Lesson summary

"Methods for solving irrational equations"

11th grade of physical and mathematical profile.

Zelenodolsky municipal district of the Republic of Tatarstan

Valieva S.Z.

Lesson topic: Methods for solving irrational equations

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

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

Lesson type: seminar.
Lesson plan:

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

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

In the previous lesson, we considered solving irrational equations containing square roots by squaring them. In this case, we obtain a consequence 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 considered solving equations using the definition of a square root. In this case, the check can be omitted. However, when solving equations, it is not always necessary to immediately proceed to the “blind” application of algorithms for solving the equation. In the tasks of the Unified State Exam, there are quite a few equations, in 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 a word.

II. Learning new material.

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 \u003d t 2;

4t 2 – 3t – 27 = 0

x 2 - 2x - 15 \u003d 0

x 2 - 2x - 6 \u003d 9;

Answer: -3; 5.

2 way. ODZ research.

solve the equation

ODZ:


x \u003d 2. By checking we make sure that x \u003d 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 \u003d 5 (-)

2=4, hence x=1. By checking we make sure that x \u003d 1 is the root of this equation.

4 way. Reduction of an equation to a system by introducing a variable.

solve the equation

Let =u,
=v.

We get the system:

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

we get x = 1.

Answer: x = 1.

5 way. Selection of a full square.

solve the equation

Let's open the modules. Because -1≤cos0.5x≤1, then -4≤cos0.5x-3≤-2, so . Likewise,

Then we get the equation

x = 4πn, nZ.

Answer: 4πn, nZ.

6 way. Assessment method

solve the equation

ODZ: x 3 - 2x 2 - 4x + 8 ≥ 0, by definition, the right side -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 the 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. Use of 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 \u003d 1 and x \u003d
.

1. 
   Consolidation.(each student is given a worksheet)

Front oral work

Find an idea for solving equations (1-10)

1.
(ODZ - )

2.
x = 2

3. x 2 - 3x +
(replacement)

4. (selection of a full square)

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

6.
(by multiplication by the adjoint expression)

7.
because
. 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 with subsequent verification

solve equations numbered 11,13,17,19

Solve Equations:

12. (x + 6) 2 -

14.

Assessment method

Using the properties of monotonicity of functions.

Use of vectors.

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

1. 
   Homework: Solve the remaining equations.

Bibliography:

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

1. 
   Didactic materials on algebra and principles of analysis for grade 11 /B.M. Ivlev, S.M. Sahakyan, S.I. Schwarzburd. – M.: Enlightenment, 2003.
2. 
   Mordkovich A. G. Algebra and the beginnings of analysis. 10 - 11 cells: Task book for general education. institutions. – M.: Mnemosyne, 2000.
3. 
   Ershova A.P., Goloborodko V.V. Independent and control work on algebra and principles of analysis for grades 10-11. – M.: Ileksa, 2004
4. 
   KIM USE 2002 - 2010

6. Algebraic simulator. A.G. Merzlyak, V.B. Polonsky, M.S. Yakir. Handbook for schoolchildren and entrants. Moscow.: "Ileksa" 2001.
7. Equations and inequalities. Non-standard solution methods. Educational - methodical manual. 10 - 11 classes. S.N. Oleinik, M.K. Potapov, P.I. Pasichenko. Moscow. "Bustard". 2001

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

Such equations are always solved in 3 steps:

1. Separate the root. In other words, if there are other numbers or functions to the left of the equal sign in addition to the root, all this must be moved to the right by changing the sign. At the same time, only the radical should remain on the left - without any coefficients.
2. 2. We square both sides of the equation. At the same time, remember that the range of the root is all non-negative numbers. Hence the function on the right irrational equation must also be non-negative: g (x) ≥ 0.
3. The third step follows logically 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, it is necessary to substitute the resulting candidate numbers into the original equation and check: is the correct numerical equality really obtained?

