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How to solve fractional rational equations. Solving integer and fractionally rational equations

How to solve fractional rational equations. Solving integer and fractionally rational equations

Today we will figure out how to solve fractional rational equations.

Let's see: from the equations

(1) 2x + 5 = 3(8 - x),

(3)

(4)

fractional rational equations are only (2) and (4), while (1) and (3) are entire equations.

I propose to solve equation (4), and then formulate the rule.

Since the equation is fractional, we need to find a common denominator. In this equation, this expression is 6 (x - 12) (x - 6). Then we multiply both sides of the equation by a common denominator:

After reduction, we get the whole equation:

6 (x - 6) 2 - 6 (x - 12) 2 \u003d 5 (x - 12) (x - 6).

Having solved this equation, it is necessary to check whether the obtained roots turn the denominators of the fractions in the original equation to zero.

Expanding the brackets:
6x 2 - 72x + 216 - 6x 2 + 144x - 864 = 5x 2 - 90x + 360, we simplify the equation: 5x 2 - 162x + 1008 = 0.

Finding the roots of the equation
D=6084, √D=78,
x 1 = (162 - 78) / 10 = 84/10 = 8.4 and x 2 = (162 + 78) / 10 = 240/10 = 24.

At x = 8.4 and 24, the common denominator is 6(x - 12)(x - 6) ≠ 0, which means that these numbers are the roots of equation (4).

Answer: 8,4; 24.

Solving the proposed equation, we arrive at the following provisions:

1) We find a common denominator.

2) Multiply both sides of the equation by a common denominator.

3) We solve the resulting whole equation.

4) We check which of the roots turn the common denominator to zero and exclude them from the solution.

Let us now look at an example of how the resulting positions work.

Solve the equation:

1) Common denominator: x 2 - 1

2) We multiply both parts of the equation by a common denominator, we get the whole equation: 6 - 2 (x + 1) \u003d 2 (x 2 - 1) - (x + 4) (x - 1)

3) We solve the equation: 6 - 2x - 2 \u003d 2x 2 - 2 - x 2 - 4x + x + 4

x 2 - x - 2 = 0

x 1 = - 1 and x 2 = 2

4) When x \u003d -1, the common denominator x 2 - 1 \u003d 0. The number -1 is not a root.

For x \u003d 2, the common denominator is x 2 - 1 ≠ 0. The number 2 is the root of the equation.

Answer: 2.

As you can see, our provisions work. Don't be afraid, you will succeed! The most important find the common denominator correctly and do the transformations carefully. We hope that when solving fractional rational equations, you will always get the correct answers. If you have any questions or want to practice solving such equations, sign up for lessons with the author of this article, tutor Valentina Galinevskaya.

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Solution of fractional rational equations

Help Guide

Rational equations are equations in which both the left and right sides are rational expressions.

(Recall: rational expressions are integer and fractional expressions without radicals, including the operations of addition, subtraction, multiplication or division - for example: 6x; (m - n) 2; x / 3y, etc.)

Fractional-rational equations, as a rule, are reduced to the form:

Where P(x) and Q(x) are polynomials.

To solve such equations, multiply both sides of the equation by Q(x), which can lead to the appearance of extraneous roots. Therefore, when solving fractional rational equations, it is necessary to check the found roots.

A rational equation is called an integer, or algebraic, if it does not have a division by an expression containing a variable.

Examples of a whole rational equation:

5x - 10 = 3(10 - x)

3x
-=2x-10
4

If in a rational equation there is a division by an expression containing the variable (x), then the equation is called fractional rational.

An example of a fractional rational equation:

15
x + - = 5x - 17
x

Fractional rational equations are usually solved as follows:

1) find a common denominator of fractions and multiply both parts of the equation by it;

2) solve the resulting whole equation;

3) exclude from its roots those that turn the common denominator of the fractions to zero.

Examples of solving integer and fractional rational equations.

Example 1. Solve the whole equation

x – 1 2x 5x
-- + -- = --.
2 3 6

Solution:

Finding the lowest common denominator. This is 6. Divide 6 by the denominator and multiply the result by the numerator of each fraction. We get an equation equivalent to this one:

3(x - 1) + 4x 5x
------ = --
6 6

Since the denominator is the same on the left and right sides, it can be omitted. Then we have a simpler equation:

3(x - 1) + 4x = 5x.

We solve it by opening brackets and reducing like terms:

3x - 3 + 4x = 5x

3x + 4x - 5x = 3

Example solved.

