Basic methods for solving equations

What is a solution to an equation?

Identity transformation. Main

types of identical transformations.

foreign root. Root loss.

Equation solution is a process consisting mainly in replacing a given equation with another equation that is equivalent to it . Such a replacement is calledidentity transformation . The main identity transformations are as follows:

1.

Replacing one expression with another, identically equal to it. For example, the equation (3 x+ 2 ) 2 = 15 x+ 10 can be replaced by the following equivalent:9 x 2 + 12 x + 4 = 15 x + 10 .

2.

The transfer of terms of the equation from one side to the other with opposite signs. So, in the previous equation, we can transfer all its members from the right side to the left side with the “-” sign: 9 x 2 + 12 x + 4 – 15 x- 10 = 0, after which we get:9 x 2 – 3 x- 6 = 0 .

3.

Multiplication or division of both sides of an equation by the same expression (number) other than zero. This is very important, becausethe new equation may not be equivalent to the previous one if the expression by which we are multiplying or dividing may be equal to zero.

EXAMPLE The equationx- 1 = 0 has a single rootx= 1.

Multiplying both sides byx- 3 , we get the equation

( x- 1)( x- 3) = 0, which has two roots:x= 1 andx = 3.

The last value is not the root of the given equation

x- 1 = 0. This is the so-calledextraneous root .

Conversely, division can lead toroot loss . So

in our case, ifx- 1 )( x- 3 ) = 0 is the original

equation, then the rootx= 3 will be lost in division

both sides of the equationx- 3 .

In the last equation (item 2), we can divide all its terms by 3 (not zero!) and finally get:

3 x 2 – x – 2 = 0 .

This equation is equivalent to the original one:

(3 x+ 2) 2 = 15 x + 10 .

4.

Canraise both sides of the equation to an odd power orextract an odd root from both sides of the equation . It must be remembered that:

a) erectioneven degree may causeto the acquisition of extraneous roots ;

b)wrong extractioneven root can lead toloss of roots .

EXAMPLES. Equation 7x = 35 has a single rootx = 5 .

By squaring both sides of this equation, we get

the equation:

49 x 2 = 1225 .

having two roots:x = 5 andx = – 5. Last value

is a foreign root.

Wrong taking the square root of both

parts of equation 49x 2 = 1225 results in 7x = 35,

and we lose the rootx = – 5.

Correct taking the square root leads to

equation: | 7x | = 35, a hence two cases:

1) 7 x = 35, thenx = 5 ; 2) – 7 x = 35, thenx = – 5 .

Therefore, whencorrect extracting a square

root we do not lose the roots of the equation.

What meansright extract root? Here we meet

with a very important conceptarithmetic root

(cm. ).

Can lead to the appearance of so-called extraneous roots. In this article, we will firstly analyze in detail what is extraneous roots. Secondly, let's talk about the reasons for their occurrence. And thirdly, using examples, we will consider the main ways of sifting out extraneous roots, that is, checking the roots for the presence of extraneous ones among them in order to exclude them from the answer.

Extraneous roots of the equation, definition, examples

Algebra school textbooks do not define an extraneous root. There, the idea of ​​an extraneous root is formed by describing the following situation: with the help of some transformations of the equation, the transition from the original equation to the consequence equation is carried out, the roots of the obtained consequence equation are found, and the found roots are checked by substitution into the original equation, which shows that some of the found roots are not the roots of the original equation, these roots are called extraneous roots for the original equation.

Based on this base, you can take for yourself the following definition of an extraneous root:

Definition

extraneous roots are the roots of the equation-consequence obtained as a result of transformations, which are not the roots of the original equation.

Let's take an example. Consider the equation and the corollary of this equation x·(x−1)=0 , obtained by replacing the expression with the expression x·(x−1) which is identically equal to it. The original equation has a single root 1 . The equation obtained as a result of the transformation has two roots 0 and 1 . So 0 is an extraneous root for the original equation.

Causes of the possible appearance of extraneous roots

If no “exotic” transformations are used to obtain the consequence equation, but only basic transformations of equations are used, then extraneous roots can arise only for two reasons:

• due to the expansion of the ODZ and
• because both sides of the equation are raised to the same even power.

