.
x 2 \u003d 4;
Received incomplete quadratic equation has two roots -2 and 2.

Let's check the obtained roots, for this we will substitute the values ​​of the variable x into the original equation.
Examination.
When x 1 \u003d -2 - true:
When x 2 \u003d -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 us find the ODZ of this equation. From the definition square root it follows that in this equation two conditions must be simultaneously satisfied:

ODZ of the given equation: 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; x2=0.
After checking, we establish that x 2 \u003d 0 is an extra root.
Answer: x 1 \u003d 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, as a result we get the equation x 2 \u003d x + 1. The roots of this equation:

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

x+10 And x0 And x 2 \u003d x + 1, from which it follows that the negative root for the irrational equation is extraneous and must be discarded.

Example 5 . Solve the equation += 7.

Let's square both sides of the equation and perform the reduction of similar terms, transfer the terms from one part of the equation to the other and multiply both parts 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 get 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 (which is also a consequence of the original one) has roots x 1 \u003d 4, x 2 \u003d 11. Both roots, as the test shows, satisfy the original equation.

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

Comment. When squaring equations, students often in equations like (*) multiply root expressions, that is, instead of equation = 12, they write the equation = 12. This does not lead to errors, since the equations are consequences of the equations. However, it should be borne in mind that in the general case, such a multiplication of radical expressions gives non-equivalent equations.

In the examples discussed above, it was possible to first transfer one of the radicals to the right side of the equation. Then one radical will remain 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 (solitude of the radical) is quite often used in solving irrational equations.

Example 6. Solve equation-= 3.

Having isolated the first radical, we obtain the equation
=+ 3, which is equivalent to the original one.

Squaring both sides of this equation, we get the equation

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

4x - 5 = 3(*). This equation is a consequence of the original equation. Squaring both sides of the equation, we arrive at the equation
16x 2 - 40x + 25 \u003d 9 (x 2 - Zx + 3), or

7x2 - 13x - 2 = 0.

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

Answer: x = 2.

Note that if we immediately, without isolating one of the radicals, were squaring both parts 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 also used. Consider an example of using the method of replacing the unknown (the method of introducing an auxiliary variable).

What will be discussed? On equations that contain a function of a variable under the sign of the radical. However, the sign of the radical can be replaced by a degree with a fractional exponent. Such equations are irrational.

Basic properties of irrational equations

1. Any root of an even degree is arithmetic, i.e. radical expressions are always non-negative and take only non-negative values.

2. Any root of an odd degree is defined for all values ​​of the radical expression and can take on any values.

3. The equation √(f(x)) = g(x) is equivalent to the system (hereinafter, under the notation √(f(x)) we mean the square root of the expression in brackets):

(f(x) = (g(x))2,
(g(x) ≥ 0.

How can irrational equations be solved?

1. Raise both sides of the equation to the same power.
2. Change of variable.
3. By multiplying both parts by the same expressions.
4. Application of the properties of the functions included in the equation.

Let us consider examples of equations solved by these methods.

Example 1

Solve the equation √ (3x 2 - 14x + 17) = 3 - 2x.

Solution.

We use property 3 of the above and get the system:

(3x 2 - 14x + 17 \u003d (3 - 2x) 2,
(3 – 2x ≥ 0.

From the first equation we get x 2 + 2x - 8 = 0. Its roots are -4 and 2. But only the number -4 satisfies the inequality of our system.

Answer: -4.

There is another way to solve this equation. Let's not write down the system. Forget inequality. We work only with the equation. But let's remember that raising both sides of the equation to an even power leads to a corollary equation. It, along with the roots of the original equation, may contain other roots, which are called extraneous. Therefore, after solving the consequence equation, it is necessary to find a way to weed out extraneous roots. This can usually be done by checking, which in this case is considered as one of the stages of the solution.

Obviously, we will again obtain the roots of the consequence equation: -4 and 2. The verification is carried out by substituting into the original equation √ (3x 2 - 14x + 17) \u003d 3 - 2x.

If x = -4, then we get √121 = 11, which is correct. For x = 2, we get √1 = -1, which is not true and the root 2 is eliminated.

Answer: x = -4.

Example 2

Solve the equation 3 √(4x + 3) - 3 √(x + 2) = 1

Solution.

