Dsr 14 fractional rational equations. Rational equations

§ 1 Integer and fractional rational equations

In this lesson we will look at concepts such as rational equation, rational expression, whole expression, fractional expression. Let's consider solving rational equations.

A rational equation is an equation in which the left and right sides are rational expressions.

Rational expressions are:

Fractional.

An integer expression is made up of numbers, variables, integer powers using the operations of addition, subtraction, multiplication, and division by a number other than zero.

For example:

IN fractional expressions there is a division by a variable or an expression with a variable. For example:

A fractional expression does not make sense for all values of the variables included in it. For example, the expression

at x = -9 it does not make sense, since at x = -9 the denominator goes to zero.

This means that a rational equation can be integer or fractional.

A whole rational equation is a rational equation in which the left and right sides are whole expressions.

For example:

A fractional rational equation is a rational equation in which either the left or right sides are fractional expressions.

For example:

§ 2 Solution of an entire rational equation

Let's consider the solution of an entire rational equation.

For example:

Multiply both sides of the equation by the smallest common denominator denominators of the fractions included in it.

For this:

1. find the common denominator for denominators 2, 3, 6. It is equal to 6;

2. find an additional factor for each fraction. To do this, divide the common denominator 6 by each denominator

additional factor for fraction

3. multiply the numerators of the fractions by their corresponding additional factors. Thus, we obtain the equation

which is equivalent to the given equation

Let's open the brackets on the left, move the right part to the left, changing the sign of the term when transferred to the opposite one.

Let us bring similar terms of the polynomial and get

We see that the equation is linear.

Having solved it, we find that x = 0.5.

§ 3 Solution of a fractional rational equation

Let's consider solving a fractional rational equation.

For example:

1.Multiply both sides of the equation by the least common denominator of the denominators of the rational fractions included in it.

Let's find the common denominator for the denominators x + 7 and x - 1.

It is equal to their product (x + 7)(x - 1).

2. Let's find an additional factor for each rational fraction.

To do this, divide the common denominator (x + 7)(x - 1) by each denominator. Additional factor for fractions

equal to x - 1,

additional factor for fraction

equals x+7.

3.Multiply the numerators of the fractions by their corresponding additional factors.

We obtain the equation (2x - 1)(x - 1) = (3x + 4)(x + 7), which is equivalent to this equation

4.Multiply the binomial by the binomial on the left and right and get the following equation

5. We move the right side to the left, changing the sign of each term when transferring to the opposite:

6. Let us present similar terms of the polynomial:

7. Both sides can be divided by -1. We get a quadratic equation:

8. Having solved it, we will find the roots

Since in Eq.

the left and right sides are fractional expressions, and in fractional expressions for some values variable denominator can go to zero, then it is necessary to check whether the common denominator does not go to zero when x1 and x2 are found.

At x = -27, the common denominator (x + 7)(x - 1) does not vanish; at x = -1, the common denominator also does not equal to zero.

Therefore, both roots -27 and -1 are roots of the equation.

When solving a fractional rational equation, it is better to immediately indicate the range of acceptable values. Eliminate those values at which the common denominator goes to zero.

Let's consider another example of solving a fractional rational equation.

For example, let's solve the equation

We factor the denominator of the fraction on the right side of the equation

We get the equation

Let's find the common denominator for the denominators (x - 5), x, x(x - 5).

It will be the expression x(x - 5).

Now let's find the range of acceptable values of the equation

To do this, we equate the common denominator to zero x(x - 5) = 0.

We obtain an equation, solving which we find that at x = 0 or at x = 5 the common denominator goes to zero.

This means that x = 0 or x = 5 cannot be the roots of our equation.

Additional multipliers can now be found.

Additional factor for rational fractions

additional factor for the fraction

will be (x - 5),

and the additional factor of the fraction

We multiply the numerators by the corresponding additional factors.

We get the equation x(x - 3) + 1(x - 5) = 1(x + 5).

Let's open the brackets on the left and right, x2 - 3x + x - 5 = x + 5.

Let's move the terms from right to left, changing the sign of the transferred terms:

X2 - 3x + x - 5 - x - 5 = 0

And after bringing similar terms, we obtain a quadratic equation x2 - 3x - 10 = 0. Having solved it, we find the roots x1 = -2; x2 = 5.

But we have already found out that at x = 5 the common denominator x(x - 5) goes to zero. Therefore, the root of our equation

will be x = -2.

§ 4 Brief summary lesson

Important to remember:

When solving fractional rational equations, proceed as follows:

1. Find the common denominator of the fractions included in the equation. Moreover, if the denominators of fractions can be factored, then factor them and then find the common denominator.

2.Multiply both sides of the equation by a common denominator: find additional factors, multiply the numerators by additional factors.

3.Solve the resulting whole equation.

4. Eliminate from its roots those that make the common denominator vanish.

List of used literature:

Makarychev Yu.N., N.G. Mindyuk, Neshkov K.I., Suvorova S.B. / Edited by Telyakovsky S.A. Algebra: textbook. for 8th grade. general education institutions. - M.: Education, 2013.
Mordkovich A.G. Algebra. 8th grade: In two parts. Part 1: Textbook. for general education institutions. - M.: Mnemosyne.
Rurukin A.N. Lesson-based developments in algebra: 8th grade. - M.: VAKO, 2010.
Algebra 8th grade: lesson plans according to the textbook by Yu.N. Makarycheva, N.G. Mindyuk, K.I. Neshkova, S.B. Suvorova / Auth.-comp. T.L. Afanasyeva, L.A. Tapilina. -Volgograd: Teacher, 2005.

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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 synthesis;
development of initiative, the ability to make decisions, and not stop there;
development critical thinking;
development of research skills.

Educating:

upbringing cognitive interest to the subject;
fostering independence in decision-making educational tasks;
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 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. (Transfer everything with the unknown to left side equations, all numbers are on the right. Lead similar terms. Find unknown factor).
What is the name of equation number 3? ( Square.) Methods for solving quadratic equations. ( Selection full square, by formulas, using Vieta’s theorem and its consequences.)
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 roots given 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.

We have already learned to solve quadratic equations. Now let's extend the studied methods to rational equations.

What is a 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 coincide with invalid values variables that were obtained by solving the second inequality, they are both solutions to this equation.

Answer:.

So, let's formulate an algorithm for solving 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, let's move all the terms to left side, 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 pedagogical ideas "Public lesson" ().
School.xvatit.com ().
Rudocs.exdat.com ().

Homework

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

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A manual for the textbook by Makarychev Yu.N. A manual for the textbook by Mordkovich A.G.

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. Move all expressions contained in the equation to the left side 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 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: $x^2+x-2=0$.
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.
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. $(x+2)(x+3)(x+4)(x+5)=3$.