Example 2. Solve a fractional rational equation

x – 3 1 x + 5
-- + - = ---.
x - 5 x x(x - 5)

We find a common denominator. This is x(x - 5). So:

x 2 – 3x x – 5 x + 5
--- + --- = ---
x(x - 5) x(x - 5) x(x - 5)

Now we get rid of the denominator again, since it is the same for all expressions. We reduce like terms, equate the equation to zero and get a quadratic equation:

x 2 - 3x + x - 5 = x + 5

x 2 - 3x + x - 5 - x - 5 = 0

x 2 - 3x - 10 = 0.

Having solved the quadratic equation, we find its roots: -2 and 5.

Let's check if these numbers are the roots of the original equation.

For x = –2, the common denominator x(x – 5) does not vanish. So -2 is the root of the original equation.

At x = 5, the common denominator vanishes, and two of the three expressions lose their meaning. So the number 5 is not the root of the original equation.

Answer: x = -2

More examples

Example 1

x 1 \u003d 6, x 2 \u003d - 2.2.

Answer: -2.2; 6.

Example 2

Presentation and lesson on the topic: "Rational equations. Algorithm and examples for solving rational equations"

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Introduction to irrational equations

Guys, we learned how to solve quadratic equations. But mathematics is not limited to them. Today we will learn how to solve rational equations. The concept of rational equations is in many ways similar to the concept of rational numbers. Only in addition to numbers, now we have introduced some variable $x$. And thus we get an expression in which there are operations of addition, subtraction, multiplication, division and raising to an integer power.

Let $r(x)$ be rational expression. Such an expression can be a simple polynomial in the variable $x$ or a ratio of polynomials (the operation of division is introduced, as for rational numbers).
The equation $r(x)=0$ is called rational equation.
Any equation of the form $p(x)=q(x)$, where $p(x)$ and $q(x)$ are rational expressions, will also be rational equation.

Consider examples of solving rational equations.

Example 1
Solve the equation: $\frac(5x-3)(x-3)=\frac(2x-3)(x)$.

Solution.
Let's move all expressions to the left side: $\frac(5x-3)(x-3)-\frac(2x-3)(x)=0$.
If ordinary numbers were represented on the left side of the equation, then we would bring two fractions to a common denominator.
Let's do this: $\frac((5x-3)*x)((x-3)*x)-\frac((2x-3)*(x-3))((x-3)*x )=\frac(5x^2-3x-(2x^2-6x-3x+9))((x-3)*x)=\frac(3x^2+6x-9)((x-3) *x)=\frac(3(x^2+2x-3))((x-3)*x)$.
We got the equation: $\frac(3(x^2+2x-3))((x-3)*x)=0$.

A fraction is zero if and only if the numerator of the fraction is zero and the denominator is non-zero. Then separately equate the numerator to zero and find the roots of the numerator.
$3(x^2+2x-3)=0$ or $x^2+2x-3=0$.
$x_(1,2)=\frac(-2±\sqrt(4-4*(-3)))(2)=\frac(-2±4)(2)=1;-3$.
Now let's check the denominator of the fraction: $(x-3)*x≠0$.
The product of two numbers is equal to zero when at least one of these numbers is equal to zero. Then: $x≠0$ or $x-3≠0$.
$x≠0$ or $x≠3$.
The roots obtained in the numerator and denominator do not match. So in response we write down both roots of the numerator.
Answer: $x=1$ or $x=-3$.

If suddenly, one of the roots of the numerator coincided with the root of the denominator, then it should be excluded. Such roots are called extraneous!

Algorithm for solving rational equations:

1. Move all expressions contained in the equation to the left of the equal sign.
2. Convert this part of the equation to algebraic fraction: $\frac(p(x))(q(x))=0$.
3. Equate the resulting numerator to zero, that is, solve the equation $p(x)=0$.
4. Equate the denominator to zero and solve the resulting equation. If the roots of the denominator coincided with the roots of the numerator, then they should be excluded from the answer.

Example 2
Solve the equation: $\frac(3x)(x-1)+\frac(4)(x+1)=\frac(6)(x^2-1)$.

