Quadratic equation - easy to solve! *Hereinafter referred to as “KU”. Friends, it would seem that there could be nothing simpler in mathematics than solving such an equation. But something told me that many people have problems with him. I decided to see how many on-demand impressions Yandex gives out per month. Here's what happened, look:
What does it mean? This means that about 70,000 people a month are looking for this information, and this is summer, and what will happen during the school year - there will be twice as many requests. This is not surprising, because those guys and girls who graduated from school a long time ago and are preparing for the Unified State Exam are looking for this information, and schoolchildren also strive to refresh their memory.
Despite the fact that there are a lot of sites that tell you how to solve this equation, I decided to also contribute and publish the material. Firstly, I want visitors to come to my site based on this request; secondly, in other articles, when the topic of “KU” comes up, I will provide a link to this article; thirdly, I’ll tell you a little more about his solution than is usually stated on other sites. Let's get started! The content of the article:
A quadratic equation is an equation of the form:
where coefficients a,band c are arbitrary numbers, with a≠0.
In the school course, the material is given in the following form - the equations are divided into three classes:
1. They have two roots.
2. *Have only one root.
3. They have no roots. It is worth especially noting here that they do not have real roots
How are roots calculated? Just!
We calculate the discriminant. Underneath this “terrible” word lies a very simple formula:
The root formulas are as follows:
*You need to know these formulas by heart.
You can immediately write down and solve:
Example:
1. If D > 0, then the equation has two roots.
2. If D = 0, then the equation has one root.
3. If D< 0, то уравнение не имеет действительных корней.
Let's look at the equation:
In this regard, when the discriminant is equal to zero, the school course says that one root is obtained, here it is equal to nine. Everything is correct, it is so, but...
This idea is somewhat incorrect. In fact, there are two roots. Yes, yes, don’t be surprised, you get two equal roots, and to be mathematically precise, then the answer should write two roots:
x 1 = 3 x 2 = 3
But this is so - a small digression. At school you can write it down and say that there is one root.
Now the next example:
As we know, the root of a negative number cannot be taken, so there is no solution in this case.
That's the whole decision process.
Quadratic function.
This shows what the solution looks like geometrically. This is extremely important to understand (in the future, in one of the articles we will analyze in detail the solution to the quadratic inequality).
This is a function of the form:
where x and y are variables
a, b, c – given numbers, with a ≠ 0
The graph is a parabola:
That is, it turns out that by solving a quadratic equation with “y” equal to zero, we find the points of intersection of the parabola with the x axis. There can be two of these points (the discriminant is positive), one (the discriminant is zero) and none (the discriminant is negative). Details about the quadratic function You can view article by Inna Feldman.
Let's look at examples:
Example 1: Solve 2x 2 +8 x–192=0
a=2 b=8 c= –192
D=b 2 –4ac = 8 2 –4∙2∙(–192) = 64+1536 = 1600
Answer: x 1 = 8 x 2 = –12
*It was possible to immediately divide the left and right sides of the equation by 2, that is, simplify it. The calculations will be easier.
Example 2: Decide x 2–22 x+121 = 0
a=1 b=–22 c=121
D = b 2 –4ac =(–22) 2 –4∙1∙121 = 484–484 = 0
We found that x 1 = 11 and x 2 = 11
It is permissible to write x = 11 in the answer.
Answer: x = 11
Example 3: Decide x 2 –8x+72 = 0
a=1 b= –8 c=72
D = b 2 –4ac =(–8) 2 –4∙1∙72 = 64–288 = –224
The discriminant is negative, there is no solution in real numbers.
Answer: no solution
The discriminant is negative. There is a solution!
Here we will talk about solving the equation in the case when a negative discriminant is obtained. Do you know anything about complex numbers? I will not go into detail here about why and where they arose and what their specific role and necessity in mathematics is; this is a topic for a large separate article.
The concept of a complex number.
A little theory.
