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9DOI 10.24412/2709-1201-2022-26-33 mastering methods for solving equations with unknowns within the framework of a modular sign in secondary schools
ibragimov firadun nadir
doctor of pedagogical sciences, professor, Sheki branch of ASPU, Azerbaijan ORCID:https://orcid.org/0000-0002-0775-1048
karimova mehriban nurmammad
heard teacher at Sheki branch of ASPU, Azerbaijan
Abstract
The article substantiates the relevance of the topic, draws attention to the types of equations that are unknown within the framework of the module attribute in teaching mathematics in a secondary school. It is emphasized that in the content of the subject of mathematics, materials related to unknown equations within the sign of the modulus are not sufficiently distributed, and the content is not enough for the development of concepts along the content line "Algebra and functions ". The content of unknown types of equations under the module sign, which is considered important for inclusion in the content of the subject of mathematics in research work, is adequately given to the development of the concept of a function and the interpretation of their effective solution. methods are presented in examples.
Keywords: modulus, modulus of a number, equation, family of equations, unknown variable, equation root, set of solutions, identity, number line, interval, method of intervals.
Relevance of the research topic. While a logical connection between the concepts of "function" and "types of equations" in the" line of the content of algebra and functions" is mainly expected, compatibility between types of equations (exponential, logarithmic, trigonometric, irrational types of equations), unknown within the module symbol is not guaranteed. This "gap" negatively affects the implementation of both the expected general results and the results of meaningful risk in mathematics. Therefore, it is necessary to justify the content of the types of equations unknown under the sign of the modulus, which are considered important for inclusion in the content of the subject of mathematics, in an adequate way for the development of the concept of a function, and to interpret their effective methods of solving. This need determines the relevance of the research topic. The methodological basis of the study was: 1) the idea of systematic analysis by L. Bertalanffy; 2) "system-structural" approach to the object of study, formed as a section of dialectics. Interpretation of research materials. As the name suggests, in these equations, the unknown is inside the sign of the absolute value. Their solution is based on the following reasoning: if the absolute value of a number x is equal to a positive number, then this number x itself is equal to either a or—a.
The content of mathematics taught in general education schools includes examples of unknown equations inside the module symbol in the following order:
|/(x)| = a; |/(x)| = |#(x)|; |/(x)| = g(x); |/(x)| ± |#(x)| ± ■■■ ± Kx)| = a, [3-5] In the IX grades of our general education school, the technology for solving equations with a variable within the framework of the module symbol is interpreted as follows.
When solving equations with a variable inside the sign of the module, two cases are considered: 1) the expression inside the sign of the module (module) is positive or equal to zero; 2) The expression inside the modulo sign is negative.
These considerations are argued in accordance with the definition of the absolute value of a
x, if x > 0,
number, namely: |x| = . ,r ^ _ J '—x,if x < 0.
In order for students to become subjects of knowledge and skills (components of knowledge and activity) aimed at solving equations with a variable within the mark of the module, it is considered appropriate to apply the following sequence of classes.
Example 1. |3x — 2| + 11 = 5 Solution: Only the modulus expression is stored on the left side of the equality: |3x — 2| = —6.The last number does not satisfy the definition of the modulus. Because the absolute value of a number must be either zero or a positive number. Answer: 0
It is important here that students understand that these types of equations do not have solutions. Example 2. |x — 3| =6
should be either 6 or -6. The resulting numbers 9 and -3 satisfy the given equation. So the equation has two roots. x — 3 Example 3. |x2 — 2x| = 3
x2 — 2x = 3, —(x2 — 2x) = 3. Case 1: x2 — 2x = 3; xx = 3 or x2 = —1. 3 and -3 are solutions to this equation. Case 2:
—(x2 — 2x) = 3; x2 — 2x + 3 = 0 The discriminant in the equation is negative. So this equation has a solution {—1; 3}. Example 4. |2x — 4| = 1 — 3x.
( 2x — 4, if x > 2,
Solution: According to the definition of absolute value: |2x — 4| =
Solution: Write this equation on the board.] f 2
[—(2x — 4), if x < 2.
Case1. If, x > 2 and |2x — 4| = 2x — 4 this equation takes the form 2x — 4 = 1 — 3x . This is
written like this: (_ x. > 2 _
<-2x — 4 = 1 — 3x.
