How to reduce to a quadratic equation. Lesson on the topic: "Equations reducible to quadratic"

MUNICIPAL EDUCATIONAL INSTITUTION TUMANOVSKAYA SECONDARY SCHOOL OF MOSKALENSKY MUNICIPAL DISTRICT OF OMSK REGION

Lesson topic: EQUATIONS REDUCABLE TO SQUARE

Developed by teacher of mathematics and physics at Tumanovskaya secondary school BIRIKH TATYANA VIKTOROVNA

2008

Purpose of the lesson: 1) consider ways to solve equations reducible to quadratic ones; teach how to solve such equations. 2) develop students’ speech and thinking, attentiveness, and logical thinking. 3) instill an interest in mathematics,

Lesson type: Lesson on learning new material

Lesson plan: 1. organizational stage
2. oral work
3. practical work
4. summing up the lesson

PROGRESS OF THE LESSON
Today in the lesson we will get acquainted with the topic “Equations reducible to quadratic”. Each student must be able to correctly and rationally solve equations, learn to apply various ways when solving the given quadratic equations.
1. Oral work 1. Which of the numbers: -3, -2, -1, 0, 1, 2, 3 are the roots of the equation: a) x 3 – x = 0; b) y 3 – 9y = 0; c) y 3 + 4y = 0? - How many solutions can a third-degree equation have? - What method did you use to solve these equations?2. Check the solution to the equation: x 3 - 3x 2 + 4x – 12 = 0 x 2 (x - 3) + 4 (x - 3) = 0(x - 3) (x 2 + 4) = 0 (x - 3) (x - 2) (x + 2) = 0 Answer: x = 3, x = -2, x = 2 Students explain the mistake they made. I summarize the oral work. So, you were able to solve the three proposed equations orally and find the mistake made when solving the fourth equation. When solving equations orally, the following two methods were used: placing the common factor outside the bracket sign and factoring. Now let's try to apply these methods when doing written work.
2. Practical work 1. One student solves the equation on the board 25x 3 – 50x 2 – x + 2 = 0 When deciding, he turns special attention to change the characters in the second bracket. He recites the entire solution and finds the roots of the equation.2. I suggest that stronger students solve the equation x 3 – x 2 – 4(x - 1) 2 = 0. When checking a solution, I draw students’ special attention to the most important points.3. Work on the board. Solve the equation (x 2 + 2x) 2 – 2(x 2 + 2x) – 3 = 0 When solving this equation, students find out that it is necessary to use a “new” method - introducing a new variable.Let us denote by the variable y = x 2 + 2x and substitute it into this equation. y 2 – 2y – 3 = 0. Let's solve the quadratic equation for the variable y. Then we find the value of the variable x.4 . Consider the equation (x 2 – x + 1) (x 2 – x - 7) = 65. Let's answer the questions:- what degree is this equation?- what solution method is most rational to use to solve it?- what new variable should be introduced? (x 2 – x + 1) (x 2 – x - 7) = 65 Let's denote y = x 2 – x (y + 1) (y – 7) = 65Next, the class solves the equation independently. We check the solutions to the equation at the board.5. For strong students, I suggest solving the equation x 6 – 3x 4 – x 2 – 3 = 0 Answer: -1, 1 6. The equation (2x 2 + 7x - 8) (2x 2 + 7x - 3) – 6 = 0 class proposes to solve as follows: the strongest students - solve independently; for the rest, one of the students on the board decides.Solve: 2x 2 + 7x = y(y - 8) (y - 3) – 6 = 0 We find: y1 = 2, y2 = 9 Substitute into our equation and find the values ​​of x, for this we solve the equations:2x 2 + 7x = 2 2x 2 + 7x = 9As a result of solving two equations, we find four values ​​of x, which are the roots of this equation.7. At the end of the lesson, I propose to solve the equation x 6 – 1 = 0 orally. When solving it is necessary to apply the difference of squares formula; we can easily find the roots.(x 3) 2 – 1 = 0 (x 3 - 1) (x 3 + 1) = 0 Answer: -1, 1.
3. Summing up the lesson Once again I draw the students’ attention to the methods that were used to solve equations reduced to quadratic equations. Students’ work in class is evaluated, I comment on the grades and point out mistakes made. We write down our homework. As a rule, the lesson proceeds at a fast pace, and the students’ performance is high. Thank you all very much for the good work.

Lesson #1

Lesson type: lesson of learning new material.

Lesson format: conversation.

Target: develop the ability to solve equations reduced to quadratic ones.

