Write down some consequence of the equation. Equivalent equations, transformation of equations. About ways to solve equations

home Allowing you to move from the equation being solved to the so-called equivalent equations And corollary equations

, from whose solutions it is possible to determine the solution to the original equation. In this article we will analyze in detail which equations are called equivalent and which are called corollary equations, give the corresponding definitions, give explanatory examples and explain how to find the roots of an equation using the known roots of an equivalent equation and a corollary equation.

Equivalent equations, definition, examples

Let us define equivalent equations.

Definition Equivalent equations

- these are equations that have the same roots or do not have roots.

Let us define equivalent equations.

Definitions that are the same in meaning, but slightly different in wording, are given in various mathematics textbooks, for example, The two equations f(x)=g(x) and r(x)=s(x) are called equivalent

Let us define equivalent equations.

, if they have the same roots (or, in particular, if both equations have no roots). Equations that have the same roots are called equivalent equations

. Equations that do not have roots are also considered equivalent.

By the same roots is meant the following: if some number is the root of one of the equivalent equations, then it is also the root of any other of these equations, and not one of the equivalent equations can have a root that is not the root of any other of them. these equations.

Let us give examples of equivalent equations. For example, three equations 4 x = 8, 2 x = 4 and x = 2 are equivalent. Indeed, each of them has a single root 2, so they are equivalent by definition. Another example: two equations x·0=0 and 2+x=x+2 are equivalent, the sets of their solutions coincide: the root of both the first and second of them is any number. The two equations x=x+5 and x 4 =−1 are also examples of equivalent equations; they both have no real solutions.

The stated definition of equivalent equations applies to both equations with one variable and equations with a large number variables. However, for equations with two, three, etc. variables, the word “roots” in the definition must be replaced with the word “solutions”. So,

Let us define equivalent equations.

Definition- these are equations that have the same solutions or do not have them.

Let's show an example of equivalent equations with several variables. x 2 +y 2 +z 2 =0 and 5 x 2 +x 2 y 4 z 8 =0 - here is an example of equivalent equations with three variables x, y and z, they both have a unique solution (0, 0, 0) . But equations with two variables x+y=5 and x·y=1 are not equivalent, since, for example, a pair of values ​​x=2, y=3 is a solution to the first equation (when substituting these values ​​into the first equation we get the correct equality 2+3=5), but is not a solution to the second (when substituting these values ​​into the second equation we get the incorrect equality 2·3=1).

Consequence equations

Here are the definitions of corollary equations from school textbooks:

Let us define equivalent equations.

If each root of the equation f(x)=g(x) is at the same time a root of the equation p(x)=h(x), then the equation p(x)=h(x) is called consequence equations f(x)=g(x) .

Let us define equivalent equations.

If all the roots of the first equation are roots of the second equation, then the second equation is called consequence first equation.

Let's give a couple of examples of corollary equations. The equation x 2 =3 2 is a consequence of the equation x−3=0. Indeed, the second equation has a single root x=3, this root is also the root of the equation x 2 =3 2, therefore, by definition, the equation x 2 =3 2 is a consequence of the equation x−3=0. Another example: the equation (x−2)·(x−3)·(x−4)=0 is a consequence of the equation , since all the roots of the second equation (there are two of them, these are 2 and 3) are obviously the roots of the first equation.

From the definition of a corollary equation it follows that absolutely any equation is a consequence of any equation that has no roots.

It is worth citing several rather obvious consequences from the definition of equivalent equations and the definition of a corollary equation:

  • If two equations are equivalent, then each of them is a consequence of the other.
  • If each of two equations is a consequence of the other, then these equations are equivalent.
  • Two equations are equivalent if and only if each of them is a consequence of the other.
  • Algebra: textbook for 8th grade. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
  • Mordkovich A. G. Algebra and beginning of mathematical analysis. Grade 11. At 2 p.m. Part 1. Textbook for students educational institutions (profile level) / A. G. Mordkovich, P. V. Semenov. - 2nd ed., erased. - M.: Mnemosyne, 2008. - 287 p.: ill. ISBN 978-5-346-01027-2.
  • 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.
  • Let two equations be given

    If each root of equation (2.1) is also a root of equation (2.2), then equation (2.2) is called a consequence of the equation(2.1). Note that the equivalence of equations means that each of the equations is a consequence of the other.

    In the process of solving an equation, it is often necessary to apply transformations that lead to an equation that is a consequence of the original one. The corollary equation is satisfied by all the roots of the original equation, but, in addition to them, the corollary equation can also have solutions that are not roots of the original equation, these are the so-called outsiders roots. To identify and weed out extraneous roots, they usually do this: all found roots of the corollary equation are checked by substitution into the original equation.

    If, when solving an equation, we replaced it with a corollary equation, then the above check is an integral part of solving the equation. Therefore, it is important to know under what transformations given equation goes into investigation.

    Consider the equation

    and multiply both of its parts by the same expression, which makes sense for all values. We get the equation

    whose roots are both the roots of equation (2.3) and the roots of equation . This means that equation (2.4) is a consequence of equation (2.3). It is clear that equations (2.3) and (2.4) are equivalent if the “extraneous” equation has no roots.

    So, if both sides of the equation are multiplied by the expression that makes sense for any values ​​of , then we get an equation that is a consequence of the original one. The resulting equation will be equivalent to the original one if the equation has no roots. Note that the inverse transformation, i.e. transition from equation (2.4) to equation (2.3) by dividing both sides of equation (2.4) by the expression is, as a rule, unacceptable, since it can lead to a loss of solutions (in this case, the roots of the equation may be “lost”). For example, an equation has two roots: 3 and 4. Dividing both sides of the equation by leads to an equation that has only one root 4, i.e. root loss has occurred.

    Let's take equation (2.3) again and square both sides. We get the equation

    the roots of which are both the roots of equation (2.3) and the roots of the “extraneous” equation, i.e. equation (2.5) is a consequence of equation (2.3).

    For example, an equation has a root of 4. If both sides of this equation are squared, you get an equation that has two roots: 4 and -2. This means that the equation is a consequence of the equation. When moving from equation to equation, an extraneous root -2 appeared.

    So, when both sides of the equation are squared (and generally to any even power), we obtain an equation that is a consequence of the original one. This means that with this transformation, the appearance of extraneous roots is possible. Note that raising both sides of the equation to the same odd power results in an equation equivalent to the given one.

    The following transformations are most often used when solving equations:

    Other transformations

    In the list presented in the previous paragraph, we deliberately did not include such transformations as raising both sides of the equation to the same natural power, logarithm, potentiating both sides of the equation, extracting the root of the same degree from both sides of the equation, freeing external function and others. The fact is that these transformations are not so general: transformations from the above list are used to solve equations of all types, and the transformations just mentioned are used to solve certain types of equations (irrational, exponential, logarithmic, etc.). They are discussed in detail within the framework of the corresponding methods for solving the corresponding types of equations. Here are links to their detailed descriptions:

    • Raising both sides of an equation to the same natural power.
    • Taking logarithms of both sides of the equation.
    • Potentiating both sides of the equation.
    • Extracting the root of the same power from both sides of an equation.
    • Replacing an expression corresponding to one of the parts of the original equation with an expression from another part of the original equation.