Here it is worth recalling that the expansion of the ODZ as a result of the transformation of the equation mainly occurs

• When reducing fractions;
• When replacing a product with one or more zero factors by zero;
• When replacing zero with a fraction with a zero numerator;
• When using some properties of powers, roots, logarithms;
• When using some trigonometric formulas;
• When multiplying both parts of the equation by the same expression, which vanishes on the ODZ for this equation;
• When released in the process of solving the signs of logarithms.

The example from the previous paragraph of the article illustrates the appearance of an extraneous root due to the expansion of the ODZ, which takes place when passing from the equation to the corollary equation x·(x−1)=0 . The ODZ for the original equation is the set of all real numbers, except for zero, the ODZ for the resulting equation is the set R, that is, the ODZ is extended by the number zero. This number eventually turns out to be an extraneous root.

We will also give an example of the appearance of an extraneous root due to the raising of both parts of the equation to the same even power. The irrational equation has a single root 4, and the consequence of this equation, obtained from it by squaring both parts of the equation, that is, the equation , has two roots 1 and 4 . From this it can be seen that squaring both sides of the equation led to the appearance of an extraneous root for the original equation.

Note that the expansion of the ODZ and the raising of both parts of the equation to the same even power does not always lead to the appearance of extraneous roots. For example, when passing from the equation to the corollary equation x=2, the ODZ expands from the set of all non-negative numbers to the set of all real numbers, but extraneous roots do not appear. 2 is the only root of both the first and second equations. Also, there is no appearance of extraneous roots during the transition from the equation to the equation-consequence. The only root of both the first and second equations is x=16 . That is why we are not talking about the causes of the appearance of extraneous roots, but about the reasons for the possible appearance of extraneous roots.

What is weeding out extraneous roots?

The term "eliminating extraneous roots" can only be called a well-established term, it is not found in all algebra textbooks, but it is intuitive, which is why it is usually used. What is meant by sifting out extraneous roots becomes clear from the following phrase: “... verification is an obligatory step in solving the equation, which will help to detect extraneous roots, if any, and discard them (usually they say “weed out”)” .

Thus,

Definition

Weeding out extraneous roots is the detection and rejection of extraneous roots.

Now you can move on to ways to weed out extraneous roots.

Methods for weeding out extraneous roots

Substitution check

The main way to weed out extraneous roots is a substitution check. It allows you to weed out extraneous roots that could arise due to the expansion of the ODZ, and due to the raising of both parts of the equation to the same even power.

The substitution check is as follows: the found roots of the consequence equation are substituted in turn into the original equation or into any equation equivalent to it, those that give the correct numerical equality are the roots of the original equation, and those that give an incorrect numerical equality or expression, meaningless are extraneous roots for the original equation.

Let's use an example to show how extraneous roots are screened out through substitution into the original equation.

In some cases, weeding out extraneous roots is more appropriate to carry out in other ways. This applies mainly to those cases where the substitution check is associated with significant computational difficulties or when the standard way of solving equations of a certain type involves a different check (for example, sifting out extraneous roots when solving fractional-rational equations is carried out according to the condition that the denominator of the fraction is not equal to zero ). Let's analyze alternative ways of sifting out extraneous roots.

According to ODZ

In contrast to the substitution check, screening out extraneous roots by ODZ is not always appropriate. The fact is that this method allows you to filter out only extraneous roots that arise due to the expansion of the ODZ, and it does not guarantee the elimination of extraneous roots that could arise for other reasons, for example, due to raising both parts of the equation to the same even power . Moreover, it is not always easy to find the ODZ for the equation being solved. Nevertheless, the method of sifting out extraneous roots by ODZ should be kept in service, since its use often requires less computational work than the use of other methods.

The sifting of extraneous roots according to the ODZ is carried out as follows: all found roots of the consequence equation are checked for belonging to the region of permissible values ​​of the variable for the original equation or any equation equivalent to it, those that belong to the ODZ are the roots of the original equation, and those of them which do not belong to the ODZ are extraneous roots for the original equation.

