Lesson on the topic: "Equations reducible to quadratic equations." Rational equations
Lesson-research on the topic "Equations reducible to quadratic equations"
“People should be valued according to the goals they set for themselves”
N.N. Miklouho-Maclay.
Goals:
Educational: bring into the system the students’ knowledge on this topic (repeat the theory, develop the ability to determine the type of equation and choose a rational way to solve this equation);
Developmental: intensive and creative thinking, desire to find a solution;
Educational: instilling interest in oral work, developing the skills of conscious assimilation of material.
show a way to solve equations by introducing a new variable.
Lesson progress
Today in class I would like to invite you to take a deeper look into the wonderful world of mathematics - into the world of equations, into the world of search, into the world of research.
To teach mathematics,
She must be loved first.
Let's see how things really stand in this regard in relation to all the equations we have studied.
But first let's check homework
Interactive whiteboard. Slide with solution (swap notebooks)
Secondary immersion in the topic
To understand mathematics,
You need to know it in detail.
As far as we know the topic in detail, we will try to understand the progress next job:
Slide. And remember, what is an equation? (“the curtain” reveals the correct answer)
(Equality containing an unknown).
What does it mean to solve an equation?
(This means finding all its roots or proving that they do not exist).
What is the root of the equation?
(The value of the variable at which the equation becomes true).
What types of equations do you know and can solve? (Linear, quadratic, fractional rational, biquadratic).
All methods of solving equations known to you can be figuratively represented in the form of “keys”. Lesson symbol – a bunch of keys –
“Linear equations”, “Quadratic equations”, “Fractional rational equations”, “Equations reducible to quadratic equations”. “Biquadratic equations”
(we hang the keys on the board)
Slide (faces) and types of equations
find the roots of the equations.
Solution on the board
x 4 – 10 x 2 + 9 = 0, biquadratic equation
(x-10) 2 -3(x-10)-4=0
Let us summarize our research work.
Conclusion: So, we solved two equations of different types using the same method - the method of introducing a new variable, where the original equation is reduced to a quadratic one.
Now let's try to create a solution algorithm
And our task is to try to “hone” this key, to learn to open the secret of the equations with such a key.
The actual creative part
To get you interested in mathematics,
You need to draw attention to it.
Let's try to see how much we can attract attention
Let's consider solving equations of higher degrees using
factorization.
Answer: -1; -0,5; 1.
To be friends with mathematics,
You need to be logical in everything.
There is no doubt that the equations that are proposed without logic are almost impossible to master. Now we will make sure of this.
What substitution can be made in each equation.
Now try to reduce this equation to a quadratic one, we have already defined the substitution (choose any equation for yourself if you wish) and check it.
Summing up.
Reflection
Today in the lesson we just tried to slightly “hone” our “key”, you still have a lot of work ahead of you to make this key work perfectly.
At home: Collection of GIA-2010 p. 151 No. 128,129, No. 130,131. Thank you for the lesson. It was interesting for me to work with you. I wish you good luck, new searches and discoveries.
4. Summing up the lesson.
What new did you learn in the lesson?
Which tasks were difficult? What do you remember?
How did the class work during the lesson?
Who did the best job?
Evaluate students' answers at the board.
Give marks for the lesson, justifying their assignment.
Lesson topic: Solving equations that reduce to quadratic.
Lesson objectives:
educational: Introduce students to the biquadratic equation, relying on students’ previous experience in solving quadratic equations, consolidate the ability to solve equations reduced to quadratic equations by substitution and determine which substitution is more rational to make.
developing: promote the development of attention, logical thinking, skills to analyze, compare and draw conclusions.
educating: developing the ability to plan work, look for rational ways to carry it out, and the ability to defend one’s opinion in a reasoned manner
Progress of the lesson.
1. Organizational moment.
Hello guys.
Among the most important sciences
There is only one most important one.
Learn algebra, it is the head of sciences,
Everyone needs it very much for life,
When you reach the heights of science,
You will know the value of your knowledge,
You will understand that algebras of beauty,
They will not be a bad treasure for life.
