We introduced the equation above in § 7. First, let us recall what a rational expression is. This is an algebraic expression made up of numbers and the variable x using the operations of addition, subtraction, multiplication, division and exponentiation with a natural exponent.
If r(x) is a rational expression, then the equation r(x) = 0 is called a rational equation.
However, in practice it is more convenient to use a slightly broader interpretation of the term “rational equation”: this is an equation of the form h(x) = q(x), where h(x) and q(x) are rational expressions.
Until now, we could not solve any rational equation, but only one that, as a result of various transformations and reasoning, was reduced to linear equation. Now our capabilities are much greater: we will be able to solve a rational equation that reduces not only to linear
mu, but also to the quadratic equation.
Let us recall how we solved rational equations before and try to formulate a solution algorithm.
Example 1. Solve the equation
Solution. Let's rewrite the equation in the form
In this case, as usual, we take advantage of the fact that the equalities A = B and A - B = 0 express the same relationship between A and B. This allowed us to move the term to the left side of the equation with the opposite sign.
Let's transform the left side of the equation. We have
Let us recall the conditions of equality fractions zero: if and only if two relations are simultaneously satisfied:
1) the numerator of the fraction is zero (a = 0); 2) the denominator of the fraction is different from zero).
Equating the numerator of the fraction on the left side of equation (1) to zero, we obtain
It remains to check the fulfillment of the second condition indicated above. The relation means for equation (1) that . The values x 1 = 2 and x 2 = 0.6 satisfy the indicated relationships and therefore serve as the roots of equation (1), and at the same time the roots of the given equation.
1) Let's transform the equation to the form
2) Let us transform the left side of this equation:
(simultaneously changed the signs in the numerator and
fractions).
Thus, the given equation takes the form
3) Solve the equation x 2 - 6x + 8 = 0. Find
4) For the found values, check the fulfillment of the condition 
. The number 4 satisfies this condition, but the number 2 does not. This means that 4 is the root of the given equation, and 2 is an extraneous root.
ANSWER: 4.
2. Solving rational equations by introducing a new variable
The method of introducing a new variable is familiar to you; we have used it more than once. Let us show with examples how it is used in solving rational equations.
Example 3. Solve the equation x 4 + x 2 - 20 = 0.
Solution. Let's introduce a new variable y = x 2 . Since x 4 = (x 2) 2 = y 2, then the given equation can be rewritten as
y 2 + y - 20 = 0.
This is a quadratic equation, the roots of which can be found using known formulas; we get y 1 = 4, y 2 = - 5.
But y = x 2, which means the problem has been reduced to solving two equations:
x 2 =4; x 2 = -5.
From the first equation we find that the second equation has no roots.
Answer: .
An equation of the form ax 4 + bx 2 +c = 0 is called a biquadratic equation (“bi” is two, i.e., a kind of “double quadratic” equation). The equation just solved was precisely biquadratic. Any biquadratic equation is solved in the same way as the equation from Example 3: introduce a new variable y = x 2, solve the resulting quadratic equation with respect to the variable y, and then return to the variable x.
Example 4. Solve the equation
Solution. Note that the same expression x 2 + 3x appears twice here. This means that it makes sense to introduce a new variable y = x 2 + 3x. This will allow us to rewrite the equation in a simpler and more pleasant form (which, in fact, is the purpose of introducing a new variable- and simplifying the recording
becomes clearer, and the structure of the equation becomes clearer):
Now let’s use the algorithm for solving a rational equation.
1) Let’s move all the terms of the equation into one part:
= 0
2) Transform the left side of the equation
So, we have transformed the given equation to the form
3) From the equation - 7y 2 + 29y -4 = 0 we find (you and I have already solved quite a lot of quadratic equations, so it’s probably not worth always giving detailed calculations in the textbook).
4) Let's check the found roots using condition 5 (y - 3) (y + 1). Both roots satisfy this condition.
So, the quadratic equation for the new variable y is solved: 
Since y = x 2 + 3x, and y, as we have established, takes two values: 4 and , we still have to solve two equations: x 2 + 3x = 4; x 2 + Zx = . The roots of the first equation are the numbers 1 and - 4, the roots of the second equation are the numbers
In the examples considered, the method of introducing a new variable was, as mathematicians like to say, adequate to the situation, that is, it corresponded well to it. Why? Yes, because the same expression clearly appeared in the equation several times and there was a reason to designate this expression with a new letter. But this does not always happen; sometimes a new variable “appears” only during the transformation process. This is exactly what will happen in the next example.
Example 5. Solve the equation
x(x-1)(x-2)(x-3) = 24.
Solution. We have
x(x - 3) = x 2 - 3x;
(x - 1)(x - 2) = x 2 -Зx+2.
This means that the given equation can be rewritten in the form
(x 2 - 3x)(x 2 + 3x + 2) = 24
Now a new variable has “appeared”: y = x 2 - 3x.
With its help, the equation can be rewritten in the form y (y + 2) = 24 and then y 2 + 2y - 24 = 0. The roots of this equation are the numbers 4 and -6.
