Fractional equations. ODZ.
Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)
We continue to master the equations. We already know how to work with linear and quadratic equations. Remained last view – fractional equations . Or they are also called much more respectably - fractional rational equations. It is the same.
Fractional equations.
As the name implies, these equations necessarily contain fractions. But not just fractions, but fractions that have unknown in denominator. At least in one. For example:
Let me remind you that if the denominators are only numbers, these are linear equations.
How to decide fractional equations? First of all, get rid of fractions! After this, the equation most often turns into linear or quadratic. And then we know what to do... In some cases it can turn into an identity, such as 5=5 or an incorrect expression, such as 7=2. But this rarely happens. I will mention this below.
But how to get rid of fractions!? Very simple. Applying the same identical transformations.
We need to multiply the entire equation by the same expression. So that all denominators are reduced! Everything will immediately become easier. Let me explain with an example. Let us need to solve the equation:
How were you taught in elementary school? We move everything to one side, bring it to a common denominator, etc. Forget how horrible dream! This is what you need to do when you add or subtract fractions. Or you work with inequalities. And in equations, we immediately multiply both sides by an expression that will give us the opportunity to reduce all the denominators (i.e., in essence, by a common denominator). And what is this expression?
On the left side, reducing the denominator requires multiplying by x+2. And on the right, multiplication by 2 is required. This means that the equation must be multiplied by 2(x+2). Multiply:
This is a common multiplication of fractions, but I’ll describe it in detail:
Please note that I am not opening the bracket yet (x + 2)! So, in its entirety, I write it:
On the left side it contracts entirely (x+2), and on the right 2. Which is what was required! After reduction we get linear the equation:
And everyone can solve this equation! x = 2.
Let's solve another example, a little more complicated:
If we remember that 3 = 3/1, and 2x = 2x/ 1, we can write:
And again we get rid of what we don’t really like - fractions.
We see that to reduce the denominator with X, we need to multiply the fraction by (x – 2). And a few are not a hindrance to us. Well, let's multiply. All left side and all right side:
Parentheses again (x – 2) I'm not revealing. I work with the bracket as a whole as if it were one number! This must always be done, otherwise nothing will be reduced.
With a feeling of deep satisfaction we reduce (x – 2) and we get an equation without any fractions, with a ruler!
Now let’s open the brackets:
We bring similar ones, move everything to the left side and get:
But before that we will learn to solve other problems. On interest. That's a rake, by the way!
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)
You can get acquainted with functions and derivatives.
T. Kosyakova,
School No. 80, Krasnodar
Solving quadratic and fractional rational equations containing parameters
Lesson 4
Lesson topic:
The purpose of the lesson: develop the ability to solve fractional rational equations containing parameters.
Lesson type: introduction of new material.
1. (Orally) Solve the equations:
Example 1. Solve the equation 
Solution. 
Let's find invalid values a: 
Answer. If 
If a = – 19
, then there are no roots.
Example 2. Solve the equation 
Solution.
Let's find invalid parameter values a :
10 – a = 5, a = 5;
10 – a = a, a = 5.
Answer. If a = 5 a № 5 , That x=10– a .
Example 3. At what parameter values b
the equation 
It has:
a) two roots; b) the only root?
Solution.
1) Find invalid parameter values b :
x = b, b 2 (b 2
– 1) – 2b 3 + b 2 = 0, b 4
– 2b 3 = 0,
b= 0 or b = 2;
x = 2, 4( b 2 – 1) – 4b 2 + b 2
= 0, b 2 – 4 = 0, (b – 2)(b + 2) = 0,
b= 2 or b = – 2.
2) Solve the equation x 2 ( b 2 – 1) – 2b 2x+ b 2 = 0:
D=4 b 4 – 4b 2 (b 2 – 1), D = 4 b 2 .
A) 
Excluding invalid parameter values b , we find that the equation has two roots if b № – 2, b № – 1, b № 0, b № 1, b № 2 .
b) 4b 2 = 0, b = 0, but this is an invalid parameter value b ; If b 2 –1=0 , i.e. b=1 or.
Answer: a) if b № –2 , b № –1, b № 0, b № 1, b № 2 , then two roots; b) if b=1 or b=–1 , then the only root.
Independent work
Option 1
Solve the equations:
Option 2
Solve the equations:
Answers
IN 1. and if a=3
, then there are no roots; If 
b) if if a
№
2
, then there are no roots.