2. Lesson motivation.
The epigraph of our lesson is the words of Galileo Galilei “Without persistent mental work, no one can advance far in mathematics. But anyone who knows the joy of knowledge, who has seen the beauty of mathematics, will not regret the effort expended.” D In order to successfully solve equations that reduce to quadratic equations, you need to have a good knowledge of the theory of solving these same quadratic equations. Therefore, we will repeat the concepts and formulas necessary in the future. I. P. Pavlov “Learn the basics of science before you climb to its heights. Never take on the next one without mastering the previous one.”
3. Updating knowledge. Frontal survey, oral work with the class.
Test “Continue the phrase” (follow-up self-test and assessment of knowledge).
A quadratic equation is an equation of the form...
The roots of a quadratic equation are found using the formula...
The number of roots of a quadratic equation depends on...
A reduced quadratic equation is an equation of the form...
Ways to solve quadratic equations: ...
What equations are called fractional rational equations?
Algorithm for solving fractional rational equations.
The main property of proportion.
When does a fraction equal 0?
Solving the equation x-8x -9 = 0 using known methods.
4.Learning new material.
Biquadratic equations
| Bi quadratic equation: ax 4 +bx 2 + c = 0 |
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| Solution algorithm |
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| Make a variable replacement: | x 2 = t |
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| It will turn out: | at 2 + bt + c = 0 |
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| Find the roots of a quadratic equation: | t 1,2
= |
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| Reverse substitution: | ||
| If tk | no roots |
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| Thus, a biquadratic equation can have from 0 to 4 solutions. |
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| Questions: Show me general view biquadratic equation. Give an algorithm for solving a biquadratic equation. How many roots can a biquadratic equation have? |
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Consider the solution to the textbook example.
Decision No. 733(1, 2, 4)
Method for introducing a new variable
Suggest ways to solve the following equation:
Drawing up an algorithm for solving equations that reduce to quadratic ones.
Solution algorithm:
Enter variable replacement
Write a quadratic equation with a new variable
Solve a new quadratic equation
Equations reduced to quadratic.
Biquadratic equations
Preliminary preparation to the lesson:
students should be able to solve biquadratic equations and equations reduced to quadratic ones by introducing a new variable;
Students prepare reports in advance about the great Italian mathematicians.
Lesson objectives:
1) educational: consideration of methods for solving equations reducible to quadratic equations;
2) educational: developing skills in group work and conscious activity of students;
3) developing: development of students’ mental activity, interaction skills between students, and the ability to generalize the facts being studied.
Equipment: crossword grid on cards, cards, poster - travel plan, notes on the board, positive code, carbon copy.
Lesson type: lesson-trip around the country “Mathematics”.
Lesson progress
I. Organizational moment
The trip plan, which lists the names of the stations, is displayed on the slide.
Today we will go on a journey through the country of Mathematics. Let's stop in the city of Equations of the third and fourth degrees, continue our acquaintance with biquadratic equations, and hear reports about Italian mathematicians.
II. Traveling around the country "Mathematics"
1. Station for crossword lovers.
The grid with answers is pre-recorded on a positive code or on back side boards.
Each of you has cards with a crossword grid and questions. Place a blank sheet of paper and carbon paper under the card. Write your answers only in nominative case. Solve the crossword puzzle, hand in the cards, and do a self-test on the sheet.
Horizontal:
4.What is the expression b 4 – 4ac for a quadratic equation with coefficients a, b, c? (Discriminant.)
6. The value of the variable at which the equation turns into a true equality. (Root.)
8. Equation of the form ax 4 + bx 2 + c = 0, where A ≠ 0. (Biquadratic.)
9. French mathematician. (Viet.)
10. An equation in which the left and right sides are integer expressions. (Whole.)
11. An equation with one variable that has the same set of roots. (Equivalent.)
Vertical:
1. The set of roots of an equation. (Solution.)
2. Solution of the equation Oh 2 = 0. (Zero.)
3. Equality containing a variable. (Equation.)
5. Quadratic equation in which one of the coefficients b or c equals 0. (Incomplete.)
7. Quadratic equation in which the first coefficient is equal to one. (Added.)
2. Station "Istoricheskaya".
Checking homework.
