Lesson topic: “The whole equation and its roots.”
Goals:
consider a way to solve an entire equation using factorization;
educational:
developing:
educational:
Class: 9
Textbook: Algebra. 9th grade: textbook for educational institutions/ [Yu.N. Makarychev, N.G. Mindyuk, K.I. Neshkov, S.B. Suvorov]; edited by S.A. Telyakovsky.- 16th ed. – M.: Education, 2010
Equipment: computer with projector, presentation “Whole Equations”
During the classes:
Organizing time.
Watch the video “Everything is in your hands.”
There are times in life when you give up and it seems like nothing will work out. Then remember the words of the sage “Everything is in your hands:” and let these words be the motto of our lesson.
2x + 6 =10, 14x = 7, x 2 – 16 = 0, x – 3 = 5 + 2x, x 2 = 0,
Message of the lesson topic, goals.
Today we will get acquainted with a new type of equations - these are whole equations. Let's learn how to solve them.
Let's write down the number in the notebook, Classwork and the topic of the lesson: “The whole equation, its roots.”
2.Updating basic knowledge.
Solve the equation:
Answers: a)x = 0; b) x =5/3; c) x = -, ; d) x = 1/6; - 1/6; e) there are no roots; e) x = 0; 5; - 5; g) 0; 1; -2; h)0; 1; - 1; i) 0.2; - 0.2; j) -3; 3.
3.Formation of new concepts.
Conversation with students:
What is an equation? (equality containing an unknown number)
What types of equations do you know? (linear, square)
3.How many roots can it have? linear equation?) (one, many and no roots)
4.How many roots can a quadratic equation have?
What determines the number of roots? (from discriminant)
In what case does a quadratic equation have 2 roots? (D0)
In what case does a quadratic equation have 1 root? (D=0)
In what case does a quadratic equation have no roots? (D0)
Whole equation is an equation of the left and right sides, which is an entire expression. (read aloud).
From the considered linear and quadratic equations, we see that the number of roots is not greater than its degree.
Do you think it is possible to determine the number of its roots without solving an equation? (possible children's answers)
Let's get acquainted with the rule for determining the degree of an entire equation?
If an equation with one variable is written as P(x)=0, where P(x) is a polynomial standard view, then the degree of this polynomial is called the degree of the equation. The degree of an arbitrary integer equation is the degree of an equivalent equation of the form P(x) = 0, where P(x) is a polynomial of standard form.
The equationn Ouch degree has no moren roots.
The whole equation can be solved in several ways:
ways to solve entire equations
factorization graphic introduction new
variable
(Write the diagram in a notebook)
Today we will look at one of them: factorization using the following equation as an example: x 3 – 8x 2 – x +8 = 0. (the teacher explains on the board, students write down the solution to the equation in a notebook)
What is the name of the method of factorization that can be used to left side Factor the equations? (grouping method). Let's factorize the left side of the equation, and to do this, group the terms on the left side of the equation.
When does the product of factors equal zero? (when at least one of the multipliers equal to zero). Let us equate each factor of the equation to zero.
Let's solve the resulting equations
How many roots did we get? (write in notebook)
x 2 (x – 8) – (x – 8) = 0
(x – 8) (x 2 – 1) = 0
(x – 8)(x – 1)(x + 1) = 0
x 1 = 8, x 2 = 1, x 3 = - 1.
Answer: 8; 1; -1.
4.Formation of skills and abilities. Practical part.
work on textbook No. 265 (write in notebook)
What is the degree of the equation and how many roots does each equation have:
Answers: a) 5, b) 6, c) 5, d) 2, e) 1, f) 1
№ 266(a)(solution at the board with explanation)
Solve the equation:
5. Lesson summary:
Consolidation theoretical material:
What equation with one variable is called an integer? Give an example.
How to find the degree of an entire equation? How many roots does an equation with one variable of the first, second, nth degree have?
6.Reflection
Evaluate your work. Raise your hand who...
1) understood the topic perfectly
2) understood the topic well
I'm still experiencing difficulties
7.Homework:
clause 12 (p. 75-77 example 1) No. 267 (a, b).
“student checklist”
Student checklist
| Stages of work Grade | Total |
|||||
| Verbal counting | Solve the equation | Solving Quadratic Equations | Solving cubic equations | |||
Student checklist
Class______ Last name First name ___________________
| Stages of work Grade | Total |
|||||
| Verbal counting | Solve the equation | What is the degree of familiar equations | Solving Quadratic Equations | Solving cubic equations | ||
Student checklist
Class______ Last name First name ___________________
| Stages of work Grade | Total |
|||||
| Verbal counting | Solve the equation | What is the degree of familiar equations | Solving Quadratic Equations | Solving cubic equations | ||
View document contents
"Handout"
1.Solve the equations:
a) x 2 = 0 e) x 3 – 25x = 0
a) x 2 = 0 e) x 3 – 25x = 0
b) 3x – 5 = 0 g) x(x – 1)(x + 2) = 0
c) x 2 –5 = 0 h) x 4 – x 2 = 0
d) x 2 = 1/36 i) x 2 –0.01 = 0.03
e) x 2 = – 25 j) 19 – c 2 = 10
3. Solve the equations:
x 2 -5x+6=0 y 2 -4y+7=0 x 2 -12x+36=0
4. Solve the equations:
I option II option III option
x 3 -1=0 x 3 - 4x=0 x 3 -12x 2 +36x=0
"test"
Hello! Now you will be offered a 4-question math test. Click on the buttons on the screen under the questions that, in your opinion, have the correct answer. Click the "next" button to start testing. Good luck!