A complex number z is a number of the form
z = a + bi
where a and b are real numbers, i is the so-called imaginary unit.
a+bi – this is a SINGLE NUMBER, not an addition.
The imaginary unit is equal to the root of minus one:
Now consider the equation:
We get two conjugate roots.
Incomplete quadratic equation.
Let's consider special cases, this is when the coefficient “b” or “c” is equal to zero (or both are equal to zero). They can be solved easily without any discriminants.
Case 1. Coefficient b = 0.
The equation becomes:
Let's transform:
Example:
4x 2 –16 = 0 => 4x 2 =16 => x 2 = 4 => x 1 = 2 x 2 = –2
Case 2. Coefficient c = 0.
The equation becomes:
Let's transform and factorize:
*The product is equal to zero when at least one of the factors is equal to zero.
Example:
9x 2 –45x = 0 => 9x (x–5) =0 => x = 0 or x–5 =0
x 1 = 0 x 2 = 5
Case 3. Coefficients b = 0 and c = 0.
Here it is clear that the solution to the equation will always be x = 0.
Useful properties and patterns of coefficients.
There are properties that allow you to solve equations with large coefficients.
Ax 2 + bx+ c=0 equality holds
a + b+ c = 0, That
- if for the coefficients of the equation Ax 2 + bx+ c=0 equality holds
a+ c =b, That
These properties help solve a certain type of equation.
Example 1: 5001 x 2 –4995 x – 6=0
The sum of the odds is 5001+( – 4995)+(– 6) = 0, which means
Example 2: 2501 x 2 +2507 x+6=0
Equality holds a+ c =b, Means
Regularities of coefficients.
1. If in the equation ax 2 + bx + c = 0 the coefficient “b” is equal to (a 2 +1), and the coefficient “c” is numerically equal to the coefficient “a”, then its roots are equal
ax 2 + (a 2 +1)∙x+ a= 0 = > x 1 = –a x 2 = –1/a.
Example. Consider the equation 6x 2 + 37x + 6 = 0.
x 1 = –6 x 2 = –1/6.
2. If in the equation ax 2 – bx + c = 0 the coefficient “b” is equal to (a 2 +1), and the coefficient “c” is numerically equal to the coefficient “a”, then its roots are equal
ax 2 – (a 2 +1)∙x+ a= 0 = > x 1 = a x 2 = 1/a.
Example. Consider the equation 15x 2 –226x +15 = 0.
x 1 = 15 x 2 = 1/15.
3. If in Eq. ax 2 + bx – c = 0 coefficient “b” is equal to (a 2 – 1), and coefficient “c” is numerically equal to the coefficient “a”, then its roots are equal
ax 2 + (a 2 –1)∙x – a= 0 = > x 1 = – a x 2 = 1/a.
Example. Consider the equation 17x 2 +288x – 17 = 0.
x 1 = – 17 x 2 = 1/17.
4. If in the equation ax 2 – bx – c = 0 the coefficient “b” is equal to (a 2 – 1), and the coefficient c is numerically equal to the coefficient “a”, then its roots are equal
ax 2 – (a 2 –1)∙x – a= 0 = > x 1 = a x 2 = – 1/a.
Example. Consider the equation 10x 2 – 99x –10 = 0.
x 1 = 10 x 2 = – 1/10
Vieta's theorem.
Vieta's theorem is named after the famous French mathematician Francois Vieta. Using Vieta's theorem, we can express the sum and product of the roots of an arbitrary KU in terms of its coefficients.
45 = 1∙45 45 = 3∙15 45 = 5∙9.
In total, the number 14 gives only 5 and 9. These are the roots. With a certain skill, using the presented theorem, you can solve many quadratic equations orally immediately.
Vieta's theorem, in addition. It is convenient in that after solving a quadratic equation in the usual way (through a discriminant), the resulting roots can be checked. I recommend doing this always.