AT 2. If a=2
, then there are no roots; If a=0
, then there are no roots; If 
b) if a=– 1
, then the equation becomes meaningless; if there are no roots;
If
Homework assignment.
Solve the equations:
Answers: a) If a № –2 , That x= a ; If a=–2 , then there are no solutions; b) if a № –2 , That x=2; If a=–2 , then there are no solutions; c) if a=–2 , That x– any number except 3 ; If a № –2 , That x=2; d) if a=–8 , then there are no roots; If a=2 , then there are no roots; If
Lesson 5
Lesson topic:"Solving fractional rational equations containing parameters."
Lesson objectives:
training in solving equations with non-standard conditions;
conscious assimilation by students of algebraic concepts and connections between them.
Lesson type: systematization and generalization.
Checking homework.
Example 1. Solve the equation 
a) relative to x; b) relative to y.
Solution.
a) Find invalid values y: y=0, x=y, y 2 =y 2 –2y,
y=0– invalid parameter value y.
If y№ 0 , That x=y–2; If y=0, then the equation becomes meaningless.
b) Find invalid parameter values x: y=x, 2x–x 2 +x 2 =0, x=0– invalid parameter value x; y(2+x–y)=0, y=0 or y=2+x;
y=0 does not satisfy the condition y(y–x)№ 0 .
Answer: a) if y=0, then the equation becomes meaningless; If y№ 0 , That x=y–2; b) if x=0 x№ 0 , That y=2+x .
Example 2. For what integer values of the parameter a are the roots of the equation 
belong to the interval 
D = (3 a + 2) 2 – 4a(a+ 1) 2 = 9 a 2 + 12a + 4 – 8a 2 – 8a,
D = ( a + 2) 2 .
If a № 0 or a № – 1 , That
Answer: 5 .
Example 3. Find relatively x integer solutions to the equation 
Answer. If y=0, then the equation does not make sense; If y=–1, That x– any integer except zero; If y№ 0, y№ – 1, then there are no solutions.
Example 4. Solve the equation 
with parameters a
And b
.
If a№
–b
, That 
Answer. If a= 0 or b= 0 , then the equation becomes meaningless; If a№ 0, b№ 0, a=–b , That x– any number except zero; If a№ 0, b№ 0, a№ –b, That x=–a, x=–b .
Example 5. Prove that for any value of the parameter n other than zero, the equation 
has a single root equal to –n
.
Solution. 
i.e. x=–n, which was what needed to be proven.
Homework assignment.
1. Find integer solutions to the equation 
2. At what parameter values c the equation 
It has:
a) two roots; b) the only root?
3. Find all the integer roots of the equation 
If a ABOUT N
.
4. Solve the equation 3xy – 5x + 5y = 7: a) relatively y; b) relatively x .
1. The equation is satisfied by any integer equal values of x and y other than zero.
2. a) When
b) at or 
3. – 12; – 9; 0
.
4. a) If then there are no roots; If 
b) if then there are no roots; If 
Test
Option 1
1. Determine the type of equation 7c(c + 3)x 2 +(c–2)x–8=0 when: a) c=–3; b) c=2 ; V) c=4 .
2. Solve the equations: a) x 2 –bx=0 ; b) cx 2 –6x+1=0; V) 
3. Solve the equation 3x–xy–2y=1:
a) relatively x ;
b) relatively y .
nx 2 – 26x + n = 0, knowing that the parameter n accepts only integer values.
5. For what values of b does the equation 
It has:
a) two roots;
b) the only root?
Option 2
1. Determine the type of equation 5c(c + 4)x 2 +(c–7)x+7=0 when: a) c=–4 ; b) c=7 ; V) c=1 .
2. Solve the equations: a) y 2 +cy=0 ; b) ny 2 –8y+2=0 ; V) 
3. Solve the equation 6x–xy+2y=5:
a) relatively x ;
b) relatively y .
4. Find the integer roots of the equation nx 2 –22x+2n=0 , knowing that the parameter n accepts only integer values.
5. For what values of the parameter a does the equation 
It has:
a) two roots;
b) the only root?
Answers
IN 1. 1. a) Linear equation;
b) incomplete quadratic equation; c) quadratic equation.
2. a) If b=0, That x=0; If b№ 0, That x=0, x=b;
b) 
If cО (9;+Ґ ), then there are no roots;
c) if a=–4
, then the equation becomes meaningless; If a№
–4
, That x=– a
.