1. Solve the equation:
3x + 6 = 0
Correct
No answer
Roots
Correct
No answer
Roots
4. Solve the equation: 0 x = - 4
Roots
A lot of
roots
View presentation content
"1"
- Solve the equation:
- ORAL WORK
Goals:
educational:
- generalize and deepen information about equations; introduce the concept of a whole equation and its degree, its roots; Consider a way to solve an entire equation using factorization.
- generalize and deepen information about equations;
- introduce the concept of a whole equation and its degree, its roots;
- Consider a way to solve an entire equation using factorization.
developing:
- development of mathematical and general outlook, logical thinking, ability to analyze, draw conclusions;
- development of mathematical and general outlook, logical thinking, ability to analyze, draw conclusions;
educational:
- cultivate independence, clarity and accuracy in actions.
- cultivate independence, clarity and accuracy in actions.
- Psychological attitude
- We continue to generalize and deepen information about equations;
- get acquainted with the concept of the whole equation,
with the concept of degree of equation;
- develop skills in solving equations;
- control the level of material assimilation;
- In class we can make mistakes, have doubts, and consult.
- Each student sets his own guidelines.
- What equations are called integers?
- What is the degree of an equation?
- How many roots does it have? equation nth degrees?
- Methods for solving equations of first, second and third degrees.
- Lesson Plan
a) x 2 = 0 e) x 3 – 25x = 0 c) x 2 –5 = 0 h) x 4 – x 2 = 0 d) x 2 = 1/36 i) x 2 –0,01 = 0,03 e) x 2 = – 25 k) 19 – s 2 = 10
Solve the equations:
For example:
X²=x³-2(x-1)
- Equations
If the equation is with one variable
written as
P(x) = 0, where P(x) is a polynomial of standard form,
then the degree of this polynomial is called
degree of this equation
2x³+2x-1=0 (5th degree)
14x²-3=0 (4th degree)
For example:
What is the degree of acquaintance equations for us?
- a) x 2 = 0 e) x 3 – 25x = 0
- b) 3x – 5 = 0 g) x(x – 1)(x + 2) = 0
- c) x 2 – 5 = 0 h) x 4 – x 2 = 0
- d) x 2 = 1/36 i) x 2 – 0,01 = 0,03
- e) x 2 = – 25 k) 19 – s 2 = 10
- Solve the equations:
- 2 ∙x + 5 =15
- 0∙x = 7
How many roots can an equation of degree 1 have?
No more than one!
0, D=-12, D x 1 =2, x 2 =3 no roots x=6. How many roots can an equation of degree I (quadratic) have? No more than two!" width="640" - Solve the equations:
- x 2 -5x+6=0 y 2 -4y+7=0 x 2 -12x+36=0
- D=1, D0, D=-12, D
x 1 =2, x 2 =3 no roots x=6.
How many roots can an equation of degree I have? (square) ?
No more than two!
Solve the equations:
- I option II option III option
x 3 -1=0x 3 - 4x=0 x 3 -12x 2 +36x=0
- x 3 =1 x(x 2 - 4)=0 x(x 2 -12x+36)=0
x=1 x=0, x=2, x= -2 x=0, x=6
1 root 3 roots 2 roots
- How many roots can an equation of degree I I I have?
No more than three!
- How many roots do you think the equation can have?
IV, V, VI, VII, n th degrees?
- No more than four, five, six, seven roots!
No more at all n roots!
ax²+bx+c=0
Quadratic equation
ax + b = 0
Linear equation
No roots
No roots
One root
Let's expand the left side of the equation
by multipliers:
x²(x-8)-(x-8)=0
Answer:=1, =-1.
- Third degree equation of the form: ax³+bx²+cx+d=0
By factorization
(8x-1)(2x-3)-(4x-1)²=38
Let's open the brackets and give
similar terms
16x²-24x-2x+3-16x²+8x-138=0
Answer: x=-2
The motto of our lesson: “The more I know, the more I can.” Epigaph:
Who doesn't notice anything
He doesn't study anything.
Who doesn't study anything
He's always whining and bored.
(poet R. Seph).
Mathematical dictation
1.Insert the missing oneswords and indicate matches
1.What is it called?
equation?
1. Find all of it... or
prove that... no.
2.What is it called?
root of the equation?
2. ……, containing
variable.
3.What does it mean to decide
the equation?
3. ……., in which
the equation is reversed
to the correct number
equality.
Solve equations orally:
a) x² = 0b) 3x – 6 = 0
c) x² – 9 = 0
d) x(x – 1)(x + 2) = 0
e) x² = – 25
Solve the equation:
x⁴-6x²+5=0The whole equation and its roots
Lesson objectives:
summarize and deepen information aboutequations
introduction to the concept of whole
the equation
introduction to the concept of degree
equations
formation of solution skills
equations
Equations
x5
2
x 1 x 1
3
x
2
x 5
x3 1 x 2 1
3x2
4
2
(x 3 1) x 2 x 3 2(x 1)
x
2x 1
x 12
whole
equations
fractional
equations
Whole equation
An entire equation with onethe variable is the equation
the left and right parts of which
whole expressions.