Returning to the original variable x, we obtain two equations x 2 - 3x = 4 and x 2 - 3x = - 6. From the first equation we find x 1 = 4, x 2 = - 1; the second equation has no roots.
ANSWER: 4, - 1.
Lesson objectives:
Educational:
- formation of the concept of fractional rational equations;
- consider various ways to solve fractional rational equations;
- consider an algorithm for solving fractional rational equations, including the condition that the fraction is equal to zero;
- teach solving fractional rational equations using an algorithm;
- checking the level of mastery of the topic by conducting a test.
Developmental:
- developing the ability to correctly operate with acquired knowledge and think logically;
- development of intellectual skills and mental operations - analysis, synthesis, comparison and generalization;
- development of initiative, the ability to make decisions, and not stop there;
- development of critical thinking;
- development of research skills.
Educating:
- fostering cognitive interest in the subject;
- fostering independence in solving educational problems;
- nurturing will and perseverance to achieve final results.
Lesson type: lesson - explanation of new material.
During the classes
1. Organizational moment.
Hello guys! There are equations written on the board, look at them carefully. Can you solve all of these equations? Which ones are not and why?
Equations in which the left and right sides are fractional rational expressions are called fractional rational equations. What do you think we will study in class today? Formulate the topic of the lesson. So, open your notebooks and write down the topic of the lesson “Solving fractional rational equations.”
2. Updating knowledge. Frontal survey, oral work with the class.
And now we will repeat the main theoretical material that we will need to study a new topic. Please answer the following questions:
- What is an equation? ( Equality with a variable or variables.)
- What is the name of equation number 1? ( Linear.) A method for solving linear equations. ( Move everything with the unknown to the left side of the equation, all numbers to the right. Give similar terms. Find unknown factor).
- What is the name of equation number 3? ( Square.) Methods for solving quadratic equations. ( Isolating a complete square using formulas using Vieta’s theorem and its corollaries.)
- What is proportion? ( Equality of two ratios.) The main property of proportion. ( If the proportion is correct, then the product of its extreme terms is equal to the product of the middle terms.)
- What properties are used when solving equations? ( 1. If you move a term in an equation from one part to another, changing its sign, you will get an equation equivalent to the given one. 2. If both sides of the equation are multiplied or divided by the same non-zero number, you get an equation equivalent to the given one.)
- When does a fraction equal zero? ( A fraction is equal to zero when the numerator is zero and the denominator is not zero..)
3. Explanation of new material.
Solve equation No. 2 in your notebooks and on the board.
Answer: 10.
What fractional rational equation can you try to solve using the basic property of proportion? (No. 5).
(x-2)(x-4) = (x+2)(x+3)
x 2 -4x-2x+8 = x 2 +3x+2x+6
x 2 -6x-x 2 -5x = 6-8
Solve equation No. 4 in your notebooks and on the board.
Answer: 1,5.
What fractional rational equation can you try to solve by multiplying both sides of the equation by the denominator? (No. 6).
x 2 -7x+12 = 0
D=1›0, x 1 =3, x 2 =4.
Answer: 3;4.
Now try to solve equation number 7 using one of the following methods.
(x 2 -2x-5)x(x-5)=x(x-5)(x+5) |
|
||
(x 2 -2x-5)x(x-5)-x(x-5)(x+5)=0 |
x 2 -2x-5=x+5 |
||
x(x-5)(x 2 -2x-5-(x+5))=0 |
x 2 -2x-5-x-5=0 |
||
x(x-5)(x 2 -3x-10)=0 |
|||
x=0 x-5=0 x 2 -3x-10=0 |
|||
x 1 =0 x 2 =5 D=49 |
|||
x 3 =5 x 4 =-2 |
x 3 =5 x 4 =-2 |
||
Answer: 0;5;-2. |
Answer: 5;-2. |
Explain why this happened? Why are there three roots in one case and two in the other? What numbers are the roots of this fractional rational equation?
Until now, students have not encountered the concept of an extraneous root; it is indeed very difficult for them to understand why this happened. If no one in the class can give a clear explanation of this situation, then the teacher asks leading questions.
- How do equations No. 2 and 4 differ from equations No. 5,6,7? ( In equations No. 2 and 4 there are numbers in the denominator, No. 5-7 are expressions with a variable.)
- What is the root of an equation? ( The value of the variable at which the equation becomes true.)
- How to find out if a number is the root of an equation? ( Make a check.)
When testing, some students notice that they have to divide by zero. They conclude that the numbers 0 and 5 are not the roots of this equation. The question arises: is there a way to solve fractional rational equations that allows us to eliminate this error? Yes, this method is based on the condition that the fraction is equal to zero.
x 2 -3x-10=0, D=49, x 1 =5, x 2 =-2.
If x=5, then x(x-5)=0, which means 5 is an extraneous root.
If x=-2, then x(x-5)≠0.
Answer: -2.
Let's try to formulate an algorithm for solving fractional rational equations in this way. Children formulate the algorithm themselves.
Algorithm for solving fractional rational equations:
- Move everything to the left side.
- Reduce fractions to a common denominator.
- Create a system: a fraction is equal to zero when the numerator is equal to zero and the denominator is not equal to zero.