3. a) If y=3, then there are no roots; If);
b) a=–3, a=1.
Additional tasks
Solve the equations:
Literature
1. Golubev V.I., Goldman A.M., Dorofeev G.V. About the parameters from the very beginning. – Tutor, No. 2/1991, p. 3–13.
2. Gronshtein P.I., Polonsky V.B., Yakir M.S. The necessary conditions in problems with parameters. – Kvant, No. 11/1991, p. 44–49.
3. Dorofeev G.V., Zatakavay V.V. Problem solving containing parameters. Part 2. – M., Perspective, 1990, p. 2–38.
4. Tynyakin S.A. Five hundred and fourteen problems with parameters. – Volgograd, 1991.
5. Yastrebinetsky G.A. Problems with parameters. – M., Education, 1986.
§ 1 Integer and fractional rational equations
In this lesson we will look at concepts such as rational equation, rational expression, whole expression, fractional expression. Let's consider solving rational equations.
A rational equation is an equation in which the left and right sides are rational expressions.
Rational expressions are:
Fractional.
An integer expression is made up of numbers, variables, integer powers using the operations of addition, subtraction, multiplication, and division by a number other than zero.
For example:
IN fractional expressions there is a division by a variable or an expression with a variable. For example:
A fractional expression does not make sense for all values of the variables included in it. For example, the expression
at x = -9 it does not make sense, since at x = -9 the denominator goes to zero.
This means that a rational equation can be integer or fractional.
A whole rational equation is a rational equation in which the left and right sides are whole expressions.
For example:
A fractional rational equation is a rational equation in which either the left or right sides are fractional expressions.
For example:
§ 2 Solution of an entire rational equation
Let's consider the solution of an entire rational equation.
For example:
Let's multiply both sides of the equation by the least common denominator of the denominators of the fractions included in it.
For this:
1. find the common denominator for denominators 2, 3, 6. It is equal to 6;
2. find an additional factor for each fraction. To do this, divide the common denominator 6 by each denominator
additional factor for fraction
additional factor for fraction
3. multiply the numerators of the fractions by their corresponding additional factors. Thus, we obtain the equation
which is equivalent to the given equation
Let's open the brackets on the left, move the right part to the left, changing the sign of the term when transferred to the opposite one.
Let us bring similar terms of the polynomial and get
We see that the equation is linear.
Having solved it, we find that x = 0.5.
§ 3 Solution of a fractional rational equation
Let's consider solving a fractional rational equation.
For example:
1.Multiply both sides of the equation by the least common denominator of the denominators of the rational fractions included in it.
Let's find the common denominator for the denominators x + 7 and x - 1.
It is equal to their product (x + 7)(x - 1).
2. Let's find an additional factor for each rational fraction.
To do this, divide the common denominator (x + 7)(x - 1) by each denominator. Additional factor for fractions
equal to x - 1,
additional factor for fraction
equals x+7.
3.Multiply the numerators of the fractions by their corresponding additional factors.
We obtain the equation (2x - 1)(x - 1) = (3x + 4)(x + 7), which is equivalent to this equation
4.Multiply the binomial by the binomial on the left and right and get the following equation
5. We move the right side to the left, changing the sign of each term when transferring to the opposite:
6. Let us present similar terms of the polynomial:
7. Both sides can be divided by -1. We get a quadratic equation:
8. Having solved it, we will find the roots
Since in Eq.
the left and right sides are fractional expressions, and in fractional expressions, for some values of the variables, the denominator can become zero, then it is necessary to check whether the common denominator does not go to zero when x1 and x2 are found.
At x = -27, the common denominator (x + 7)(x - 1) does not vanish; at x = -1, the common denominator also does not equal to zero.
Therefore, both roots -27 and -1 are roots of the equation.
When solving a fractional rational equation, it is better to immediately indicate the region acceptable values. Eliminate those values at which the common denominator goes to zero.
Let's consider another example of solving a fractional rational equation.
For example, let's solve the equation
We factor the denominator of the fraction on the right side of the equation
We get the equation
Let's find the common denominator for the denominators (x - 5), x, x(x - 5).
It will be the expression x(x - 5).
Now let's find the range of acceptable values of the equation
To do this, we equate the common denominator to zero x(x - 5) = 0.
We obtain an equation, solving which we find that at x = 0 or at x = 5 the common denominator goes to zero.