We are with you at the Istoricheskaya station. We will hear student reports about the great Italian mathematicians. Listen carefully. For interesting addition You can also get a “5”.
Student. Italian mathematicians of the 16th century made a great contribution to the problem of solving equations of 3rd and 4th degrees. N. Tartaglia, A. Fiore, D. Cardano, L. Ferrari and others. In 1535, a scientific duel took place between A. Fiore and N. Tartaglia, in which the latter won a brilliant victory. In 2 hours he solved 30 problems proposed by A. Fiore, and A. Fiore himself could not solve a single one given to him by N. Tartaglia.
Teacher. Are there any extras? Who else has prepared reports about Italian mathematicians?
Messages prepared by students are heard. Each message will take 2-3 minutes.
Teacher. So, N. Tartaglia solved 30 problems in 2 hours. How many equations can you solve? What solutions will you choose?
3. City of Equations (oral part)
This is not just a city of Equations, but equations of the third and fourth degrees. You have to answer all the questions. Only by answering them can you move on.
Task 1. How would you solve the equations for each group?
1) X 3 – X = 0, X 3 + 9X = 0, X 4 – 4X 2 = 0, at 4 – 16 = 0.
2) 9at 3 - 18at 2 – y + 2 = 0, x 3 – 5X 2 + 16X – 80 = 0, 6at 4 – 3at 3 + 12at 2 – 6at = 0.
3) (at 2 – at + 1)(at 2 – at – 7) = 65, (X 2 + 2X) 2 – 2(X 2 + 2X) – 3 = 0,
(X 2 + X – 1)(X 2 + X + 2) = 40.
Answers:
Examples of group 1) are best solved by factoring by taking the common factor out of brackets or using abbreviated multiplication formulas.
Examples of group 2) are better solved by grouping and factorization.
Examples of group 3) are better solved by introducing a new variable and moving to a quadratic equation.
Task 2. What factor would you put out of brackets in the examples of group 1) task 1?
Answers: X(X 2 – 1) = 0,
X(X 2 + 9) = 0,
X 2 (X 2 – 4) = 0.
Task 3. How would you group the terms in the examples of group 2) of task 1?
Answers: (9at 3 – 18at 2) – (at – 2) = 0,
(X 3 – 5X 2) + (16X – 80) = 0,
(6at 4 – 3at 3) + (12at 2 – 6at) = 0.
Task 4. What would you denote by a new variable in the examples of group 3) of task 1?
Answers: at 2 – at = t,
x 2 + 2x = t,
x 2 + x = t.
Task 5. How can you factor a polynomial? at 4 – 16 = 0?
Answer: (at 2 – 4)(at 2 + 4) = (at – 2)(at + 2)(at 2 + 4) = 0.
4. City of Equations. Practical part.
You have completed the oral work in the city of Equations, and we are going to travel further on this interesting city and continue to get acquainted with interesting equations.
Task 6.
Two students complete tasks at the board at the same time.
A) The first student solves at the board with an explanation.
9X 3 – 18X 2 – x + 2 = 0.
b) The second student solves the equation silently, then explains the solution, the class listens and asks questions if something is not clear.
X 3 + X 2 – 4(X + 1) 2 = 0.
Task 7. Solve the equation (see appendix.)
The task is completed independently according to the options. Previously, together with the teacher, they consider possible replacements for introducing a new variable. Checked orally.
OptionI.
(X 2 + 2X) 2 – 2(X 2 + 2X) – 3 = 0.
X 2 + 2X = t.
OptionII.
(X 2 – X + 1)(X 2 – X – 7) = 0.
Substitution to introduce a new variable X 2 - X = t.
Task 8.
Additional task for those who can cope with the previous equations earlier.
(2X 2 + X – 1)(2X 2 + X – 4) + 2 = 0.
Substitution for introducing a new variable 2 X 2 + X = t.
Task 9. Solve the equation.
Students comment on the progress of the solution from their seats.
X 4 (X + 1) – 6X 2 (X + 1) + 5(X + 1) = 0.
