Let's continue talking about solving equations. In this article we will go into detail about rational equations and principles of solution 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, consider solutions typical examples with all necessary explanations.
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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 points 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 parts are integers 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 left side With opposite sign to get zero on the right side;
- after this, on the left side of the equation the resulting standard view.
The result is algebraic equation, which is equivalent to the original integer equation. So in the most simple cases solving entire equations is reduced to solving linear or quadratic equations, and in general case– to solve 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 true numerical equality, 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 higher than the second is associated with significant difficulties, and for equations of degree higher than the fourth there is no general formulas roots. Therefore, to solve entire equations of the third, fourth and more high degrees Often you have 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 by known formulas roots through the discriminant is not difficult, the roots are equal. They are the desired roots of the original equation.
Answer:
Also useful for solving entire rational equations method for introducing a new variable. In some cases, it allows you to move to equations whose degree is lower than the degree of the original whole equation.
Example.
Find the real roots of a rational equation (x 2 +3 x+1) 2 +10=−2 (x 2 +3 x−4).
Solution.
Reducing this entire rational equation to an algebraic equation is, to put it mildly, not a very good idea, since in this case we will come to the need to solve a fourth-degree equation that does not have rational roots. Therefore, you will have to look for another solution.
Here it is easy to see that you can introduce a new variable y and replace the expression x 2 +3·x with it. This replacement leads us to the whole equation (y+1) 2 +10=−2·(y−4) , which, after moving the expression −2·(y−4) to the left side and subsequent transformation of the expression formed there, is reduced to a quadratic equation y 2 +4·y+3=0. The roots of this equation y=−1 and y=−3 are easy to find, for example, they can be selected based on the theorem inverse to Vieta’s theorem.
Now we move on to the second part of the method of introducing a new variable, that is, to performing a reverse replacement. After performing the reverse substitution, we obtain two equations x 2 +3 x=−1 and x 2 +3 x=−3, which can be rewritten as x 2 +3 x+1=0 and x 2 +3 x+3 =0 . Using the formula for the roots of a quadratic equation, we find the roots of the first equation. And the second quadratic equation doesn't have 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 non-standard method or artificial reception to solve 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 of the approaches to solving the equation is based on the following statement: 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 equal to zero, that 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 area acceptable values, - they are the desired roots of the original fractional rational equation.
For example, let's solve a fractional rational equation using this algorithm.
Example.
Solve the equation.
Solution.
First, we solve the quadratic equation x 2 −2·x−11=0. Its roots can be calculated using the root formula for the even second coefficient, we have D 1 =(−1) 2 −1·(−11)=12, And .
Secondly, we find the ODZ of the variable x for the original equation. It consists of all numbers for which x 2 +3·x≠0, which is the same as x·(x+3)≠0, whence x≠0, x≠−3.
It remains to check whether the roots found in the first step are included in the ODZ. Obviously yes. Therefore, the original fractional rational equation has two roots.
Answer:
Note that this approach is more profitable than the first if the ODZ is easy to find, and is especially beneficial if the roots of the equation p(x) = 0 are irrational, for example, or rational, but with a rather large numerator and/or denominator, for example, 127/1101 and −31/59. This is due to the fact that in such cases, checking the condition q(x)≠0 will require significant computational effort, and it is easier to exclude extraneous roots using the ODZ.
In other cases, when solving the equation, especially when the roots of the equation p(x) = 0 are integers, it is more profitable to use the first of the given algorithms. That is, it is advisable to immediately find the roots of the entire equation p(x)=0, and then check whether the condition q(x)≠0 is satisfied for them, rather than finding the ODZ, and then solving the equation p(x)=0 on this ODZ . This is due to the fact that in such cases it is usually easier to check than to find DZ.
Let us consider the solution of two examples to illustrate the specified nuances.
Example.
Find the roots of the equation.
Solution.
First, let's find the roots of the whole equation (2 x−1) (x−6) (x 2 −5 x+14) (x+1)=0, composed using the numerator of the fraction. The left side of this equation is a product, and the right side is zero, therefore, according to the method of solving equations through factorization, this equation is equivalent to a set of four equations 2 x−1=0 , x−6=0 , x 2 −5 x+ 14=0 , x+1=0 . Three of these equations are linear and one is quadratic; we can solve them. From the first equation we find x=1/2, from the second - x=6, from the third - x=7, x=−2, from the fourth - x=−1.