This means that x = 0 or x = 5 cannot be the roots of our equation.
Additional multipliers can now be found.
An additional factor for rational fractions
additional factor for the fraction
will be (x - 5),
and the additional factor of the fraction
We multiply the numerators by the corresponding additional factors.
We get the equation x(x - 3) + 1(x - 5) = 1(x + 5).
Let's open the brackets on the left and right, x2 - 3x + x - 5 = x + 5.
Let's move the terms from right to left, changing the sign of the transferred terms:
X2 - 3x + x - 5 - x - 5 = 0
And after bringing similar terms, we obtain a quadratic equation x2 - 3x - 10 = 0. Having solved it, we find the roots x1 = -2; x2 = 5.
But we have already found out that at x = 5 the common denominator x(x - 5) goes to zero. Therefore, the root of our equation
will be x = -2.
§ 4 Brief summary lesson
Important to remember:
When solving fractional rational equations, proceed as follows:
1. Find the common denominator of the fractions included in the equation. Moreover, if the denominators of fractions can be factored, then factor them and then find the common denominator.
2.Multiply both sides of the equation by a common denominator: find additional factors, multiply the numerators by additional factors.
3.Solve the resulting whole equation.
4. Eliminate from its roots those that make the common denominator vanish.
List of used literature:
- Makarychev Yu.N., N.G. Mindyuk, Neshkov K.I., Suvorova S.B. / Edited by Telyakovsky S.A. Algebra: textbook. for 8th grade. general education institutions. - M.: Education, 2013.
- Mordkovich A.G. Algebra. 8th grade: In two parts. Part 1: Textbook. for general education institutions. - M.: Mnemosyne.
- Rurukin A.N. Lesson developments in algebra: 8th grade. - M.: VAKO, 2010.
- Algebra 8th grade: lesson plans based on the textbook by Yu.N. Makarycheva, N.G. Mindyuk, K.I. Neshkova, S.B. Suvorova / Auth.-comp. T.L. Afanasyeva, L.A. Tapilina. -Volgograd: Teacher, 2005.
Let's get acquainted with rational and fractional rational equations, give their definition, give examples, and also analyze the most common types of problems.
Yandex.RTB R-A-339285-1
Rational equation: definition and examples
Acquaintance with rational expressions begins in the 8th grade of school. At this time, in algebra lessons, students increasingly begin to encounter assignments with equations that contain rational expressions in their notes. Let's refresh our memory on what it is.
Definition 1
Rational equation is an equation in which both sides contain rational expressions.
In various manuals you can find another formulation.
Definition 2
Rational equation- this is an equation, the left side of which contains a rational expression, and the right side contains zero.
The definitions that we have given for rational equations are equivalent, since they say the same thing. The correctness of our words is confirmed by the fact that for any rational expressions P And Q equations P = Q And P − Q = 0 will be equivalent expressions.
Now let's look at the examples.
Example 1
Rational equations:
x = 1 , 2 x − 12 x 2 y z 3 = 0 , x x 2 + 3 x - 1 = 2 + 2 7 x - a (x + 2) , 1 2 + 3 4 - 12 x - 1 = 3 .
Rational equations, just like equations of other types, can contain any number of variables from 1 to several. First we'll look at simple examples, in which the equations will contain only one variable. And then we will begin to gradually complicate the task.
Rational equations are divided into two large groups: integer and fractional. Let's see what equations will apply to each of the groups.
Definition 3
A rational equation will be integer if its left and right sides contain entire rational expressions.
Definition 4
A rational equation will be fractional if one or both of its parts contain a fraction.
Fractional rational equations necessarily contain division by a variable or the variable is present in the denominator. There is no such division in the writing of whole equations.
Example 2
3 x + 2 = 0 And (x + y) · (3 · x 2 − 1) + x = − y + 0, 5– entire rational equations. Here both sides of the equation are represented by integer expressions.
1 x - 1 = x 3 and x: (5 x 3 + y 2) = 3: (x − 1) : 5 are fractional rational equations.
Whole rational equations include linear and quadratic equations.
Solving whole equations
Solving such equations usually comes down to converting them into equivalent algebraic equations. This can be achieved by carrying out equivalent transformations of equations in accordance with the following algorithm:
- first we get zero on the right side of the equation; to do this, we need to move the expression that is on the right side of the equation to its left side and change the sign;
- then we transform the expression on the left side of the equation into a polynomial standard view.