- Solve the equation.
- Check inequality to exclude extraneous roots.
- Write down the answer.
Discussion: how to formalize the solution if you use the basic property of proportion and multiplying both sides of the equation by a common denominator. (Add to the solution: exclude from its roots those that make the common denominator vanish).
4. Initial comprehension of new material.
Work in pairs. Students choose how to solve the equation themselves depending on the type of equation. Assignments from the textbook “Algebra 8”, Yu.N. Makarychev, 2007: No. 600(b,c,i); No. 601(a,e,g). The teacher monitors the completion of the task, answers any questions that arise, and provides assistance to low-performing students. Self-test: answers are written on the board.
b) 2 – extraneous root. Answer: 3.
c) 2 – extraneous root. Answer: 1.5.
a) Answer: -12.5.
g) Answer: 1;1.5.
5. Setting homework.
- Read paragraph 25 from the textbook, analyze examples 1-3.
- Learn an algorithm for solving fractional rational equations.
- Solve in notebooks No. 600 (a, d, e); No. 601(g,h).
- Try to solve No. 696(a) (optional).
6. Completing a control task on the topic studied.
The work is done on pieces of paper.
Example task:
A) Which of the equations are fractional rational?
B) A fraction is equal to zero when the numerator is ______________________ and the denominator is _______________________.
Q) Is the number -3 the root of equation number 6?
D) Solve equation No. 7.
Assessment criteria for the assignment:
- “5” is given if the student completed more than 90% of the task correctly.
- "4" - 75%-89%
- "3" - 50%-74%
- “2” is given to a student who has completed less than 50% of the task.
- A rating of 2 is not given in the journal, 3 is optional.
7. Reflection.
On the independent work sheets, write:
- 1 – if the lesson was interesting and understandable to you;
- 2 – interesting, but not clear;
- 3 – not interesting, but understandable;
- 4 – not interesting, not clear.
8. Summing up the lesson.
So, today in the lesson we got acquainted with fractional rational equations, learned to solve these equations in various ways, and tested our knowledge with the help of independent educational work. You will learn the results of your independent work in the next lesson, and at home you will have the opportunity to consolidate your knowledge.
Which method of solving fractional rational equations, in your opinion, is easier, more accessible, and more rational? Regardless of the method for solving fractional rational equations, what should you remember? What is the “cunning” of fractional rational equations?
Thanks everyone, lesson is over.
In this article I will show you algorithms for solving seven types of rational equations, which can be reduced to quadratic by changing variables. In most cases, the transformations that lead to replacement are very non-trivial, and it is quite difficult to guess about them on your own.
For each type of equation, I will explain how to make a change of variable in it, and then show a detailed solution in the corresponding video tutorial.
You have the opportunity to continue solving the equations yourself, and then check your solution with the video lesson.
So, let's begin.
1 . (x-1)(x-7)(x-4)(x+2)=40
Note that on the left side of the equation there is a product of four brackets, and on the right side there is a number.
1. Let's group the brackets by two so that the sum of the free terms is the same.
2. Multiply them.
3. Let's introduce a change of variable.
In our equation, we will group the first bracket with the third, and the second with the fourth, since (-1)+(-4)=(-7)+2:
At this point the variable replacement becomes obvious:
We get the equation 
Answer: 
2 .
An equation of this type is similar to the previous one with one difference: on the right side of the equation is the product of the number and . And it is solved in a completely different way:
1. We group the brackets by two so that the product of the free terms is the same.
2. Multiply each pair of brackets.
3. We take x out of each factor.
4. Divide both sides of the equation by .
5. We introduce a change of variable.
In this equation, we group the first bracket with the fourth, and the second with the third, since:
Note that in each bracket the coefficient at and the free term are the same. Let's take a factor out of each bracket:
Since x=0 is not a root of the original equation, we divide both sides of the equation by . We get:
We get the equation: 
Answer: 
3
. 
Note that the denominators of both fractions contain quadratic trinomials, in which the leading coefficient and the free term are the same. Let us take x out of the bracket, as in the equation of the second type. We get:
Divide the numerator and denominator of each fraction by x:
Now we can introduce a variable replacement:
We obtain an equation for the variable t:
4 .
Note that the coefficients of the equation are symmetrical with respect to the central one. This equation is called returnable .
To solve it,
1. Divide both sides of the equation by (We can do this since x=0 is not a root of the equation.) We get:
2. Let’s group the terms in this way:
3. In each group, let’s take the common factor out of brackets:
4. Let's introduce the replacement:
5. Express through t the expression:
From here 
We get the equation for t:
Answer:
5. Homogeneous equations.
Equations that have a homogeneous structure can be encountered when solving exponential, logarithmic and trigonometric equations, so you need to be able to recognize it.
Homogeneous equations have the following structure:
In this equality, A, B and C are numbers, and the square and circle denote identical expressions. That is, on the left side of a homogeneous equation there is a sum of monomials having the same degree (in this case, the degree of monomials is 2), and there is no free term.