An analysis of the information provided leads to the conclusion that it is advisable to screen out extraneous roots according to the ODZ if at the same time:

• it is easy to find the ODZ for the original equation,
• extraneous roots could arise only due to the expansion of the ODZ,
• substitution verification is associated with significant computational difficulties.

We will show how weeding out extraneous roots is carried out in practice.

Under the terms of the ODZ

As we said in the previous paragraph, if extraneous roots could arise only due to the expansion of the ODZ, then they can be filtered out according to the ODZ for the original equation. But it is not always easy to find ODZ in the form of a numerical set. In such cases, it is possible to screen out extraneous roots not according to the ODZ, but according to the conditions that determine the ODZ. Let us explain how the screening of extraneous roots is carried out according to the conditions of the ODZ.

The found roots are substituted in turn into the conditions that determine the ODZ for the original equation or any equation equivalent to it. Those of them that satisfy all conditions are the roots of the equation. And those of them that do not satisfy at least one condition or give an expression that does not make sense are extraneous roots for the original equation.

Let us give an example of screening out extraneous roots according to the conditions of the ODZ.

Screening out extraneous roots arising from raising both sides of the equation to an even power

It is clear that weeding out extraneous roots arising from raising both parts of the equation to the same even power can be done by substituting into the original equation or into any equation equivalent to it. But such verification can be associated with significant computational difficulties. In this case, it is worth knowing an alternative way to weed out extraneous roots, which we will talk about now.

Screening out extraneous roots that may arise when both parts of irrational equations of the form are raised to the same even power , where n is some even number, can be carried out according to the condition g(x)≥0 . This follows from the definition of an even root: an even root n is a non-negative number whose nth power is equal to the root number, whence . Thus, the voiced approach is a kind of symbiosis of the method of raising both parts of the equation to the same degree and the method of solving irrational equations by determining the root. That is, the equation , where n is an even number, is solved by raising both parts of the equation to the same even power, and sifting out extraneous roots is performed according to the condition g(x)≥0 taken from the method for solving irrational equations to determine the root.

Loss of roots and extraneous roots when solving equations

MOU "Secondary School No. 2 with in-depth study of individual subjects" of the city of Vsevolozhsk. The research work was prepared by a student of class 11 B: Vasilyev Vasily. Project leader: Egorova Lyudmila Alekseevna.

Equation To begin with, consider various ways to solve this equation sinx+cosx =- 1

Solution #1 sinx+cosx =-1 i Y x 0 1 sin(x+)=- 1 sin(x+)=- x+ =- +2 x+ = +2 + x=- +2 x= +2 Answer: +2

Solution No. 2 sinx + cosx \u003d - 1 i Answer: +2 y x 0 1 2sin cos + - + + \u003d 0 sin cos + \u003d 0 cos (cos + sin) \u003d 0 cos \u003d 0 cos + sin \u003d 1 \u003d + m tg =-1 = + m =- + x=- +2 x= +2

Solution #3 i y x 0 1 sinx+cosx =- 1 2 = x= x+ x sin2x=0 2x= x= Answer:

sinx+cosx =-1 Solution #4 i y x 0 1 + =- 1 2tg +1- =-1- 2tg =- 2 =- + n x= - + 2 n Answer: - + 2 n

We compare solutions Correct solutions Let's figure out in which cases extraneous roots may appear and why #2 Answer: +2 #3 Answer: #4 Answer: + 2 n #1 Answer: +2

Verification of the solution Do I need to do a verification? Check the roots just in case, for reliability? This is of course useful when it is easy to substitute, but mathematicians are rational people and do not do unnecessary actions. Consider different cases and remember when verification is really needed.

1. The simplest ready-made formulas c osx =a x=a =a s inx =a t gx =a However, when using such formulas, one should be aware of the conditions under which they can be applied. For example, the formula = can be used under the condition a 0, -4ac 0 And the grossest mistake is the answer x= arccos2+2 for the equation cosx =2, since the formula x= arccos a +2 can only be used for the roots of the equation cosx = a, where | a | one

2. Transformations Often, when solving equations, you have to carry out many transformations. If the equation is replaced by a new one, having all the roots of the previous one, and transformed so that there is no loss or acquisition of roots, then such equations are called equivalent. 1. When transferring the components of the equation from one part to another. 2. By adding the same number to both parts. 3 . When multiplying both sides of the equation by the same non-zero number. 4 . When applying identities that are true on the set of all real numbers. In this case, verification is not required!