Tasks:

• introduce students to one of the ways to solve equations;
• develop skills in solving such equations;
• create conditions for the formation of interest in the subject and the development of logical thinking;
• to ensure personal and humane relationships between participants in the educational process.

Lesson plan:

1. Organizational moment.

3. Studying new material.
4. Consolidation of new material.
5. Homework.
6. Lesson summary.

PROGRESS OF THE LESSON

1. Organizational moment

Teacher:“Guys, today we begin to study the important and interesting topic"Equations Reducible to Quadratic Equations." You know the concept of a quadratic equation. Let's remember what we know on this topic."

Schoolchildren are given instructions:

• Remember the definitions associated with this topic.
• Remember the methods for solving known equations.
• Remember your difficulties when completing tasks on topics that are “close” to this one.
• Remember ways to overcome difficulties.
• Think through possible research tasks and ways to complete them.
• Remember where previously solved problems were applied.

Students recall the form of a complete quadratic equation, an incomplete quadratic equation, conditions for solving a complete quadratic equation, methods for solving incomplete quadratic equations, the concept of a whole equation, the concept of a degree.

The teacher suggests solving the following equations (work in pairs):

a) x 2 – 10x + 21 = 0
b) 3x 2 + 6x + 8 = 0
c) x (x – 1) + x 2 (x – 1) = 0

One of the students comments on the solution to these equations.

3. Learning new material

The teacher suggests considering and solving the following equation (problem problem):

(x 2 – 5x + 4) (x 2 – 5x + 6) = 120

Students talk about the power of a given equation and suggest multiplying these factors. But there are students who notice the same terms in this equation. What solution method can be applied here?
The teacher invites the students to turn to the textbook (Yu. N. Makarychev “Algebra-9”, paragraph 11, p. 63) and figure out the solution to this equation. The class is divided into two groups. Those students who understand the solution method perform the following tasks:

a) (x 2 + 2x) (x 2 +2x + 2) = –1
b) (x 2 – 7) 2 – 4 (x 2 – 7) – 45 = 0,

the rest are solution algorithm such equations and analyze the solution to the next equation together with the teacher.

(2x 2 + 3) 2 – 12(2x 2 + 3) + 11 = 0.

Algorithm:

– enter a new variable;
– create an equation containing this variable;
– solve the equation;
– substitute the found roots into the substitution;
– solve the equation with the initial variable;
– check the roots found, write down the answer.

4. Consolidation of new material

Work in pairs: “strong” explains, “weak” repeats, decides.

Solve the equation:

a) 9x 3 – 27x 2 = 0
b) x 4 – 13x 2 + 36 = 0

Teacher:“Let's remember where else we used solving quadratic equations?”

Students:“When solving inequalities; when finding the domain of definition of a function; when solving equations with a parameter.”
The teacher offers optional assignments. The class is divided into 4 groups. Each group explains the solution to their task.

a) Solve the equation:
b) Find the domain of definition of the function:
c) At what values A the equation has no roots:
d) Solve the equation: x + – 20 = 0.

5. Homework

No. 221(a, b, c), No. 222(a, b, c).

The teacher suggests preparing messages:

1. “Historical information about the creation of these equations” (based on materials from the Internet).
2. Methods for solving equations on the pages of the Kvant magazine.

Creative tasks can be completed at will in separate notebooks:

a) x 6 + 2x 4 – 3x 2 = 0
b) (x 2 + x) / (x 2 + x – 2) – (x 2 + x – 5) / (x 2 + x – 4) = 1

6. Lesson summary

The guys tell us what they learned new in the lesson, what tasks caused difficulties, where they applied them, and how they evaluate their performance.

Lesson #2

Lesson type: lesson on consolidating skills and abilities.

Lesson format: lesson workshop.

Target: consolidate acquired knowledge, develop the ability to solve equations on this topic.

Tasks:

• develop the ability to solve equations reduced to quadratic ones;
• develop independent thinking skills;
• develop the ability to conduct analysis and search for missing information;
• cultivate activity, independence, discipline.

Lesson plan:

1. Organizational moment.
2. Updating the subjective experience of students.
3. Problem solving.
4. Independent work.
5. Homework.
6. Lesson summary.

PROGRESS OF THE LESSON

1. Organizational moment

Teacher:“In the last lesson we learned about equations that can be reduced to quadratic equations. Which mathematician contributed to the solution of equations of the third and fourth degrees?”

The student who prepared the message talks about Italian mathematicians of the 16th century.