With the roots found, it is quite easy to check whether the denominator of the fraction on the left side of the original equation vanishes, but determining the ODZ, on the contrary, is not so simple, since for this you will have to solve an algebraic equation of the fifth degree. Therefore, let's give up finding ODZ in favor of checking the roots. To do this, we substitute them one by one instead of the variable x in the expression x 5 −15 x 4 +57 x 3 −13 x 2 +26 x+112, obtained after substitution, and compare them with zero: (1/2) 5 −15·(1/2) 4 + 57·(1/2) 3 −13·(1/2) 2 +26·(1/2)+112= 1/32−15/16+57/8−13/4+13+112=
122+1/32≠0
;
6 5 −15·6 4 +57·6 3 −13·6 2 +26·6+112= 448≠0
;
7 5 −15·7 4 +57·7 3 −13·7 2 +26·7+112=0;
(−2) 5 −15·(−2) 4 +57·(−2) 3 −13·(−2) 2 + 26·(−2)+112=−720≠0 ;
(−1) 5 −15·(−1) 4 +57·(−1) 3 −13·(−1) 2 + 26·(−1)+112=0 .
Thus, 1/2, 6 and −2 are the desired roots of the original fractional rational equation, and 7 and −1 are extraneous roots.
Answer:
1/2 , 6 , −2 .
Example.
Find the roots of a fractional rational equation.
Solution.
First, let's find the roots of the equation (5 x 2 −7 x−1) (x−2)=0. This equation is equivalent to a set of two equations: square 5 x 2 −7 x−1=0 and linear x−2=0. Using the formula for the roots of a quadratic equation, we find two roots, and from the second equation we have x=2.
Checking whether the denominator goes to zero at the found values of x is quite unpleasant. And determining the range of permissible values of the variable x in the original equation is quite simple. Therefore, we will act through ODZ.
In our case, the ODZ of the variable x of the original fractional rational equation consists of all numbers except those for which the condition x 2 +5·x−14=0 is satisfied. The roots of this quadratic equation are x=−7 and x=2, from which we draw a conclusion about the ODZ: it consists of all x such that .
It remains to check whether the found roots and x=2 belong to the range of acceptable values. The roots belong, therefore, they are roots of the original equation, and x=2 does not belong, therefore, it is an extraneous root.
Answer:
It will also be useful to separately dwell on the cases when in a fractional rational equation of the form there is a number in the numerator, that is, when p(x) is represented by some number. Wherein
- if this number is non-zero, then the equation has no roots, since a fraction is equal to zero if and only if its numerator is equal to zero;
- if this number is zero, then the root of the equation is any number from the ODZ.
Example.
Solution.
Since the numerator of the fraction on the left side of the equation contains a non-zero number, then for any x the value of this fraction cannot be equal to zero. Therefore, this equation has no roots.
Answer:
no roots.
Example.
Solve the equation.
Solution.
The numerator of the fraction on the left side of this fractional rational equation contains zero, so the value of this fraction is zero for any x for which it makes sense. In other words, the solution to this equation is any value of x from the ODZ of this variable.
It remains to determine this range of acceptable values. It includes all values of x for which x 4 +5 x 3 ≠0. The solutions to the equation x 4 +5 x 3 =0 are 0 and −5, since this equation is equivalent to the equation x 3 (x+5)=0, and it in turn is equivalent to the combination of two equations x 3 =0 and x +5=0, from where these roots are visible. Therefore, the desired range of acceptable values is any x except x=0 and x=−5.
Thus, a fractional rational equation has infinitely many solutions, which are any numbers except zero and minus five.
Answer:
Finally, it's time to talk about solving fractional rational equations arbitrary type. 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 perform a cast rational fractions To common denominator and simplify the resulting expression: . So we come to the equation.
On next stage we need to solve the equation −2·x−1=0. We find x=−1/2.
It remains to check whether the found number −1/2 is not an extraneous root of the original equation. To do this, you can check or find the VA of the variable x of the original equation. Let's demonstrate both approaches.
Let's start with checking. We substitute the number −1/2 into the original equation instead of the variable x, and we get the same thing, −1=−1. The substitution gives the correct numerical equality, so x=−1/2 is the root of the original equation.
Now we will show how the last point of the algorithm is performed through ODZ. The range of permissible values of the original equation is the set of all numbers except −1 and 0 (at x=−1 and x=0 the denominators of the fractions vanish). The root x=−1/2 found in the previous step belongs to the ODZ, therefore, x=−1/2 is the root of the original equation.
Answer:
−1/2 .
Let's look at another example.