2x — 4 = 1 — 3x from the equation it is found that this value 5x = 5, x = 1, does not satisfy x > 2 the condition. That is, in this case, the equation has no solution.
Case 2. If , x < 2 |2x — 4| = —(2x — 4)and this equation takes the form—(2x — 4) = 1 — 3x. In ( x < 2
this case, the |_(2x — 4)1 — 3x system is taken. —(2x — 4) = 1 — 3x from the equation it is
found that this value x = —3 satisfies the x < 2 condition. So this equation has one root {3}. [2; 91-94]
In other classes of secondary schools
|/(x)| = #(x); |/(x)| ± |£(x)| ± ■■■ ± Kx)| = a equations in the form are used as educational tasks, students must perform cognitive operations that are adequate to the requirements of the tasks. When solving these equations, students are often involved in solving methods that are carried out by dividing the numerical axis into segments in which the signs of expressions under the module sign are preserved. [3-5]
This solution method is called the "interval method". Such a division is carried out by points at which at least one of the expressions under the modulus sign vanishes. Then the equation is solved on each interval and the root included in this interval is extracted.
It should be noted that modular equations of this type can be solved by taking various combinations of expression signs inside the module, but this approach is inconvenient if there are many modular signs in the equation.
Let us give two examples of the application of the interval method. Example 5. |x + 1| + |x — 2| = 3.
Solution: There are two expressions|x + 1| and |x — 2| in this equation, entered under the absolute value sign: ; it becomes zero when the first is present x = —1, and zero when the second is present x = 2 . If we place these two points (abscissa—1 and abscissa 2) on a number line, we divide it into three intervals: the first infinite interval is the points to the left of the abscissa —1 point, the
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second interval is the —1abscissa and 2 all points in between, including the abscissa points, and the third the infinite interval includes the points to the right of the point whose abscissa is 2. The expressions in each of these three intervals can be written as follows |x + 1| and |x — 2|without using the sign of the absolute value:
—x — 1, in the first interval
I _
1 1 {x + 1, in the first and second intervals
T ^ • , f—x + 2, in the first and second intervals,
Like this |x — 2| = -j i.i.
x — 2 in the third interval
Therefore, the left side of the equation in example 5 looks like this:
In the first interval: (—x — 1) + (—x + 2) = —2x + 1;
In the second interval: (x + 1) + (—x + 2) = 3;
In the third interval: (x + 1) + (x — 2) = 2x — 1.
Now it is easy to find all the points of the equation in example 5.
It is clear that any number in the second—1 < x < 2 interval is a root of this equation, because for each value in taken from this interval, the left and right sides of the equation in example 5 take the numerical value 3. If there are roots of the equation in the first interval, then it is obvious that they must be equal to the roots of the equation—2x + 1 = 3 . This equation has only one (one) root x =
— 1 : (this root is defined above). Note that this root is not included in the first interval (its x =
— 1value is in the second interval). Thus, the equation in example 5 does not have a root in the first interval. Similarly, it can be determined that the equation has no roots in the third interval (we consider it useful to present this to students as an independent work).
Thus, the equation in example 5 has infinitely many roots. Each number in the interval
— 1 < x < 2 is the root of the equation. This equation has no other roots [12; 19-20] Example 6. |x + 2| + |x — 1| — |x — 3|=4
Solution: Equation x + 2 = 0, x — 1 = 0, x — 3 = 0 roots are divided into intervals number axes (—ro;2], (—2; 1], (1;3], (3 + ro). For each of these intervals, the following equation is solved:
1. Since for x e —2] in this interval the equation x + 2 < 0, x — 1<0, x — 3<0 —(x + 2) — (x — 1) + (x — 3) = 4 is replaced by the equation and its root —8 e (—ro; —2].
2. Since the x e (—2; 1], equation x + 2>0, x—1<0, x — 3 < 0 replaced by an equation
x + 2 — (x — 1) + (x — 3) = 4. Its root is x = 4 , but this root does not lie in the interval (-2; 1], therefore, the equation does not have a root in the interval (—2; 1].
3. When therefore x e (1; 3], in this interval the equation x + 2>0,x—1<0, x — 3 < 0 replaced by an equation x + 2 + (x — 1) + (x — 3) =4. Its root x = 2e(1; 3].