10. Power equation
If an equation with onevariable is written as P(x)=0,
where P(x) is the standard polynomial
form, then the degree of this polynomial
is called the degree of the equation, i.e.
greatest of degrees
monomials.
Examples: x⁵-2x³+2x-1=05th
degree
4th
x⁴-14x²-3=0
degree
11. What is the degree of the equation?
5a) 2x²-6x⁵+1=0
2
d) (x+8)(x-7)=0
6
b) x⁶-4x²-3=0
1 5
x 0
7
V)
5x(x²+4)=17
d)
x x
5
2 4
5
1
3
e) 5x-
12. Let's repeat
linear equationaх+b=0
aх2 + bx + c = 0
a bunch of
roots
no roots
one root
quadratic equation
D=0
one root
D>0
two roots
D<0
no roots
13. First degree equation
14. Third degree equation
Solve the equationx3 8x 2 x 8 0
Solution: expand the left side
factoring equations 2
x (x 8) (x 8) 0
(x 8)(x 2 1) 0
x 8 0
x2 1 0
x1 8, x2answer
1, x3 1
15. Solve the equation:
(8x-1)(2x-3)-(4x-1)²=38Solution: Let's open the brackets and give
similar terms
16x²-24x-2x+3-16x²+8x-1-38=0
-18x-36=0
CHECK YOURSELF!
x+2=0
x=-2
Answer: x=-2
16. Let's solve the biquadratic equation:
X⁴ - 5 x² - 36 = 0Let's make a replacement: x² = a, a≥ 0
a² - 5a -36 =0
D=169
a1= -4 (not suitable, because a≥0)
a2 = 9
X² = 9
x1 = 3 and x2 = -3
Answer: 3 and -3.
17. Solve the equation:
x⁴-6x²+5=0Answer: 1, -1, V5, - V5
18. Establish the correspondence: Equation method.
Sample textSecond level
Third level
Fourth level
Fifth level
19. Test
1) Determine the degree of the equation(x 2 3) 5 x(x 1) 15
a) 2
b) 3
in 1
2) Which numbers are roots
x(x 1)(x 2) 0?
equations
a) -1
b) 0
at 2
3) Solve the equation 9 x 3 27 x 2 0
a) 0;-3
b) -3;0;3
c) 0;3
20.
1)What equation is called
whole and how to distinguish it from
fractional?
2)
What is the degree of an equation?
3)
What are the roots of an equation?
4)
5)
How many roots can it have?
1st degree equation?
How many roots can it have?
2nd degree equation?
21. Homework:
Think and answer the question: “How muchroots can have a whole equation with
one variable of the 2nd, 3rd, 4th, third degree?
Consider the equation.
31x 3 – 10x = (x – 5) 2 + 6x 2
Both the left and right sides of the equation are integer expressions.
Recall that such equations are called entire equations.
Let's return to our original equation and open the brackets using the squared difference formula.
Let's move all the terms of the equation to the left side and present similar terms.
The expressions “minus ten x” and “plus ten x” cancel each other out.
After bringing similar terms, we obtain an equation, on the left side of which there is a polynomial of the standard form (in general terms we will call it “Pe from x”), and on the right side there is zero.
To determine the degree of an entire equation, it is necessary to reduce it to the form pe from x equals zero, that is, to an equation in which the left side contains a polynomial of the standard form, and the right side contains zero.
After this, it is necessary to determine the degree of the polynomial pe from x. This will be the degree of the equation.
Let's look at an example. Let's try to determine the degree of this equation.
Let's open the brackets using the formula for the square of the sum.
Next, we move all the terms of the equation to the left side and present similar terms.
So, we have obtained an equation, on the left side of which is a polynomial of the standard form of the second degree, and on the right side is zero. This means that the degree of this equation is second.
The degree of the equation determines how many roots it has.
It can be proven that an equation of the first degree has one root, an equation of the second degree has no more than two roots, an equation of the third degree has no more than three roots, and so on.
The degree of an equation also tells us how the equation can be solved.
For example, we reduce an equation of the first degree to the form a x plus be equals ce, where a is not equal to zero.
We reduce an equation of the second degree to an equivalent equation, on the left side of which there is a square trinomial, and on the right side there is a zero. Such an equation is solved using the formula for the roots of a quadratic equation or Vieta's theorem.
There is no universal method for solving equations of higher degrees, but there are basic methods that we will consider with examples.
Let's solve the equation of the third power x to the third power minus eight x to the second power minus x plus eight equals zero.
To solve this equation, we factorize its left side using the grouping method and using the difference of squares formula.
Next, you need to remember that the product is equal to zeros when one of the factors is equal to zero. Based on this, we conclude that either x minus 8 is equal to zero, or x minus 1 is equal to zero, or x plus one is equal to zero. Therefore, the roots of the equation will be the numbers minus one, one and eight.
Sometimes, to solve equations of degree higher than two, it is convenient to introduce a new variable.
Let's look at a similar example.
If we open the brackets, move all the terms of the equation to the left side, bring similar terms and present the left side of the equation in the form of a polynomial of a standard form, then none of the methods known to us will help solve this equation. In this case, it is worth paying attention to the fact that both brackets contain the same expressions.
It is this expression that we will denote as the new variable igrik.
Then our equation will be reduced to an equation with the variable ig...
Next, we’ll simply open the brackets and move all the terms of the equation to the left side.
