Basic methods for solving equations

What is the solution to an equation?

Identical transformation. Basic

types of identity transformations.

Foreign root. Root loss.

Solving the equation is a process consisting mainly of replacing a given equation with another equation that is equivalent to it . This replacement is calledidentical transformation . The main identity transformations are as follows:

1.

Replacing one expression with another that is 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.

Transferring terms of an equation from one side to the other with reverse signs. So, in the previous equation we can transfer all its terms from the right side to the left 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.

Multiplying or dividing 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 we are multiplying or dividing by 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, if (x – 1 )( x – 3 ) = 0 is the original

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

both sides of the equation onx – 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 the odd root from both sides of the equation . It must be remembered that:

a) construction ineven degree may causeto the acquisition of foreign 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 an extraneous root.

Incorrect taking the square root of both

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

and we are losing our rootsx = – 5.

Correct taking the square root results in

equation: | 7x | = 35, A hence to two cases:

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

Therefore, whencorrect extracting square

roots we do not lose the roots of the equation.

What meansRight extract the root? This is where we meet

with a very important conceptarithmetic root

(cm. ).

In the last lesson, we used three steps to solve equations.

The first stage is technical. Using a chain of transformations from the original equation, we arrive at a fairly simple one, which we solve and find the roots.

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

The third stage is verification. Checking all 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 distinguished. But all these stages need to be “kept in mind” and carried out in one form or another. It is imperative to analyze the equivalence of transformations. And if the analysis shows that a check needs to be performed, then it is mandatory. 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, the ODZ (permissible value range) is very often used. If it is difficult to check using the 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.

Solution

The ODZ of the equation is determined 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.

Let's square both sides of the equation, move the terms from one side of the equation to the other, add similar terms, and get a quadratic equation x squared equals two. Its roots are

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

Examination

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

the equality is true, which means that x equals 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.

Solution

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 find 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.

Solution

Finding the ODZ in this equation is quite difficult.

Let's carry out the transformation: square both sides of this equation,

Let's move all the terms to the left side of the equation and bring like terms, write two roots under one, get similar radicals, bring like ones, divide by the coefficient minus 12, and factor the radical expression, we get an equation in the form of a product of two factors equal to zero. Having solved it, we find the roots:

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

Since we raised both sides 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 equals one is the root of the equation.

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

This means that x equal to zero is an extraneous root.

Answer: one.

Task 4

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

Solution

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

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

Now let's start finding the roots of the equation. Given that three is equal to the logarithm of eight to base two, we write the equation as follows: the logarithm of the expression x square plus five x plus two to base two is equal to the logarithm of eight to base two. Let us potentiate the equation, obtain and solve a quadratic equation.

The discriminant is forty-nine.

Calculate the roots:

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

Examination

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

Answer: minus six; one.

In the last lesson we looked at the issue of the appearance of extraneous roots. We can detect them through verification. Is it possible to lose roots when solving an equation and how to prevent this?

When performing such actions on an equation, such as, firstly, dividing both sides 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 definition of the equation) ;

secondly, narrowing the OD of the equation during the solution process can lead to the loss of the roots of the equation.

Remember!

The equation written as

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

you need to factorize by putting the common factor out of brackets;

then, equate each factor to zero, thereby obtaining two equations.

We calculate their roots.

Exercise 1

Solve the equation x cube equals x.

First way

Divide both sides of this equation by x, we get x square equals one, having roots x first equals 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, and we get: x multiplied by x squared minus one equals 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 the first method, we lost one root - x equals zero.

Answer: minus one; zero; one.

Remember! Reducing both sides of the equation by a factor containing the unknown can result in lost roots.

Task 2

Solve the equation: the decimal logarithm of x squared is equal to two.

Solution

First way

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

Its roots: x first equals ten; X second is equal to minus ten.

Second way

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

Its root - x is equal to ten

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

Remember! When solving an equation, use the available formulas wisely.