To solve a homogeneous equation, divide both sides by
Attention! When dividing the right and left sides of an equation by an expression containing an unknown, you can lose roots. Therefore, it is necessary to check whether the roots of the expression by which we divide both sides of the equation are the roots of the original equation.
Let's go the first way. We get the equation:
Now we introduce variable replacement:
Let us simplify the expression and obtain a biquadratic equation for t:
Answer: or
7
. 
This equation has the following structure:
To solve it, you need to select a complete square on the left side of the equation.
To select a full square, you need to add or subtract twice the product. Then we get the square of the sum or difference. This is crucial for successful variable replacement.
Let's start by finding twice the product. This will be the key to replacing the variable. In our equation, twice the product is equal to
Now let's figure out what is more convenient for us to have - the square of the sum or the difference. Let's first consider the sum of expressions:
Great! This expression is exactly equal to twice the product. Then, in order to get the square of the sum in brackets, you need to add and subtract the double product:
Smirnova Anastasia Yurievna
Lesson type: lesson of learning new material.
Form of organization of educational activities: frontal, individual.
The purpose of the lesson: to introduce a new type of equations - fractional rational equations, to give an idea of the algorithm for solving fractional rational equations.
Lesson objectives.
Educational:
- formation of the concept of a fractional rational equation;
- consider an algorithm for solving fractional rational equations, including the condition that the fraction is equal to zero;
- teach solving fractional rational equations using an algorithm.
Developmental:
- create conditions for developing skills in applying acquired knowledge;
- promote the development of students’ cognitive interest in the subject;
- developing students’ ability to analyze, compare and draw conclusions;
- development of skills of mutual control and self-control, attention, memory, oral and written speech, independence.
Educating:
- fostering cognitive interest in the subject;
- fostering independence in solving educational problems;
- nurturing will and perseverance to achieve final results.
Equipment: textbook, blackboard, crayons.
Textbook "Algebra 8". Yu.N. Makarychev, N.G. Mindyuk, K.I. Neshkov, S.B. Suvorova, edited by S.A. Telyakovsky. Moscow "Enlightenment". 2010
Five hours are allocated for this topic. This is the first lesson. The main thing is to study the algorithm for solving fractional rational equations and practice this algorithm in exercises.
During the classes
1. Organizational moment.
Hello guys! Today I would like to start our lesson with a quatrain:
To make life easier for everyone,
What would be decided, what would be possible,
Smile, good luck to everyone,
So that there are no problems,
We smiled at each other, created a good mood and started working.
There are equations written on the board, look at them carefully. Can you solve all of these equations? Which ones are not and why?
Equations in which the left and right sides are fractional rational expressions are called fractional rational equations. What do you think we will study in class today? Formulate the topic of the lesson. So, open your notebooks and write down the topic of the lesson “Solving fractional rational equations.”
2. Updating knowledge. Frontal survey, oral work with the class.
And now we will repeat the main theoretical material that we will need to study a new topic. Please answer the following questions:
- What is an equation? ( Equality with a variable or variables.)
- What is the name of equation number 1? ( Linear.) A method for solving linear equations. ( Move everything with the unknown to the left side of the equation, all numbers to the right. Give similar terms. Find unknown factor).
- What is the name of equation number 3? ( Square.) Methods for solving quadratic equations. (P about formulas)
- What is proportion? ( Equality of two ratios.) The main property of proportion. ( If the proportion is correct, then the product of its extreme terms is equal to the product of the middle terms.)
- What properties are used when solving equations? ( 1. If you move a term in an equation from one part to another, changing its sign, you will get an equation equivalent to the given one. 2. If both sides of the equation are multiplied or divided by the same non-zero number, you get an equation equivalent to the given one.)
- When does a fraction equal zero? ( A fraction is equal to zero when the numerator is zero and the denominator is not zero..)
3. Explanation of new material.
Solve equation No. 2 in your notebooks and on the board.
Answer: 10.
What fractional rational equation can you try to solve using the basic property of proportion? (No. 5).
(x-2)(x-4) = (x+2)(x+3)
x 2 -4x-2x+8 = x 2 +3x+2x+6
x 2 -6x-x 2 -5x = 6-8
Solve equation No. 4 in your notebooks and on the board.
Answer: 1,5.
What fractional rational equation can you try to solve by multiplying both sides of the equation by the denominator? (No. 6).
x 2 -7x+12 = 0
D=1›0, x 1 =3, x 2 =4.
Answer: 3;4.
We will look at solving equations like equation No. 7 in the following lessons.
Explain why this happened? Why are there three roots in one case and two in the other? What numbers are the roots of this fractional rational equation?
Until now, students have not encountered the concept of an extraneous root; it is indeed very difficult for them to understand why this happened. If no one in the class can give a clear explanation of this situation, then the teacher asks leading questions.
- How do equations No. 2 and 4 differ from equations No. 5 and 6? ( In equations No. 2 and 4 there are numbers in the denominator, No. 5-6 - expressions with a variable.)
- What is the root of an equation? ( The value of the variable at which the equation becomes true.)
- How to find out if a number is the root of an equation? ( Make a check.)