Let us bring similar terms and get the quadratic equation already familiar to us.
It is not difficult to find the roots of this equation. Game one is equal to six, game two is equal to minus sixteen.
Now let's return to the original equation by performing the reverse substitution.
Initially, we took the expression two x squared minus x as the game. And since we have two values for the variable y, we get two equations. In each equation we move all terms to the left side and solve the resulting two quadratic equations. The roots of the first equation are the numbers minus one point five and two, and the second equation has no roots, since its discriminant is less than zero.
So, the solution to this fourth-degree equation is the numbers minus one point five and two.
A special place in the classification of entire equations has an equation of the form a x to the fourth power plus be x to the second power plus ce equals zero. Equations of this type are called biquadratic equations.
Such equations can be solved using a change of variable.
Let's look at an example.
In this equation, let's denote x square by igrik. It is worth noting that the iGrik variable cannot take negative values.
We obtain a quadratic equation whose roots are the numbers one twenty-fifth and one.
Let's do the reverse replacement.
The roots of the first equation are one-fifth and minus one-fifth, and the roots of the second are one and minus one.
Thus, we have found the four roots of the original biquadratic equation.
Let's get acquainted with rational and fractional rational equations, give their definition, give examples, and also analyze the most common types of problems.
Yandex.RTB R-A-339285-1
Rational equation: definition and examples
Acquaintance with rational expressions begins in the 8th grade of school. At this time, in algebra lessons, students increasingly begin to encounter assignments with equations that contain rational expressions in their notes. Let's refresh our memory on what it is.
Definition 1
Rational equation is an equation in which both sides contain rational expressions.
In various manuals you can find another formulation.
Definition 2
Rational equation- this is an equation, the left side of which contains a rational expression, and the right side contains zero.
The definitions that we gave for rational equations are equivalent, since they talk about the same thing. The correctness of our words is confirmed by the fact that for any rational expressions P And Q equations P = Q And P − Q = 0 will be equivalent expressions.
Now let's look at the examples.
Example 1
Rational equations:
x = 1 , 2 x − 12 x 2 y z 3 = 0 , x x 2 + 3 x - 1 = 2 + 2 7 x - a (x + 2) , 1 2 + 3 4 - 12 x - 1 = 3 .
Rational equations, just like equations of other types, can contain any number of variables from 1 to several. To begin with, we will look at simple examples in which the equations will contain only one variable. And then we will begin to gradually complicate the task.
Rational equations are divided into two large groups: integer and fractional. Let's see what equations will apply to each of the groups.
Definition 3
A rational equation will be integer if its left and right sides contain entire rational expressions.
Definition 4
A rational equation will be fractional if one or both of its parts contain a fraction.
Fractional rational equations necessarily contain division by a variable or the variable is present in the denominator. There is no such division in the writing of whole equations.
Example 2
3 x + 2 = 0 And (x + y) · (3 · x 2 − 1) + x = − y + 0, 5– entire rational equations. Here both sides of the equation are represented by integer expressions.
1 x - 1 = x 3 and x: (5 x 3 + y 2) = 3: (x − 1) : 5 are fractionally rational equations.
Whole rational equations include linear and quadratic equations.
Solving whole equations
Solving such equations usually comes down to converting them into equivalent algebraic equations. This can be achieved by carrying out equivalent transformations of equations in accordance with the following algorithm:
- first we get zero on the right side of the equation; to do this, we need to move the expression that is on the right side of the equation to its left side and change the sign;
- then we transform the expression on the left side of the equation into a polynomial of standard form.
We must obtain an algebraic equation. This equation will be equivalent to the original equation. Easy cases allow us to reduce the whole equation to a linear or quadratic one to solve the problem. In general, we solve an algebraic equation of degree n.
Example 3
It is necessary to find the roots of the whole equation 3 (x + 1) (x − 3) = x (2 x − 1) − 3.
Solution
Let us transform the original expression in order to obtain an equivalent algebraic equation. To do this, we will transfer the expression contained on the right side of the equation to the left side and replace the sign with the opposite one. As a result we get: 3 (x + 1) (x − 3) − x (2 x − 1) + 3 = 0.
Now let's transform the expression that is on the left side into a standard form polynomial and perform the necessary actions with this polynomial:
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
We managed to reduce the solution of the original equation to the solution of a quadratic equation of the form x 2 − 5 x − 6 = 0. The discriminant of this equation is positive: D = (− 5) 2 − 4 · 1 · (− 6) = 25 + 24 = 49 . This means there will be two real roots. Let's find them using the formula for the roots of a quadratic equation:
x = - - 5 ± 49 2 1,
x 1 = 5 + 7 2 or x 2 = 5 - 7 2,
x 1 = 6 or x 2 = - 1
Let's check the correctness of the roots of the equation that we found during the solution. For this, we substitute the numbers we received into the original equation: 3 (6 + 1) (6 − 3) = 6 (2 6 − 1) − 3 And 3 · (− 1 + 1) · (− 1 − 3) = (− 1) · (2 · (− 1) − 1) − 3. In the first case 63 = 63 , in the second 0 = 0 . Roots x=6 And x = − 1 are indeed the roots of the equation given in the example condition.
Answer: 6 , − 1 .
Let's look at what "degree of an entire equation" means. We will often encounter this term in cases where we need to represent an entire equation in algebraic form. Let's define the concept.