Raise both sides of the equation to the third power

(3 √(4x + 3) - 3 √(x + 2))3 = 13.

We get (4x + 3) - (x + 2) - 3(3 √(4x + 3) 3 √(x + 2))(3 √(4x + 3) - 3 √(x + 2)) = 1

Or (4x + 3) - (x + 2) - 3 3 √ ((4x + 3) (x + 2)) (3 √ (4x + 3) - 3 √ (x + 2)) \u003d 1.

Given the initial condition, the equation will take the form

(4x + 3) - (x + 2) - 3 3 √ ((4x + 3) (x + 2)) = 1. After performing simple transformations, we get

3x - 3 3 √((4x + 3)(x + 2)) = 0,

x = 3 √((4x + 3)(x + 2)).

To solve this equation, it is necessary to cube again.

Having done it, we will have

x 3 \u003d 4x 2 + 11x + 6,

x 3 - 4x 2 - 11x - 6 = 0.

Find the selection method one root of the equation. This number is -1.

Dividing the polynomial x 3 - 4x 2 - 11x - 6 by x + 1 with a corner, we get the trinomial x 2 - 5x - 6.

The roots of the equation x 2 - 5x - 6 \u003d 0 - numbers: -1; 6.

Hence, the roots of the equation x 3 - 4x 2 - 11x - 6 \u003d 0 will be the numbers -1; 6.

Substituting the numbers -1; 6 into the original equation, make sure that the root of the equation is the number 6.

Answer: 6.

Example 3

Solve the equation x 2 - x √ (4x + 5) = 8x + 10

Solution.
Note that 8x + 10 = 2(√(4x + 5)) 2 . By checking we make sure that x \u003d 0 is not the root of this equation. So, dividing by x 2 both parts of this equation, we get it equivalent:

1 – √(4x + 5)/x = 2(√(4x + 5)/x) 2

Replace √(4x + 5)/x = t and solve the resulting quadratic equation 1 – t = 2t2.

We get t 1 = -1 and t 2 = 1/2. Let's go back to the original variable x and get 2 equations

1) √(4x + 5)/x = -1,

2) √(4x + 5)/x = 1/2

From the first equation x = -1. (x = 5 has to be discarded after verification).

From the second -x = 8 ± 2√21. To filter out extraneous roots, it is easier to analyze the condition here than to do a substitution. After all, the equation is easily transformed to the form √(4x + 5) = 0.5x, which is equivalent to the system

(4x + 5 \u003d 0.25x 2,
(0.5x ≥ 0.

It is now obvious that x = 8 + 2√21 fits. and general

answer: x = -1 and x = 8 + 2√21.

Example 4

Solve the equation √(8x + 1) + √(3x - 5) = √(7x + 4) + √(2x - 2).

Solution.

Let's use the formula √а + √b = (a – b) / (√а – √b), which is valid for a ≥ 0; b ≥ 0; a ≠ b.

Taking into account the ODZ (x ≥ 1 2/3), this formula can be applied to the expressions on the left and right sides of the equation.

And we get: (5x + 6) / (√ (8x + 1) - √ (3x - 5)) \u003d (5x + 6) / (√ (7x + 4) - √ (2x - 2))

or (5x + 6)((√(8x + 1) - √(3x - 5)) - (√(7x + 4) - √(2x - 2)) = 0

It is equivalent to a set of 2 equations:

1) (5x + 6) = 0 and

2) √(8x + 1) - √(3x - 5) = √(7x + 4) - √(2x - 2)

From the first we get x = -1.2. But this value is not included in the ODZ.

Let's compare the second equation with the original one. When adding these equations, we get:

2√(8x + 1) = 2√(7x + 4).

x = 3.

Answer: 3.

It is impossible to describe all the ways to solve irrational equations in one article. It is unlikely that there will be a source with such full content. Yes, you don't need it. For successful preparation for the exam, As well as the training of any specialist in general, it is important not to memorize a theory or methods and reproduce them in similar cases, but, more importantly, to master them and apply them in an unfamiliar situation. That is, some basic stock of knowledge must be learned to apply creatively. Then you yourself will be able to invent new ways, that is, to make discoveries.

I wish you success. And share your findings with your friends. This can also be done through comments on blog articles.

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Methods for solving irrational equations.

Preliminary preparation for the lesson: students should be able to solve irrational equations in a variety of ways.