2. Actualization of subjective experience

1) Checking homework

A student is called to the board and solves equations similar to home ones:

a) (x 2 – 10) 2 – 3 (x 2 – 10) – 4 = 0
b) x 4 – 10 x 2 + 9 = 0

At this time, to bridge gaps in knowledge, “weak” students receive cards. The “weak” student comments the solution to the “strong” student, the “strong” student marks the solution with “+” or “–” signs.

2) Repetition of theoretical material

Students are asked to fill out a table like:

Students fill out the third column at the end of the lesson.
The task completed on the board is checked. A sample solution remains on the board.

3. Problem solving

The teacher offers a choice of two groups of equations. The class is divided into two groups. One performs tasks according to the model, the other looks for new methods for solving equations. If decisions cause difficulties, then students can turn to a model - reasoning.

a) (2x 2 + 3) 2 – 12 (2x 2 + 3) + 11 = 0 a) (5x – 63) (5 x – 18) = 550
b) x 4 – 4x 2 + 4 = 0 b) 2x 3 – 7 x 2 + 9 = 0

The first group comments on their solution, the second group checks the solution through an overhead projector and comments on their solution methods.

Teacher: Guys, let's look at one interesting equation: (x 2 – 6 x – 9) 2 = x (x 2 – 4 x – 9).

– What method do you propose to solve it?

Students begin to discuss the problem problem in groups. They propose to open the brackets, bring similar terms, obtain a whole algebraic equation of the fourth degree, and find whole roots, if any, among the divisors of the free term; then factorize and find the roots of this equation.
The teacher approves the solution algorithm and suggests considering another solution method.

Let us denote x 2 – 4x – 9 = t, then x 2 – 6x – 9 = t – 2x. We obtain the equation t 2 – 5tx + 4x 2 = 0 and solve it for t.

The original equation breaks down into a set of two equations:

x 2 – 4 x – 9 = 4x x = – 1
x 2 – 4 x – 9 = x x = 9
x = (5 + 61)/2 x = (5 – 61)/2

4. Independent work

Students are offered the following equations to choose from:

a) x 4 – 6 x 2 + 5 = 0 a) (1 – y 2) + 7 (1 – y 2) + 12 = 0
b) (x 2 + x) 2 – 8 (x 2 + x) + 12 = 0 b) x 4 + 4 x 2 – 18 x 2 – 12 x + 9 = 0
c) x 6 + 27 x 4 – 28 = 0

The teacher comments on the equations of each group, drawing attention to the fact that the equation under point c) allows students to deepen their knowledge and skills.
Independent work is done on paper using carbon paper.
Students check solutions through an overhead projector after exchanging notebooks.

5. Homework

No. 223(g, e, f), No. 224(a, b) or No. 225, No. 226.

Creative task.

Determine the degree of the equation and derive the Vieta formula for this equation:

6. Lesson summary

Students return to filling out the “I learned” column of the table.

Lesson #3

Lesson type: lesson of review and systematization of knowledge.

Lesson format: lesson is a competition.

Objective of the lesson: learn to correctly evaluate your knowledge and skills, correctly correlate your capabilities with the tasks offered.

Tasks:

• teach you how to apply your knowledge comprehensively;
• identify the depth and strength of skills and abilities;
• promote rational organization of labor;
• foster activity and independence.

Lesson plan:

1. Organizational moment.
2. Updating the subjective experience of students.
3. Problem solving.
4. Independent work.
5. Homework.
6. Lesson summary.

PROGRESS OF THE LESSON

1. Organizational moment

Teacher:“Today we will conduct an unusual lesson, a competition lesson. You are already familiar with the Italian mathematicians Fiori, N. Tartaglia, L. Ferrari, D. Cardano from the last lesson.

On February 12, 1535, a scientific duel took place between Fiori and N. Tartaglia, in which Tartaglia won a brilliant victory. In two hours he solved all thirty problems proposed by Fiori, while Fiori did not solve a single problem of Tartaglia.
How many equations can you solve in a lesson? What methods should you choose? Italian mathematicians offer you their equations.”

2. Actualization of subjective experience

Oral work

1) Which of the numbers: – 3, – 2, – 1, 0, 1, 2, 3 are the roots of the equation:

a) x 3 – x = 0 b) y 3 – 9 y = 0 c) y 3 + 4 y = 0?

– How many solutions can a third-degree equation have?
– What method will you use to solve these equations?

2) Check the solution to the equation. Find the mistake you made.

x 3 – 3x 2 + 4x – 12 = 0
x 2 (x – 3) + 4 (x – 3) = 0
(x – 3)(x 2 + 4) = 0
(x – 3)(x + 2)(x – 2) = 0
x = 3, x = – 2, x = 2.

Work in pairs. Students explain how to solve equations and the mistake they made.