Solving an irrational equation

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

2x 2 - 14x + 13 = (5 - x) 2
2x2 - 14x + 13 = 25 - 10x + x2
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 \u003d -2

It remains only to substitute these numbers in the original equation, i.e. perform a check. But even here you can do the right thing to simplify the final decision.

How to simplify the decision

Let's think: why do we even check at the end of solving an irrational equation? We want to make sure that when substituting our roots, there will be a non-negative number to the right of the equal sign. After all, we already know for sure that it is a non-negative number on the left, because the arithmetic square root (because of which 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) \u003d g (6) \u003d 5 - 6 \u003d -1< 0
g (x 2) = g (−2) = 5 − (−2) = 5 + 2 = 7 > 0

From the values ​​obtained, it follows that the root x 1 = 6 does not suit us, since when substituting into the right side of the original equation, we get a negative number. But the root x 2 \u003d −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 the root x 2 = −2 is substituted, turns into a positive number, i.e. the range of the arithmetic root is not violated.

That's the whole algorithm! As you can see, solving equations with radicals is not so difficult. The main thing is not to forget to check the received roots, otherwise it is very likely to get extra answers.

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Municipal educational institution

"Kudinskaya secondary school No. 2"

Ways to solve irrational equations

Completed by: Egorova Olga,

Supervisor:

Teacher

mathematics,

higher qualification

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

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

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

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

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

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

Introduction

Mathematical education received in a general education school is an essential component of general education and the general culture of a modern person. Almost everything that surrounds a modern person is all connected in one way or another with mathematics. And the latest advances in physics, engineering and information technology leave no doubt that in the future the state of affairs will remain the same. Therefore, the solution of many practical problems is reduced to solving various types of equations that need to be learned to solve. One of these types are irrational equations.

Irrational equations

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

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

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

Types of irrational equations

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

2..

Let's consider the first of them.

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

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

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

1. - where g(x)≥ 0 and

2. - where g(x) ≤ 0.

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

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

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

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

Consider the second kind of irrational equations.

. Let the equation . His ODZ:

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

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

Section 1. Methods for solving irrational equations

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

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

Solve the equation:

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

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

Solve the equation:

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

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

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

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

Answer: x = 1.

2 method. Replacing an adjacent system of conditions

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

Solve the equation:

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

This equation is equivalent to the system

Answer: the equation has no solutions.

3 method. Using the properties of the nth root

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

2.

3.

4.

5.

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

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

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

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

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

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

Answer: -1,-2.

Solve the equation: .

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

Answer: x = 4.25.

4 method. Introduction of new variables

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

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

1. Find the ODZ of the original equation.

2. Go from the equation to its corollary.

3. Find the roots of the resulting equation.

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

The check is as follows:

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

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

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

Such a solution method with the indicated method of verification makes it possible to avoid cumbersome calculations in the case of direct substitution of each of the found roots of the last equation into the original one.

Solve the irrational equation:

.

The set of admissible values ​​of this equation:

Setting , after substitution we obtain the equation

or its equivalent equation

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

.

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

, .

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

, .

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

Answer: x = 2.

Solve the irrational equation:

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

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

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

5 method. Identity Equation Transformation

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

Solve the equation:

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

.

We get:

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

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

when the equation has a solution

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

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

EXAMPLE 10:

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

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

Notes.

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

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

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

Answer: x = 4.

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

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

Answer: x = 3.

8 method. Application of the derivative in solving irrational equations

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

EXAMPLE 15:

Solve the equation: (1)

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

Answer:

EXAMPLE 16:

Solve the irrational equation:

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

Answer: x = 3.

9 method. Functional

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

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

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

Answer: x = 3.

EXAMPLE 18:

Solve the irrational equation: (1)

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

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

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

Solution: This equation is equivalent to a mixed system

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 index is an 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;
The resulting 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 of the square root it follows that in this equation two conditions must be satisfied simultaneously:

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. AT this case an irrational equation is equivalent to a 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 = 4, x 2 = 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).