When testing, some students notice that they have to divide by zero. They conclude that the numbers 0 and 5 are not the roots of this equation. The question arises: is there a way to solve fractional rational equations that allows us to eliminate this error? Yes, this method is based on the condition that the fraction is equal to zero.
Let's try to formulate an algorithm for solving fractional rational equations in this way. Children formulate the algorithm themselves.
Algorithm for solving fractional rational equations:
- Move everything to the left side.
- Reduce fractions to a common denominator.
- Create a system: a fraction is equal to zero when the numerator is equal to zero and the denominator is not equal to zero.
- Solve the equation.
- Check inequality to exclude extraneous roots.
- Write down the answer.
4. Initial comprehension of new material.
Work in pairs. Students choose how to solve the equation themselves depending on the type of equation. Assignments from the textbook “Algebra 8”, Yu.N. Makarychev, 2007: No. 600(b,c); No. 601(a,e). The teacher monitors the completion of the task, answers any questions that arise, and provides assistance to low-performing students. Self-test: answers are written on the board.
b) 2 - extraneous root. Answer: 3.
c) 2 - extraneous root. Answer: 1.5.
a) Answer: -12.5.
5. Setting homework.
- Read paragraph 25 from the textbook, analyze examples 1-3.
- Learn an algorithm for solving fractional rational equations.
- Solve in notebooks No. 600 (d, d); No. 601(g,h).
6. Summing up the lesson.
So, today in the lesson we got acquainted with fractional rational equations and learned to solve these equations in various ways. Regardless of how you solve fractional rational equations, what should you keep in mind? What is the “cunning” of fractional rational equations?
Thanks everyone, lesson is over.
Let's continue talking about solving equations. In this article we will go into detail about rational equations and principles of solving rational equations with one variable. First, let's figure out what type of equations are called rational, give a definition of whole rational and fractional rational equations, and give examples. Next, we will obtain algorithms for solving rational equations, and, of course, we will consider solutions to typical examples with all the necessary explanations.
Page navigation.
Based on the stated definitions, we give several examples of rational equations. For example, x=1, 2·x−12·x 2 ·y·z 3 =0, , are all rational equations.
From the examples shown, it is clear that rational equations, as well as equations of other types, can be with one variable, or with two, three, etc. variables. In the following paragraphs we will talk about solving rational equations with one variable. Solving equations in two variables and their large number deserve special attention.
In addition to dividing rational equations by the number of unknown variables, they are also divided into integer and fractional ones. Let us give the corresponding definitions.
Definition.
The rational equation is called whole, if both its left and right sides are integer rational expressions.
Definition.
If at least one of the parts of a rational equation is a fractional expression, then such an equation is called fractionally rational(or fractional rational).
It is clear that whole equations do not contain division by a variable; on the contrary, fractional rational equations necessarily contain division by a variable (or a variable in the denominator). So 3 x+2=0 and (x+y)·(3·x 2 −1)+x=−y+0.5– these are whole rational equations, both of their parts are whole expressions. A and x:(5 x 3 +y 2)=3:(x−1):5 are examples of fractional rational equations.
Concluding this point, let us pay attention to the fact that the linear equations and quadratic equations known to this point are entire rational equations.
Solving whole equations
One of the main approaches to solving entire equations is to reduce them to equivalent ones algebraic equations. This can always be done by performing the following equivalent transformations of the equation:
- first, the expression from the right side of the original integer equation is transferred to the left side with the opposite sign to obtain zero on the right side;
- after this, on the left side of the equation the resulting standard form.
The result is an algebraic equation that is equivalent to the original integer equation. Thus, in the simplest cases, solving entire equations is reduced to solving linear or quadratic equations, and in the general case, to solving an algebraic equation of degree n. For clarity, let's look at the solution to the example.
Example.
Find the roots of the whole equation 3·(x+1)·(x−3)=x·(2·x−1)−3.
Solution.
Let us reduce the solution of this entire equation to the solution of an equivalent algebraic equation. To do this, firstly, we transfer the expression from the right side to the left, as a result we arrive at the equation 3·(x+1)·(x−3)−x·(2·x−1)+3=0. And, secondly, we transform the expression formed on the left side into a standard form polynomial by completing the necessary: 3·(x+1)·(x−3)−x·(2·x−1)+3= (3 x+3) (x−3)−2 x 2 +x+3= 3 x 2 −9 x+3 x−9−2 x 2 +x+3=x 2 −5 x−6. Thus, solving the original integer equation is reduced to solving the quadratic equation x 2 −5·x−6=0.
We calculate its discriminant D=(−5) 2 −4·1·(−6)=25+24=49, it is positive, which means that the equation has two real roots, which we find using the formula for the roots of a quadratic equation:
To be completely sure, let's do it checking the found roots of the equation. First we check the root 6, substitute it instead of the variable x in the original integer equation: 3·(6+1)·(6−3)=6·(2·6−1)−3, which is the same, 63=63. This is a valid numerical equation, therefore x=6 is indeed the root of the equation. Now we check the root −1, we have 3·(−1+1)·(−1−3)=(−1)·(2·(−1)−1)−3, from where, 0=0 . When x=−1, the original equation also turns into a correct numerical equality, therefore, x=−1 is also a root of the equation.
