\(\bullet\) 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 equation, the ODZ 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's find ODZ given equation 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 specialized level of the exam must be able to cope with such tasks. 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.
How to prepare for the exam using the Shkolkovo educational portal?
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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.
At x=5 the common denominator x*(x-5) becomes zero. Therefore, this number is not the root of the original fractional rational equation, since there will be a division by zero.
We have already learned how to solve quadratic equations. Now let's extend the studied methods to rational equations.
What's happened rational expression? We have already encountered this concept. Rational expressions are expressions made up of numbers, variables, their powers and symbols of mathematical operations.
Accordingly, rational equations are equations of the form: , where 
- rational expressions.
Previously, we considered only those rational equations that can be reduced to linear ones. Now let's look at those rational equations that can be reduced to quadratic equations.
Example 1
Solve the equation: .
Solution:
A fraction is equal to 0 if and only if its numerator is equal to 0 and its denominator is not equal to 0.
We get the following system:
The first equation of the system is a quadratic equation. Before solving it, let's divide all its coefficients by 3. We get:
We get two roots: ; .
Since 2 never equals 0, two conditions must be met: 
. Since none of the roots of the equation obtained above coincides with acceptable values variables that were obtained by solving the second inequality, they are both solutions to this equation.
Answer:.
So, let's formulate a solution algorithm rational equations:
1. Move all terms to the left side so that the right side ends up with 0.
2. Transform and simplify the left side, bring all fractions to a common denominator.
3. Equate the resulting fraction to 0 using the following algorithm: 
.
4. Write down those roots that were obtained in the first equation and satisfy the second inequality in the answer.
Let's look at another example.
Example 2
Solve the equation: 
.
Solution
At the very beginning, we move all the terms to the left so that 0 remains on the right. We get:
Now let's bring the left side of the equation to a common denominator:
This equation is equivalent to the system:
The first equation of the system is a quadratic equation.
Coefficients of this equation: . We calculate the discriminant:
We get two roots: ; .
Now let's solve the second inequality: the product of factors is not equal to 0 if and only if none of the factors is equal to 0.
Two conditions must be met: 
. We find that of the two roots of the first equation, only one is suitable - 3.
Answer:.
In this lesson, we remembered what a rational expression is, and also learned how to solve rational equations, which reduce to quadratic equations.
In the next lesson we will look at rational equations as models of real situations, and also look at motion problems.
Bibliography
- Bashmakov M.I. Algebra, 8th grade. - M.: Education, 2004.
- Dorofeev G.V., Suvorova S.B., Bunimovich E.A. and others. Algebra, 8. 5th ed. - M.: Education, 2010.
- Nikolsky S.M., Potapov M.A., Reshetnikov N.N., Shevkin A.V. Algebra, 8th grade. Tutorial for educational institutions. - M.: Education, 2006.
- Festival of Pedagogical Ideas" Public lesson" ().
- School.xvatit.com ().
- Rudocs.exdat.com ().
Homework
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 encountered 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.
Lesson objectives:
Educational:
- formation of the concept of fractional rational equations;
- consider various ways to solve fractional rational equations;
- consider an algorithm for solving fractional rational equations, including the condition that the fraction is equal to zero;
- teach solving fractional rational equations using an algorithm;
- checking the level of mastery of the topic by conducting a test.
Developmental:
- developing the ability to correctly operate with acquired knowledge and think logically;
- development of intellectual skills and mental operations - analysis, synthesis, comparison and generalization;
- development of initiative, the ability to make decisions, and not stop there;
- development of critical thinking;
- development of research skills.
Educating:
- fostering cognitive interest in the subject;
- fostering independence in solving educational problems;
- nurturing will and perseverance to achieve final results.
Lesson type: lesson - explanation of new material.
During the classes
1. Organizational moment.
Hello guys! There are equations written on the board, look at them carefully. Can you solve all of these equations? Which ones are not and why?
Equations in which the left and right sides are fractional rational expressions are called fractional rational equations. What do you think we will study in class today? Formulate the topic of the lesson. So, open your notebooks and write down the topic of the lesson “Solving fractional rational equations.”
2. Updating knowledge. Frontal survey, oral work with the class.
And now we will repeat the main theoretical material that we need to study new topic. Please answer the following questions:
- What is an equation? ( Equality with a variable or variables.)
- What is the name of equation number 1? ( Linear.) Solution linear equations. (Move everything with the unknown to the left side of the equation, all numbers to the right. Give similar terms. Find unknown factor).
- What is the name of equation number 3? ( Square.) Solutions quadratic equations. (Isolating a complete square using formulas using Vieta’s theorem and its corollaries.)
- What is proportion? ( Equality of two ratios.) The main property of proportion. ( If the proportion is correct, then the product of its extreme terms is equal to the product of the middle terms.)
- What properties are used when solving equations? ( 1. If you move a term in an equation from one part to another, changing its sign, you will get an equation equivalent to the given one. 2. If both sides of the equation are multiplied or divided by the same non-zero number, you get an equation equivalent to the given one.)
- When does a fraction equal zero? ( A fraction is equal to zero when the numerator is zero and the denominator is not zero..)