However, not every equation can be solved by equivalent transformations. More often it is necessary to apply unequal transformations. Often such transformations are based on the use of formulas that are not true for all real values. In this case, in particular, the domain of definition of the equation changes. This error is in solution #4. We will analyze the error, but first we will look again at the solution number 4. sinx+cosx=-1 + =-1 2tg +1- =-1- 2tg =-2 =- + n x = - + 2 n The error lies in the formula sin2x= You can use this formula, but you should additionally check whether they are roots numbers of the form + for which tg is not defined. Now it is clear that there is a loss of roots in the solution. Let's bring it to the end.

Solution #4 i y x 0 1 Check the numbers = + n by substitution: x= + 2 n sin(+ 2 n)+ cos (+ 2 n)=sin + cos =0+(-1)=- 1 So x= +2 n is the root of the equation Answer: +2 sinx+cosx =-1 + =- 1 2tg +1- =-1- 2tg =- 2 =- + n x= - + 2 n

We considered one of the ways to lose roots, there are a great many of them in mathematics, so you need to decide carefully, remembering all the rules. Just as you can lose the roots of an equation, you can also gain extra ones in the course of solving it. Let's consider solution #3, which made such a mistake.

Solution #3 i y x 0 1 2 2 and extra roots! Extraneous roots could appear when both sides of the equation were squared. In this case, you need to check. For n=2k we have sin k+cos k=-1; cos k=-1 for k=2m-1 , Then n=2(2m+1)=4m+2 , x= = +2 m , Answer: +2 For n=2k+1 we have sin + cos =- 1 sin(+ k)+ cos (+ k)=- 1 cos k-sin k=- 1 cos k=-1 for k=2m+1 n=2(2m+1)+ 1=2m+3 x= ( 4m+3)= +2 m=- +2 sinx+cosx =- 1 = x= x+ x sin2x=0 2x= x=

So, we have considered a couple of possible cases, of which there are a great many. Try not to waste your time and not make stupid mistakes.

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In the last lesson, when solving equations, we used three stages.

The first stage is technical. With the help of a chain of transformations from the original equation, we come to a fairly simple one, which we solve and find the roots.

The second stage is the analysis of the solution. We analyze the transformations that we performed and find out if they are equivalent.

The third stage is verification. Checking all the found roots by substituting them into the original equation is mandatory when performing transformations that can lead to a corollary equation

Is it always necessary to distinguish three stages when solving an equation?

Of course not. As, for example, in solving this equation. In everyday life, they are usually not isolated. But all these stages must be “keep in mind” and performed in one form or another. Be sure to analyze the equivalence of transformations. And if the analysis showed that it is necessary to perform a check, then it is required. Otherwise, the equation cannot be considered solved correctly.

Is it always possible to check the roots of an equation only by substitution?

If equivalent transformations were used when solving the equation, then verification is not required. When checking the roots of an equation, ODZ (range of acceptable values) is very often used. If it is difficult to check for ODZ, then it is performed by substituting it into the original equation.

Exercise 1

Solve the equation square root of two x plus three equals one plus x.

Decision

The ODZ equation is defined by a system of two inequalities: two x plus three is greater than or equal to zero and one plus x is greater than or equal to zero. The solution is x greater than or equal to minus one.

We square both sides of the equation, transfer the terms from one side of the equation to the other, add like terms, we get the quadratic equation x squared equals two. Its roots are

x first, second equals plus or minus the square root of two.

Examination

The x value of the first equals the square root of two is the root of the equation, since it is included in the DPV.
The value of x second is minus the square root of two is not the root of the equation, because it is not included in the ODZ.
Let's check the x-root is equal to the square root of two, substituting it into the original equality, we get

true equality, so x equal to the square root of two is the root of the equation.

Answer: square root of two.

Task 2

Solve the equation square root of x minus eight equals five minus x.