Teacher:“You guys are great! You have completed the first task of Italian mathematicians.”

3. Problem solving

Two students at the blackboard:

a) Find the coordinates of the points of intersection with the coordinate axes of the graph of the function:

b) Solve the equation:

Students in the class choose to complete one or two tasks. Students at the board consistently comment on their actions.

4. “End-to-end” independent work

The set of cards is compiled according to difficulty level and with answer options.

1) x 4 – x 2 – 12 = 0
2) 16 x 3 – 32 x 2 – x + 2 = 0
3) (x 2 + 2 x) 2 – 7 (x 2 + 2 x) – 8 = 0
4) (x 2 + 3 x + 1) (x 2 + 3 x + 3) = – 1
5) x 4 + x 3 – 4 x 2 + x + 1 = 0

Possible answers:

1) a) – 2; 2 b) – 3; 3 c) no solution
2) a) – 1/4; 1/4 b) – 1/4; 1/4; 2 c) 1/4; 2
3) a) – 4; 1; 2 b) –1; 1; – 4; 2 c) – 4; 2
4) a) – 2; – 1; b) – 2; – 1; 1 c) 1; 2
5) a) – 1; (– 3 + 5) /2 b) 1; (– 3 – 5) /2 c) 1; (– 3 – 5)/2; (–3 + 5) /2.

5. Homework

Collection of tasks for a written exam in algebra: No. 72, No. 73 or No. 76, No. 78.

Additional task. Determine the value of the parameter a for which the equation x 4 + (a 2 – a + 1) x 2 – a 3 – a = 0

a) has a single root;
b) has two different roots;
c) has no roots.

Quadratic equation or an equation of the second degree with one unknown is an equation that, after transformations, can be reduced to the following form:

ax 2 + bx + c = 0 - quadratic equation

Where x- this is the unknown, but a, b And c- coefficients of the equation. In quadratic equations a called the first coefficient ( a ≠ 0), b is called the second coefficient, and c called a known or free member.

Equation:

ax 2 + bx + c = 0

called complete quadratic equation. If one of the coefficients b or c is equal to zero, or both of these coefficients are equal to zero, then the equation is presented in the form of an incomplete quadratic equation.

Reduced quadratic equation

The complete quadratic equation can be reduced to more convenient view, dividing all its members into a, that is, for the first coefficient:

Equation x 2 + px + q= 0 is called a reduced quadratic equation. Therefore, any quadratic equation in which the first coefficient is equal to 1 can be called reduced.

For example, the equation:

x 2 + 10x - 5 = 0

is reduced, and the equation:

3x 2 + 9x - 12 = 0

can be replaced by the above equation, dividing all its terms by -3:

x 2 - 3x + 4 = 0

Solving Quadratic Equations

To solve a quadratic equation, you need to reduce it to one of the following forms:

ax 2 + bx + c = 0

ax 2 + 2kx + c = 0

x 2 + px + q = 0

For each type of equation there is its own formula for finding roots:

Notice the equation:

ax 2 + 2kx + c = 0

this is the transformed equation ax 2 + bx + c= 0, in which the coefficient b- even, which allows you to replace it with type 2 k. Therefore, the formula for finding the roots for this equation can be simplified by substituting 2 into it k instead of b:

Example 1. Solve the equation:

3x 2 + 7x + 2 = 0

Since in the equation the second coefficient is not even number, and the first coefficient is not equal to unity, then we will look for roots using the very first formula, called general formula finding the roots of a quadratic equation. At first

a = 3, b = 7, c = 2

Now, to find the roots of the equation, we simply substitute the values ​​of the coefficients into the formula:

x 1 =	-2	= -	1	, x 2 =	-12	= -2
	6		3		6

Answer: -	1	, -2.
	3

Example 2:

x 2 - 4x - 60 = 0

Let's determine what the coefficients are:

a = 1, b = -4, c = -60

Since the second coefficient in the equation is an even number, we will use the formula for quadratic equations with an even second coefficient:

x 1 = 2 + 8 = 10, x 2 = 2 - 8 = -6

Answer: 10, -6.

Example 3.

y 2 + 11y = y - 25

Let's reduce the equation to general appearance:

y 2 + 11y = y - 25

y 2 + 11y - y + 25 = 0

y 2 + 10y + 25 = 0

Let's determine what the coefficients are:

a = 1, p = 10, q = 25

Since the first coefficient is equal to 1, we will look for roots using the formula for the above equations with an even second coefficient:

Answer: -5.