    The links provided contain comprehensive information on the listed transformations. Therefore, we will no longer dwell on them in this article. All subsequent information applies to transformations from the list of basic transformations.

    What happens as a result of transforming the equation?

    Carrying out all the above transformations can give either an equation that has the same roots as the original equation, or an equation whose roots contain all the roots of the original equation, but which may also have other roots, or an equation whose roots will not include all roots of the transformed equation. In the following paragraphs we will analyze which of these transformations, under which conditions, lead to which equations. This is extremely important to know for successfully solving equations.

    Equivalent transformations of equations

    Of particular interest are transformations of equations that result in equivalent equations, that is, equations that have the same set of roots as the original equation. Such transformations are called equivalent transformations. IN school textbooks the corresponding definition is not given explicitly, but it is easy to read from the context:

    Let us define equivalent equations.

    Equivalent transformations of equations are transformations that give equivalent equations.

    So why are equivalent transformations interesting? The fact is that if with their help it is possible to come from the equation being solved to a fairly simple equivalent equation, then solving this equation will give the desired solution to the original equation.

    Of the transformations listed in the previous paragraph, not all are always equivalent. Some transformations are equivalent only under certain conditions. Let's make a list of statements that determine which transformations and under what conditions are equivalent transformations of the equation. To do this, we will take the above list as a basis, and to the transformations that are not always equivalent, we will add conditions that give them equivalence. Here is the list:

    • Replacing an expression on the left or right side of an equation with an expression that does not change the variables for the equation is an equivalent transformation of the equation.

    Let us explain why this is so. To do this, we take an equation with one variable (similar reasoning can be carried out for equations with several variables) of the form A(x)=B(x), we denoted the expressions on its left and right sides as A(x) and B(x), respectively . Let the expression C(x) be identically equal to the expression A(x), and the ODZ of the variable x of the equation C(x)=B(x) coincides with the ODZ of the variable x for the original equation. Let us prove that the transformation of the equation A(x)=B(x) into the equation C(x)=B(x) is an equivalent transformation, that is, we will prove that the equations A(x)=B(x) and C(x) =B(x) are equivalent.

    To do this, it is enough to show that any root of the original equation is a root of the equation C(x)=B(x), and any root of the equation C(x)=B(x) is a root of the original equation.

    Let's start with the first part. Let q be the root of the equation A(x)=B(x), then when we substitute it for x we ​​will get the correct numerical equality A(q)=B(q). Since the expressions A(x) and C(x) are identically equal and the expression C(q) makes sense (this follows from the condition that the OD for the equation C(x)=B(x) coincides with the OD for the original equation) , then the numerical equality A(q)=C(q) is true. Next we use the properties of numerical equalities. Due to the symmetry property, the equality A(q)=C(q) can be rewritten as C(q)=A(q) . Then, due to the transitivity property, the equalities C(q)=A(q) and A(q)=B(q) imply the equality C(q)=B(q). This proves that q is the root of the equation C(x)=B(x) .

    The second part, and with it the entire statement as a whole, is proved in an absolutely analogous way.

    The essence of the analyzed equivalent transformation is as follows: it allows you to work separately with expressions on the left and right sides of the equations, replacing them identically in equal terms on the original ODZ variables.

    The most common example: we can replace the sum of numbers on the right side of the equation x=2+1 with its value, which will result in an equivalent equation of the form x=3. Indeed, we replaced the expression 2+1 with the identically equal expression 3, and the ODZ of the equation did not change. Another example: on the left side of the equation 3·(x+2)=7·x−2·x+4−1 we can, and on the right – , which will lead us to the equivalent equation 3·x+6=5·x+ 3. The resulting equation is indeed equivalent, since we replaced the expressions with identically equal expressions and at the same time obtained an equation that has an OD that coincides with the OD for the original equation.

    • Adding the same number to both sides of an equation or subtracting the same number from both sides of an equation is an equivalent transformation of the equation.

    Let us prove that adding the same number c to both sides of the equation A(x)=B(x) gives the equivalent equation A(x)+c=B(x)+c and that subtracting from both sides of the equation A(x) =B(x) of the same number c gives the equivalent equation A(x)−c=B(x)−c.

    Let q be the root of the equation A(x)=B(x), then the equality A(q)=B(q) is true. The properties of numerical equalities allow us to add to both sides of a true numerical equality or subtract the same number from its parts. Let us denote this number as c, then the equalities A(q)+c=B(q)+c and A(q)−c=B(q)−c are valid. From these equalities it follows that q is the root of the equation A(x)+c=B(x)+c and the equation A(x)−c=B(x)−c.

    Now back. Let q be the root of the equation A(x)+c=B(x)+c and the equation A(x)−c=B(x)−c, then A(q)+c=B(q)+c and A (q)−c=B(q)−c . We know that subtracting the same number from both sides of a true numerical equality produces a true numerical equality. We also know that adding the correct numerical equality to both sides gives the correct numerical equality. Let us subtract the number c from both sides of the correct numerical equality A(q)+c=B(q)+c, and add the number c to both sides of the equality A(x)−c=B(x)−c. This will give us the correct numerical equalities A(q)+c−c=B(q)+c−c and A(q)−c+c=B(q)+c−c, from which we conclude that A(q) =B(q) . From the last equality it follows that q is the root of the equation A(x)=B(x) .

    This proves the original statement as a whole.

    Let us give an example of such a transformation of equations. Let's take the equation x−3=1, and transform it by adding the number 3 to both sides, after which we get the equation x−3+3=1+3, which is equivalent to the original one. It is clear that in the resulting equation you can perform operations with numbers, as we discussed in the previous item on the list, as a result we have the equation x=4. So, performing equivalent transformations, we accidentally solved the equation x−3=1, its root is the number 4. The considered equivalent transformation is very often used to get rid of identical numerical terms located in different parts equations For example, in both the left and right sides of the equation x 2 +1=x+1 there is the same term 1, subtracting the number 1 from both sides of the equation allows us to move on to the equivalent equation x 2 +1−1=x+1−1 and further to equivalent equation x 2 =x, and thereby get rid of these identical terms.

    • Adding to both sides of the equation or subtracting from both sides of the equation an expression for which the ODZ is not narrower than the ODZ for the original equation is an equivalent transformation.

    Let's prove this statement. That is, we prove that the equations A(x)=B(x) and A(x)+C(x)=B(x)+C(x) are equivalent, provided that the ODZ for the expression C(x) is not already , than ODZ for the equation A(x)=B(x) .

    First we prove one auxiliary point. Let us prove that, under the specified conditions, the OD equations before and after the transformation are the same. Indeed, the ODZ for the equation A(x)+C(x)=B(x)+C(x) can be considered as the intersection of the ODZ for the equation A(x)=B(x) and the ODZ for the expression C(x) . From this and from the fact that the ODZ for the expression C(x) is not narrower by condition than the ODZ for the equation A(x)=B(x), it follows that the ODZ for the equations A(x)=B(x) and A (x)+C(x)=B(x)+C(x) are the same.