Solution.
We will solve according to the points of the algorithm.
1. $\frac(3x)(x-1)+\frac(4)(x+1)-\frac(6)(x^2-1)=0$.
2. $\frac(3x)(x-1)+\frac(4)(x+1)-\frac(6)(x^2-1)=\frac(3x)(x-1)+\ frac(4)(x+1)-\frac(6)((x-1)(x+1))= \frac(3x(x+1)+4(x-1)-6)((x -1)(x+1))=$ $=\frac(3x^2+3x+4x-4-6)((x-1)(x+1))=\frac(3x^2+7x- 10)((x-1)(x+1))$.
$\frac(3x^2+7x-10)((x-1)(x+1))=0$.
3. Equate the numerator to zero: $3x^2+7x-10=0$.
$x_(1,2)=\frac(-7±\sqrt(49-4*3*(-10)))(6)=\frac(-7±13)(6)=-3\frac( 1)(3);1$.
4. Equate the denominator to zero:
$(x-1)(x+1)=0$.
$x=1$ and $x=-1$.
One of the roots $x=1$ coincided with the root of the numerator, then we do not write it down in response.
Answer: $x=-1$.

It is convenient to solve rational equations using the change of variables method. Let's demonstrate it.

Example 3
Solve the equation: $x^4+12x^2-64=0$.

Solution.
We introduce a replacement: $t=x^2$.
Then our equation will take the form:
$t^2+12t-64=0$ is an ordinary quadratic equation.
$t_(1,2)=\frac(-12±\sqrt(12^2-4*(-64)))(2)=\frac(-12±20)(2)=-16; 4$.
Let's introduce an inverse replacement: $x^2=4$ or $x^2=-16$.
The roots of the first equation are a pair of numbers $x=±2$. The second one has no roots.
Answer: $x=±2$.

Example 4
Solve the equation: $x^2+x+1=\frac(15)(x^2+x+3)$.
Solution.
Let's introduce a new variable: $t=x^2+x+1$.
Then the equation will take the form: $t=\frac(15)(t+2)$.
Next, we will act according to the algorithm.
1. $t-\frac(15)(t+2)=0$.
2. $\frac(t^2+2t-15)(t+2)=0$.
3. $t^2+2t-15=0$.
$t_(1,2)=\frac(-2±\sqrt(4-4*(-15)))(2)=\frac(-2±\sqrt(64))(2)=\frac( -2±8)(2)=-5; 3$.
4. $t≠-2$ - the roots do not match.
We introduce a reverse substitution.
$x^2+x+1=-5$.
$x^2+x+1=3$.
Let's solve each equation separately:
$x^2+x+6=0$.
$x_(1,2)=\frac(-1±\sqrt(1-4*(-6)))(2)=\frac(-1±\sqrt(-23))(2)$ - no roots.
And the second equation: $x^2+x-2=0$.
The roots of this equation will be the numbers $x=-2$ and $x=1$.
Answer: $x=-2$ and $x=1$.

Example 5
Solve the equation: $x^2+\frac(1)(x^2) +x+\frac(1)(x)=4$.

Solution.
We introduce a replacement: $t=x+\frac(1)(x)$.
Then:
$t^2=x^2+2+\frac(1)(x^2)$ or $x^2+\frac(1)(x^2)=t^2-2$.
We got the equation: $t^2-2+t=4$.
$t^2+t-6=0$.
The roots of this equation are the pair:
$t=-3$ and $t=2$.
Let's introduce the reverse substitution:
$x+\frac(1)(x)=-3$.
$x+\frac(1)(x)=2$.
We will decide separately.
$x+\frac(1)(x)+3=0$.
$\frac(x^2+3x+1)(x)=0$.
$x_(1,2)=\frac(-3±\sqrt(9-4))(2)=\frac(-3±\sqrt(5))(2)$.
Let's solve the second equation:
$x+\frac(1)(x)-2=0$.
$\frac(x^2-2x+1)(x)=0$.
$\frac((x-1)^2)(x)=0$.
The root of this equation is the number $x=1$.
Answer: $x=\frac(-3±\sqrt(5))(2)$, $x=1$.

Tasks for independent solution

Solve Equations:

1. $\frac(3x+2)(x)=\frac(2x+3)(x+2)$.

2. $\frac(5x)(x+2)-\frac(20)(x^2+2x)=\frac(4)(x)$.
3. $x^4-7x^2-18=0$.
4. $2x^2+x+2=\frac(8)(2x^2+x+4)$.
5. $(x+2)(x+3)(x+4)(x+5)=3$.

Fractional equations. ODZ.

Attention!
There are additional
material in Special Section 555.
For those who strongly "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. The last view remains fractional equations. Or they are also called much more solid - fractional rational equations. This is the same.

Fractional equations.