We must obtain an algebraic equation. This equation will be equivalent to the original equation. Easy cases allow us to reduce the whole equation to a linear or quadratic one to solve the problem. In general, we solve an algebraic equation of degree n.
Example 3
It is necessary to find the roots of the whole equation 3 (x + 1) (x − 3) = x (2 x − 1) − 3.
Solution
Let us transform the original expression in order to obtain an equivalent algebraic equation. To do this, we will transfer the expression contained on the right side of the equation to the left side and replace the sign with the opposite one. As a result we get: 3 (x + 1) (x − 3) − x (2 x − 1) + 3 = 0.
Now let's transform the expression that is on the left side into a standard form polynomial and perform the necessary actions with this polynomial:
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
We managed to reduce the solution to the original equation to the solution quadratic equation kind x 2 − 5 x − 6 = 0. The discriminant of this equation is positive: D = (− 5) 2 − 4 · 1 · (− 6) = 25 + 24 = 49 . This means there will be two real roots. Let's find them using the formula for the roots of a quadratic equation:
x = - - 5 ± 49 2 1,
x 1 = 5 + 7 2 or x 2 = 5 - 7 2,
x 1 = 6 or x 2 = - 1
Let's check the correctness of the roots of the equation that we found during the solution. For this, we substitute the numbers we received into the original equation: 3 (6 + 1) (6 − 3) = 6 (2 6 − 1) − 3 And 3 · (− 1 + 1) · (− 1 − 3) = (− 1) · (2 · (− 1) − 1) − 3. In the first case 63 = 63 , in the second 0 = 0 . Roots x = 6 And x = − 1 are indeed the roots of the equation given in the example condition.
Answer: 6 , − 1 .
Let's look at what "degree of an entire equation" means. We will often encounter this term in cases where we need to represent an entire equation in the form of an algebraic one. Let's define the concept.
Definition 5
Degree of the whole equation- this is the degree algebraic equation, equivalent to the original integer equation.
If you look at the equations from the example above, you can establish: the degree of this whole equation is second.
If our course was limited to solving equations of the second degree, then the discussion of the topic could end there. But it's not that simple. Solving equations of the third degree is fraught with difficulties. And for equations higher than the fourth degree there is no general formulas roots In this regard, solving entire equations of the third, fourth and other degrees requires us to use a number of other techniques and methods.
The most commonly used approach to solving entire rational equations is based on the factorization method. The algorithm of actions in this case is as follows:
- we move the expression from the right side to the left so that zero remains on the right side of the record;
- We represent the expression on the left side as a product of factors, and then move on to a set of several simpler equations.
Find the solution to the equation (x 2 − 1) · (x 2 − 10 · x + 13) = 2 · x · (x 2 − 10 · x + 13) .
Solution
Move the expression from the right side of the record to the left with opposite sign: (x 2 − 1) · (x 2 − 10 · x + 13) − 2 · x · (x 2 − 10 · x + 13) = 0. Converting the left-hand side to a polynomial of the standard form is inappropriate due to the fact that this will give us an algebraic equation of the fourth degree: x 4 − 12 x 3 + 32 x 2 − 16 x − 13 = 0. The ease of conversion does not justify all the difficulties in solving such an equation.
It’s much easier to go the other way: let’s take the common factor out of brackets x 2 − 10 x + 13 . So we arrive at an equation of the form (x 2 − 10 x + 13) (x 2 − 2 x − 1) = 0. Now we replace the resulting equation with a set of two quadratic equations x 2 − 10 x + 13 = 0 And x 2 − 2 x − 1 = 0 and find their roots through the discriminant: 5 + 2 3, 5 - 2 3, 1 + 2, 1 - 2.
Answer: 5 + 2 3, 5 - 2 3, 1 + 2, 1 - 2.
In the same way, we can use the method of introducing a new variable. This method allows us to move to equivalent equations with degrees lower than the degrees in the original integer equation.
Example 5
Does the equation have roots? (x 2 + 3 x + 1) 2 + 10 = − 2 (x 2 + 3 x − 4)?
Solution
If we now try to reduce an entire rational equation to an algebraic one, we will get an equation of degree 4 that has no rational roots. Therefore, it will be easier for us to go the other way: introduce a new variable y, which will replace the expression in the equation x 2 + 3 x.