    Now we will prove the equivalence of the equations A(x)=B(x) and A(x)+C(x)=B(x)+C(x), provided that the ranges of acceptable values ​​for these equations are the same. We will not give a proof of the equivalence of the equations A(x)=B(x) and A(x)−C(x)=B(x)−C(x) under the specified condition, since it is similar.

    Let q be the root of the equation A(x)=B(x), then the numerical equality A(q)=B(q) is true. Since the ODZ of the equations A(x)=B(x) and A(x)+C(x)=B(x)+C(x) are the same, then the expression C(x) makes sense at x=q, which means C(q) is some number. If we add C(q) to both sides of the correct numerical equality A(q)=B(q) , this will give the correct numerical inequality A(q)+C(q)=B(q)+C(q) , from which it follows that q is the root of the equation A(x)+C(x)=B(x)+C(x) .

    Back. Let q be the root of the equation A(x)+C(x)=B(x)+C(x), then A(q)+C(q)=B(q)+C(q) is a true numerical equality. We know that subtracting the same number from both sides of a true numerical equality produces a true numerical equality. Subtract C(q) from both sides of the equality A(q)+C(q)=B(q)+C(q) , this gives A(q)+C(q)−C(q)=B(q)+C(q)−C(q) and further A(q)=B(q) . Therefore, q is the root of the equation A(x)=B(x) .

    Thus, the statement in question is completely proven.

    Let's give an example of this transformation. Let's take the equation 2 x+1=5 x+2. We can add to both sides, for example, the expression −x−1. Adding this expression will not change the ODZ, which means that such a transformation is equivalent. As a result of this, we obtain the equivalent equation 2 x+1+(−x−1)=5 x+2+(−x−1). This equation can be transformed further: open the brackets and reduce similar terms on its left and right sides (see the first item in the list). After performing these actions, we obtain the equivalent equation x=4·x+1. The transformation of equations under consideration is often used to get rid of identical terms that are simultaneously on the left and right sides of the equation.

    • If you move a term in an equation from one part to another, changing the sign of this term to the opposite, you will get an equation equivalent to the given one.

    This statement is a consequence of the previous ones.

    Let us show how this equivalent transformation of the equation is carried out. Let's take the equation 3·x−1=2·x+3. Let's move the term, for example, 2 x from the right side to the left, changing its sign. In this case, we obtain the equivalent equation 3·x−1−2·x=3. You can also move minus one from the left side of the equation to the right, changing the sign to plus: 3 x−2 x=3+1. Finally, bringing similar terms leads us to the equivalent equation x=4.

    • Multiplying or dividing both sides of an equation by the same non-zero number is an equivalent transformation.

    Let's give a proof.

    Let A(x)=B(x) be some equation and c be some number different from zero. Let us prove that multiplying or dividing both sides of the equation A(x)=B(x) by the number c is an equivalent transformation of the equation. To do this, we prove that the equations A(x)=B(x) and A(x) c=B(x) c, as well as the equations A(x)=B(x) and A(x):c= B(x):c - equivalent. This can be done this way: prove that any root of the equation A(x)=B(x) is a root of the equation A(x) c=B(x) c and a root of the equation A(x):c=B(x) :c , and then prove that any root of the equation A(x) c=B(x) c , like any root of the equation A(x):c=B(x):c, is a root of the equation A(x) =B(x) . Let's do it.

    Let q be the root of the equation A(x)=B(x) . Then the numerical equality A(q)=B(q) is true. Having studied the properties of numerical equalities, we learned that multiplying or dividing both sides of a true numerical equality by the same number other than zero leads to a true numerical equality. Multiplying both sides of the equality A(q)=B(q) by c, we obtain the correct numerical equality A(q) c=B(q) c, from which it follows that q is the root of the equation A(x) c= B(x)·c . And dividing both sides of the equality A(q)=B(q) by c, we obtain the correct numerical equality A(q):c=B(q):c, from which it follows that q is the root of the equation A(x):c =B(x):c .

    Now in the other direction. Let q be the root of the equation A(x) c=B(x) c. Then A(q)·c=B(q)·c is a true numerical equality. Dividing both parts of it by a non-zero number c, we obtain the correct numerical equality A(q)·c:c=B(q)·c:c and further A(q)=B(q) . It follows that q is the root of the equation A(x)=B(x) . If q is the root of the equation A(x):c=B(x):c . Then A(q):c=B(q):c is a true numerical equality. Multiplying both parts of it by a non-zero number c, we obtain the correct numerical equality A(q):c·c=B(q):c·c and further A(q)=B(q) . It follows that q is the root of the equation A(x)=B(x) .

    The statement has been proven.

    Let's give an example of this transformation. With its help, you can, for example, get rid of fractions in the equation. To do this, you can multiply both sides of the equation by 12. The result is an equivalent equation of the form , which can then be transformed into the equivalent equation 7 x−3=10, which does not contain fractions in its notation.

    • Multiplying or dividing both sides of an equation by the same expression, the OD for which is not narrower than the OD for the original equation and does not vanish by the OD for the original equation, is an equivalent transformation.

    Let's prove this statement. To do this, we prove that if the ODZ for the expression C(x) is not narrower than the ODZ for the equation A(x)=B(x), and C(x) does not vanish on the ODZ for the equation A(x)=B( x) , then the equations A(x)=B(x) and A(x) C(x)=B(x) C(x), as well as the equations A(x)=B(x) and A( x):C(x)=B(x):C(x) - equivalent.

    Let q be the root of the equation A(x)=B(x) . Then A(q)=B(q) is a true numerical equality. From the fact that the ODZ for the expression C(x) is not the same ODZ for the equation A(x)=B(x), it follows that the expression C(x) makes sense when x=q. This means that C(q) is some number. Moreover, C(q) is nonzero, which follows from the condition that the expression C(x) does not vanish. If we multiply both sides of the equality A(q)=B(q) by a non-zero number C(q), this will give the correct numerical equality A(q)·C(q)=B(q)·C(q) , from which it follows that q is the root of the equation A(x)·C(x)=B(x)·C(x) . If we divide both sides of the equality A(q)=B(q) by a non-zero number C(q), this will give the correct numerical equality A(q):C(q)=B(q):C(q) , from which it follows that q is the root of the equation A(x):C(x)=B(x):C(x) .

    Back. Let q be the root of the equation A(x)·C(x)=B(x)·C(x) . Then A(q)·C(q)=B(q)·C(q) is a true numerical equality. Note that the ODZ for the equation A(x) C(x)=B(x) C(x) is the same as the ODZ for the equation A(x)=B(x) (we justified this in one of the previous paragraphs current list). Since C(x) by condition does not vanish on the ODZ for the equation A(x)=B(x), then C(q) is a nonzero number. Dividing both sides of the equality A(q) C(q)=B(q) C(q) by a non-zero number C(q) we obtain the correct numerical equality A(q)·C(q):C(q)=B(q)·C(q):C(q) and further A(q)=B(q) . It follows that q is the root of the equation A(x)=B(x) . If q is the root of the equation A(x):C(x)=B(x):C(x) . Then A(q):C(q)=B(q):C(q) is a true numerical equality. Multiplying both sides of the equality A(q):C(q)=B(q):C(q) by a non-zero number C(q) we obtain the correct numerical equality A(q):C(q)·C(q)=B(q):C(q)·C(q) and further A(q)=B(q) . It follows that q is the root of the equation A(x)=B(x) .