3. Explanation of new material.
Solve equation No. 2 in your notebooks and on the board.
Answer: 10.
Which fractional rational equation Can you try to solve using the basic property of proportion? (No. 5).
(x-2)(x-4) = (x+2)(x+3)
x 2 -4x-2x+8 = x 2 +3x+2x+6
x 2 -6x-x 2 -5x = 6-8
Solve equation No. 4 in your notebooks and on the board.
Answer: 1,5.
What fractional rational equation can you try to solve by multiplying both sides of the equation by the denominator? (No. 6).
x 2 -7x+12 = 0
D=1›0, x 1 =3, x 2 =4.
Answer: 3;4.
Now try to solve equation number 7 using one of the following methods.
(x 2 -2x-5)x(x-5)=x(x-5)(x+5) |
|
||
(x 2 -2x-5)x(x-5)-x(x-5)(x+5)=0 |
x 2 -2x-5=x+5 |
||
x(x-5)(x 2 -2x-5-(x+5))=0 |
x 2 -2x-5-x-5=0 |
||
x(x-5)(x 2 -3x-10)=0 |
|||
x=0 x-5=0 x 2 -3x-10=0 |
|||
x 1 =0 x 2 =5 D=49 |
|||
x 3 =5 x 4 =-2 |
x 3 =5 x 4 =-2 |
||
Answer: 0;5;-2. |
Answer: 5;-2. |
Explain why this happened? Why are there three roots in one case and two in the other? What numbers are the roots of this fractional rational equation?
Until now, students have not encountered the concept of an extraneous root; it is indeed very difficult for them to understand why this happened. If no one in the class can give a clear explanation of this situation, then the teacher asks leading questions.
- How do equations No. 2 and 4 differ from equations No. 5,6,7? ( In equations No. 2 and 4 there are numbers in the denominator, No. 5-7 are expressions with a variable.)
- What is the root of an equation? ( The value of the variable at which the equation becomes true.)
- How to find out if a number is the root of an equation? ( Make a check.)
When testing, some students notice that they have to divide by zero. They conclude that the numbers 0 and 5 are not the roots of this equation. The question arises: is there a way to solve fractional rational equations that allows us to eliminate this error? Yes, this method is based on the condition that the fraction is equal to zero.
x 2 -3x-10=0, D=49, x 1 =5, x 2 =-2.
If x=5, then x(x-5)=0, which means 5 is an extraneous root.
If x=-2, then x(x-5)≠0.
Answer: -2.
Let's try to formulate an algorithm for solving fractional rational equations in this way. Children formulate the algorithm themselves.
Algorithm for solving fractional rational equations:
- Move everything to the left side.
- Reduce fractions to a common denominator.
- Create a system: a fraction is equal to zero when the numerator is equal to zero and the denominator is not equal to zero.
- Solve the equation.
- Check inequality to exclude extraneous roots.
- Write down the answer.
Discussion: how to formalize the solution if you use the basic property of proportion and multiplying both sides of the equation by a common denominator. (Add to the solution: exclude from its roots those that make the common denominator vanish).
4. Initial comprehension of new material.
Work in pairs. Students choose how to solve the equation themselves depending on the type of equation. Assignments from the textbook “Algebra 8”, Yu.N. Makarychev, 2007: No. 600(b,c,i); No. 601(a,e,g). The teacher monitors the completion of the task, answers any questions that arise, and provides assistance to low-performing students. Self-test: answers are written on the board.
b) 2 – extraneous root. Answer: 3.
c) 2 – extraneous root. Answer: 1.5.
a) Answer: -12.5.
g) Answer: 1;1.5.
5. Setting homework.
- Read paragraph 25 from the textbook, analyze examples 1-3.
- Learn an algorithm for solving fractional rational equations.
- Solve in notebooks No. 600 (a, d, e); No. 601(g,h).
- Try to solve No. 696(a) (optional).
6. Completing a control task on the topic studied.
The work is done on pieces of paper.
Example task:
A) Which of the equations are fractional rational?
B) A fraction is equal to zero when the numerator is ______________________ and the denominator is _______________________.
Q) Is the number -3 the root of equation number 6?
D) Solve equation No. 7.
Assessment criteria for the assignment:
- “5” is given if the student completed more than 90% of the task correctly.
- "4" - 75%-89%
- "3" - 50%-74%
- “2” is given to a student who has completed less than 50% of the task.
- A rating of 2 is not given in the journal, 3 is optional.
7. Reflection.
On the independent work sheets, write:
- 1 – if the lesson was interesting and understandable to you;
- 2 – interesting, but not clear;
- 3 – not interesting, but understandable;
- 4 – not interesting, not clear.
8. Summing up the lesson.
So, today in the lesson we got acquainted with fractional rational equations, learned how to solve these equations different ways, tested their knowledge with the help of a training independent work. You will learn the results of your independent work in the next lesson, and at home you will have the opportunity to consolidate your knowledge.
Which method of solving fractional rational equations, in your opinion, is easier, more accessible, and more rational? Regardless of the method for solving fractional rational equations, what should you remember? What is the “cunning” of fractional rational equations?
Thanks everyone, lesson is over.