Now we will work with the whole equation (y + 1) 2 + 10 = − 2 (y − 4). Let's move the right side of the equation to the left with the opposite sign and carry out the necessary transformations. We get: y 2 + 4 y + 3 = 0. Let's find the roots of the quadratic equation: y = − 1 And y = − 3.
Now let's do the reverse replacement. We get two equations x 2 + 3 x = − 1 And x 2 + 3 · x = − 3 . Let's rewrite them as x 2 + 3 x + 1 = 0 and x 2 + 3 x + 3 = 0. We use the formula for the roots of a quadratic equation in order to find the roots of the first equation from those obtained: - 3 ± 5 2. The discriminant of the second equation is negative. This means that the second equation has no real roots.
Answer:- 3 ± 5 2
Whole equations high degrees come across in tasks quite often. There is no need to be afraid of them. You need to be ready to apply non-standard method their solutions, including a number of artificial transformations.
Solving fractional rational equations
We will begin our consideration of this subtopic with an algorithm for solving fractionally rational equations of the form p (x) q (x) = 0, where p(x) And q(x)– whole rational expressions. The solution of other fractionally rational equations can always be reduced to the solution of equations of the indicated type.
The most commonly used method for solving the equations p (x) q (x) = 0 is based on the following statement: numerical fraction u v, Where v- this is a number that is different from zero, equal to zero only in those cases when the numerator of the fraction is equal to zero. Following the logic of the above statement, we can claim that the solution to the equation p (x) q (x) = 0 can be reduced to fulfilling two conditions: p(x)=0 And q(x) ≠ 0. This is the basis for constructing an algorithm for solving fractional rational equations of the form p (x) q (x) = 0:
- find the solution to the whole rational equation p(x)=0;
- we check whether the condition is satisfied for the roots found during the solution q(x) ≠ 0.
If this condition is met, then the found root. If not, then the root is not a solution to the problem.
Example 6
Let's find the roots of the equation 3 · x - 2 5 · x 2 - 2 = 0 .
Solution
We are dealing with a fractional rational equation of the form p (x) q (x) = 0, in which p (x) = 3 x − 2, q (x) = 5 x 2 − 2 = 0. Let's start solving the linear equation 3 x − 2 = 0. The root of this equation will be x = 2 3.
Let's check the found root to see if it satisfies the condition 5 x 2 − 2 ≠ 0. To do this, substitute a numerical value into the expression. We get: 5 · 2 3 2 - 2 = 5 · 4 9 - 2 = 20 9 - 2 = 2 9 ≠ 0.
The condition is met. It means that x = 2 3 is the root of the original equation.
Answer: 2 3 .
There is another option for solving fractional rational equations p (x) q (x) = 0. Recall that this equation is equivalent to the whole equation p(x)=0 on the range of permissible values of the variable x of the original equation. This allows us to use the following algorithm in solving the equations p (x) q (x) = 0:
- solve the equation p(x)=0;
- find the range of permissible values of the variable x;
- we take the roots that lie in the range of permissible values of the variable x as the desired roots of the original fractional rational equation.
Solve the equation x 2 - 2 x - 11 x 2 + 3 x = 0.
Solution
First, let's solve the quadratic equation x 2 − 2 x − 11 = 0. To calculate its roots, we use the roots formula for the even second coefficient. We get D 1 = (− 1) 2 − 1 · (− 11) = 12, and x = 1 ± 2 3 .
Now we can find the ODZ of variable x for the original equation. These are all the numbers for which x 2 + 3 x ≠ 0. It's the same as x (x + 3) ≠ 0, from where x ≠ 0, x ≠ − 3.
Now let’s check whether the roots x = 1 ± 2 3 obtained at the first stage of the solution are within the range of permissible values of the variable x. We see them coming in. This means that the original fractional rational equation has two roots x = 1 ± 2 3.
Answer: x = 1 ± 2 3
The second solution method described is simpler than the first in cases where the range of permissible values of the variable x is easily found, and the roots of the equation p(x)=0 irrational. For example, 7 ± 4 · 26 9. The roots can be rational, but with a large numerator or denominator. For example, 127 1101 And − 31 59 . This saves time on checking the condition q(x) ≠ 0: It is much easier to exclude roots that are not suitable according to the ODZ.
In cases where the roots of the equation p(x)=0 are integers, it is more expedient to use the first of the described algorithms for solving equations of the form p (x) q (x) = 0. Find the roots of an entire equation faster p(x)=0, and then check whether the condition is satisfied for them q(x) ≠ 0, 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.