Definition 5
Degree of the whole equation is the degree of an algebraic equation equivalent to the original integer equation.
If you look at the equations from the example above, you can establish: the degree of this whole equation is second.
If our course was limited to solving equations of the second degree, then the discussion of the topic could end there. But it's not that simple. Solving equations of the third degree is fraught with difficulties. And for equations above the fourth degree there are no general root formulas at all. In this regard, solving entire equations of the third, fourth and other degrees requires us to use a number of other techniques and methods.
The most commonly used approach to solving entire rational equations is based on the factorization method. The algorithm of actions in this case is as follows:
- we move the expression from the right side to the left so that zero remains on the right side of the record;
- We represent the expression on the left side as a product of factors, and then move on to a set of several simpler equations.
Find the solution to the equation (x 2 − 1) · (x 2 − 10 · x + 13) = 2 · x · (x 2 − 10 · x + 13) .
Solution
We move the expression from the right side of the record to the left with the opposite sign: (x 2 − 1) · (x 2 − 10 · x + 13) − 2 · x · (x 2 − 10 · x + 13) = 0. Converting the left-hand side to a polynomial of the standard form is inappropriate due to the fact that this will give us an algebraic equation of the fourth degree: x 4 − 12 x 3 + 32 x 2 − 16 x − 13 = 0. The ease of conversion does not justify all the difficulties in solving such an equation.
It’s much easier to go the other way: let’s take the common factor out of brackets x 2 − 10 x + 13 . So we arrive at an equation of the form (x 2 − 10 x + 13) (x 2 − 2 x − 1) = 0. Now we replace the resulting equation with a set of two quadratic equations x 2 − 10 x + 13 = 0 And x 2 − 2 x − 1 = 0 and find their roots through the discriminant: 5 + 2 3, 5 - 2 3, 1 + 2, 1 - 2.
Answer: 5 + 2 3, 5 - 2 3, 1 + 2, 1 - 2.
In the same way, we can use the method of introducing a new variable. This method allows us to move to equivalent equations with degrees lower than the degrees in the original integer equation.
Example 5
Does the equation have roots? (x 2 + 3 x + 1) 2 + 10 = − 2 (x 2 + 3 x − 4)?
Solution
If we now try to reduce an entire rational equation to an algebraic one, we will get an equation of degree 4 that has no rational roots. Therefore, it will be easier for us to go the other way: introduce a new variable y, which will replace the expression in the equation x 2 + 3 x.
Now we will work with the whole equation (y + 1) 2 + 10 = − 2 · (y − 4). Let's move the right side of the equation to the left with the opposite sign and carry out the necessary transformations. We get: y 2 + 4 y + 3 = 0. Let's find the roots of the quadratic equation: y = − 1 And y = − 3.
Now let's do the reverse replacement. We get two equations x 2 + 3 x = − 1 And x 2 + 3 · x = − 3 . Let's rewrite them as x 2 + 3 x + 1 = 0 and x 2 + 3 x + 3 = 0. We use the formula for the roots of a quadratic equation in order to find the roots of the first equation from those obtained: - 3 ± 5 2. The discriminant of the second equation is negative. This means that the second equation has no real roots.
Answer:- 3 ± 5 2
Entire equations of high degrees appear in problems quite often. There is no need to be afraid of them. You need to be ready to use a non-standard method for solving them, including a number of artificial transformations.
Solving fractional rational equations
We will begin our consideration of this subtopic with an algorithm for solving fractionally rational equations of the form p (x) q (x) = 0, where p(x) And q(x)– whole rational expressions. The solution of other fractionally rational equations can always be reduced to the solution of equations of the indicated type.
The most commonly used method for solving the equations p (x) q (x) = 0 is based on the following statement: numerical fraction u v, Where v- this is a number that is different from zero, equal to zero only in those cases when the numerator of the fraction is equal to zero. Following the logic of the above statement, we can claim that the solution to the equation p (x) q (x) = 0 can be reduced to fulfilling two conditions: p(x)=0 And q(x) ≠ 0. This is the basis for constructing an algorithm for solving fractional rational equations of the form p (x) q (x) = 0:
- find the solution to the whole rational equation p(x)=0;
- we check whether the condition is satisfied for the roots found during the solution q(x) ≠ 0.
If this condition is met, then the found root. If not, then the root is not a solution to the problem.
Example 6
Let's find the roots of the equation 3 · x - 2 5 · x 2 - 2 = 0 .
Solution
We are dealing with a fractional rational equation of the form p (x) q (x) = 0, in which p (x) = 3 x − 2, q (x) = 5 x 2 − 2 = 0. Let's start solving the linear equation 3 x − 2 = 0. The root of this equation will be x = 2 3.
Let's check the found root to see if it satisfies the condition 5 x 2 − 2 ≠ 0. To do this, substitute a numerical value into the expression. We get: 5 · 2 3 2 - 2 = 5 · 4 9 - 2 = 20 9 - 2 = 2 9 ≠ 0.
The condition is met. It means that x = 2 3 is the root of the original equation.
Answer: 2 3 .
There is another option for solving fractional rational equations p (x) q (x) = 0. Recall that this equation is equivalent to the whole equation p(x)=0 on the range of permissible values of the variable x of the original equation. This allows us to use the following algorithm in solving the equations p (x) q (x) = 0:
- solve the equation p(x)=0;
- find the range of permissible values of the variable x;
- we take the roots that lie in the range of permissible values of the variable x as the desired roots of the original fractional rational equation.