As the name implies, these equations necessarily contain fractions. But not just fractions, but fractions that have unknown in the denominator. At least in one. For example:

Let me remind you, if in the denominators only numbers, these are linear equations.

How to decide fractional equations? First of all, get rid of the fractions! After that, the equation, most often, turns into a linear or quadratic one. And then we know what to do... In some cases, it can turn into an identity, like 5=5 or an incorrect expression, like 7=2. But this rarely happens. Below I will mention it.

But how to get rid of fractions!? Very simple. Applying all the same identical transformations.

We need to multiply the whole equation by the same expression. So that all denominators decrease! Everything will immediately become easier. I explain with an example. Let's say we need to solve the equation:

How were they taught in elementary school? We transfer everything in one direction, reduce it to a common denominator, etc. Forget how bad dream! This is what you need to do when you add or subtract fractional expressions. Or work with inequalities. And in equations, we immediately multiply both parts by an expression that will give us the opportunity to reduce all denominators (ie, in essence, by a common denominator). And what is this expression?

On the left side, to reduce the denominator, you need to multiply by x+2. And on the right, multiplication by 2 is required. So, the equation must be multiplied by 2(x+2). We multiply:

This is the usual multiplication of fractions, but I will write in detail:

Please note that I am not opening the parenthesis yet. (x + 2)! So, in its entirety, I write it:

On the left side, it is reduced entirely (x+2), and in the right 2. As required! After reduction we get linear the equation:

Anyone 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 can be written:

And again we get rid of what we don’t really like - from fractions.

We see that to reduce the denominator with x, it is necessary to multiply the fraction by (x - 2). And units are not a hindrance to us. Well, let's multiply. All left side and all right side:

Brackets again (x - 2) I don't reveal. 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 cut (x - 2) and we get the equation without any fractions, in a ruler!

And now we open the brackets:

We give similar ones, transfer everything to the left side and get:

But before that, we will learn to solve other problems. For interest. Those rakes, 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. Learning - with interest!)

you can get acquainted with functions and derivatives.

We continue talking about solution of equations. In this article, we will focus on rational equations and principles for solving rational equations with one variable. First, let's figure out what kind of equations are called rational, give a definition of integer rational and fractional rational equations, and give examples. Further, we will obtain algorithms for solving rational equations, and, of course, consider the solutions of typical examples with all the necessary explanations.

Page navigation.

Based on the sounded 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 can be seen that rational equations, as well as equations of other types, can be either with one variable, or with two, three, etc. variables. In the following paragraphs, we will talk about solving rational equations in one variable. Solving equations with two variables and their large number deserve special attention.

In addition to dividing rational equations by the number of unknown variables, they are also divided into integer and fractional. Let us give the corresponding definitions.

Definition.

The rational equation is called whole, if both its left and right parts 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 integer 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 are entire rational equations, both of their parts are integer expressions. A and x:(5 x 3 +y 2)=3:(x−1):5 are examples of fractional rational equations.

Concluding this paragraph, let us pay attention to the fact that linear equations and quadratic equations known by this moment are entire rational equations.

Solving entire equations

One of the main approaches to solving entire equations is their reduction to equivalent 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 get zero on the right side;
after that, on the left side of the equation, the resulting standard form.

The result is an algebraic equation that is equivalent to the original whole equation. So in the simplest cases, the solution of entire equations is reduced to the solution of linear or quadratic equations, and in the general case - to the solution of an algebraic equation of degree n. For clarity, let's analyze the solution of 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 whole 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 polynomial of the standard form by doing 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, the solution of the original integer equation is reduced to the solution of the quadratic equation x 2 −5·x−6=0 .

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 by the formula of the roots of the quadratic equation:

To be completely sure, let's do 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, so 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, whence, 0=0 . For x=−1, the original equation also turned into a true numerical equality, therefore, x=−1 is also the root of the equation.

Answer:

6 , −1 .

Here it should also be noted that the term “power of an entire equation” is associated with the representation of an entire equation in the form of an algebraic equation. We give the corresponding definition:

Definition.

The degree of the whole equation call the degree of an algebraic equation equivalent to it.

According to this definition, the whole equation from the previous example has the second degree.

On this one could finish with the solution of entire rational equations, if not for one but .... As is known, the solution of algebraic equations of degree higher than the second is associated with significant difficulties, and for equations of degree higher than the fourth, there are no general formulas for roots at all. Therefore, to solve entire equations of the third, fourth and more high degrees often have to resort to other methods of solution.