    The statement has been proven.

    For clarity, we give an example of carrying out a disassembled transformation. Let's divide both sides of the equation x 3 ·(x 2 +1)=8·(x 2 +1) by the expression x 2 +1. This transformation is equivalent, since the expression x 2 +1 does not vanish on the OD for the original equation and the OD of this expression is not narrower than the OD for the original equation. As a result of this transformation, we obtain the equivalent equation x 3 ·(x 2 +1):(x 2 +1)=8·(x 2 +1):(x 2 +1), which can be further transformed to the equivalent equation x 3 =8.

    Transformations leading to corollary equations

    In the previous paragraph, we examined which transformations from the list of basic transformations and under what conditions are equivalent. Now let's see which of these transformations and under what conditions lead to corollary equations, that is, to equations that contain all the roots of the transformed equation, but in addition to them may also have other roots - extraneous roots for the original equation.

    Transformations leading to corollary equations are in demand no less than equivalent transformations. If with their help it is possible to obtain an equation that is quite simple in terms of solution, then its solution and subsequent elimination of extraneous roots will give a solution to the original equation.

    Note that all equivalent transformations can be considered special cases of transformations that lead to corollary equations. This is understandable, because an equivalent equation is a special case of a corollary equation. But from a practical point of view, it is more useful to know that the transformation under consideration is precisely equivalent, and not leading to a corollary equation. Let us explain why this is so. If we know that the transformation is equivalent, then the resulting equation will definitely not have roots extraneous to the original equation. And the transformation leading to the corollary equation may be the cause of the appearance of extraneous roots, which obliges us in the future to carry out an additional action - sifting out extraneous roots. Therefore, in this section of the article we will focus on transformations, as a result of which extraneous roots may appear for the original equation. And it is really important to be able to distinguish such transformations from equivalent transformations in order to clearly understand when it is necessary to filter out extraneous roots, and when this is not necessary.

    Let's analyze the entire list of basic transformations of equations given in the second paragraph of this article in order to search for transformations, as a result of which extraneous roots may appear.

    • Replacing expressions on the left and right sides of the equation with identically equal expressions.

    We have proven that this transformation is equivalent if its implementation does not change the OD. And if the DL changes, what will happen? The narrowing of the ODZ can lead to the loss of roots; this will be discussed in more detail in the next paragraph. And with the expansion of the ODZ, extraneous roots may appear. It is not difficult to justify this. Let us present the corresponding reasoning.

    Let the expression C(x) be such that it is identically equal to the expression A(x) and the OD for the equation C(x)=B(x) is wider than the OD for the equation A(x)=B(x). Let us prove that the equation C(x)=B(x) is a consequence of the equation A(x)=B(x), and that among the roots of the equation C(x)=B(x) there may be roots that are foreign to the equation A( x)=B(x) .

    Let q be the root of the equation A(x)=B(x) . Then A(q)=B(q) is a true numerical equality. Since the ODZ for the equation C(x)=B(x) is wider than the ODZ for the equation A(x)=B(x), then the expression C(x) is defined at x=q. Then, taking into account the identical equality of the expressions C(x) and A(x) , we conclude that C(q)=A(q) . From the equalities C(q)=A(q) and A(q)=B(q), due to the transitivity property, the equality C(q)=B(q) follows. From this equality it follows that q is the root of the equation C(x)=B(x) . This proves that under the specified conditions the equation C(x)=B(x) is a consequence of the equation A(x)=B(x) .

    It remains to prove that the equation C(x)=B(x) can have roots different from the roots of the equation A(x)=B(x). Let us prove that any root of the equation C(x)=B(x) from the ODZ for the equation A(x)=B(x) is a root of the equation A(x)=B(x). Path p is the root of the equation C(x)=B(x), belonging to the ODZ for the equation A(x)=B(x). Then C(p)=B(p) is a true numerical equality. Since p belongs to the ODZ for the equation A(x)=B(x), then the expression A(x) is defined for x=p. From this and from the identical equality of the expressions A(x) and C(x) it follows that A(p)=C(p) . From the equalities A(p)=C(p) and C(p)=B(p), due to the transitivity property, it follows that A(p)=B(p), which means p is the root of the equation A(x)= B(x) . This proves that any root of the equation C(x)=B(x) from the ODZ for the equation A(x)=B(x) is a root of the equation A(x)=B(x). In other words, on the ODZ for the equation A(x)=B(x) there cannot be roots of the equation C(x)=B(x), which are extraneous roots for the equation A(x)=B(x). But according to the condition, the ODZ for the equation C(x)=B(x) is wider than the ODZ for the equation A(x)=B(x). And this allows the existence of a number r that belongs to the ODZ for the equation C(x)=B(x) and does not belong to the ODZ for the equation A(x)=B(x), which is the root of the equation C(x)=B(x). That is, the equation C(x)=B(x) may have roots that are foreign to the equation A(x)=B(x), and all of them will belong to the set to which the ODZ for the equation A(x)=B is extended (x) when replacing the expression A(x) in it with the identically equal expression C(x).

    So, replacing the expressions on the left and right sides of the equation with identically equal expressions, as a result of which the ODZ is expanded, in the general case leads to a corollary equation (that is, it can lead to the appearance of extraneous roots) and only in a particular case leads to equivalent equation (in the event that the resulting equation does not have roots foreign to the original equation).

    Let us give an example of carrying out a parsed transformation. Replacing the expression on the left side of the equation identically equal to it by the expression x·(x−1) leads to the equation x·(x−1)=0, in this case the expansion of the ODZ occurs - the number 0 is added to it. The resulting equation has two roots 0 and 1, and substituting these roots into the original equation shows that 0 is an extraneous root for the original equation, and 1 is the root of the original equation. Indeed, substituting zero into the original equation gives the meaningless expression , since it contains division by zero, and substituting one gives the correct numerical equality , which is the same as 0=0 .

    Note that a similar transformation of a similar equation into the equation (x−1)·(x−2)=0, as a result of which the ODZ also expands, does not lead to the appearance of extraneous roots. Indeed, both roots of the resulting equation (x−1)·(x−2)=0 - numbers 1 and 2, are roots of the original equation, which is easy to verify by checking by substitution. With these examples, we once again wanted to emphasize that replacing an expression on the left or right side of the equation with an identically equal expression, which expands the ODZ, does not necessarily lead to the appearance of extraneous roots. But it can also lead to their appearance. So, if such a transformation took place in the process of solving the equation, then it is necessary to carry out a check in order to identify and filter out extraneous roots.

    Most often, the ODZ of an equation can expand and extraneous roots may appear due to the replacement by zero of the difference of identical expressions or the sum of expressions with opposite signs, due to the replacement by zero of products with one or more zero factors, due to the reduction of fractions and due to the use of properties roots, powers, logarithms, etc.