Decision

The ODZ of an irrational equation is determined by a system of two inequalities: x minus eight is greater than or equal to zero and five minus x is greater than or equal to zero. Solving it, we get that this system has no solutions. The root of the equation cannot be any of the values ​​of the variable x.

Answer: no roots.

Task 3

Solve the equation square root of x cubed plus four x minus one minus eight square roots of x to the fourth power minus x equals square root of x cubed minus one plus two square roots of x.

Decision

Finding the ODZ in this equation is quite difficult.

Let's perform transformations: let's square both sides of this equation,

we transfer all the terms to the left side of the equation and bring like terms, write two roots under one, get like radicals, give like ones, divide by a factor of minus 12, and decompose the root expression into factors, we get an equation in the form of a product of two factors equal to zero. Solving it, we find the roots:

x the first is equal to one, x the second is equal to zero.

Since we raised both parts of the equation to an even power, checking the roots is mandatory.

Examination

If x is equal to one, then

we get the correct equality, which means that x equal to one is the root of the equation.

If x is zero, then the square root of minus one is undefined.

Hence, x equal to zero is an extraneous root.

Answer: one.

Task 4

Solve the equation for the logarithm of x squared plus five x plus two base two equals three.

Decision

Let's find the ODZ equation. To do this, we solve the inequality x square plus five x plus two greater than zero.

We solve the inequality by the method of intervals. To do this, we decompose its left side into factors, having previously solved the quadratic equation, and taking into account the inequality sign, we determine the ODZ. ODZ is equal to the union of open rays from minus infinity to minus fraction five plus the square root of seventeen divided by two, and from minus fraction five minus the square root of seventeen divided by two to plus infinity.

Now let's start looking for the roots of the equation. Given that three is equal to the logarithm of eight to the base of two, we write the equation in the following form: the logarithm of the expression x squared plus five x plus two to the base two is equal to the logarithm of eight to the base two. We potentiate the equation, we obtain and solve the quadratic equation.

The discriminant is forty nine.

We calculate the roots:

x first equals minus six; X second is equal to one.

Examination

Minus six belongs to the ODZ, one belongs to the ODZ, which means that both numbers are the roots of the equation.

Answer: minus six; one.

In the last lesson, we considered the issue of the appearance of extraneous roots. We can detect them by checking. Is it possible to lose roots when solving an equation and how to prevent this?

When performing such actions on the equation, as, firstly, dividing both parts of the equation by the same expression ax from x (except for those cases when it is known for sure that ax from x is not equal to zero for any x from the domain of the equation) ;

secondly, the narrowing of the ODZ equation in the process of solving can lead to the loss of the roots of the equation.

Remember!

The equation written in the form

ef from x multiplied by ash from x is equal to zhe from x multiplied by ash from x is solved in this way:

it is necessary to factorize by taking the common factor out of brackets;

then, each factor is equated to zero, thereby obtaining two equations.

We calculate their roots.

Exercise 1

Solve the equation x cube equals x.

First way

We divide both sides of this equation by x, we get x squared equal to one, having roots x first equal to one,

X second is equal to minus one.

Second way

x cube equals x. Let's move x to the left side of the equation, take x out of brackets, we get: x times x squared, minus one is equal to zero.

Let's calculate its roots:

X first is equal to zero, x second is equal to one, x third is equal to minus one.

The equation has three roots.

When solving in the first way, we lost one root - x is equal to zero.

Answer: minus one; zero; one.

Remember! Reducing both sides of the equation by a factor containing the unknown can lead to the loss of roots.

Task 2

Solve the equation decimal logarithm of x squared is two.

Decision

First way

By definition of the logarithm, we get the quadratic equation x squared equals one hundred.

Its roots: x first equals ten; x second equals minus ten.

Second way

By the property of the logarithm, we have two decimal logarithms x equals two.

Its root - x is equal to ten

In the second method, there was a loss of the x-root equal to minus ten. And the reason is that they applied the wrong formula, narrowing the scope of the equation. The expression decimal logarithm of x squared is defined for all x except x equal to zero. The expression decimal logarithm x is for x greater than zero. The correct formula is decimal logarithm x square is equal to two decimal logarithms modulo x.

Remember! When solving an equation, correctly apply the available formulas.