Example 8
Find the roots of the equation (2 x - 1) (x - 6) (x 2 - 5 x + 14) (x + 1) x 5 - 15 x 4 + 57 x 3 - 13 x 2 + 26 x + 112 = 0.
Solution
Let's start by looking at the whole equation (2 x − 1) (x − 6) (x 2 − 5 x + 14) (x + 1) = 0 and finding its roots. To do this, we apply the method of solving equations through factorization. It turns out that the original 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, of which three are linear and one is quadratic. Finding roots: from the first equation x = 1 2, from the second – x = 6, from the third – x = 7 , x = − 2 , from the fourth – x = − 1.
Let's check the obtained roots. It is difficult for us to determine the ODZ in this case, since for this we will have to solve an algebraic equation of the fifth degree. It will be easier to check the condition according to which the denominator of the fraction, which is on the left side of the equation, should not go to zero.
Let’s take turns substituting the roots for the variable x in the expression x 5 − 15 x 4 + 57 x 3 − 13 x 2 + 26 x + 112 and calculate its value:
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 .
The verification carried out allows us to establish that the roots of the original fractional rational equation are 1 2, 6 and − 2 .
Answer: 1 2 , 6 , - 2
Example 9
Find the roots of the fractional rational equation 5 x 2 - 7 x - 1 x - 2 x 2 + 5 x - 14 = 0.
Solution
Let's start working with the equation (5 x 2 − 7 x − 1) (x − 2) = 0. Let's find its roots. It’s easier for us to imagine this equation as a combination of quadratic and linear equations 5 x 2 − 7 x − 1 = 0 And x − 2 = 0.
We use the formula for the roots of a quadratic equation to find the roots. We obtain from the first equation two roots x = 7 ± 69 10, and from the second x = 2.
It will be quite difficult for us to substitute the value of the roots into the original equation to check the conditions. It will be easier to determine the ODZ of the variable x. In this case, the ODZ of the variable x is all numbers except those for which the condition is met x 2 + 5 x − 14 = 0. We get: x ∈ - ∞, - 7 ∪ - 7, 2 ∪ 2, + ∞.
Now let's check whether the roots we found belong to the range of permissible values of the variable x.
The roots x = 7 ± 69 10 belong, therefore, they are the roots of the original equation, and x = 2- does not belong, therefore, it is an extraneous root.
Answer: x = 7 ± 69 10.
Let us examine separately the cases when the numerator of a fractional rational equation of the form p (x) q (x) = 0 contains a number. In such cases, if the numerator contains a number other than zero, then the equation will have no roots. If this number is equal to zero, then the root of the equation will be any number from the ODZ.
Example 10
Solve the fractional rational equation - 3, 2 x 3 + 27 = 0.
Solution
This equation will not have roots, since the numerator of the fraction on the left side of the equation contains a non-zero number. This means that at no value of x will the value of the fraction given in the problem statement be equal to zero.
Answer: no roots.
Example 11
Solve the equation 0 x 4 + 5 x 3 = 0.
Solution
Since the numerator of the fraction contains zero, the solution to the equation will be any value x from the ODZ of the variable x.
Now let's define the ODZ. It will include all values of x for which x 4 + 5 x 3 ≠ 0. 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 this in turn is equivalent to the combination of two equations x 3 = 0 and x + 5 = 0, where these roots are visible. We come to the conclusion that the desired range of acceptable values are any x except x = 0 And x = − 5.
It turns out that the fractional rational equation 0 x 4 + 5 · x 3 = 0 has an infinite number of solutions, which are any numbers other than zero and - 5.
Answer: - ∞ , - 5 ∪ (- 5 , 0 ∪ 0 , + ∞
Now let's talk about fractional rational equations arbitrary type and methods for solving them. They can be written as r(x) = s(x), Where r(x) And s(x)– rational expressions, and at least one of them is fractional. Solving such equations reduces to solving equations of the form p (x) q (x) = 0.
We already know that we can obtain an equivalent equation by transferring an expression from the right side of the equation to the left with the opposite sign. This means that the equation r(x) = s(x) is equivalent to the equation r (x) − s (x) = 0. We have also already discussed ways to convert a rational expression into a rational fraction. Thanks to this, we can easily transform the equation r (x) − s (x) = 0 into an identical rational fraction of the form p (x) q (x) .