Example 4.

x 2 - 7x + 6 = 0

Let's determine what the coefficients are:

a = 1, p = -7, q = 6

Since the first coefficient is equal to 1, we will look for roots using the formula for the above equations with an odd second coefficient:

x 1 = (7 + 5) : 2 = 6, x 2 = (7 - 5) : 2 = 1

General theory of problem solving using equations

Before moving on to specific types of problems, we first present general theory to solve various problems using equations. First of all, problems in such disciplines as economics, geometry, physics and many others are reduced to equations. General procedure to solve problems using equations is as follows:

• All the quantities we are looking for from the problem conditions, as well as any auxiliary ones, are denoted by variables convenient for us. Most often these variables are last letters Latin alphabet.
• Using the numerical values ​​given in the problem, as well as verbal relationships, one or more equations are compiled (depending on the conditions of the problem).
• They solve the resulting equation or their system and throw out “illogical” solutions. For example, if you need to find the area, then a negative number will obviously be an extraneous root.
• We get the final answer.

Example problem in algebra

Here we give an example of a problem that reduces to quadratic equation without relying on any specific area.

Example 1

Find two such irrational numbers, when adding the squares, the result will be five, and when they are added together in the usual way, three will be obtained.

Let's denote these numbers by the letters $x$ and $y$. According to the conditions of the problem, it is quite easy to create two equations $x^2+y^2=5$ and $x+y=3$. We see that one of them is square. To find a solution you need to solve the system:

$\cases(x^2+y^2=5,\\x+y=3.)$

First we express from the second $x$

Substituting into the first and performing elementary transformations

$(3-y)^2 +y^2=5$

$9-6y+y^2+y^2=5$

We moved on to solving the quadratic equation. Let's do this using formulas. Let's find the discriminant:

First root

$y=\frac(3+\sqrt(17))(2)$

Second root

$y=\frac(3-\sqrt(17))(2)$

Let's find the second variable.

For the first root:

$x=3-\frac(3+\sqrt(17))(2)=\frac(3-\sqrt(17))(2)$

For the second root:

$x=3-\frac(3-\sqrt(17))(2)=\frac(3+\sqrt(17))(2)$

Since the sequence of numbers is not important to us, we get one pair of numbers.

Answer: $\frac(3-\sqrt(17))(2)$ and $\frac(3+\sqrt(17))(2)$.

Example of a problem in physics

Let's consider an example of a problem leading to the solution of a quadratic equation in physics.

Example 2

A helicopter flying uniformly in calm weather has a speed of $250$ km/h. He needs to fly from his base to the place of the fire, which is located $70$ km away and return back. At this time, the wind was blowing towards the base, slowing down the helicopter’s movement towards the forest. Because of this, he got back to the base 1 hour earlier. Find the wind speed.

Let us denote the wind speed by $v$. Then we get that the helicopter will fly towards the forest with a real speed equal to $250-v$, and back its real speed will be $250+v$. Let's calculate the time for the journey there and the journey back.

$t_1=\frac(70)(250-v)$

$t_2=\frac(70)(250+v)$

Since the helicopter got back to the base $1$ hour earlier, we will have

$\frac(70)(250-v)-\frac(70)(250+v)=1$

Let's bring the left side to a common denominator, apply the rule of proportion and perform elementary transformations:

$\frac(17500+70v-17500+70v)((250-v)(250+v))=1$

$140v=62500-v^2$

$v^2+140v-62500=0$

We obtained a quadratic equation to solve this problem. Let's solve it.

We will solve it using a discriminant:

$D=19600+250000=269600≈519^2$

The equation has two roots:

$v=\frac(-140-519)(2)=-329.5$ and $v=\frac(-140+519)(2)=189.5$

Since we were looking for speed (which cannot be negative), it is obvious that the first root is superfluous.

Answer: $189.5$

Example problem in geometry

Let's consider an example of a problem leading to the solution of a quadratic equation in geometry.

Example 3

Find the area of ​​a right triangle that satisfies the following conditions: its hypotenuse is equal to $25$, and its legs are in the ratio of $4$ to $3$.

In order to find the required area we need to find the legs. Let us mark one part of the leg through $x$. Then, expressing the legs through this variable, we find that their lengths are equal to $4x$ and $3x$. Thus, from the Pythagorean theorem we can form the following quadratic equation:

$(4x)^2+(3x)^2=625$

(the root $x=-5$ can be ignored, since the leg cannot be negative)

We found that the legs are equal to $20$ and $15$, respectively, which means the area

$S=\frac(1)(2)\cdot 20\cdot 15=150$

There are several classes of equations that can be solved by reducing them to quadratic equations. One of these equations 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 the common denominator of all fractions that are included in 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).

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.