How to solve rational equations. Video lesson “Rational equations

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 for and free member the same. Let's take a factor out of each bracket:

Since x=0 is not a root original equation, 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 denote 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:

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

Additional materials
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Educational aids and simulators in the Integral online store for grade 8
A manual for the textbook by Makarychev Yu.N. A manual for the textbook by Mordkovich A.G.

Introduction to Irrational Equations

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.

Solution.
Let's move all expressions to 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 it this way: $\frac((5x-3)*x)((x-3)*x)-\frac((2x-3)*(x-3))((x-3)*x )=\frac(5x^2-3x-(2x^2-6x-3x+9))((x-3)*x)=\frac(3x^2+6x-9)((x-3) *x)=\frac(3(x^2+2x-3))((x-3)*x)$.
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 equal to zero, and the denominator is different from 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:

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

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. $(x+2)(x+3)(x+4)(x+5)=3$.

We have already learned how 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, by to 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 real situations, and also consider movement tasks.

Bibliography

1. Bashmakov M.I. Algebra, 8th grade. - M.: Education, 2004.
2. Dorofeev G.V., Suvorova S.B., Bunimovich E.A. and others. Algebra, 8. 5th ed. - M.: Education, 2010.
3. Nikolsky S.M., Potapov M.A., Reshetnikov N.N., Shevkin A.V. Algebra, 8th grade. Tutorial for educational institutions. - M.: Education, 2006.

1. Festival pedagogical ideas "Public lesson" ().
2. School.xvatit.com ().
3. Rudocs.exdat.com ().

Homework

Least common denominator is used to simplify this equation. This method is used when you cannot write the given equation with one rational expression on each side of the equation (and use the crisscross method of multiplication). This method is used when you are given a rational equation with 3 or more fractions (in the case of two fractions, it is better to use criss-cross multiplication).

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:

1. What is an equation? ( Equality with a variable or variables.)
2. 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. Lead similar terms. Find unknown factor).
3. 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.)
4. 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.)
5. 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.)
6. 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.

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?

Students still have the concept extraneous root haven’t met, it’s really 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:

1. Move everything to the left side.
2. Reduce fractions to a common denominator.
3. Create a system: a fraction is equal to zero when the numerator is equal to zero and the denominator is not equal to zero.
4. Solve the equation.
5. Check inequality to exclude extraneous roots.
6. 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.

1. Read paragraph 25 from the textbook, analyze examples 1-3.
2. Learn an algorithm for solving fractional rational equations.
3. Solve in notebooks No. 600 (a, d, e); No. 601(g,h).
4. 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.