In such cases, sometimes the approach to solving entire rational equations based on factorization method. At the same time, the following algorithm is followed:

first they seek to have zero on the right side of the equation, for 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 you to go to a set of several simpler equations.

The above algorithm for solving the whole 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 . It is quite obvious here that it is not advisable to transform the left 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, whose solution is difficult.

On the other hand, it is obvious that x 2 −10·x+13 can be found on the left side of the resulting equation, thereby representing 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 the known root formulas through the discriminant is not difficult, the roots are equal. They are the desired roots of the original equation.

Answer:

It is also useful for solving entire rational equations. method for introducing a new variable. In some cases, it allows one to pass to equations whose degree is lower than the degree of the original integer 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.

It is easy to see here that you can introduce a new variable y and replace the expression x 2 +3 x with it. Such a replacement leads us to the whole equation (y+1) 2 +10=−2 (y−4) , which after transferring the expression −2 (y−4) to the left side and subsequent transformation of the expression formed there, reduces to 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 found based on the inverse theorem of Vieta's theorem.

Now let's move on to the second part of the method of introducing a new variable, that is, to making a reverse substitution. 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 . According to the formula of the roots of the 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 ready to look for a non-standard method or an artificial technique for solving them.

Solution of fractionally rational equations

First, it will be useful to understand how to solve fractionally rational equations of the form , where p(x) and q(x) are rational integer expressions. And then we will show how to reduce the solution of the remaining fractionally rational equations to the solution of equations of the indicated form.

One of the approaches 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 not defined), 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, the solution of the equation is reduced to the fulfillment of two conditions p(x)=0 and q(x)≠0 .

This conclusion is consistent with the following algorithm for solving a fractionally rational equation. To solve a fractional rational equation of the form

solve the whole rational equation p(x)=0 ;
and check whether the condition q(x)≠0 is satisfied for each found root, while

if true, then this root is the root of the original equation;
if not, then this root is extraneous, that is, it is not the root of the original equation.

Let's analyze an example of using the voiced algorithm when solving a fractional rational equation.

Example.

Find the roots of the equation.

Solution.

This is a fractionally rational equation of the form , where p(x)=3 x−2 , q(x)=5 x 2 −2=0 .

According to the algorithm for solving fractionally rational equations of this kind, we first need to solve the equation 3·x−2=0 . it linear equation, whose root is x=2/3 .

It remains to check for this root, that is, to check whether it satisfies the condition 5·x 2 −2≠0 . We substitute the number 2/3 instead of x into the expression 5 x 2 −2, we get . The condition is met, so x=2/3 is the root of the original equation.

Answer:

2/3 .

The solution of a fractional rational equation can be approached from a slightly different position. This equation is equivalent to the whole equation p(x)=0 on the variable x of the original equation. That is, you can follow this algorithm for solving a fractionally rational equation :

solve the equation p(x)=0 ;
find ODZ variable x ;
take the roots belonging to the region of admissible 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 an 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 x (x+3)≠0 , whence x≠0 , x≠−3 .

It remains to check whether the roots found at the first step are included in the ODZ. Obviously yes. Therefore, the original fractionally rational equation has two roots.

Answer:

Note that this approach is more profitable than the first one if the ODZ is easily found, and it 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 efforts, and it is easier to exclude extraneous roots from the ODZ.

In other cases, when solving the equation, especially when the roots of the equation p(x)=0 are integers, it is more advantageous to use the first of the above algorithms. That is, it is advisable to immediately find the roots of the whole equation p(x)=0 , and then check whether the condition q(x)≠0 is satisfied for them, and not find the ODZ, and then solve the equation p(x)=0 on this ODZ . This is due to the fact that in such cases it is usually easier to make a check than to find the ODZ.

Consider the solution of two examples to illustrate the stipulated nuances.

Example.

Find the roots of the equation.

Solution.

First we find the roots of the whole equation (2 x−1) (x−6) (x 2 −5 x+14) (x+1)=0, compiled 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 the 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 them to see if the denominator of the fraction located on the left side of the original equation does not vanish, and it is not so easy to determine the ODZ, since this will have to solve an algebraic equation of the fifth degree. Therefore, we will refuse to find the ODZ in favor of checking the roots. To do this, we substitute them in turn 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 fractionally 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 we find the roots of the equation (5x2 −7x−1)(x−2)=0. This equation is equivalent to a set of two equations: the square 5·x 2 −7·x−1=0 and the linear x−2=0 . According to the formula of the roots of the quadratic equation, we find two roots, and from the second equation we have x=2.