Example.
Find the roots of the equation.
Solution.
We need to solve a fractional rational equation, let's go through all the steps of the algorithm.
First, we move the term from the right side to the left, we get .
Secondly, we transform the expression formed on the left side: . As a result, we arrive at the equation x=0.
Its root is obvious - it is zero.
At the fourth step, it remains to find out whether the found root is extraneous to the original fractional rational equation. When it is substituted into the original equation, the expression is obtained. Obviously, it doesn't make sense because it contains division by zero. Whence we conclude that 0 is an extraneous root. Therefore, the original equation has no roots.
7, which leads to Eq. From this we can conclude that the expression in the denominator of the left side must be equal to that of the right side, that is, . Now we subtract from both sides of the triple: . By analogy, from where, and further.
The check shows that both roots found are roots of the original fractional rational equation.
Answer:
Bibliography.
- Algebra: textbook for 8th grade. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
- Mordkovich A. G. Algebra. 8th grade. At 2 p.m. Part 1. Textbook for students educational 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.
Presentation and lesson on the topic: "Rational equations. Algorithm and examples of 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 only. Today we will learn how to solve rational equations. The concept of rational equations is in many ways similar to the concept rational numbers. Only in addition to numbers, now we have introduced some variable $x$. And thus we get an expression in which the operations of addition, subtraction, multiplication, division and raising to an integer power are present.Let $r(x)$ be rational expression. Such an expression can be a simple polynomial in the variable $x$ or a ratio of polynomials (a division operation 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.
Let's look at examples of solving rational equations.
Example 1.Solve the equation: $\frac(5x-3)(x-3)=\frac(2x-3)(x)$.
Solution.
Let's move all the expressions to the left side: $\frac(5x-3)(x-3)-\frac(2x-3)(x)=0$.
If the left side of the equation were represented regular numbers, 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 equal to zero if and only if the numerator of the fraction is zero and the denominator is non-zero. Then we 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 coincide. So we write down both roots of the numerator in the answer.
Answer: $x=1$ or $x=-3$.
If suddenly one of the roots of the numerator coincides with the root of the denominator, then it should be excluded. Such roots are called extraneous!
Algorithm for solving rational equations:
1. Transfer all expressions contained in the equation to left side from 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 coincide 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.
Let's 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 $x=1$ coincides with the root of the numerator, then we do not write it down in the answer.
Answer: $x=-1$.
It is convenient to solve rational equations using the change of variables method. Let's demonstrate this.
Example 3.
Solve the equation: $x^4+12x^2-64=0$.
Solution.
Let's introduce the replacement: $t=x^2$.
Then our equation will take the form:
$t^2+12t-64=0$ - ordinary quadratic equation.
$t_(1,2)=\frac(-12±\sqrt(12^2-4*(-64)))(2)=\frac(-12±20)(2)=-16; $4.
Let's introduce the reverse substitution: $x^2=4$ or $x^2=-16$.
The roots of the first equation are a pair of numbers $x=±2$. The second thing is that it 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 proceed 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 coincide.
Let's 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: $x^2+x-2=0$.
Roots given equation there will be 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.
Let's introduce the 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'll 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: $x=\frac(-3±\sqrt(5))(2)$, $x=1$.
Problems to solve independently
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. $(x+2)(x+3)(x+4)(x+5)=3$.
In this article I will show you algorithms for solving seven types of rational equations, which can be reduced to quadratic by changing variables. In most cases, the transformations that lead to replacement are very non-trivial, and it is quite difficult to guess about them on your own.
For each type of equation, I will explain how to make a change of variable in it, and then show a detailed solution in the corresponding video tutorial.
You have the opportunity to continue solving the equations yourself, and then check your solution with the video lesson.
So, let's begin.
1 . (x-1)(x-7)(x-4)(x+2)=40
Note that on the left side of the equation there is a product of four brackets, and on the right side there is a number.
1. Let's group the brackets by two so that the sum of the free terms is the same.
2. Multiply them.
3. Let's introduce a change of variable.
In our equation, we will group the first bracket with the third, and the second with the fourth, since (-1)+(-4)=(-7)+2:
At this point the variable replacement becomes obvious:
We get the equation 
Answer: 
2 .
