How to solve rational equations. The simplest rational equations

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 of 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 by ordinary numbers, then we would reduce the 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 an 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. $(x+2)(x+3)(x+4)(x+5)=3$.

"Solving fractional rational equations"

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 will need to study a 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 No. 1? ( Linear.) A method for solving 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).

3. What is the name of equation No. 3? ( Square.) Methods for solving quadratic equations. ( Isolating a complete square using formulas using Vieta’s theorem and its corollaries.)

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.

What fractional rational equation can you try to solve using the basic property of proportion? (No. 5).

(x-2)(x-4) = (x+2)(x+3)

x2-4x-2x+8 = x2+3x+2x+6

x2-6x-x2-5x = 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).

D=1›0, x1=3, x2=4.

Answer: 3;4.

Now try to solve equation number 7 using one of the following methods.


(x2-2x-5)x(x-5)=x(x-5)(x+5)
(x2-2x-5)x(x-5)-x(x-5)(x+5)=0
x(x-5)(x2-2x-5-(x+5))=0			x2-2x-5-x-5=0
x(x-5)(x2-3x-10)=0
x=0 x-5=0 x2-3x-10=0
x1=0 x2=5 D=49

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.

In equations No. 2 and 4 there are numbers in the denominator, No. 5-7 are expressions with a variable

The value of the variable at which the equation becomes true

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.

x2-3x-10=0, D=49, x1=5, x2=-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 the 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”, 2007: No. 000 (b, c, i); No. 000(a, d, 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.

2. Learn the algorithm for solving fractional rational equations.

3. Solve in notebooks No. 000 (a, d, e); No. 000(g, h).

4. Try to solve No. 000(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 to solve these equations in various ways, and tested our knowledge with the help of independent educational 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.

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?

2. Updating knowledge. Frontal survey, oral work with the class.

And now we will repeat the main theoretical material that we will need to study a 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.) A method for solving 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.) Methods for solving 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.

What 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?

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.)

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.

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.

Thanks everyone, lesson is over.

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Solving fractional rational equations

Reference Guide

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

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

Fractional rational equations are usually reduced to the form:

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

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

A rational equation is called whole, or algebraic, if it does not divide by an expression containing a variable.

Examples of a whole rational equation:

5x – 10 = 3(10 – x)

3x
- = 2x – 10
4

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

Example of a fractional rational equation:

15
x + - = 5x – 17
x

Fractional rational equations are usually solved as follows:

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

2) solve the resulting whole equation;

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

Examples of solving integer and fractional rational equations.

Example 1. Let's solve the whole equation

x – 1 2x 5x
-- + -- = --.
2 3 6

Solution:

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

3(x – 1) + 4x 5x
------ = --
6 6

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

3(x – 1) + 4x = 5x.

We solve it by opening the brackets and combining similar terms:

3x – 3 + 4x = 5x

3x + 4x – 5x = 3

The example is solved.

Example 2. Solve a fractional rational equation

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

Finding a common denominator. This is x(x – 5). So:

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

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

x 2 – 3x + x – 5 = x + 5

x 2 – 3x + x – 5 – x – 5 = 0

x 2 – 3x – 10 = 0.

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

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

At x = –2, the common denominator x(x – 5) does not vanish. This means –2 is the root of the original equation.

At x = 5, the common denominator goes to zero, and two out of three expressions become meaningless. This means that the number 5 is not the root of the original equation.

Answer: x = –2

More examples

Example 1.

x 1 =6, x 2 = - 2.2.

Answer: -2,2;6.

Example 2.