4. Since the x e (3; equation x + 2>0,x — 1>0, x — 3 > 0 replaced by an equation
x + 2 + (x — 1) — (x — 3) = 4x Its root exists x = 0 , but this root is not in the range (3; +ro). That is, there is no root of the equation in this interval. So, the root of the equation is equal to —8 and 2. [10; 137-138] There are two intermediate results worth noting here:
From the analysis of the research materials, it was concluded that the implementation of this method on the presented examples is easily mastered by students.
Our research leads us to the conclusion that the materials presented in this and other lessons related to unknown equations within the sign of the modulus are not sufficiently distributed and insufficient for the development of the concepts of the subject area "Algebra and functions".
It can be seen from the analysis of scientific sources that this is precisely the position of the majority of scientists, mathematicians, and methodologists. They point out that the general way to solve equations with an unknown (variable) within the sign of the modulus is to replace the given
equation by defining the modulus of the number with an equivalent equation or system of equations or system of equations, and they also emphasize that the simplest |/(x)| = b is an equation; where /(x) is a given function and b is a given number. When solving such an equation, the following situations are possible:
1) b < 0.
In this case, due to the property of the modulus, the equation has no solution.
2) b = 0.
In this case, the equation is identical to the equation /(x) = 0 3) b > 0.
r/(x) = — b
In this case, the equation|/(x)| = b coincides with the system of equations
N , , and the set
L/(x) = ь
of solutions of these equations becomes the set of solutions of the union equation |/(x)| = ¿.[8;134] Based on the analysis of scientific sources and our many years of experience, we come to the conclusion that it is possible (and useful) to include a wider content of equations within the sign of the module in the content of mathematics as a general educational subject, and these equations should take their rightful place in the content of general mathematical education for the development of the concept functions. [5-7]
Let's visualize our idea with examples. To do this, we start with a simple type of equations in modular notation, corresponding to the development of the concept of a function in the course of general mathematical education.
|F(x)| = a equations in the form. As emphasized above, students' attention is focused on two cases:
1) since the a < 0 left side of the equation is non-negative, the right side cannot be negative, in which case the equation loses its meaning;
2) a > 0 the equation is reduced to solving two equations. That is: 1) F(x) = a , F(x) > 0 ; 2) —F(x) = —a, or F(x) = —a,. F(x) < 0 [8; 83]
Let us visualize our reasoning with concrete examples. Example 7. |2x — 1| = 3
1 1
Solution: 1) if 2x — 1 > 0or x > -, then we get 2x — 1 = 3, x1 = 2x ; 2) 2x — 1 < 0 or x < -
when, —(2x — 1) = 3 and hence. x2 = —1
Example 8. At what value m does the equation|x2 — 6x| + 1 = m have three different real roots? Solution: |x2 — 6x| + 1 = m ^ |x2 — 6x| = m — 1. This shows m < 1 that the equation does not have three real roots. That's why it should be m > 1. Then we'll get it
|x2 — 6x| = m — 1 ^
x2 — 6x = —(m — 1) 1 x2 — 6x = m — 1
rx2 — 6x + (m — 1) = 0
Ix2 — 6x — (m — 1) = 0 since m > 1 the discriminant of the equation x2 — 6x — (m — 1) = 0 x2 is positive, it has two different roots. if there is a discriminant D = 36 — 4(m — 1) = 0 ^ m = 10 of an equation x2 — 6x — (m — 1) = 0, then this equationhas two equal roots. Thus, if m = 10 two real versions of the equation x2 — 6x — (m — 1) = 0 and the equation x2 — 6x — (m — 1) = 0 has two real equal roots. Since two of the four roots taken are the same, the equation has three different roots. [11; 111-112] Example 9. log2(x + 1)2 + log2|x + 1| = 6
Solution: If we write the equation in the form 2 log2 |x + 1| + log2 |x + 1| = 6, we get
log2|x + 1| = 2; |x + 1|=4
1) when x > —1, we find x + 1 = 4, Xj=3;
2) when x < —1, becomes —(x + 1) = 4 and x2 = —5. Example 10. 3|x2-4x-x| = 81
Solution: If we equate the bases, it is possible3|%2-4x-x| = 34. From here we get the equation |x2 — 4x — 1| = 4. There will be two cases here:
1)x2 - 4x - 1 = 4,x1 = —1, x2 = 5;
2) x2 — 4x — 1 = —4, x3 = 3, x4 = 1. Example 11.|t^x + ct^x| = 2V2.