Answer:
6 , −1 .
Here it should also be noted that the term “degree of the whole equation” is associated with the representation of an entire equation in the form of an algebraic equation. Let us give the corresponding definition:
Definition.
The power of the whole equation is called the degree of an equivalent algebraic equation.
According to this definition, the entire equation from the previous example has the second degree.
This could have been the end of solving entire rational equations, if not for one thing…. As is known, solving algebraic equations of degree above the second is associated with significant difficulties, and for equations of degree above the fourth there are no general root formulas at all. Therefore, to solve entire equations of the third, fourth and higher degrees, it is often necessary to resort to other solution methods.
In such cases, an approach to solving entire rational equations based on factorization method. In this case, the following algorithm is adhered to:
- first, they ensure that there is a zero on the right side of the equation; to do this, they transfer the expression from the right side of the whole equation to the left;
- then, the resulting expression on the left side is presented as a product of several factors, which allows us to move on to a set of several simpler equations.
The given algorithm for solving an entire equation through factorization requires a detailed explanation using an example.
Example.
Solve the whole equation (x 2 −1)·(x 2 −10·x+13)= 2 x (x 2 −10 x+13) .
Solution.
First, as usual, we transfer the expression from the right side to the left side of the equation, not forgetting to change the sign, we get (x 2 −1)·(x 2 −10·x+13)− 2 x (x 2 −10 x+13)=0 . Here it is quite obvious that it is not advisable to transform the left-hand side of the resulting equation into a polynomial of the standard form, since this will give an algebraic equation of the fourth degree of the form x 4 −12 x 3 +32 x 2 −16 x−13=0, the solution of which is difficult.
On the other hand, it is obvious that on the left side of the resulting equation we can x 2 −10 x+13 , thereby presenting it as a product. We have (x 2 −10 x+13) (x 2 −2 x−1)=0. The resulting equation is equivalent to the original whole equation, and it, in turn, can be replaced by a set of two quadratic equations x 2 −10·x+13=0 and x 2 −2·x−1=0. Finding their roots using known root formulas through a discriminant is not difficult; the roots are equal. They are the desired roots of the original equation.
Answer:
Also useful for solving entire rational equations method for introducing a new variable. In some cases, it allows you to move to equations whose degree is lower than the degree of the original whole equation.
Example.
Find the real roots of a rational equation (x 2 +3 x+1) 2 +10=−2 (x 2 +3 x−4).
Solution.
Reducing this entire rational equation to an algebraic equation is, to put it mildly, not a very good idea, since in this case we will come to the need to solve a fourth-degree equation that does not have rational roots. Therefore, you will have to look for another solution.
Here it is easy to see that you can introduce a new variable y and replace the expression x 2 +3·x with it. This replacement leads us to the whole equation (y+1) 2 +10=−2·(y−4) , which, after moving the expression −2·(y−4) to the left side and subsequent transformation of the expression formed there, is reduced to a quadratic equation y 2 +4·y+3=0. The roots of this equation y=−1 and y=−3 are easy to find, for example, they can be selected based on the theorem inverse to Vieta's theorem.
Now we move on to the second part of the method of introducing a new variable, that is, to performing a reverse replacement. After performing the reverse substitution, we obtain two equations x 2 +3 x=−1 and x 2 +3 x=−3, which can be rewritten as x 2 +3 x+1=0 and x 2 +3 x+3 =0 . Using the formula for the roots of a quadratic equation, we find the roots of the first equation. And the second quadratic equation has no real roots, since its discriminant is negative (D=3 2 −4·3=9−12=−3 ).
Answer:
In general, when we are dealing with entire equations of high degrees, we must always be prepared to search for a non-standard method or an artificial technique for solving them.
Solving fractional rational equations
First, it will be useful to understand how to solve fractional rational equations of the form , where p(x) and q(x) are integer rational expressions. And then we will show how to reduce the solution of other fractionally rational equations to the solution of equations of the indicated type.
One approach to solving the equation is based on the following statement: the numerical fraction u/v, where v is a non-zero number (otherwise we will encounter , which is undefined), is equal to zero if and only if its numerator is equal to zero, then is, if and only if u=0 . By virtue of this statement, solving the equation is reduced to fulfilling two conditions p(x)=0 and q(x)≠0.
This conclusion corresponds to the following algorithm for solving a fractional rational equation. To solve a fractional rational equation of the form , you need
- solve the whole rational equation p(x)=0 ;
- and check whether the condition q(x)≠0 is satisfied for each root found, while
- if true, then this root is the root of the original equation;
- if it is not satisfied, then this root is extraneous, that is, it is not the root of the original equation.
Let's look at an example of using the announced algorithm when solving a fractional rational equation.
Example.
Find the roots of the equation.
Solution.
This is a fractional rational equation, and of the form , where p(x)=3·x−2, q(x)=5·x 2 −2=0.
According to the algorithm for solving fractional rational equations of this type, we first need to solve the equation 3 x−2=0. This is a linear equation whose root is x=2/3.