TRANSPORTATION METHOD
With this method, the coefficient “a” is multiplied by the free term, as if “thrown” to it, which is why it is called "transfer" method. This method is used when the roots of the equation can be easily found using Vieta's theorem and, most importantly, when the discriminant is an exact square.
If A± b+c≠ 0, then the transfer technique is used, for example:
2X 2 – 11x+ 5 = 0 (1) => X 2 – 11x+ 10 = 0 (2)
Using Vieta's theorem in equation (2), it is easy to determine that x 1 = 10 x 2 = 1
The resulting roots of the equation must be divided by 2 (since the two were “thrown” from x 2), we get
x 1 = 5 x 2 = 0.5.
What is the rationale? Look what's happening.
The discriminants of equations (1) and (2) are equal:
If you look at the roots of the equations, you only get different denominators, and the result depends precisely on the coefficient of x 2:
The second (modified) one has roots that are 2 times larger.
Therefore, we divide the result by 2.
*If we reroll the three, we will divide the result by 3, etc.
Answer: x 1 = 5 x 2 = 0.5
Sq. ur-ie and Unified State Examination.
I’ll tell you briefly about its importance - YOU MUST BE ABLE TO DECIDE quickly and without thinking, you need to know the formulas of roots and discriminants by heart. Many of the problems included in the Unified State Examination tasks boil down to solving a quadratic equation (geometric ones included).
Something worth noting!
1. The form of writing an equation can be “implicit”. For example, the following entry is possible:
15+ 9x 2 - 45x = 0 or 15x+42+9x 2 - 45x=0 or 15 -5x+10x 2 = 0.
You need to bring it to a standard form (so as not to get confused when solving).
2. Remember that x is an unknown quantity and it can be denoted by any other letter - t, q, p, h and others.
Goals:
- Introduce the concept of a reduced quadratic equation;
- “discover” the relationship between the roots and coefficients of the given quadratic equation;
- develop an interest in mathematics, showing through the example of Viet’s life that mathematics can be a hobby.
1. Checking homework
No. 309(g) x 1 =7, x 2 =
No. 311(g) x 1 =2, x 2 =-1
No. 312 (d) no roots
2. Repetition of learned material
Everyone has a table on their table. Find the correspondence between the left and right columns of the table.
| Verbal formulation | Literal expression |
| 1. Square trinomial | A. ah 2 =0 |
| 2. Discriminant | B. ax 2 +c=0, s< 0 |
| 3. Incomplete quadratic equation having one root equal to 0. | IN. D > 0 |
| 4. An incomplete quadratic equation, one root of which is 0 and the other is not equal to 0. | G. D< 0 |
| 5. Not a complete quadratic equation, the roots of which are equal in magnitude but opposite in sign. | D. akh 2 +in+c=0 |
| 6. Not a complete quadratic equation that has no real roots. | E. D=v 2 +4ac |
| 7. General view of a quadratic equation. | AND. x 2 +px+q=0 |
| 8. Condition under which a quadratic equation has two roots | Z. ah 2 +in+s |
| 9. Condition under which a quadratic equation has no roots | AND. ax 2 +c=0, c > 0 |
| 10. Condition under which a quadratic equation has two equal roots | TO. akh 2 +in=0 |
| 11. Reduced quadratic equation. | L. D = 0 |
Enter the correct answers in the table.
1-Z; 2-E; 3-A; 4-K; 5 B; 6-I; 7-D; 8-B; 9-G; 10-L; 11-F.