    • Adding the same number to both sides of an equation or subtracting the same number from both sides of an equation.

    We showed above that this transformation is always equivalent, that is, leading to an equivalent equation. Go ahead.

    • Adding the same expression to both sides of an equation or subtracting the same expression from both sides of an equation.

    In the previous paragraph, we added a condition that the ODZ for the expression being added or subtracted should not be narrower than the ODZ for the equation being transformed. This condition made the transformation in question equivalent. Here there are arguments similar to those given at the beginning of this paragraph of the article regarding the fact that an equivalent equation is a special case of a corollary equation and that knowledge about the equivalence of a transformation is practically more useful than knowledge about the same transformation, but from the standpoint of the fact that it leads to corollary equation.

    Is it possible, as a result of adding the same expression or subtracting the same expression from both sides of an equation, to obtain an equation that, in addition to all the roots of the original equation, will have some other roots? No, he can not. If the ODZ for the expression being added or subtracted is not narrower than the ODZ for the original equation, then as a result of the addition or subtraction an equivalent equation will be obtained. If the ODZ for the expression being added or subtracted is narrower than the ODZ for the original equation, then this can lead to the loss of roots, and not to the appearance of extraneous roots. We'll talk more about this in the next paragraph.

    • Transferring a term from one part of the equation to another with the sign changed to the opposite.

    This transformation of the equation is always equivalent. Therefore, it makes no sense to consider it as a transformation leading to an equation-consequence, for the reasons stated above.

    • Multiplying or dividing both sides of an equation by the same number.

    In the previous paragraph, we proved that if the multiplication or division of both sides of the equation is carried out by a non-zero number, then this is an equivalent transformation of the equation. Therefore, again, there is no point in talking about it as a transformation leading to a corollary equation.

    But here it is worth paying attention to the reservation about the difference from zero of the number by which both sides of the equation are multiplied or divided. For division this clause is clear - with primary classes we realized that You can't divide by zero. Why this clause for multiplication? Let's think about what multiplying both sides of the equation by zero results in. For clarity, let's take a specific equation, for example, 2 x+1=x+5. This is a linear equation that has a single root, which is the number 4. Let's write down the equation that will be obtained by multiplying both sides of this equation by zero: (2 x+1) 0=(x+5) 0. Obviously, the root of this equation is any number, because when you substitute any number into this equation instead of the variable x, you get the correct numerical equality 0=0. That is, in our example, multiplying both sides of the equation by zero led to a corollary equation, which caused the appearance of an infinite number of extraneous roots for the original equation. Moreover, it is worth noting that in this case the usual methods of screening out extraneous roots do not cope with their task. This means that the transformation performed is useless for solving the original equation. And this is a typical situation for the transformation under consideration. This is why a transformation such as multiplying both sides of an equation by zero is not used to solve equations. We still have to look at this transformation and other transformations that should not be used to solve equations in the last paragraph.

    • Multiplying or dividing both sides of an equation by the same expression.

    In the previous paragraph, we proved that this transformation is equivalent if two conditions are met. Let's remind them. The first condition: the OD for this expression should not be narrower than the OD for the original equation. The second condition: the expression by which the multiplication or division is carried out must not vanish on the ODZ for the original equation.

    Let's change the first condition, that is, we will assume that the OD for the expression by which we plan to multiply or divide both sides of the equation is narrower than the OD for the original equation. As a result of such a transformation, an equation will be obtained for which the ODZ will be narrower than the ODZ for the original equation. Such transformations can lead to the loss of roots; we will talk about them in the next paragraph.

    What will happen if we remove the second condition about the non-zero values ​​of the expression by which both sides of the equation are multiplied or divided by the ODZ for the original equation?

    Dividing both sides of the equation by the same expression, which vanishes by the OD for the original equation, will result in an equation whose OD is narrower than the OD for the original equation. Indeed, numbers will fall out of it, turning the expression by which the division was carried out to zero. This can lead to root loss.

    What about multiplying both sides of the equation by the same expression, which vanishes on the ODZ for the original equation? It can be shown that when both sides of the equation A(x)=B(x) are multiplied by the expression C(x), for which the ODZ is not narrower than the ODZ for the original equation, and which vanishes by the ODZ for the original equation, the equation is obtained is a consequence that, in addition to all the roots of the equation A(x)=B(x), it can also have other roots. Let's do this, especially since this paragraph of the article is precisely devoted to transformations leading to corollary equations.

    Let the expression C(x) be such that the ODZ for it is not narrower than the ODZ for the equation A(x)=B(x), and it vanishes on the ODZ for the equation A(x)=B(x) . Let us prove that in this case the equation A(x)·C(x)=B(x)·C(x) is a consequence of the equation A(x)=B(x) .

    Let q be the root of the equation A(x)=B(x) . Then A(q)=B(q) is a true numerical equality. Since the ODZ for the expression C(x) is not narrower than the ODZ for the equation A(x)=B(x), then the expression C(x) is defined at x=q, which means that C(q) is a certain number. Multiplying both sides of a true numerical equality by any number gives a true numerical equality, therefore, A(q)·C(q)=B(q)·C(q) is a true numerical equality. This means q is the root of the equation A(x)·C(x)=B(x)·C(x) . This proves that any root of the equation A(x)=B(x) is a root of the equation A(x) C(x)=B(x) C(x), which means that the equation A(x) C (x)=B(x)·C(x) is a consequence of the equation A(x)=B(x) .

    Note that under the specified conditions, the equation A(x)·C(x)=B(x)·C(x) may have roots that are foreign to the original equation A(x)=B(x). They are all numbers from the ODZ for the original equation that turn the expression C(x) to zero (all numbers that turn the expression C(x) to zero are the roots of the equation A(x) C(x)=B(x) C(x) , since their substitution into the indicated equation gives the correct numerical equality 0=0 ), but which are not roots of the equation A(x)=B(x) . The equations A(x)=B(x) and A(x)·C(x)=B(x)·C(x) under the specified conditions will be equivalent when all numbers from the ODZ for the equation A(x)=B (x) , which make the expression C(x) vanish, are the roots of the equation A(x)=B(x) .

    So, multiplying both sides of the equation by the same expression, the ODZ for which is not narrower than the ODZ for the original equation, and which vanishes by the ODZ for the original equation, in the general case leads to a corollary equation, that is, it can lead to the appearance of foreign roots.

    Let's give an example to illustrate. Let's take the equation x+3=4. Its only root is the number 1. Let's multiply both sides of this equation by the same expression, which vanishes by the ODZ for the original equation, for example, by x·(x−1) . This expression vanishes at x=0 and x=1. Multiplying both sides of the equation by this expression gives us the equation (x+3) x (x−1)=4 x (x−1). The resulting equation has two roots: 1 and 0. The number 0 is an extraneous root for the original equation that appeared as a result of the transformation.

    Transformations that may lead to loss of roots

    Some conversions from under certain conditions can lead to loss of roots. For example, when dividing both sides of the equation x·(x−2)=x−2 by the same expression x−2, the root is lost. Indeed, as a result of such a transformation, the equation x=1 is obtained with a single root, which is the number 1, and the original equation has two roots 1 and 2.