Solution. Let's do some simple transformations first:
sm^ cosx
= 2V2;
2
stn2x к
= 2V2. From here: |sin2x| =
1)sm2x = tt,^ = (—1)fc • ~ + k • ^ ;
2)sin2x = — — ; x = —(—1)fc • - + k • - (fc = 0; ±1; ±2;...).
2 8 2
Reply:x = + (fc = 0; ±1; ±2;...).
Equations of the form F(|x|)= a. It is permissible to draw students' attention to the following explanation for solving this type of equations. According to the definition of the modulus, this
equation is written in the form of the following two systems: = U'
|—x| = |x| it happens because it is F(|—x|) = F(|x|). So the function y = F(x) — a is even. Therefore, if this x1is the root of this equation, then this —x1 is the root of this equation. From this judgment it is clear that to solve this equation it is enough to solve one of the above systems. Then the obtained roots and their opposite numbers will be the roots of this equation. Example 12.cos|x| = sin|x|
Solution:^ |x| = 1, |x| + = + k = 0,±1,±2,...).
Example 13. 2logs*2 — 21+l°gs|x| + 2log5|x|-1 = 1. Solution: The equation can be written like this:
1
22iogsM _ 2 • 2iogsM + - • 2iogsM = 1,
2 • 22l°gsW _ 3 • 2log5|x| -2 = 0.
From here, if we accept 2logs|x| = у (1), it becomes 2y2 _ 3y _ 2 = 0, y1 = 2, y2 = -1 .
If we consider the value y1 in (1), we will find its roots 2logs|x| = 2, log5|x| = 1, |x| = 5, x1 = 5, x2 = _5. Since y2 < 0 it does not satisfy
equation (1).
Equations of the form |F(|x|)| = a(a > 0). We present the solution of two examples of equations in this form.
Example 14. |2 • |x| _ 3| = 5.
Solution: First write equations 2- |x| _ 3 = 5 and 2 • |x| _ 3 = _5. If you solve the first equation 2 • |x| = 8, |x| = 4,x1 = 4, x2 = _4. Obtained by solving the second equation 2 • |x| = _2, |x| = _1. Since the left side of this equation is positive and the right
side is negative, equality is impossible. Example 15.|log3|x|| = 1.
Solution: We write the following system from this equation:{|log3 x| = 1
x > 0.
We consider two cases:
1) when x > 1, since the |log3 x| = log3 x get equation log3 x = 1 . This is happening here x1 = 3.
2) when 0 < x < 1 occurs |log3 x| = _ log3 x and _ log3 x = 1 the equation is obtained , whence x2 =1.
1
The roots opposite to these roots are x3 = _3 and x4 = _ -. [9; 86]
Equations in the form |F(x)| = ^(x). Students should know that here ^(x) > 0 and the given equation is same power as the following two systems of equations:
iF(x) = ^(x) (-F(x) = ^(x), { <p(x) >0 { <p(x) > 0. Example 16. x2 + 2x — 3 • |x + 1| — 3 = 0
Solution: The equation can be written like this: 3 • |x + 1| = x2 + 2x — 3. The right side of this equation must be x2 + 2x — 3 > 0 or (x — 1) • (x + 3) > 0. This happens when x < —3 or x > 1 . In this case, if we solve the equations 3(x + 1) = x2 + 2x — 3 and 3(x + 1) = —x2 — 2x + 3 , their roots will be respectively x = —2, x = 3 and x = 0, x = —5 . Of these, the roots x1 = 3 and x2 = —5 satisfy the conditions x < —3 and x > 1 . These roots are the roots of the given equation .
Example 17.|sinx| = sinx + 2
Solution: Here the right side x is positive for all values of. There are two cases to consider:
fsinx = stnx + 2, , (—stnx = stnx + 2, i ■ ^ r. and i . „ (. stnx >0 (. stnx < 0.
The first system has no solutions, the second has 2 stnx = —2; stnx = —1,
n
* = —2 )fc = 0; ±1, ±2,...). [9; 88]
Equations of the form |F(x)| =F(x). Students should be clear that equations are a special case of equation the form of |F(x)| = ^(x). To solve the equation, you need to look at the following
(F(x) = F(x), system: { F(x) > 0.