An equation of this type is similar to the previous one with one difference: on the right side of the equation is the product of the number and . And it is solved in a completely different way:
1. We group the brackets by two so that the product of the free terms is the same.
2. Multiply each pair of brackets.
3. We take x out of each factor.
4. Divide both sides of the equation by .
5. We introduce a change of variable.
In this equation, we group the first bracket with the fourth, and the second with the third, since:
Note that in each bracket the coefficient at and the free term are the same. Let's take a factor out of each bracket:
Since x=0 is not a root of the original equation, we divide both sides of the equation by . We get:
We get the equation: 
Answer: 
3
. 
Note that the denominators of both fractions are square trinomials, for 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 indicate identical expressions. That is, on the left side of a homogeneous equation there is a sum of monomials having same degree(V in this case the degree of the monomials is 2), and there is no free term.
To solve 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 get biquadratic equation relative to 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:
An integer expression is a mathematical expression made up of numbers and literal variables using the operations of addition, subtraction and multiplication. Integers also include expressions that involve division by any number other than zero.
The concept of a fractional rational expression
A fractional expression is a mathematical expression that, in addition to the operations of addition, subtraction and multiplication performed with numbers and letter variables, as well as division by a number not equal to zero, also contains division into expressions with letter variables.
Rational expressions are all whole and fractional expressions. Rational equations are equations in which the left and right sides are rational expressions. If in a rational equation the left and right sides are integer expressions, then such a rational equation is called an integer.
If in a rational equation the left or right sides are fractional expressions, then such a rational equation is called fractional.
Examples of fractional rational expressions
1. x-3/x = -6*x+19
2. (x-4)/(2*x+5) = (x+7)/(x-2)
3. (x-3)/(x-5) + 1/x = (x+5)/(x*(x-5))
Scheme for solving a fractional rational equation
1. Find the common denominator of all fractions that are included in the equation.
2. Multiply both sides of the equation by a common denominator.
3. Solve the resulting whole equation.
4. Check the roots and exclude those that make the common denominator vanish.
Since we are solving fractional rational equations, there will be variables in the denominators of the fractions. This means that they will be a common denominator. And in the second point of the algorithm we multiply by a common denominator, then extraneous roots may appear. At which the common denominator will be equal to zero, which means multiplying by it will be meaningless. Therefore, at the end it is necessary to check the obtained roots.
Let's look at an example:
Solve the fractional rational equation: (x-3)/(x-5) + 1/x = (x+5)/(x*(x-5)).
We will stick to general scheme: Let's first find the common denominator of all fractions. We get x*(x-5).
Multiply each fraction by a common denominator and write the resulting whole equation.
(x-3)/(x-5) * (x*(x-5))= x*(x+3);
1/x * (x*(x-5)) = (x-5);
(x+5)/(x*(x-5)) * (x*(x-5)) = (x+5);
x*(x+3) + (x-5) = (x+5);
Let us simplify the resulting equation. We get:
x^2+3*x + x-5 - x - 5 =0;
x^2+3*x-10=0;
We get a simple reduced quadratic equation. We solve it with any of known methods, we get the roots x=-2 and x=5.
Now we check the obtained solutions:
Substitute the numbers -2 and 5 into the common denominator. At x=-2 the common denominator x*(x-5) does not vanish, -2*(-2-5)=14. This means that the number -2 will be the root of the original fractional rational equation.
When x=5 the common denominator x*(x-5) becomes equal to zero. Therefore, this number is not the root of the original fractional rational equation, since there will be a division by zero.
\(\bullet\) A rational equation is an equation represented in the form \[\dfrac(P(x))(Q(x))=0\] where \(P(x), \Q(x)\) - polynomials (the sum of “X’s” in various powers, multiplied by various numbers).
The expression on the left side of the equation is called a rational expression.
The EA (range of acceptable values) of a rational equation is all the values of \(x\) at which the denominator does NOT vanish, that is, \(Q(x)\ne 0\) .
\(\bullet\) For example, equations \[\dfrac(x+2)(x-3)=0,\qquad \dfrac 2(x^2-1)=3, \qquad x^5-3x=2\] are rational equations.