It remains to check for this root, that is, check whether it satisfies the condition 5 x 2 −2≠0. We substitute the number 2/3 into the expression 5 x 2 −2 instead of x, and we get . The condition is met, so x=2/3 is the root of the original equation.
Answer:
2/3 .
You can approach solving a fractional rational equation from a slightly different position. This equation is equivalent to the integer equation p(x)=0 on the variable x of the original equation. That is, you can stick to this algorithm for solving a fractional rational equation :
- solve the equation p(x)=0 ;
- find the ODZ of variable x;
- take roots belonging to the region of acceptable values - they are the desired roots of the original fractional rational equation.
For example, let's solve a fractional rational equation using this algorithm.
Example.
Solve the equation.
Solution.
First, we solve the quadratic equation x 2 −2·x−11=0. Its roots can be calculated using the root formula for the even second coefficient, we have D 1 =(−1) 2 −1·(−11)=12, And .
Secondly, we find the ODZ of the variable x for the original equation. It consists of all numbers for which x 2 +3·x≠0, which is the same as x·(x+3)≠0, whence x≠0, x≠−3.
It remains to check whether the roots found in the first step are included in the ODZ. Obviously yes. Therefore, the original fractional rational equation has two roots.
Answer:
Note that this approach is more profitable than the first if the ODZ is easy to find, and is especially beneficial if the roots of the equation p(x) = 0 are irrational, for example, or rational, but with a rather large numerator and/or denominator, for example, 127/1101 and −31/59. This is due to the fact that in such cases, checking the condition q(x)≠0 will require significant computational effort, and it is easier to exclude extraneous roots using the ODZ.
In other cases, when solving the equation, especially when the roots of the equation p(x) = 0 are integers, it is more profitable to use the first of the given algorithms. That is, it is advisable to immediately find the roots of the entire equation p(x)=0, and then check whether the condition q(x)≠0 is satisfied for them, rather than finding the ODZ, and then solving the equation p(x)=0 on this ODZ . This is due to the fact that in such cases it is usually easier to check than to find DZ.
Let us consider the solution of two examples to illustrate the specified nuances.
Example.
Find the roots of the equation.
Solution.
First, let's find the roots of the whole equation (2 x−1) (x−6) (x 2 −5 x+14) (x+1)=0, composed using the numerator of the fraction. The left side of this equation is a product, and the right side is zero, therefore, according to the method of solving equations through factorization, this equation is equivalent to a set of four equations 2 x−1=0 , x−6=0 , x 2 −5 x+ 14=0 , x+1=0 . Three of these equations are linear and one is quadratic; we can solve them. From the first equation we find x=1/2, from the second - x=6, from the third - x=7, x=−2, from the fourth - x=−1.
With the roots found, it is quite easy to check whether the denominator of the fraction on the left side of the original equation vanishes, but determining the ODZ, on the contrary, is not so simple, since for this you will have to solve an algebraic equation of the fifth degree. Therefore, we will abandon finding the ODZ in favor of checking the roots. To do this, we substitute them one by one instead of the variable x in the expression x 5 −15 x 4 +57 x 3 −13 x 2 +26 x+112, obtained after substitution, and compare them with zero: (1/2) 5 −15·(1/2) 4 + 57·(1/2) 3 −13·(1/2) 2 +26·(1/2)+112= 1/32−15/16+57/8−13/4+13+112=
122+1/32≠0
;
6 5 −15·6 4 +57·6 3 −13·6 2 +26·6+112= 448≠0
;
7 5 −15·7 4 +57·7 3 −13·7 2 +26·7+112=0;
(−2) 5 −15·(−2) 4 +57·(−2) 3 −13·(−2) 2 + 26·(−2)+112=−720≠0 ;
(−1) 5 −15·(−1) 4 +57·(−1) 3 −13·(−1) 2 + 26·(−1)+112=0 .
Thus, 1/2, 6 and −2 are the desired roots of the original fractional rational equation, and 7 and −1 are extraneous roots.
Answer:
1/2 , 6 , −2 .
Example.
Find the roots of a fractional rational equation.
Solution.
First, let's find the roots of the equation (5 x 2 −7 x−1) (x−2)=0. This equation is equivalent to a set of two equations: square 5 x 2 −7 x−1=0 and linear x−2=0. Using the formula for the roots of a quadratic equation, we find two roots, and from the second equation we have x=2.
Checking whether the denominator goes to zero at the found values of x is quite unpleasant. And determining the range of permissible values of the variable x in the original equation is quite simple. Therefore, we will act through ODZ.
In our case, the ODZ of the variable x of the original fractional rational equation consists of all numbers except those for which the condition x 2 +5·x−14=0 is satisfied. The roots of this quadratic equation are x=−7 and x=2, from which we draw a conclusion about the ODZ: it consists of all x such that .
It remains to check whether the found roots and x=2 belong to the range of acceptable values. The roots belong, therefore, they are roots of the original equation, and x=2 does not belong, therefore, it is an extraneous root.
Answer:
It will also be useful to separately dwell on the cases when in a fractional rational equation of the form there is a number in the numerator, that is, when p(x) is represented by some number. Wherein
- if this number is non-zero, then the equation has no roots, since a fraction is equal to zero if and only if its numerator is equal to zero;
- if this number is zero, then the root of the equation is any number from the ODZ.