This shows that the solution of the inequality F(x) >0 is also the solution of this equation.
i
cosx —
2 i
Solution: Set of solutions to the inequality cosx — - >0 is
n n
— — + < x < — + 2rcn( n = 0, ±1, ±2,...)
Equations in the form|F(x)| = |^(x)|. Students should be aware that the following two situations are possible when|a| = |£|:
Example 18.
i
cosx — 2
1) a = b becomes when there are (^ > 0 or (^ < 0
' U > 0 U < 0
2) a = — b becomes when there are ( ^ > 0 or (^ < 0 7 t- b < 0 lb > 0
Based on these relations, the given equation can be solved.
— x —
Example 19. Zff ^^^ =0. [1;47]
Solution: Domain of the equation: x 2 — x — 1^0 or x^
2 , n -u n ^ -1±V29 .
x2+x — 7^0 or x^ —-— are prices
Can be written the equation as Iff |*2+*_1| = 0, Zff|x2 — x — 1| = Zff|x2 + x — 7|,
|x2 — x — 1| = |x2 + x — 7|
Here we consider two cases:
1) if we solve the x2 — x — 1 = —(x2 + x — 7) equation, we get x1 = —2, x2 = 2
2) if we solve the x2 — x — 1 = (x2 + x — 7) equation, we get x3 = 3
Equations in the form |F1(x) + a1| ± — ± |Fw(x) + an|= a. As mentioned above, the interval method is used to solve such equations. The intervals are determined based on the values of the variable that make the summation on the left side equal to zero. Example 20.|log^ x — 2| — |log3 x — 2| = 2
Solution: Substituting in log3 x = y, we get the |2y — 2| — |y — 2| =2 equation. The roots of this
1
equation are equal toy1 = —2 and y2 = 2. Thus, we obtain x1 = - and x2 = 9 from the equations
log3 x = —2 and log3 x = 2 .
Equations in the form |F1(x)| + |F2(x)| = a. Since the left side is not negative, it must be a > 0. Here, it is intended to look at such equations that the functions F1(x) and F2(x) statisty the condition |F1(x) — F2(x)| = a. [9; 95] Example 21. |x2 — 4| + |x2 — 9| = 5
Solution: It is clear, that since the sum of two non-negative expressions is equal to 5, then each of them should not be more than 5, that is: |x2 — 4| < 5 and |x2 — 9| <5 inequalities are obtained. This is where the —1 < x2 <9 and 4 < x2 < 14 inequalities come from. Comparing these two inequalities shows that this is how it should be 4 < x2 < 9 . If we write this 2 < |x| < 3 inequality in the form: 2 < |x|and |x| <3 or x< —2, x > 2 and — 3 < x < 3 and it becomes. Thus, the set of roots of this equation is equal to [-2;-3] and [2;3].
Scientific novelty and theoretical significance of the work. 1) In the "Meaningful series of algebra and functions" of the subject of mathematics studied in secondary schools, with a logical connection between the concepts of "function" and "types of equations", it is mainly assumed that there are types of equations (superscript, logarithmic, trigonometric, irrational types of equations), unknown under the module sign, compatibility between them is established; 2) The existence of a theoretical "gap" is put in the spotlight, which negatively affects the implementation of both general results and the results expected in the subject of mathematics along meaningful lines; 3) The content of the types of equations that are considered important for inclusion in the content of the subject of mathematics and unknown under the module sign is adequately justified for the development of the concept of a function;
The practical significance of the work. Based on the results of the study, it was concluded that in general education schoolsIt is important to include in the content of the subject of mathematics the content of the types of equations unknown under the sign of the modulus, so that they are adequately substantiated for the development of the concept of a function, and methods for solving each type.identification will have a positive impact on the formation of an environment for eliminating shortcomings that may occur in the cognitive development of students.
Result.1) While the logical connection between the concepts of "function" and "types of equations" is assumed in the "line of the content of algebra and functions" of the subject of mathematics studied in high school, types of equations that are unknown under the modulus symbol (superscript, logarithmic, trigonometric, irrational types of equations) compatibility between them is not ensured; 2) The content of the types of equations that are considered important for inclusion in the content of the subject of mathematics, unknown under the module sign, should be adequate to the development of the concept of a function; 4) There are special methods for solving unknown types of equations under the module symbol.
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