3. Consolidation of the studied material
Solve the equations:
a) -5x 2 + 8x -3=0;
Solution:
D=64 – 4(-5)(-3) = 4,
x 1 = x 2 = = a + b + c = -5+8-3=0
b) 2 x 2 +6x – 8 = 0;
Solution:
D=36 – 4 2 (-8)= 100,
x 1 = = x 2 = a + b + c = 2+6-8=0
c) 2009 x 2 +x – 2010 =0
Solution:
a + b + c = 2009+1 + (-2010) =0, then x 1 =1 x 2 =
4. Expansion of the school course
ax 2 +in+c=0, if a+b+c=0, then x 1 =1 x 2 =Let's consider solving the equations
a) 2x 2 + 5x +3 = 0
Solution:
D = 25 -24 = 1 x 1 = x 2 = a – b + c = 2-5 + 3 = 0
b) -4x 2 -5x -1 =0
Solution:
D = 25 – 16 = 9 x 1 = – 1 x 2 = a – b + c = -4-(-5) – 1 = 0
c)1150x 2 +1135x -15 = 0
Solution:
a – b+c = 1150-1135 +(-15) = 0 x 1 = – 1 x 2 =
ax 2 +in+c=0, if a-b+c=0, then x 1 = – 1 x 2 =
5. New theme
Let's check your completion of the first task. What new concepts did you encounter? 11 – f, i.e.
The given quadratic equation is x 2 + px + q = 0.
The topic of our lesson.
Let's fill out the following table.
The left column is in the notebooks and one student is at the blackboard.
Solving the equation akh 2 +in+c=0
Right column, more prepared student at the blackboard
Solving the equation x 2 + px + q = 0, with a = 1, b = p, c = q
The teacher (if necessary) helps, the rest are in notebooks.
6. Practical part
X 2 – 6 X + 8 = 0,
D = 9 – 8 = 1,
x 1 = 3 – 1 = 2
x 2 = 3 + 1 = 4
X 2 + 6 X + 8 = 0,
D = 9 – 8 = 0,
x 1 = -3 – 1 = -4
x 2 = -3 + 1 = -2
X 2 + 20 X + 51 = 0,
D = 100 – 51 = 49
x 1 = 10 – 7 = 3
x 2 = 10 + 7 = 17
X 2 – 20 X – 69 = 0,
D = 100 – 69 = 31
Based on the results of our calculations, we will fill out the table.
Equation no. R x 1+ x 2 q x 1 x 2 1 -6 6 8 8
Let's compare the results obtained with the coefficients of quadratic equations.
What conclusion can be drawn?
7. Historical background
The relationship between the roots and coefficients of a quadratic equation was first established by the famous French scientist Francois Viète (1540–1603).
François Viète was a lawyer by profession and worked as an adviser to the king for many years. And although mathematics was his hobby, or as they say, a hobby, thanks to hard work he achieved great results in it. Viet in 1591 introduced letter notation for unknowns and coefficients of equations. This made it possible to write roots and other properties of the equation using general formulas.
The disadvantage of Vieta's algebra was that it only recognized positive numbers. To avoid negative solutions, he replaced equations or looked for artificial solutions, which took a lot of time, complicated the solution and often led to errors.
Viète made many different discoveries, but he himself most valued the establishment of the relationship between the roots and coefficients of a quadratic equation, that is, the relationship called “Viète’s theorem.”
We will consider this theorem in the next lesson.
8. Generalization of knowledge
Questions:
- Which equation is called a reduced quadratic equation?
- What formula can be used to find the roots of the given quadratic equation?
- What determines the number of roots of the given quadratic equation?
- What is the discriminant of a reduced quadratic equation?
- How are the roots of the above quadratic equation and its coefficients related?
- Who made this connection?
9. Homework
clause 4.5, No. 321(b,f) No.322(a,d,g,h)
Fill the table.
The equation Roots Sum of roots Product of roots X 2 – 8x + 7 = 0 1 and 7 8 7
Literature
CM. Nikolsky and others, “Algebra 8” textbook of the “MSU-School” series - M.: Prosveshchenie, 2007.
Bibliographic description: Gasanov A. R., Kuramshin A. A., Elkov A. A., Shilnenkov N. V., Ulanov D. D., Shmeleva O. V. Methods for solving quadratic equations // Young scientist. 2016. No. 6.1. P. 17-20..02.2019).
Our project is about ways to solve quadratic equations. Goal of the project: learn to solve quadratic equations in ways not included in the school curriculum. Task: find all possible ways to solve quadratic equations and learn how to use them yourself and introduce these methods to your classmates.