May 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 methods of filtering 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

School algebra textbooks do not provide a definition of 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, a transition is made from the original equation to the corollary equation, the roots of the resulting corollary equation are found, and the found roots are checked by substituting into the original equation, which shows that some of the found ones roots are not roots of the original equation, these roots are called extraneous roots for the original equation.

Starting from this base, you can accept for yourself the following definition of an extraneous root:

Definition

Extraneous roots- these are the roots of the corollary equation obtained as a result of transformations, which are not the roots of the original equation.

Let's give an example. Let's consider the equation and the consequence of this equation x·(x−1)=0, obtained by replacing the expression with the identically equal expression x·(x−1) . The original equation has a single root 1. The equation obtained as a result of the transformation has two roots 0 and 1. This means 0 is an extraneous root for the original equation.

Reasons for the possible appearance of foreign roots

If to obtain the corollary equation you do not use any “exotic” transformations, but use only basic transformations of equations, then extraneous roots can arise only for two reasons:

• due to the expansion of ODZ and
• due to raising both sides of the equation to the same even power.

It is worth recalling here that the expansion of the ODZ as a result of transforming the equation mainly occurs

• When reducing fractions;
• When replacing a product with one or more zero factors by zero;
• When replacing a fraction with a zero numerator with zero;
• When using some properties of powers, roots, logarithms;
• When using some trigonometric formulas;
• When both sides of an equation are multiplied by the same expression, it vanishes by the ODZ for that equation;
• When freeing from logarithm signs in the solution process.

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

We will also give an example of the appearance of an extraneous root due to raising both sides 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 sides of the equation, that is, the equation , has two roots 1 and 4. From this it is clear that squaring both sides of the equation led to the appearance of an extraneous root for the original equation.

Note that expanding the ODZ and raising both sides of the equation to the same even power does not always lead to the appearance of extraneous roots. For example, when moving 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 no extraneous roots appear. 2 is the only root of both the first and second equations. Also, no extraneous roots appear when moving from an equation to a corollary equation. The only root of both the first and second equations is x=16. That is why we are not talking about the reasons for the appearance of extraneous roots, but about the reasons for the possible appearance of extraneous roots.

What is screening out extraneous roots?

The term “sifting out extraneous roots” can only with a stretch be called established; 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 a mandatory step in solving an equation, which will help to detect extraneous roots, if any, and discard them (usually they say “weed out”).”

Thus,

Definition

Screening out extraneous roots- this is the detection and discarding of extraneous roots.

Now you can move on to methods of screening out extraneous roots.

Methods for screening out extraneous roots

Substitution check

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

The substitution test is as follows: the found roots of the corollary 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 the incorrect numerical equality or expression are the roots of the original equation. meaningless, are extraneous roots for the original equation.

Let us show with an example how to filter out extraneous roots through substitution into the original equation.

In some cases, it is more expedient to filter out extraneous roots using other methods. This applies mainly to those cases when checking by substitution is associated with significant computational difficulties or when the standard method of solving equations of a certain type requires another check (for example, screening 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 look at alternative ways to weed out extraneous roots.

According to DL

Unlike testing by substitution, filtering out extraneous roots using 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 sifting out of extraneous roots that could arise for other reasons, for example, due to raising both sides of the equation to the same even power . Moreover, it is not always easy to find the OD for the equation being solved. Nevertheless, the method of sifting out extraneous roots using ODZ is worth keeping in service, since its use often requires less computational work than the use of other methods.

Weeding out extraneous roots according to the ODZ is carried out as follows: all found roots of the corollary equation are checked to see if they belong to the range of permissible values ​​of the variable for the original equation or any equation equivalent to it, those that belong to the ODZ are roots of the original equation, and those that belong to the ODZ are roots of the original equation, and those that which do not belong to the ODZ are extraneous roots for the original equation.

Analysis of the information provided leads to the conclusion that it is advisable to sift out extraneous roots using ODZ if at the same time:

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

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

According to the terms of the DL

As we said in the previous paragraph, if extraneous roots could arise only due to the expansion of the ODZ, then they can be eliminated using 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 weeding out extraneous roots is carried out under the conditions of ODZ.