Solution. Let's take out the common factor:
(X+ 1)(X 4 – 6X 2 + 5) = 0, whence X+ 1 = 0 or X 4 – 6X 2 + 5 = 0, i.e. or X= -1, or
X 4 – 6X 2 + 5 = 0. The last equation is biquadratic:
X 2 = t,
t 2 - 6 t + 5 = 0.
By the theorem inverse to Vieta’s theorem t 1 + t 2 = 6, t 1 · t 2 = 5. Hence t 1 =1, t 2 = 5. So X 2 = 1, or X 2 = 5, whence X 1.2 = ± 1, X 3.4 = ±.
Answer:- 1, 1, -, .
Task 10. Solve the equation.
First, the teacher discusses the solution with the class. The student then solves part of the example at the board.
(X + 1)(X + 2)(X + 3)(X + 4) = 360.
Solution. Let's group the factors first:
((X + 1)(X+ 4)) · (( X + 2)(X + 3)) = 360,
(X 2 + 5X + 4)(X 2 + 5X + 6) = 360,
Let X 2 + 5X= t, Then ( t + 4) ( t + 6) = 360.
t 2 + 10t + 24 – 360 = 0,
t + 10t – 336 = 0,
D= 100 + 4 336 = 1444 = 38 2.
Where t 1 = = 14, t 2 = = - 24.
Means, X 2 + 5X= 14 or X 2 + 5X= -24, i.e. X 2 + 5X– 14 = 0 or X 2 + 5X + 24 = 0.
In the second case D= 25 – 4 24 = -71
In the first case there are two roots X 1 = -7, X 2 = 2.
Answer: - 7; 2.
Task 11. Solve the equation. (See appendix.)
The one who correctly solves the most biquadratic equations in 10 minutes will receive a “5”. Students work independently followed by peer review.
A) X 4 – 5X 2 – 36 = 0,
b) at 4 – 6at 2 + 8 = 0,
c) 4 X 4 – 5X 2 + 1 = 0,
G) X 4 – 25X 2 + 144 = 0,
e) 5 at 4 – 5at 2 + 2 = 0,
e) t 4 – 2t 2 – 3 = 0.
Task 12. At what values A equation t 2 + at+ 9 = 0, has no roots? (See appendix.)
This example is for repetition.
5. Home station
You have arrived at Domashnyaya station. Get homework.
Task 13. Solve the equation of Italian mathematicians:
(3X 2 + X – 4) 2 + 3X 2 + X= 4. (see appendix.)
Task 14. Find and solve 3-4 equations proposed by A. Fiore and N. Tartaglia.
III. Summing up the lesson.
Our journey is over. So, count how many equations each of you solved.
In 2 lessons the whole class solved... equations. Lesson grades...
Application
Solutions
Task 6.
A) Solution.
9X 2 (X – 2) – (X – 2) = 0,
(X – 2)(9X 2 – 1) = 0,
X– 2 = 0, or 9 X 2 – 1 = 0,
X= 2 or X 2 = , i.e. X 1.2 = ±.
Answer: - ; ; 2.
b) Solution.
X 2 (X + 1) – 4(X + 1) 2 = 0,
(X + 1)(X 2 – 4X – 4) = 0,
X+ 1 = 0 or X 2 – 4X – 4 = 0,
X= - 1, or X 1,2 = = 2 .
Answer: - 1; 2 - 2; 2 + 2.
Task 7.
OptionI.
Solution. Replacement X 2 + 2X = t, Then:
t 2 – 2t – 3 = (t + 1)(t – 3) = 0.
X 2 + 2X= - 1 or X 2 + 2X= 3,
X 2 + 2X+ 1 = 0 or X 2 + 2X – 3 = 0,
(X+ 1) 2 = 0 or ( X + 3)(X– 1) = 0.
Answer: - 3; - 1, 1.
OptionII.
Solution. Replacement
Open algebra lesson in 9th grade.
Subject: Equations reduced to quadratic.
Lesson objectives: 1) generalization and deepening of students’ knowledge of solving quadratic equations;
2) promote the formation of skills to apply various methods of solving equations;
3) develop creativity students by solving tasks containing modules and parameters.
Lesson progress:
Introductory conversation.
When solving equations, students often make a number of transformations that lead to erroneous conclusions.