Solve the equation x 2 - 2 x - 11 x 2 + 3 x = 0.
Solution
First, let's solve the quadratic equation x 2 − 2 x − 11 = 0. To calculate its roots, we use the roots formula for the even second coefficient. We get D 1 = (− 1) 2 − 1 · (− 11) = 12, and x = 1 ± 2 3 .
Now we can find the ODZ of variable x for the original equation. These are all the numbers for which x 2 + 3 x ≠ 0. It's the same as x (x + 3) ≠ 0, from where x ≠ 0, x ≠ − 3.
Now let’s check whether the roots x = 1 ± 2 3 obtained at the first stage of the solution are within the range of permissible values of the variable x. We see them coming in. This means that the original fractional rational equation has two roots x = 1 ± 2 3.
Answer: x = 1 ± 2 3
The second solution method described is simpler than the first in cases where the range of permissible values of the variable x is easily found, and the roots of the equation p(x)=0 irrational. For example, 7 ± 4 · 26 9. The roots can be rational, but with a large numerator or denominator. For example, 127 1101 And − 31 59 . This saves time on checking the condition q(x) ≠ 0: It is much easier to exclude roots that are not suitable according to the ODZ.
In cases where the roots of the equation p(x)=0 are integers, it is more expedient to use the first of the described algorithms for solving equations of the form p (x) q (x) = 0. Find the roots of an entire equation faster p(x)=0, and then check whether the condition is satisfied for them q(x) ≠ 0, 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.
Example 8
Find the roots of the equation (2 x - 1) (x - 6) (x 2 - 5 x + 14) (x + 1) x 5 - 15 x 4 + 57 x 3 - 13 x 2 + 26 x + 112 = 0.
Solution
Let's start by looking at the whole equation (2 x − 1) (x − 6) (x 2 − 5 x + 14) (x + 1) = 0 and finding its roots. To do this, we apply the method of solving equations through factorization. It turns out that the original 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, of which three are linear and one is quadratic. Finding roots: from the first equation x = 1 2, from the second – x=6, from the third – x = 7 , x = − 2 , from the fourth – x = − 1.
Let's check the obtained roots. It is difficult for us to determine the ODZ in this case, since for this we will have to solve an algebraic equation of the fifth degree. It will be easier to check the condition according to which the denominator of the fraction, which is on the left side of the equation, should not go to zero.
Let’s take turns substituting the roots for the variable x in the expression x 5 − 15 x 4 + 57 x 3 − 13 x 2 + 26 x + 112 and calculate its value:
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 .
The verification carried out allows us to establish that the roots of the original fractional rational equation are 1 2, 6 and − 2 .
Answer: 1 2 , 6 , - 2
Example 9
Find the roots of the fractional rational equation 5 x 2 - 7 x - 1 x - 2 x 2 + 5 x - 14 = 0.
Solution
Let's start working with the equation (5 x 2 − 7 x − 1) (x − 2) = 0. Let's find its roots. It’s easier for us to imagine this equation as a set of quadratic and linear equations 5 x 2 − 7 x − 1 = 0 And x − 2 = 0.
We use the formula for the roots of a quadratic equation to find the roots. We obtain from the first equation two roots x = 7 ± 69 10, and from the second x = 2.
It will be quite difficult for us to substitute the value of the roots into the original equation to check the conditions. It will be easier to determine the ODZ of the variable x. In this case, the ODZ of the variable x is all numbers except those for which the condition is met x 2 + 5 x − 14 = 0. We get: x ∈ - ∞, - 7 ∪ - 7, 2 ∪ 2, + ∞.
Now let's check whether the roots we found belong to the range of permissible values of the variable x.
The roots x = 7 ± 69 10 belong, therefore, they are the roots of the original equation, and x = 2- does not belong, therefore, it is an extraneous root.
Answer: x = 7 ± 69 10 .
Let us examine separately the cases when the numerator of a fractional rational equation of the form p (x) q (x) = 0 contains a number. In such cases, if the numerator contains a number other than zero, then the equation will have no roots. If this number is equal to zero, then the root of the equation will be any number from the ODZ.
Example 10
Solve the fractional rational equation - 3, 2 x 3 + 27 = 0.
Solution
This equation will not have roots, since the numerator of the fraction on the left side of the equation contains a non-zero number. This means that at no value of x will the value of the fraction given in the problem statement be equal to zero.
Answer: no roots.
Example 11
Solve the equation 0 x 4 + 5 x 3 = 0.
Solution
Since the numerator of the fraction contains zero, the solution to the equation will be any value x from the ODZ of the variable x.
Now let's define the ODZ. It will include all values of x for which x 4 + 5 x 3 ≠ 0. 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 this in turn is equivalent to the combination of two equations x 3 = 0 and x + 5 = 0, where these roots are visible. We come to the conclusion that the desired range of acceptable values are any x except x = 0 And x = − 5.
It turns out that the fractional rational equation 0 x 4 + 5 · x 3 = 0 has an infinite number of solutions, which are any numbers other than zero and - 5.
Answer: - ∞ , - 5 ∪ (- 5 , 0 ∪ 0 , + ∞
Now let's talk about fractional rational equations of arbitrary form and methods for solving them. They can be written as r(x) = s(x), Where r(x) And s(x)– rational expressions, and at least one of them is fractional. Solving such equations reduces to solving equations of the form p (x) q (x) = 0.