In the first ODZ equation– these are all \(x\) such that \(x\ne 3\) (write \(x\in (-\infty;3)\cup(3;+\infty)\)); in the second equation – these are all \(x\) such that \(x\ne -1; x\ne 1\) (write \(x\in (-\infty;-1)\cup(-1;1)\cup(1;+\infty)\)); and in the third equation there are no restrictions on the ODZ, that is, the ODZ is all \(x\) (they write \(x\in\mathbb(R)\)). \(\bullet\) Theorems:
1) The product of two factors is equal to zero if and only if one of them is equal to zero, and the other does not lose meaning, therefore, the equation \(f(x)\cdot g(x)=0\) is equivalent to the system \[\begin(cases) \left[ \begin(gathered)\begin(aligned) &f(x)=0\\ &g(x)=0 \end(aligned) \end(gathered) \right.\\ \ text(ODZ equations)\end(cases)\] 2) A fraction is equal to zero if and only if the numerator is equal to zero and the denominator is not equal to zero, therefore, the equation \(\dfrac(f(x))(g(x))=0\) is equivalent to a system of equations \[\begin(cases) f(x)=0\\ g(x)\ne 0 \end(cases)\]\(\bullet\) Let's look at a few examples.
1) Solve the equation \(x+1=\dfrac 2x\) . Let us find the ODZ of this equation - this is \(x\ne 0\) (since \(x\) is in the denominator).
This means that the ODZ can be written as follows: .
Let's move all the terms into one part and bring them to a common denominator: \[\dfrac((x+1)\cdot x)x-\dfrac 2x=0\quad\Leftrightarrow\quad \dfrac(x^2+x-2)x=0\quad\Leftrightarrow\quad \begin( cases) x^2+x-2=0\\x\ne 0\end(cases)\] The solution to the first equation of the system will be \(x=-2, x=1\) . We see that both roots are non-zero. Therefore, the answer is: \(x\in \(-2;1\)\) .
2) Solve the equation \(\left(\dfrac4x - 2\right)\cdot (x^2-x)=0\). Let's find the ODZ of this equation. We see that the only value of \(x\) for which the left side does not make sense is \(x=0\) . So, the ODZ can be written like this: \(x\in (-\infty;0)\cup(0;+\infty)\).
Thus, this equation is equivalent to the system:
\[\begin(cases) \left[ \begin(gathered)\begin(aligned) &\dfrac 4x-2=0\\ &x^2-x=0 \end(aligned) \end(gathered) \right. \\ x\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) \left[ \begin(gathered)\begin(aligned) &\dfrac 4x=2\\ &x(x-1)= 0 \end(aligned) \end(gathered) \right.\\ x\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) \left[ \begin(gathered)\begin(aligned) &x =2\\ &x=1\\ &x=0 \end(aligned) \end(gathered) \right.\\ x\ne 0 \end(cases) \quad \Leftrightarrow \quad \left[ \begin(gathered) \begin(aligned) &x=2\\ &x=1 \end(aligned) \end(gathered) \right.\] Indeed, despite the fact that \(x=0\) is the root of the second factor, if you substitute \(x=0\) into the original equation, then it will not make sense, because expression \(\dfrac 40\) is not defined.
Thus, the solution to this equation is \(x\in \(1;2\)\) .
3) Solve the equation \[\dfrac(x^2+4x)(4x^2-1)=\dfrac(3-x-x^2)(4x^2-1)\] In our equation \(4x^2-1\ne 0\) , from which \((2x-1)(2x+1)\ne 0\) , that is, \(x\ne -\frac12; \frac12\) .
Let's move all the terms to the left side and bring them to a common denominator:
\(\dfrac(x^2+4x)(4x^2-1)=\dfrac(3-x-x^2)(4x^2-1) \quad \Leftrightarrow \quad \dfrac(x^2+4x- 3+x+x^2)(4x^2-1)=0\quad \Leftrightarrow \quad \dfrac(2x^2+5x-3)(4x^2-1)=0 \quad \Leftrightarrow\)
\(\Leftrightarrow \quad \begin(cases) 2x^2+5x-3=0\\ 4x^2-1\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) (2x-1 )(x+3)=0\\ (2x-1)(2x+1)\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) \left[ \begin(gathered) \begin( aligned) &x=\dfrac12\\ &x=-3 \end(aligned)\end(gathered) \right.\\ x\ne \dfrac 12\\ x\ne -\dfrac 12 \end(cases) \quad \ Leftrightarrow \quad x=-3\)
Answer: \(x\in \(-3\)\) .
Comment. If the answer consists of a finite set of numbers, then they can be written separated by semicolons in curly braces, as shown in the previous examples.
Problems that require solving rational equations are encountered every year in the Unified State Examination in mathematics, so when preparing to pass the certification test, graduates should definitely repeat the theory on this topic on their own. Graduates taking both the basic and profile level exam. Having mastered the theory and dealt with practical exercises on the topic “Rational Equations”, students will be able to solve problems with any number of actions and count on receiving competitive scores based on the results of passing the Unified State Exam.
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