What are “quadratic equations”?
Quadratic equation- equation of the form ax2 + bx + c = 0, Where a, b, c- some numbers ( a ≠ 0), x- unknown.
The numbers a, b, c are called the coefficients of the quadratic equation.
- a is called the first coefficient;
- b is called the second coefficient;
- c - free member.
Who was the first to “invent” quadratic equations?
Some algebraic techniques for solving linear and quadratic equations were known 4000 years ago in Ancient Babylon. The discovery of ancient Babylonian clay tablets, dating from somewhere between 1800 and 1600 BC, provides the earliest evidence of the study of quadratic equations. The same tablets contain methods for solving certain types of quadratic equations.
The need to solve equations not only of the first, but also of the second degree, even in ancient times, was caused by the need to solve problems related to finding the areas of land plots and with excavation work of a military nature, as well as with the development of astronomy and mathematics itself.
The rule for solving these equations, set out in the Babylonian texts, essentially coincides with the modern one, but it is not known how the Babylonians arrived at this rule. Almost all cuneiform texts found so far provide only problems with solutions laid out in the form of recipes, with no indication as to how they were found. Despite the high level of development of algebra in Babylon, the cuneiform texts lack the concept of a negative number and general methods for solving quadratic equations.
Babylonian mathematicians from about the 4th century BC. used the square's complement method to solve equations with positive roots. Around 300 BC Euclid came up with a more general geometric solution method. The first mathematician who found solutions to equations with negative roots in the form of an algebraic formula was an Indian scientist Brahmagupta(India, 7th century AD).
Brahmagupta laid out a general rule for solving quadratic equations reduced to a single canonical form:
ax2 + bx = c, a>0
The coefficients in this equation can also be negative. Brahmagupta's rule is essentially the same as ours.
Public competitions in solving difficult problems were common in India. One of the old Indian books says the following about such competitions: “As the sun outshines the stars with its brilliance, so a learned man will outshine his glory in public assemblies by proposing and solving algebraic problems.” Problems were often presented in poetic form.
In an algebraic treatise Al-Khwarizmi a classification of linear and quadratic equations is given. The author counts 6 types of equations, expressing them as follows:
1) “Squares are equal to roots,” i.e. ax2 = bx.
2) “Squares are equal to numbers,” i.e. ax2 = c.
3) “The roots are equal to the number,” i.e. ax2 = c.
4) “Squares and numbers are equal to roots,” i.e. ax2 + c = bx.
5) “Squares and roots are equal to the number,” i.e. ax2 + bx = c.
6) “Roots and numbers are equal to squares,” i.e. bx + c == ax2.
For Al-Khwarizmi, who avoided the use of negative numbers, the terms of each of these equations are addends and not subtractables. In this case, equations that do not have positive solutions are obviously not taken into account. The author sets out methods for solving these equations using the techniques of al-jabr and al-mukabal. His decision, of course, does not completely coincide with ours. Not to mention that it is purely rhetorical, it should be noted, for example, that when solving an incomplete quadratic equation of the first type, Al-Khorezmi, like all mathematicians until the 17th century, does not take into account the zero solution, probably because in specific practical it doesn't matter in tasks. When solving complete quadratic equations, Al-Khwarizmi sets out the rules for solving them using particular numerical examples, and then their geometric proofs.
Forms for solving quadratic equations following the model of Al-Khwarizmi in Europe were first set forth in the “Book of the Abacus,” written in 1202. Italian mathematician Leonard Fibonacci. The author independently developed some new algebraic examples of solving problems and was the first in Europe to approach the introduction of negative numbers.
This book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many problems from this book were used in almost all European textbooks of the 14th-17th centuries. The general rule for solving quadratic equations reduced to a single canonical form x2 + bх = с for all possible combinations of signs and coefficients b, c was formulated in Europe in 1544. M. Stiefel.