We already know that we can obtain an equivalent equation by transferring an expression from the right side of the equation to the left with the opposite sign. This means that the equation r(x) = s(x) is equivalent to the equation r (x) − s (x) = 0. We have also already discussed ways to convert a rational expression into a rational fraction. Thanks to this, we can easily transform the equation r (x) − s (x) = 0 into an identical rational fraction of the form p (x) q (x) .
So we move from the original fractional rational equation r(x) = s(x) to an equation of the form p (x) q (x) = 0, which we have already learned to solve.
It should be taken into account that when making transitions from r (x) − s (x) = 0 to p(x)q(x) = 0 and then to p(x)=0 we may not take into account the expansion of the range of permissible values of the variable x.
It is quite possible that the original equation r(x) = s(x) and equation p(x)=0 as a result of the transformations they will cease to be equivalent. Then the solution to the equation p(x)=0 can give us roots that will be foreign to r(x) = s(x). In this regard, in each case it is necessary to carry out verification using any of the methods described above.
To make it easier for you to study the topic, we have summarized all the information into an algorithm for solving a fractional rational equation of the form r(x) = s(x):
- we transfer the expression from the right side with the opposite sign and get zero on the right;
- transform the original expression into a rational fraction p (x) q (x) , sequentially performing operations with fractions and polynomials;
- solve the equation p(x)=0;
- We identify extraneous roots by checking their belonging to the ODZ or by substitution into the original equation.
Visually, the chain of actions will look like this:
r (x) = s (x) → r (x) - s (x) = 0 → p (x) q (x) = 0 → p (x) = 0 → elimination EXTERNAL ROOTS
Example 12
Solve the fractional rational equation x x + 1 = 1 x + 1 .
Solution
Let's move on to the equation x x + 1 - 1 x + 1 = 0. Let's transform the fractional rational expression on the left side of the equation to the form p (x) q (x) .
To do this, we will have to reduce rational fractions to a common denominator and simplify the expression:
x x + 1 - 1 x - 1 = x x - 1 (x + 1) - 1 x (x + 1) x (x + 1) = = x 2 - x - 1 - x 2 - x x · (x + 1) = - 2 · x - 1 x · (x + 1)
In order to find the roots of the equation - 2 x - 1 x (x + 1) = 0, we need to solve the equation − 2 x − 1 = 0. We get one root x = - 1 2.
All we have to do is check using any of the methods. Let's look at both of them.
Let's substitute the resulting value into the original equation. We get - 1 2 - 1 2 + 1 = 1 - 1 2 + 1. We have arrived at the correct numerical equality − 1 = − 1 . It means that x = − 1 2 is the root of the original equation.
Now let's check through the ODZ. Let us determine the range of permissible values of the variable x. This will be the entire set of numbers, with the exception of − 1 and 0 (at x = − 1 and x = 0, the denominators of the fractions vanish). The root we obtained x = − 1 2 belongs to ODZ. This means that it is the root of the original equation.
Answer: − 1 2 .
Example 13
Find the roots of the equation x 1 x + 3 - 1 x = - 2 3 · x.
Solution
We are dealing with a fractional rational equation. Therefore, we will act according to the algorithm.
Let's move the expression from the right side to the left with the opposite sign: x 1 x + 3 - 1 x + 2 3 x = 0
Let's carry out the necessary transformations: x 1 x + 3 - 1 x + 2 3 · x = x 3 + 2 · x 3 = 3 · x 3 = x.
We arrive at the equation x = 0. The root of this equation is zero.
Let's check whether this root is extraneous to the original equation. Let's substitute the value into the original equation: 0 1 0 + 3 - 1 0 = - 2 3 · 0. As you can see, the resulting equation makes no sense. This means that 0 is an extraneous root, and the original fractional rational equation has no roots.
Answer: no roots.
If we have not included other equivalent transformations in the algorithm, this does not mean that they cannot be used. The algorithm is universal, but it is designed to help, not limit.
Example 14
Solve the equation 7 + 1 3 + 1 2 + 1 5 - x 2 = 7 7 24
Solution
The easiest way is to solve the given fractional rational equation according to the algorithm. But there is another way. Let's consider it.
Subtract 7 from the right and left sides, we get: 1 3 + 1 2 + 1 5 - x 2 = 7 24.
From this we can conclude that the expression in the denominator on the left side must be equal to the reciprocal of the number on the right side, that is, 3 + 1 2 + 1 5 - x 2 = 24 7.
Subtract 3 from both sides: 1 2 + 1 5 - x 2 = 3 7. By analogy, 2 + 1 5 - x 2 = 7 3, from where 1 5 - x 2 = 1 3, and then 5 - x 2 = 3, x 2 = 2, x = ± 2
Let us carry out a check to determine whether the roots found are the roots of the original equation.
Answer: x = ± 2
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School: Branch of Municipal Educational Institution Secondary School with. Svyatoslavka in the village. Vozdvizhenka
Subject: mathematics.
Curriculum – 5 hours per week (of which 3 hours are algebra, 2 hours are geometry)
Topic: The whole equation and its roots. Solving entire equations.
Lesson type: improving skills and abilities.