    It is necessary to clearly understand when roots are lost as a result of transformations, so as not to lose roots when solving equations. Let's figure this out.

    As a result of these transformations, loss of roots can occur if and only if the ODZ for the transformed equation turns out to be narrower than the ODZ for the original equation.

    To prove this statement, two points need to be substantiated. First, it is necessary to prove that if, as a result of the indicated transformations of the equation, the ODZ is narrowed, then a loss of roots may occur. And, secondly, it is necessary to justify that if, as a result of these transformations, the roots are lost, then the ODZ for the resulting equation is narrower than the ODZ for the original equation.

    If the ODZ for the equation obtained as a result of the transformation is narrower than the ODZ for the original equation, then, naturally, not a single root of the original equation located outside the ODZ for the resulting equation can be the root of the equation obtained as a result of the transformation. This means that all these roots will be lost when moving from the original equation to an equation for which the ODZ is narrower than the ODZ for the original equation.

    Now back. Let us prove that if, as a result of these transformations, the roots are lost, then the ODZ for the resulting equation is narrower than the ODZ for the original equation. This can be done by the opposite method. The assumption that as a result of these transformations, the roots are lost, but the ODZ is not narrowed, contradicts the statements proven in the previous paragraphs. Indeed, from these statements it follows that if, when carrying out the indicated transformations, the ODZ is not narrowed, then either equivalent equations or corollary equations are obtained, which means that loss of roots cannot occur.

    So, the reason for the possible loss of roots when carrying out basic transformations of equations is the narrowing of the ODZ. It is clear that when solving equations, we should not lose roots. Here, naturally, the question arises: “What should we do to avoid losing roots when transforming equations?” We will answer it in the next paragraph. Now let's go through the list of basic transformations of equations to see in more detail which transformations can lead to the loss of roots.

    • Replacing expressions on the left and right sides of the equation with identically equal expressions.

    If you replace the expression on the left or right side of the equation with an identically equal expression, the OD for which is narrower than the OD for the original equation, this will lead to a narrowing of the OD, and because of this, roots may be lost. Most often, replacement of expressions on the left or right side of equations with identically equal expressions, carried out on the basis of some properties of roots, powers, logarithms and some trigonometric formulas, leads to a narrowing of the ODZ and, as a consequence, to the possible loss of roots. For example, replacing the expression on the left side of the equation with an identically equal expression narrows the ODZ and leads to the loss of the root −16. Similarly, replacing the expression on the left side of the equation with an identically equal expression leads to an equation for which the ODZ is narrower than the ODZ for the original equation, which entails the loss of the root −3.

    • Adding the same number to both sides of an equation or subtracting the same number from both sides of an equation.

    This transformation is equivalent, therefore, roots cannot be lost during its implementation.

    • Adding the same expression to both sides of an equation or subtracting the same expression from both sides of an equation.

    If you add or subtract an expression whose OD is narrower than the OD for the original equation, this will lead to a narrowing of the OD and, as a consequence, to a possible loss of roots. It's worth keeping this in mind. But here it is worth noting that in practice it is usually necessary to resort to adding or subtracting expressions that are present in the recording of the original equation, which does not lead to a change in the ODZ and does not entail the loss of roots.

    • Transferring a term from one part of the equation to another with the sign changed to the opposite.

    This transformation of the equation is equivalent, therefore, as a result of its implementation, the roots are not lost.

    • Multiplying or dividing both sides of an equation by the same number other than zero.

    This transformation is also equivalent, and because of it, the loss of roots does not occur.

    • Multiplying or dividing both sides of an equation by the same expression.

    This transformation can lead to a narrowing of the OD in two cases: when the OD for the expression by which the multiplication or division is carried out is narrower than the OD for the original equation, and when division is carried out by an expression that becomes zero on the OD for the original equation. Note that in practice it is usually not necessary to resort to multiplying and dividing both sides of the equation by an expression with a narrower VA. But you have to deal with division by an expression that turns into zero for the original equation. There is a method that allows you to cope with the loss of roots during such division, we will talk about it in the next paragraph of this article.

    How to avoid root loss?

    If you use only transformations from to transform equations and at the same time do not allow narrowing of the ODZ, then the loss of roots will not occur.

    Does this mean that no other transformations of the equations can be made? No, it doesn't mean that. If you come up with some other transformation of the equation and fully describe it, that is, indicate when it leads to equivalent equations, when to corollary equations, and when it can lead to the loss of roots, then it could well be adopted.

    Should we completely abandon reforms that would narrow DPD? Should not be doing that. It would not hurt to keep in your arsenal transformations in which a finite number of numbers drop out of the ODZ for the original equation. Why shouldn’t such transformations be abandoned? Because there is a method to avoid root loss in such cases. It consists of a separate check of the numbers falling out of the ODZ to see if there are roots of the original equation among them. You can check this by substituting these numbers into the original equation. Those of them that, when substituted, give the correct numerical equality, are the roots of the original equation. They need to be included in the answer. After such a check, you can safely carry out the planned transformation without fear of losing your roots.

    A typical transformation in which the ODZ for an equation is narrowed down to several numbers is to divide both sides of the equation by the same expression, which becomes zero at several points from the ODZ for the original equation. This transformation is the basis of the solution method reciprocal equations. But it is also used to solve other types of equations. Let's give an example.

    The equation can be solved by introducing a new variable. To introduce a new variable, you need to divide both sides of the equation by 1+x. But with such a division, a loss of root may occur, since although the ODZ for the expression 1+x is not narrower than the ODZ for the original equation, the expression 1+x becomes zero at x=−1, and this number belongs to the ODZ for the original equation. This means that the root −1 may be lost. To eliminate the loss of a root, you should separately check whether −1 is a root of the original equation. To do this, you can substitute −1 into the original equation and see what equality you get. In our case, the substitution gives the equality, which is the same as 4=0. This equality is false, which means −1 is not the root of the original equation. After such a check, you can carry out the intended division of both sides of the equation by 1 + x, without fear that loss of roots may occur.

    At the end of this paragraph, let us once again turn to the equations from the previous paragraph and. Transformation of these equations based on identities and leads to a narrowing of the ODZ, and this entails the loss of roots. At this point, we said that in order not to lose our roots, we need to abandon reforms that narrow the DZ. This means that these transformations must be abandoned. But what should we do? It is possible to carry out transformations not based on identities and , due to which the ODZ is narrowed, and on the basis of identities and . As a result of the transition from the original equations and to the equations and there is no narrowing of the ODZ, which means that the roots will not be lost.

    Here we especially note that when replacing expressions with identically equal expressions, you must carefully ensure that the expressions are exactly identically equal. For example, in Eq. it is impossible to replace the expression x+3 with an expression in order to simplify the appearance of the left side to , since the expressions x+3 and are not identically equal, because their values ​​do not coincide at x+3<0 . В нашем примере такое преобразование приводит к потере корня. А в общем случае замена выражения не тождественно равным выражением приводит к уравнению, которое не позволяет получить решение исходного уравнения.

    Transformations of equations that should not be used

    The transformations mentioned in this article are usually sufficient for practical needs. That is, you shouldn’t be too bothered about coming up with any other transformations; it’s better to focus on the correct use of the already proven ones.

