Let's get acquainted with rational and fractional rational equations, give their definition, give examples, and also analyze the most common types of problems.
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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 we have given for rational equations, are equivalent because they talk about 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: integers and fractions. 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 fractionally 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 of standard form.
We have to get 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 type 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 is the degree of an 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 above the fourth degree there are no general root formulas at all. 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
We move the expression from the right side of the record to the left with the 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 reschedule right side equations 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
Entire equations of high degrees appear in problems quite often. There is no need to be afraid of them. You need to be ready to use a non-standard method for solving them, 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. This 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 easier 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 the 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. Determine ADL in in this case It’s difficult for us, 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 set 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 of arbitrary form 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 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 . This 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 on the left side must be equal to the reciprocal of the number on 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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So far we have only solved integer equations with respect to the unknown, that is, equations in which the denominators (if any) did not contain the unknown.
Often you have to solve equations that contain an unknown in the denominators: such equations are called fractional equations.
To solve this equation, we multiply both sides by that is, by the polynomial containing the unknown. Will the new equation be equivalent to this one? To answer the question, let's solve this equation.
Multiplying both sides by , we get:
Solving this equation of the first degree, we find:
So, equation (2) has a single root
Substituting it into equation (1), we get:
This means that it is also a root of equation (1).
Equation (1) has no other roots. In our example, this can be seen, for example, from the fact that in equation (1)
How the unknown divisor must be equal to the dividend 1 divided by the quotient 2, that is
So, equations (1) and (2) have a single root. This means they are equivalent.
2. Let us now solve the following equation:
The simplest common denominator: ; multiply all terms of the equation by it:
After reduction we get:
Let's expand the brackets:
Bringing similar terms, we have:
Solving this equation, we find:
Substituting into equation (1), we get:
On the left side we received expressions that do not make sense.
This means that equation (1) is not a root. It follows that equations (1) and are not equivalent.
In this case, they say that equation (1) has acquired an extraneous root.
Let us compare the solution of equation (1) with the solution of the equations we considered earlier (see § 51). In solving this equation, we had to perform two operations that had not been seen before: first, we multiplied both sides of the equation by an expression containing the unknown (common denominator), and second, we reduced algebraic fractions by factors containing the unknown .
Comparing equation (1) with equation (2), we see that not all values of x that are valid for equation (2) are valid for equation (1).
It is the numbers 1 and 3 that are not acceptable values of the unknown for equation (1), but as a result of the transformation they became acceptable for equation (2). One of these numbers turned out to be a solution to equation (2), but, of course, it cannot be a solution to equation (1). Equation (1) has no solutions.
This example shows that when you multiply both sides of an equation by a factor containing the unknown and cancel algebraic fractions an equation may be obtained that is not equivalent to this one, namely: extraneous roots may appear.
From here we draw the following conclusion. When solving an equation containing an unknown in the denominator, the resulting roots must be checked by substitution into the original equation. Extraneous roots must be discarded.
T. Kosyakova,
School No. 80, Krasnodar
Solving quadratic and fractional rational equations containing parameters
Lesson 4
Lesson topic:
Objective 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
equation 
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
B-1. a) If a=3
, then there are no roots; If 
b) if if a
№
2
, then there are no roots.
B-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 equation 
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 only accepts integer values.
5. For what values of b does the equation 
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 only accepts integer values.
5. For what values of the parameter a does the equation 
has:
a) two roots;
b) the only root?
Answers
B-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. Prerequisites 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.
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 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:
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 them a 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 whole 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 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 the 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 easy, 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 dwell separately 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. At the same time
- 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:
References.
- 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.