So we move from the original fractional rational equation r(x) = s(x) to an equation of the form p (x) q (x) = 0, which we have already learned to solve.
It should be taken into account that when making transitions from r (x) − s (x) = 0 to p(x)q(x) = 0 and then to p(x)=0 we may not take into account the expansion of the range of permissible values of the variable x.
It is quite possible that the original equation r(x) = s(x) and equation p(x)=0 as a result of the transformations they will cease to be equivalent. Then the solution to the equation p(x)=0 can give us roots that will be foreign to r(x) = s(x). In this regard, in each case it is necessary to carry out verification using any of the methods described above.
To make it easier for you to study the topic, we have summarized all the information into an algorithm for solving a fractional rational equation of the form r(x) = s(x):
- we transfer the expression from the right side with the opposite sign and get zero on the right;
- transform the original expression into a rational fraction p (x) q (x) , sequentially performing operations with fractions and polynomials;
- solve the equation p(x)=0;
- we identify extraneous roots by checking their belonging to the ODZ or by substitution into the original equation.
Visually, the chain of actions will look like this:
r (x) = s (x) → r (x) - s (x) = 0 → p (x) q (x) = 0 → p (x) = 0 → elimination EXTERNAL ROOTS
Example 12
Solve the fractional rational equation x x + 1 = 1 x + 1 .
Solution
Let's move on to the equation x x + 1 - 1 x + 1 = 0. Let's transform the fractional rational expression on the left side of the equation to the form p (x) q (x) .
To do this, we will have to reduce rational fractions to a common denominator and simplify the expression:
x x + 1 - 1 x - 1 = x x - 1 (x + 1) - 1 x (x + 1) x (x + 1) = = x 2 - x - 1 - x 2 - x x · (x + 1) = - 2 · x - 1 x · (x + 1)
In order to find the roots of the equation - 2 x - 1 x (x + 1) = 0, we need to solve the equation − 2 x − 1 = 0. We get one root x = - 1 2.
All we have to do is check using any of the methods. Let's look at both of them.
Let's substitute the resulting value into the original equation. We get - 1 2 - 1 2 + 1 = 1 - 1 2 + 1. We have arrived at the correct numerical equality − 1 = − 1 . It means that x = − 1 2 is the root of the original equation.
Now let's check through the ODZ. Let us determine the range of permissible values of the variable x. This will be the entire set of numbers, with the exception of − 1 and 0 (at x = − 1 and x = 0, the denominators of the fractions vanish). The root we obtained x = − 1 2 belongs to ODZ. This means that it is the root of the original equation.
Answer: − 1 2 .
Example 13
Find the roots of the equation x 1 x + 3 - 1 x = - 2 3 · x.
Solution
We are dealing with a fractional rational equation. Therefore, we will act according to the algorithm.
Let's move the expression from the right side to the left with the opposite sign: x 1 x + 3 - 1 x + 2 3 x = 0
Let's carry out the necessary transformations: x 1 x + 3 - 1 x + 2 3 · x = x 3 + 2 · x 3 = 3 · x 3 = x.
We arrive at the equation x = 0. The root of this equation is zero.
Let's check whether this root is extraneous to the original equation. Let's substitute the value into the original equation: 0 1 0 + 3 - 1 0 = - 2 3 · 0. As you can see, the resulting equation makes no sense. This means that 0 is an extraneous root, and the original fractional rational equation has no roots.
Answer: no roots.
If we have not included other equivalent transformations in the algorithm, this does not mean that they cannot be used. The algorithm is universal, but it is designed to help, not limit.
Example 14
Solve the equation 7 + 1 3 + 1 2 + 1 5 - x 2 = 7 7 24
Solution
The easiest way is to solve the given fractional rational equation according to the algorithm. But there is another way. Let's consider it.
Subtract 7 from the right and left sides, we get: 1 3 + 1 2 + 1 5 - x 2 = 7 24.
From this we can conclude that the expression in the denominator of the left side must be equal to the number reciprocal number from the right side, that is, 3 + 1 2 + 1 5 - x 2 = 24 7.
Subtract 3 from both sides: 1 2 + 1 5 - x 2 = 3 7. By analogy, 2 + 1 5 - x 2 = 7 3, from where 1 5 - x 2 = 1 3, and then 5 - x 2 = 3, x 2 = 2, x = ± 2
Let us carry out a check to determine whether the roots found are the roots of the original equation.
Answer: x = ± 2
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