For example:
1. RU x(x+3)=2x
Divide both sides of the equation by x:
With this solution, the root X=0 was lost. What's wrong?
Divided by X, and the variable X can be equal to 0. But you cannot divide by zero.
Answer: -1; 0.
Because the denominators of both sides are the same, then
With this solution, an extraneous root X=1 appeared. Where is the error?
The common denominator cannot be 0.
Answer: X=2.
To avoid such mistakes, you need to know the rules of equivalent transitions when solving equations.
Oral survey.
What equations are called 1st degree equations?
How to solve linear equations?
How many solutions can a linear equation have?
Which equation is called an equation of the second degree?
Reduced quadratic equation?
How are quadratic equations solved?
How many roots can a quadratic equation have?
if D 0, then the quadratic equation has 2 roots.
if D = 0, then one root.
10.How to factor a quadratic trinomial?
3. Explanation of a new topic.
Today we will solve equations that can be reduced to quadratic equations and equations of degrees 3 and 4. Italian mathematicians of the 16th century made a major contribution to their solution.
Scipione Dal Ferro(1465-1526) and his student Fiori
Nicolo Tartaglia(1499-1557)
Historical information about these scientists.
Let's consider one of the equations of Italian mathematicians:
This equation can be solved using the Cardano formula for solving equations of the form , which can involve complex calculations.
Can be solved by factoring the left side of the equation.
Answer: 1; -4; 3.
Let's solve this equation in various ways:
factorization method.
both values satisfy the condition
Does not satisfy the condition
Answer: 0; -2; 2.
graphic method
Building a graph of a function
and look for the abscissa of the points of intersection of the graph with the ox axis.
3) Method of introducing a new variable.
Let it be then
This equation can also be solved in several ways.
Let's introduce a new variable
There are several classes of equations that can be solved by reducing them to quadratic equations. One such equation is biquadratic equations.
Biquadratic equations
Biquadratic equations are equations of the form a*x^4 + b*x^2 + c = 0, where a is not equal to 0.
Biquadratic equations are solved using the substitution x^2 =t. After such a substitution, we obtain a quadratic equation for t. a*t^2+b*t+c=0. We solve the resulting equation, and in the general case we have t1 and t2. If at this stage a negative root is obtained, it can be excluded from the solution, since we took t=x^2, and the square of any number is a positive number.
Returning to the original variables, we have x^2 =t1, x^2=t2.
x1,2 = ±√(t1), x3,4=±√(t2).
Let's look at a small example:
9*x^4+5*x^2 - 4 = 0.
Let's introduce the replacement t=x^2. Then the original equation will take the following form:
We solve this quadratic equation using any of the known methods and find:
The root -1 is not suitable, since the equation x^2 = -1 does not make sense.
The second root 4/9 remains. Moving on to the initial variables, we have the following equation:
x1=-2/3, x2=2/3.
This will be the solution to the equation.
Answer: x1=-2/3, x2=2/3.
Another type of equation that can be reduced to quadratic equations is fractional rational equations. Rational equations are equations whose left and right sides are rational expressions. If in a rational equation the left or right sides are fractional expressions, then such a rational equation is called fractional.
Scheme for solving a fractional rational equation
1. Find common denominator all the fractions that go into the equation.
2. Multiply both sides of the equation by a common denominator.
3. Solve the resulting whole equation.
4. Check the roots and exclude those that make the common denominator vanish.
Let's look at an example:
Solve the fractional rational equation: (x-3)/(x-5) + 1/x = (x+5)/(x*(x-5)).
We will stick to general scheme. Let's first find the common denominator of all fractions.
We get x*(x-5).
Let's multiply each fraction by a common denominator and write the resulting whole equation.
x*(x+3) + (x-5) = (x+5);
Let us simplify the resulting equation. We get,
x^2+3*x + x-5 - x - 5 =0;
Received simple reduced quadratic equation. We solve it by any of the known methods, we get the roots x=-2 and x=5. Now we check the obtained solutions. Substitute the numbers -2 and 5 into the common denominator.
At x=-2, the common denominator x*(x-5) does not vanish, -2*(-2-5)=14. This means that the number -2 will be the root of the original fractional rational equation.