Three weeks before this session, students receive homework #1: solve various irrational equations. (Students independently find 6 different irrational equations and solve them in pairs.)

One week before this lesson, students receive homework #2, which they complete individually.

1. Solve the equationdifferent ways.

2. Assess the advantages and disadvantages of each method.

3. Make a record of the conclusions in the form of a table.

Lesson Objectives:

Educational:generalization of students' knowledge on this topic, demonstration of various methods for solving irrational equations, students' ability to approach solving equations from research positions.

Educational:education of independence, the ability to listen to others and communicate in groups, increasing interest in the subject.

Developing:development logical thinking, algorithmic culture, self-education skills, self-organization, work in pairs when doing homework, the ability to analyze, compare, generalize, draw conclusions.

Equipment: computer, projector, screen, table "Rules for solving irrational equations", a poster with a quote from M.V. Lomonosov “Mathematics should be taught later that it puts the mind in order”, cards.

Rules for solving irrational equations.

Lesson type: lesson-seminar (work in groups of 5-6 people, each group must have strong students).

During the classes

I . Organizing time

(Message of the topic and objectives of the lesson)

II . Presentation research work"Methods for solving irrational equations"

(The work is presented by the student who conducted it.)

III . Analysis of methods for solving homework

(One student from each group writes down on the board their proposed solutions. Each group analyzes one of the solutions, evaluates the advantages and disadvantages, draws conclusions. The students of the groups supplement, if necessary. The analysis and conclusions of the group are evaluated. Answers must be clear and complete.)

The first way: raising both sides of the equation to the same power, followed by verification.

Solution.

Let's square both sides of the equation again:

From here

Examination:

1. Ifx=42 then, which means the number42 is not the root of the equation.

2. Ifx=2, then, which means the number2 is the root of the equation.

Answer:2.

Conclusion. When solving irrational equations by raising both parts of the equation to the same power, it is necessary to keep a verbal record, which makes the solution understandable and accessible. However, mandatory verification is sometimes complex and time-consuming. This method can be used to solve simple irrational equations containing 1-2 radicals.

The second way: equivalent transformations.

Solution:Let's square both sides of the equation:

Answer:2.

Conclusion. When solving irrational equations by the method of equivalent transitions, you need to clearly know when to put the sign of the system, and when - the aggregate. The cumbersome notation, various combinations of signs of the system and the totality often lead to errors. However, a sequence of equivalent transitions, a clear logical record without a verbal description that does not require verification, are the indisputable advantages of this method.

The third way: functional-graphic.

Solution.

Consider the functionsAnd.

1. Functionpower; is increasing, because the exponent is a positive (not integer) number.

D(f).

Let's make a table of valuesxAndf( x).

2. Functionpower; is decreasing.

Find the domain of the functionD( g).

Let's make a table of valuesxAndg( x).

Let's build these graphs of functions in one coordinate system.

Function graphs intersect at a point with an abscissaBecause functionf( x) increases, and the functiong( x) decreases, then there is only one solution to the equation.

Answer: 2.

Conclusion. The functional-graphical method is illustrative, allows you to find the number of solutions, but it is better to use it when you can easily build graphs of the functions under consideration and get an accurate answer. If the answer is approximate, then it is better to use another method.

Fourth way: introduction of a new variable.

Solution.We introduce new variables, denotingWe get the first equation of the system

Let us compose the second equation of the system.

For a variable:

For a variable

That's why

We obtain a system of two rational equations, with respect toAnd

Returning to the variable, we get

Simplification - obtaining a system of equations that do not contain radicals

1. The need to track the LPV of new variables

2. The need to return to the original variable

Conclusion. This method is best used for irrational equations containing radicals of various degrees, or the same polynomials under the root sign and behind the root sign, or mutually inverse expressions under the root sign.

- So, guys, for each irrational equation, you need to choose the most convenient way to solve it: understandable. Accessible, logical and well-designed. Raise your hand, which of you would give preference to solving this equation:

1) the method of raising both parts of the equation to the same power with verification;

2) the method of equivalent transformations;

3) functional-graphic method;

4) the method of introducing a new variable.