Checking if the denominator does not vanish at the found values of x is rather unpleasant. And to determine the range of acceptable values of the variable x in the original equation is quite simple. Therefore, we will act through the ODZ.

In our case, the ODZ of the variable x of the original fractional rational equation is made up of all numbers, except for 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 conclude about the ODZ: it is made up of all x such that .

It remains to check whether the found roots and x=2 belong to the region of admissible values. The roots - belong, therefore, they are the 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 cases where a fractional rational equation of the form contains a number in the numerator, that is, when p (x) is represented by some number. Wherein

if this number is different from zero, then the equation has no roots, since the fraction is zero if and only if its numerator is zero;
if this number is zero, then the root of the equation is any number from the ODZ.

Example.

Solution.

Since there is a non-zero number in the numerator of the fraction on the left side of the equation, for no x can the value of this fraction 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 is 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 DPV of this variable.

It remains to determine this range of acceptable values. It includes all such values x for which x 4 +5 x 3 ≠0. The solutions of the equation x 4 +5 x 3 \u003d 0 are 0 and −5, since this equation is equivalent to the equation x 3 (x + 5) \u003d 0, and it, in turn, is equivalent to the combination of two equations x 3 \u003d 0 and x +5=0 , from where these roots are visible. Therefore, the desired range of acceptable values are any x , except for x=0 and x=−5 .

Thus, a fractionally rational equation has infinitely many solutions, which are any numbers except zero and minus five.

Answer:

Finally, it's time to talk about solving arbitrary fractional rational equations. 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, we say that their solution is reduced to solving equations of the form already familiar to us.

It is known that the transfer of a term from one part of the equation to another with the opposite sign leads to an equivalent equation, so the equation r(x)=s(x) is equivalent to the equation r(x)−s(x)=0 .

We also know that any can be identically equal to this expression. 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 go 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 allowable values of the variable x may expand.

Therefore, the original equation r(x)=s(x) and the equation p(x)=0 , which we arrived at, may not be equivalent, 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) . It is possible to identify and not include extraneous roots in the answer, either by checking, or by checking their belonging to the ODZ of the original equation.

We summarize this information in algorithm for solving a fractional rational equation r(x)=s(x). To solve the fractional rational equation r(x)=s(x) , one must

Get zero on the right by moving the expression from the right side with the opposite sign.
Perform actions with fractions and polynomials on the left side of the equation, thereby converting it into a rational fraction of the form.
Solve the equation p(x)=0 .
Identify and exclude 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 go through the solutions of several examples with a detailed explanation of the solution in order to clarify the given block of information.

Example.

Solve a fractional rational equation.

Solution.

We will act in accordance with the just obtained solution algorithm. And first we transfer the terms from the right side of the equation to the left side, as a result we pass 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 perform the reduction of 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 . Find x=−1/2 .

It remains to check whether the found number −1/2 is an extraneous root of the original equation. To do this, you can check or find the ODZ variable x of the original equation. Let's demonstrate both approaches.

Let's start with a check. We substitute the number −1/2 instead of the variable x into the original equation, we get , which is the same, −1=−1. The substitution gives the correct numerical equality, therefore, x=−1/2 is the root of the original equation.

Now we will show how the last step of the algorithm is performed through the ODZ. The range of admissible values of the original equation is the set of all numbers except −1 and 0 (when x=−1 and x=0, the denominators of fractions vanish). The root x=−1/2 found at the previous step belongs to the ODZ, therefore, x=−1/2 is the root of the original equation.

Answer:

−1/2 .

Let's consider another example.

Example.

Find the roots of the equation.

Solution.

We need to solve a fractionally rational equation, let's go through all the steps of the algorithm.

First, we transfer 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 if the root found is not an outside one for the original fractionally rational equation. When it is substituted into the original equation, the expression is obtained. Obviously, it does not make sense, since 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 the equation . From this we can conclude that the expression in the denominator of the left side must be equal to from the right side, that is, . Now we subtract from both parts of the triple: . By analogy, from where, and further.

The check shows that both found roots are the roots of the original fractional rational equation.

Answer:

Bibliography.

Algebra: textbook for 8 cells. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
Mordkovich A. G. Algebra. 8th grade. At 2 pm Part 1. A textbook for students of educational institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemozina, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
Algebra: Grade 9: textbook. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.