    Literature

    1. Mordkovich A. G. Algebra and beginning of mathematical analysis. Grade 11. 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.
    2. 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.

    Class: 11

    Duration: 2 lessons.

    The purpose of the lesson:

    • (for teacher) formation in students of a holistic understanding of methods for solving irrational equations.
    • (for students) Development of the ability to observe, compare, generalize, and analyze mathematical situations (slide 2). Preparation for the Unified State Exam.

    First lesson plan(slide 3)

    1. Updating knowledge
    2. Analysis of the theory: Raising an equation to an even power
    3. Workshop on solving equations

    Second lesson plan

    1. Differentiated independent work in groups “Irrational equations on the Unified State Exam”
    2. Summary of lessons
    3. Homework

    Progress of lessons

    I. Updating knowledge

    Target: repeat the concepts necessary to successfully master the lesson topic.

    Frontal survey.

    – Which two equations are called equivalent?

    – What transformations of an equation are called equivalent?

    – Replace this equation with an equivalent one with an explanation of the applied transformation: (slide 4)

    a) x+ 2x +1; b) 5 = 5; c) 12x = -3; d) x = 32; d) = -4.

    – What equation is called the corollary equation of the original equation?

    – Can a corollary equation have a root that is not the root of the original equation? What are these roots called?

    – What transformations of the equation lead to corollary equations?

    – What is called an arithmetic square root?

    Today we will dwell in more detail on the transformation “Raising an equation to an even power”.

    II. Analysis of the theory: Raising an equation to an even power

    Teacher's explanation with active participation of students:

    Let 2m(mN) is a fixed even natural number. Then the consequence of the equationf(x) =g(x) is the equation (f(x)) = (g(x)).

    Very often this statement is used when solving irrational equations.

    Definition. An equation containing an unknown under the root sign is called irrational.

    When solving irrational equations, the following methods are used: (slide 5)

    Attention! Methods 2 and 3 require mandatory checks.

    ODZ does not always help eliminate extraneous roots.

    Conclusion: When solving irrational equations, it is important to go through three stages: technical, solution analysis, verification (slide 6).

    III. Workshop on solving equations

    Solve the equation:

    After discussing how to solve an equation by squaring, solve by going to an equivalent system.

    Conclusion: The simplest equations with integer roots can be solved by any familiar method.

    b) = x – 2

    By solving by raising both sides of the equation to the same power, students obtain roots x = 0, x = 3 -, x = 3 +, which are difficult and time-consuming to check by substitution. (Slide 7). Transition to an equivalent system

    allows you to quickly get rid of foreign roots. The condition x ≥ 2 is satisfied only by x.

    Answer: 3 +

    Conclusion: It is better to check irrational roots by moving to an equivalent system.

    c) = x – 3

    In the process of solving this equation, we obtain two roots: 1 and 4. Both roots satisfy the left side of the equation, but when x = 1 the definition of an arithmetic square root is violated. The ODZ equation does not help eliminate extraneous roots. The transition to an equivalent system gives the correct answer.

    Conclusion:good knowledge and understanding of all the conditions for determining the arithmetic square root helps to move on toperforming equivalent transformations.

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

    x + 13 - 8 + 16 = 3 + 2x - x, placing the radical on the right side, we get

    26 – x + x = 8. Application of further actions to square both sides of the equation will lead to an equation of the 4th degree. The transition to the ODZ equation gives a good result:

    let's find the ODZ equation:

    x = 3.

    Check: - 4 = , 0 = 0 correct.

    Conclusion:Sometimes it is possible to solve using the definition of the ODZ equation, but be sure to check.

    Solution: ODZ equation: -2 – x ≥ 0 x ≤ -2.

    For x ≤ -2,< 0, а ≥ 0.

    Therefore, the left side of the equation is negative, and the right side is non-negative; therefore the original equation has no roots.

    Answer: no roots.

    Conclusion:Having made the correct reasoning on the limitation in the condition of the equation, you can easily find the roots of the equation, or establish that they do not exist.

    Using the example of solving this equation, show the double squaring of the equation, explain the meaning of the phrase “solitude of radicals” and the need to check the roots found.

    h) + = 1.

    The solution of these equations is carried out using the variable replacement method until the return to the original variable. Offer the solution to those who complete the tasks of the next stage earlier.

    Control questions

    • How to solve the simplest irrational equations?
    • What do you need to remember when raising an equation to an even power? ( foreign roots may appear)
    • What is the best way to check irrational roots? ( using ODZ and conditions for the coincidence of the signs of both sides of the equation)
    • Why is it necessary to be able to analyze mathematical situations when solving irrational equations? ( For the correct and quick choice of how to solve the equation).

    IV. Differentiated independent work in groups “Irrational equations on the Unified State Exam”

    The class is divided into groups (2-3 people) according to levels of training, each group chooses an option with a task, discusses and solves the selected tasks. If necessary, seek advice from the teacher. After completing all the tasks in their version and checking the answers by the teacher, group members individually finish solving equations g) and h) of the previous stage of the lesson. For options 4 and 5 (after checking the answers and solutions by the teacher), additional tasks are written on the board and are completed individually.

    All individual solutions are submitted to the teacher for verification at the end of the lessons.

    Option 1

    Solve the equations:

    a) = 6;
    b) = 2;
    c) = 2 – x;
    d) (x + 1) (5 – x) (+ 2 = 4.

    Option 5

    1. Solve the equation:

    a) = ;
    b) = 3 – 2x;

    2. Solve the system of equations:

    Additional tasks:

    V. Summary of lessons

    What difficulties did you experience when completing USE tasks? What is needed to overcome these difficulties?

    VI. Homework

    Repeat the theory of solving irrational equations, read paragraph 8.2 in the textbook (pay attention to example 3).

    Solve No. 8.8 (a, c), No. 8.9 (a, c), No. 8.10 (a).

    Literature:

    1. Nikolsky S.M., Potapov M.K., N.N. Reshetnikov N.N., Shevkin A.V. Algebra and beginning of mathematical analysis , textbook for 11th grade of general education institutions, M.: Prosveshchenie, 2009.
    2. Mordkovich A.G. On some methodological issues related to solving equations. Mathematics at school. -2006. -No. 3.
    3. M. Shabunin. Equations. Lectures for high school students and applicants. Moscow, “Chistye Prudy”, 2005. (library “First of September”)
    4. E.N. Balayan. Problem solving workshop. Irrational equations, inequalities and systems. Rostov-on-Don, “Phoenix”, 2006.
    5. Mathematics. Preparation for the Unified State Exam 2011. Edited by F.F. Lysenko, S.Yu. Kulabukhova Legion-M, Rostov-on-Don, 2010.
    Development of an algebra lesson in the 11th specialized class

    The lesson was taught by mathematics teacher MBOU Secondary School No. 6 Tupitsyna O.V.