IV . Practical part

(Group work. Each group of students receives a card with an equation and solves it in notebooks. At this time, one representative from the group solves an example on the board. Students of each group solve the same example as a member of their group and monitor the correct execution tasks on the board.If the person answering at the blackboard makes mistakes, then the one who notices them raises his hand and helps to correct.During the lesson, each student, in addition to the example solved by his group, must write down in a notebook and others proposed to the groups, and solve them at home .)

Group 1.

Group 2

Group 3.

V . Independent work

(In groups, first there is a discussion, and then the students begin to complete the task. The correct solution prepared by the teacher is displayed on the screen.)

VI . Summing up the lesson

Now you know that solving irrational equations requires good theoretical knowledge, the ability to apply them in practice, attention, diligence, ingenuity.

Homework

Solve the equations proposed to the groups during the lesson.

If the variable in the equation is contained under the sign of the square root, then the equation is called irrational.

Sometimes mathematical model real situation is an irrational equation. Therefore, we should learn to solve at least the simplest irrational equations.

Consider the irrational equation 2 x + 1 = 3 .

Pay attention!

The method of squaring both sides of an equation is the main method for solving irrational equations.

However, this is understandable: how else to get rid of the sign of the square root?

From the equation \(2x + 1 = 9\) we find \(x = 4\). This is the root of both the equation \(2x + 1 \u003d 9\), and the given irrational equation.

The squaring method is technically simple, but sometimes leads to trouble.

Consider, for example, the irrational equation 2 x − 5 = 4 x − 7 .

By squaring both sides, we get

2 x - 5 2 = 4x - 7 2 2 x - 5 = 4 x - 7

But the value \(x = 1\), although it is the root of the rational equation \(2x - 5 = 4x - 7\), is not the root of the given irrational equation. Why? Substituting \(1\) instead of \(x\) into the given irrational equation, we get − 3 = − 3 .

How can we talk about the fulfillment of numerical equality, if both its left and right parts contain expressions that do not make sense?

In such cases they say: \(x = 1\) - extraneous root for a given irrational equation. It turns out that the given irrational equation has no roots.

An extraneous root is not a new concept for you, extraneous roots have already been encountered when solving rational equations, checking helps to detect them.

For irrational equations check - mandatory step solution of the equation, which will help to detect extraneous roots, if any, and discard them (usually they say "weed out").

Pay attention!

So, an irrational equation is solved by squaring both its parts; having solved the resulting rational equation, it is necessary to check and weed out possible extraneous roots.

Using this derivation, let's look at an example.

Example:

solve the equation 5 x − 16 = x − 2 .

Let's square both sides of the equation 5 x - 16 = x - 2: 5 x - 16 2 = x - 2 2 .

Transform and get:

5 x - 16 \u003d x 2 - 4 x 4; − x 2 9 x − 20 = 0 ; x 2 − 9 x 20 = 0; x 1 \u003d 5; x 2 = 4 .

Examination. Substituting \(x \u003d 5\) into the equation 5 x - 16 \u003d x - 2, we get 9 \u003d 3 - true equality. Substituting \(x \u003d 4\) into the equation 5 x - 16 \u003d x - 2, we get 4 \u003d 2 - the correct equality. Hence, both found values ​​are the roots of the equation 5 x − 16 = x − 2 .

You have already gained some experience in solving various equations: linear, square, rational, irrational. You know that when solving equations, various transformations are performed, for example: a member of the equation is transferred from one part of the equation to another with opposite sign; both sides of the equation are multiplied or divided by the same non-zero number; get rid of the denominator, i.e., replace the equation p x q x \u003d 0 with the equation \(p (x) \u003d 0 \); Both sides of the equation are squared.

Of course, you noticed that as a result of some transformations, extraneous roots could appear, and therefore you had to be vigilant: check all the roots found. So we will now try to comprehend all this from a theoretical point of view.

Two equations \(f (x) = g(x)\) and \(r(x) = s(x)\) are called equivalent if they have the same roots (or, in particular, if both equations have no roots) .

Usually, when solving an equation, they try to replace this equation with a simpler one, but equivalent to it. Such a change is called an equivalent transformation of the equation.

Equivalent transformations of the equation are the following transformations:

For example, replacing the equation \(2x + 5 = 7x - 8\) with the equation \(2x - 7x = - 8 - 5\) is an equivalent transformation of the equation. This means that the equations \(2x + 5 = 7x -8\) and \(2x - 7x = -8 - 5\) are equivalent.