Lesson objectives:
didactic : systematization and generalization, expansion and deepening of students’ knowledge of solving entire equations with one variable above the second degree; preparing students to apply knowledge in non-standard situations, for the Unified State Exam.
developing : development of the student’s personality through independent creative work, development of student initiative; provide a stable motivational environment, interest in the topic being studied; develop the ability to generalize, correctly select methods for solving an equation;
educational: developing interest in studying mathematics, preparing students to apply knowledge in non-standard situations; cultivate the will and perseverance to achieve final results
Lesson steps | Time | Form | Teacher activities | Student activities | Note |
1.1.Org. Moment (Introductory and motivational part, in order to enhance the activities of students) |
(Annex 1) | Defines student readiness. Focuses students' attention. Quotes the lesson motto and epigraph to the lesson. | Listen, answer questions, draw conclusions, | ||
1.2. Checking homework Updating of reference knowledge |
Oral survey (Appendix 2-4) | Coordinates student activities | Give the definition of an equation, the roots of an equation, the concept of solving an equation They solve equations orally and isolate whole equations from them. |
formation of cognitive competence |
|
1.3. Goal setting and motivation | Planning | Motivates students Communicates lesson objectives | Name and write down topic of the lesson, set their own lesson goal. | formation of communicative competence |
|
2.1. Systematization of knowledge. Goals : teach short rational writing, practice the ability to draw conclusions and generalizations |
(Appendix 5) | Gives examples of entire equations of various types. | They listen, answer questions, draw conclusions, and explain how to solve entire equations. Compile and write a supporting summary for the lesson in a notebook. | formation of cognitive, communicative and social competencies |
|
2.2. physical education minute | Commenting | Comments on a set of eye exercises | Students repeat the exercises. | ||
2.3. Consolidation. Solving whole equations Goal: teach to operate with knowledge, develop flexibility in using knowledge | Practical activities (Appendix 6) | Organizes and controls student activities. Indicates different solutions | They solve entire equations in their notebooks, show the solution on the board, and check them. Draw conclusions | Consolidation formation of information and cognitive competencies |
|
3.1. Summing up the lesson | Reflection (Appendix 7) | Motivates students to summarize the lesson Gives grades. | Summarize the material studied. They draw a conclusion. Write down homework. Evaluate their work | Complete equations |
(Annex 1)
1.Organizational moment– goals and objectives of the lesson are set.
Guys! You will have a final certification in mathematics in the form of State Examination and Unified State Examination. To successfully pass the State Exam and the Unified State Exam, you must know mathematics not only at a minimum level, but also apply your knowledge in non-standard situations. In parts B and C of the Unified State Exam, equations of higher degrees are often found. Our task: systematization and generalization, expansion and deepening of knowledge on solving entire equations with one variable above the second degree; preparation for applying knowledge in non-standard situations, for the State Examination and the Unified State Exam.
Mottoour lesson: “The more I know, the more I can.”
Epigaph:
Who doesn't notice anything
He doesn't study anything.
Who doesn't study anything
He's always whining and bored.
(poet R. Seph).
An equation is the simplest and most common mathematical problem. You have accumulated some experience in solving various equations and we need to put our knowledge in order and understand the techniques for solving non-standard equations.
U the equations themselves are of interest for study. The earliest manuscripts indicate that techniques for solving linear equations were known in ancient Babylon and ancient Egypt. Quadratic equations could be solved about 2000 years ago BC. e. Babylonians.
Standard techniques and methods for solving elementary algebraic equations are an integral part of solving all types of equations.
In the simplest cases, solving an equation with one unknown is divided into two steps: transforming the equation to a standard one and solving the standard equation. It is impossible to completely algorithmize the process of solving equations, but it is useful to remember the most common techniques that are common to all types of equations. Many equations when using non-standard techniques are solved much shorter and easier.
We will focus our attention on them.
(Appendix 2)
Updating knowledge.
For homework you were given the task to repeat the topic of equations and how to solve them.
Ø What is the equation called? ( An equation containing a variable is called an equation with one variable)
Ø What is the root of an equation?(The value of the variable at which the equation turns into the correct numerical
equality.)
Ø What does it mean to solve an equation?(Find all its roots or prove that there are no roots.)
I suggest you solve several equations orally:
a) x2 = 0 e) x3 – 25x = 0
b) 3x – 6 = 0 g) x(x – 1)(x + 2) = 0
c) x2 – 9 = 0 h) x4 – x2 = 0
d) x2 = 1/36 i) x2 – 0.01 = 0.03
e) x2 = – 25 j) 19 – c2 = 10
Tell me, what unites these equations?(single variable, whole equations, etc.)
Ø What is an entire equation with one variable called? ( Equations in which the left and right sides are integers
expressions
Ø What is the degree of an entire equation called?(The degree of an equivalent equation of the form P(x) = 0, Where P(x) – polynomial
standard type)
Ø How many roots can a whole equation have with one variable 2nd, 3rd, 4th, P th degree(no more than 2, 3, 4, P)
Do I know methods for solving entire equations?
Do I know how to apply these methods?
Will I be able to solve equations on my own?
Did you feel comfortable during the lesson?
6. On “3” - table No. 1 + 1 equation from the remaining tables.
On “4” - table No. 1 + 1 equation from any two tables
On “5” - Table No. 1 + 1 equation from each remaining
tables
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Summarizing:
Filling out the self-assessment table
Grading
At home: complete the remaining unsolved equations from all tables.