    Topic and lesson number in the topic:“Application of several transformations leading to an equation-consequence”, lesson No. 7, 8 in the topic: “Equation-consequence”

    Academic subject:Algebra and the beginnings of mathematical analysis – 11th grade (profile training according to the textbook by S.M. Nikolsky)

    Lesson type: “systematization and generalization of knowledge and skills”

    Lesson type: workshop

    The role of the teacher: direct the cognitive activity of students to develop the ability to independently apply knowledge in a complex to select the desired method or methods of transformation, leading to an equation - a consequence and application of the method in solving the equation, in new conditions.

    Required technical equipment:multimedia equipment, webcam.

    Used during the lesson:

    1. didactic teaching model- creating a problematic situation,
    2. pedagogical means– sheets indicating training modules, a selection of tasks for solving equations,
    3. type of student activity– group (groups are formed during lessons - “discovery” of new knowledge, lessons No. 1 and 2 from students with varying degrees of training and learning ability), joint or individual problem solving,
    4. personally-oriented educational technologies: modular learning, problem-based learning, search and research methods, collective dialogue, activity method, working with a textbook and various sources,
    5. health-saving technologies- exercise is performed to relieve tension,
    6. competencies:

    - educational and cognitive at a basic level- students know the concept of an equation - a consequence, a root of an equation and methods of transformation leading to an equation - a consequence, they are able to find the roots of equations and check them at a productive level;

    - at an advanced level– students can solve equations using known transformation methods, check the roots of the equations, using the range of permissible values ​​of the equations; calculate logarithms using properties based on research; informational – students independently search, extract and select the information necessary to solve educational problems in sources of various types.

    Didactic goal:

    creating conditions for:

    Formation of ideas about equations - consequences, roots and methods of transformations;

    Forming the experience of meaning-making on the basis of a logical consequence of previously studied methods of transforming equations: raising an equation to an even power, potentiating logarithmic equations, freeing an equation from denominators, bringing similar terms;

    Consolidating skills in determining the choice of transformation method, further solving the equation and choosing the roots of the equation;

    Mastering the skills of setting a problem based on known and learned information, forming requests to find out what is not yet known;

    Formation of cognitive interests, intellectual and creative abilities of students;

    Development of logical thinking, creative activity of students, design skills, ability to express their thoughts;

    Formation of a sense of tolerance and mutual assistance when working in a group;

    Awakening interest in solving equations independently;

    Tasks:

    Organize repetition and systematization of knowledge about ways to transform equations;

    - ensure mastery of methods for solving equations and checking their roots;

    - promote the development of analytical and critical thinking of students; compare and select optimal methods for solving equations;

    - create conditions for the development of research skills and group work skills;

    Motivate students to use the studied material to prepare for the Unified State Exam;

    Analyze and evaluate your work and the work of your comrades in performing this work.

    Planned results:

    *personal:

    Skills in formulating a problem based on known and learned information, forming requests to find out what is not yet known;

    Ability to select sources of information necessary to solve a problem; development of cognitive interests, intellectual and creative abilities of students;

    Development of logical thinking, creative activity, skills to express one’s thoughts, ability to build an argument;

    Self-assessment of performance results;

    Skill to work in team;

    *metasubject:

    The ability to highlight the main thing, compare, generalize, draw analogies, apply inductive methods of reasoning, put forward hypotheses when solving equations,

    Ability to interpret and apply acquired knowledge in preparation for the Unified State Exam;

    *subject:

    Knowledge of ways to transform equations,

    The ability to establish a pattern associated with various types of equations and use it when solving and selecting roots,

    Integrating lesson goals:

    1. (for the teacher) Formation in students of a holistic understanding of the methods of transforming equations and methods of solving them;
    2. (for students) Development of the ability to observe, compare, generalize, and analyze mathematical situations associated with types of equations containing various functions. Preparation for the Unified State Exam.

    Stage 1 of the lesson:

    Updating knowledge to increase motivation in the application of various methods of transforming equations (input diagnostics)

    Knowledge updating stagecarried out in the form of a test with self-test. Developmental tasks are offered, based on the knowledge acquired in previous lessons, requiring active mental activity from students and necessary to complete the task in this lesson.

    Verification work

    1. Select equations that require limiting the unknowns on the set of all real numbers:

    a) = X-2; b)3 = X-2; c) =1;

    d) ( = (; e) = ; f) +6 =5 ;

    g) = ; h) = .

    (2) Indicate the range of acceptable values ​​of each equation where there are restrictions.

    (3) Choose an example of an equation where the transformation may result in the loss of a root (use materials from previous lessons on this topic).

    Everyone checks their answers independently against the ready ones displayed on the screen. The most complex tasks are analyzed and students pay special attention to examples a, c, g, h, where restrictions exist.

    Conclusions are drawn that when solving equations, it is necessary to determine the range of values ​​​​allowed by the equation or check the roots in order to avoid extraneous values. Previously studied methods of transforming equations leading to a corollary equation are repeated. That is, students are thereby motivated to search for the correctly chosen method of solving the equation proposed to them in further work.

    Stage II of the lesson:

    Practical application of your knowledge, skills and abilities in solving equations.

    The groups are given sheets with a module compiled on the questions of this topic. The module includes five learning elements, each of which is aimed at performing specific tasks. Students with different degrees of training and learning ability independently determine the scope of their activities in the lesson, but since everyone works in groups, there is a continuous process of adjusting knowledge and skills, bringing those lagging behind to compulsory levels, others to advanced and creative levels.

    In the middle of the lesson there is a mandatory physical exercise.

    No. of educational element

    Training element indicating tasks

    Guide to mastering educational material

    UE-1

    Goal: Identify and justify the basic methods for solving equations based on the properties of functions.

    1. Exercise:

    Specify the transformation method to solve the following equations:

    A) )= -8);

    b) =

    c) ( = (

    d) ctg +x 2 -2x = ctg +24;

    d) = ;

    e) = sin x.

    2) Task:

    Solve at least two of the given equations.

    Describe what methods were used in the solved equations.

    Clause 7.3 p.212

    P.7.4 p.214

    P. 7.5 p.217

    P.7.2 p. 210

    UE-2

    Goal: Master rational techniques and methods of solution

    Exercise:

    Give examples from the above or independently selected (use materials from previous lessons) equations, in solving which you can use rational solution methods, what do they consist of? (emphasis on how to check the roots of an equation)

    UE-3

    Goal: Using the acquired knowledge when solving equations of a high level of complexity

    Exercise:

    = ( or

    ( = (

    P.7.5

    UE-4

    Set the level of mastery of the topic:

    low – solution of no more than 2 equations;

    Medium – solving no more than 4 equations;

    high – solution of no more than 5 equations

    UE-5

    Output control:

    Make a table in which you present all the methods of converting equations you use and for each method write down examples of equations you have solved, starting from lesson 1 of the topic: “Equations - consequences”

    Notes in notebooks

    Stage III of the lesson:

    The final diagnostic work represents the reflection of students, which will show readiness not only for writing a test, but also readiness for the Unified State Exam for this section.

    At the end of the lesson, all students, without exception, evaluate themselves, followed by teacher evaluation. If disagreements arise between the teacher and the student, the teacher can offer the student to complete an additional task in order to be able to objectively evaluate it. Homeworkis aimed at repeating the material before the test.




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