The found roots are in turn substituted into the conditions that determine the ODZ for the original equation or any equation equivalent to it. Those that satisfy all the 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 ODZ.

Weeding 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 sides of the equation to the same even power can be done by substituting it into the original equation or into any equation equivalent to it. But such a check may involve significant computational difficulties. In this case, it is worth knowing an alternative method of sifting out extraneous roots, which we will talk about now.

Screening out extraneous roots that may arise when raising both sides of irrational equations of the form 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 a root of an even degree: a root of an even degree n is a non-negative number, the nth power of which is equal to the radical number, whence . Thus, the approach voiced is a kind of symbiosis of the method of raising both sides of the equation to the same power 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 sides of the equation to the same even power, and eliminating extraneous roots is carried out according to the condition g(x)≥0, taken from the method of solving irrational equations by determining the root.

Loss of roots and extraneous roots when solving equations

Municipal educational institution "Secondary school No. 2 with in-depth study of individual subjects" in the city of Vsevolozhsk. The research work was prepared by a student of grade 11 B: Vasilyev Vasily. Project manager: Egorova Lyudmila Alekseevna.

Equation First, let's look at different ways to solve this equation sinx+cosx =- 1

Solution No. 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 =- 1st Answer: +2 y x 0 1 2sin cos + - + + = 0 sin cos + = 0 cos (cos + sin)= 0 cos =0 cos + sin =1 = + m tg =-1 = + m =- + x=- +2 x= +2

Solution No. 3 I y x 0 1 sinx+cosx =- 1 2 = x= x+ x sin2x=0 2x= x= Answer:

sinx+cosx =-1 Solution No. 4 i y x 0 1 + =- 1 2tg +1- =-1- 2tg =- 2 =- + n x= - + 2 n Answer: - + 2 n

Let's compare solutions Correct solutions Let's figure out in what cases extraneous roots may appear and why No. 2 Answer: +2 No. 3 Answer: No. 4 Answer: + 2 n No. 1 Answer: +2

Checking the solution Is it necessary to check? Should I check the roots just in case, to be on the safe side? This is of course useful when it’s easy to substitute, but mathematicians are rational people and don’t do unnecessary things. Let's look at 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 In cases where the roots are found using the simplest, ready-made formulas, the check does not need to be done. However, when using such formulas, you should remember the conditions under which they can be used. For example, the formula = can be used under the condition a 0, -4ac 0 And the answer x= arccos2+2 for the equation cosx =2 is considered a gross error, since the formula x= arccos a +2 can only be used for the roots of the equation cosx =a, where | a | 1

2. Transformations More often, when solving equations, you have to carry out a lot of transformations. If an equation is replaced by a new one that has all the roots of the previous one, and it is transformed so that no loss or acquisition of roots occurs, then such equations are called equivalent. 1. When transferring the components of an equation from one part to another. 2. When adding the same number to both sides. 3. When both sides of an equation are multiplied by the same non-zero number. 4 . When applying identities that are valid on the set of all real numbers. However, 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 valid for all real values. In this case, in particular, the domain of definition of the equation changes. This error is found in solution #4. Let's look at the error, but first let's look again at solution No. 4. sinx+cosx=-1 + =-1 2tg +1- =-1- 2tg =-2 =- + n x = - + 2 n The error lies in the formula sin2x= This formula can be used, but you should additionally check whether the roots are numbers of the form + for which tg is not defined. Now it is clear that the solution is the loss of roots. Let's see it through to the end.

Solution No. 4 i y x 0 1 Let's 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 looked at one of the ways to lose roots; there are a great many of them in mathematics, so you need to solve carefully, remembering all the rules. Just as you can lose the roots of an equation, you can also acquire extra ones in the course of solving it. Let's consider solution No. 3 in which such an error was made.

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, it is necessary 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 with 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 looked at a couple of possible cases, of which there are a great many. Try not to waste your time and make stupid mistakes.