The derivation of the formula for solving a quadratic equation in general form is available from Vieth, but Vieth recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli among the first in the 16th century. In addition to positive ones, negative roots are also taken into account. Only in the 17th century. thanks to the efforts Girard, Descartes, Newton and other scientists, the method of solving quadratic equations takes a modern form.
Let's look at several ways to solve quadratic equations.
Standard methods for solving quadratic equations from the school curriculum:
- Factoring the left side of the equation.
- Method for selecting a complete square.
- Solving quadratic equations using the formula.
- Graphical solution of a quadratic equation.
- Solving equations using Vieta's theorem.
Let us dwell in more detail on the solution of reduced and unreduced quadratic equations using Vieta’s theorem.
Recall that to solve the above quadratic equations, it is enough to find two numbers whose product is equal to the free term, and whose sum is equal to the second coefficient with the opposite sign.
Example.x 2 -5x+6=0
You need to find numbers whose product is 6 and whose sum is 5. These numbers will be 3 and 2.
Answer: x 1 =2, x 2 =3.
But you can also use this method for equations with the first coefficient not equal to one.
Example.3x 2 +2x-5=0
Take the first coefficient and multiply it by the free term: x 2 +2x-15=0
The roots of this equation will be numbers whose product is equal to - 15, and whose sum is equal to - 2. These numbers are 5 and 3. To find the roots of the original equation, divide the resulting roots by the first coefficient.
Answer: x 1 =-5/3, x 2 =1
6. Solving equations using the "throw" method.
Consider the quadratic equation ax 2 + bx + c = 0, where a≠0.
Multiplying both sides by a, we obtain the equation a 2 x 2 + abx + ac = 0.
Let ax = y, whence x = y/a; then we arrive at the equation y 2 + by + ac = 0, equivalent to the given one. We find its roots for 1 and 2 using Vieta’s theorem.
We finally get x 1 = y 1 /a and x 2 = y 2 /a.
With this method, the coefficient a is multiplied by the free term, as if “thrown” to it, which is why it is called the “throw” method. This method is used when the roots of the equation can be easily found using Vieta's theorem and, most importantly, when the discriminant is an exact square.
Example.2x 2 - 11x + 15 = 0.
Let’s “throw” the coefficient 2 to the free term and make a substitution and get the equation y 2 - 11y + 30 = 0.
According to Vieta's inverse theorem
y 1 = 5, x 1 = 5/2, x 1 = 2.5; y 2 = 6, x 2 = 6/2, x 2 = 3.
Answer: x 1 =2.5; X 2 = 3.
7. Properties of coefficients of a quadratic equation.
Let the quadratic equation ax 2 + bx + c = 0, a ≠ 0 be given.
1. If a+ b + c = 0 (i.e. the sum of the coefficients of the equation is zero), then x 1 = 1.
2. If a - b + c = 0, or b = a + c, then x 1 = - 1.
Example.345x 2 - 137x - 208 = 0.
Since a + b + c = 0 (345 - 137 - 208 = 0), then x 1 = 1, x 2 = -208/345.
Answer: x 1 =1; X 2 = -208/345 .
Example.132x 2 + 247x + 115 = 0
Because a-b+c = 0 (132 - 247 +115=0), then x 1 = - 1, x 2 = - 115/132
Answer: x 1 = - 1; X 2 =- 115/132
There are other properties of the coefficients of a quadratic equation. but their use is more complex.
8. Solving quadratic equations using a nomogram.
Fig 1. Nomogram
This is an old and currently forgotten method of solving quadratic equations, placed on p. 83 of the collection: Bradis V.M. Four-digit math tables. - M., Education, 1990.
Table XXII. Nomogram for solving the equation z 2 + pz + q = 0. This nomogram allows, without solving a quadratic equation, to determine the roots of the equation from its coefficients.