Example.
Solution.
Since the numerator of the fraction on the left side of the equation contains a non-zero number, then for any x the value of this fraction cannot be equal to zero. Therefore, this equation has no roots.
Answer:
no roots.
Example.
Solve the equation.
Solution.
The numerator of the fraction on the left side of this fractional rational equation contains zero, so the value of this fraction is zero for any x for which it makes sense. In other words, the solution to this equation is any value of x from the ODZ of this variable.
It remains to determine this range of acceptable values. It includes all values of x for which x 4 +5 x 3 ≠0. The solutions to the equation x 4 +5 x 3 =0 are 0 and −5, since this equation is equivalent to the equation x 3 (x+5)=0, and it in turn is equivalent to the combination of two equations x 3 =0 and x +5=0, from where these roots are visible. Therefore, the desired range of acceptable values is any x except x=0 and x=−5.
Thus, a fractional rational equation has infinitely many solutions, which are any numbers except zero and minus five.
Answer:
Finally, it's time to talk about solving fractional rational equations of arbitrary form. They can be written as r(x)=s(x), where r(x) and s(x) are rational expressions, and at least one of them is fractional. Looking ahead, let's say that their solution comes down to solving equations of the form already familiar to us.
It is known that transferring a term from one part of the equation to another with the opposite sign leads to an equivalent equation, therefore the equation r(x)=s(x) is equivalent to the equation r(x)−s(x)=0.
We also know that any , identically equal to this expression, is possible. Thus, we can always transform the rational expression on the left side of the equation r(x)−s(x)=0 into an identically equal rational fraction of the form .
So we move from the original fractional rational equation r(x)=s(x) to the equation, and its solution, as we found out above, reduces to solving the equation p(x)=0.
But here it is necessary to take into account the fact that when replacing r(x)−s(x)=0 with , and then with p(x)=0, the range of permissible values of the variable x may expand.
Consequently, the original equation r(x)=s(x) and the equation p(x)=0 that we arrived at may turn out to be unequal, and by solving the equation p(x)=0, we can get roots that will be extraneous roots of the original equation r(x)=s(x) . You can identify and not include extraneous roots in the answer either by performing a check or by checking that they belong to the ODZ of the original equation.
Let's summarize this information in algorithm for solving fractional rational equation r(x)=s(x). To solve the fractional rational equation r(x)=s(x) , you need
- Get zero on the right by moving the expression from the right side with the opposite sign.
- Perform operations with fractions and polynomials on the left side of the equation, thereby transforming it into a rational fraction of the form.
- Solve the equation p(x)=0.
- Identify and eliminate extraneous roots, which is done by substituting them into the original equation or by checking their belonging to the ODZ of the original equation.
For greater clarity, we will show the entire chain of solving fractional rational equations:
.
Let's look at the solutions of several examples with a detailed explanation of the solution process in order to clarify the given block of information.
Example.
Solve a fractional rational equation.
Solution.
We will act in accordance with the solution algorithm just obtained. And first we move the terms from the right side of the equation to the left, as a result we move on to the equation.
In the second step, we need to convert the fractional rational expression on the left side of the resulting equation to the form of a fraction. To do this, we reduce rational fractions to a common denominator and simplify the resulting expression: . So we come to the equation.
In the next step, we need to solve the equation −2·x−1=0. We find x=−1/2.
It remains to check whether the found number −1/2 is not an extraneous root of the original equation. To do this, you can check or find the VA of the variable x of the original equation. Let's demonstrate both approaches.
Let's start with checking. We substitute the number −1/2 into the original equation instead of the variable x, and we get the same thing, −1=−1. The substitution gives the correct numerical equality, so x=−1/2 is the root of the original equation.
Now we will show how the last point of the algorithm is performed through ODZ. The range of permissible values of the original equation is the set of all numbers except −1 and 0 (at x=−1 and x=0 the denominators of the fractions vanish). The root x=−1/2 found in the previous step belongs to the ODZ, therefore, x=−1/2 is the root of the original equation.
Answer:
−1/2 .
Let's look at another example.
Example.
Find the roots of the equation.
Solution.
We need to solve a fractional rational equation, let's go through all the steps of the algorithm.
First, we move the term from the right side to the left, we get .
Secondly, we transform the expression formed on the left side: . As a result, we arrive at the equation x=0.
Its root is obvious - it is zero.
At the fourth step, it remains to find out whether the found root is extraneous to the original fractional rational equation. When it is substituted into the original equation, the expression is obtained. Obviously, it doesn't make sense because it contains division by zero. Whence we conclude that 0 is an extraneous root. Therefore, the original equation has no roots.
7, which leads to Eq. From this we can conclude that the expression in the denominator of the left side must be equal to that of the right side, that is, . Now we subtract from both sides of the triple: . By analogy, from where, and further.
The check shows that both roots found are roots of the original fractional rational equation.
Answer:
Bibliography.
- 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. 8th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemosyne, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
- Algebra: 9th grade: educational. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.
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