The curvilinear scale of the nomogram is built according to the formulas (Fig. 1):
Believing OS = p, ED = q, OE = a(all in cm), from Fig. 1 similarities of triangles SAN And CDF we get the proportion
which, after substitutions and simplifications, yields the equation z 2 + pz + q = 0, and the letter z means the mark of any point on a curved scale.
Rice. 2 Solving quadratic equations using a nomogram
Examples.
1) For the equation z 2 - 9z + 8 = 0 the nomogram gives the roots z 1 = 8.0 and z 2 = 1.0
Answer:8.0; 1.0.
2) Using a nomogram, we solve the equation
2z 2 - 9z + 2 = 0.
Divide the coefficients of this equation by 2, we get the equation z 2 - 4.5z + 1 = 0.
The nomogram gives roots z 1 = 4 and z 2 = 0.5.
Answer: 4; 0.5.
9. Geometric method for solving quadratic equations.
Example.X 2 + 10x = 39.
In the original, this problem is formulated as follows: “The square and ten roots are equal to 39.”
Consider a square with side x, rectangles are constructed on its sides so that the other side of each of them is 2.5, therefore the area of each is 2.5x. The resulting figure is then supplemented to a new square ABCD, building four equal squares in the corners, the side of each of them is 2.5, and the area is 6.25
Rice. 3 Graphical method for solving the equation x 2 + 10x = 39
The area S of square ABCD can be represented as the sum of the areas of: the original square x 2, four rectangles (4∙2.5x = 10x) and four additional squares (6.25∙4 = 25), i.e. S = x 2 + 10x = 25. Replacing x 2 + 10x with the number 39, we get that S = 39 + 25 = 64, which means that the side of the square is ABCD, i.e. segment AB = 8. For the required side x of the original square we obtain
10. Solving equations using Bezout's theorem.
Bezout's theorem. The remainder of dividing the polynomial P(x) by the binomial x - α is equal to P(α) (that is, the value of P(x) at x = α).
If the number α is the root of the polynomial P(x), then this polynomial is divisible by x -α without a remainder.
Example.x²-4x+3=0
Р(x)= x²-4x+3, α: ±1,±3, α =1, 1-4+3=0. Divide P(x) by (x-1): (x²-4x+3)/(x-1)=x-3
x²-4x+3=(x-1)(x-3), (x-1)(x-3)=0
x-1=0; x=1, or x-3=0, x=3; Answer: x1 =2, x2 =3.
Conclusion: The ability to quickly and rationally solve quadratic equations is essential for solving more complex equations, such as fractional rational equations, higher power equations, biquadratic equations, and, in high school, trigonometric, exponential, and logarithmic equations. Having studied all the found methods for solving quadratic equations, we can advise our classmates, in addition to the standard methods, to solve by the transfer method (6) and solve equations using the property of coefficients (7), since they are more accessible to understanding.
Literature:
- Bradis V.M. Four-digit math tables. - M., Education, 1990.
- Algebra 8th grade: textbook for 8th grade. general education institutions Makarychev Yu. N., Mindyuk N. G., Neshkov K. I., Suvorova S. B. ed. S. A. Telyakovsky 15th ed., revised. - M.: Education, 2015
- https://ru.wikipedia.org/wiki/%D0%9A%D0%B2%D0%B0%D0%B4%D1%80%D0%B0%D1%82%D0%BD%D0%BE%D0 %B5_%D1%83%D1%80%D0%B0%D0%B2%D0%BD%D0%B5%D0%BD%D0%B8%D0%B5
- Glazer G.I. History of mathematics at school. Manual for teachers. / Ed. V.N. Younger. - M.: Education, 1964.
Quadratic equations. General information.
IN quadratic equation there must be an x squared (that’s why it’s called
"square") In addition to it, the equation may (or may not!) contain simply X (to the first power) and
just a number (free member). And there should be no X's to a power greater than two.
Algebraic equation of general form.
Where x- free variable, a, b, c— coefficients, and a≠0 .
For example: 
Expression 
called quadratic trinomial.
The elements of a quadratic equation have their own names:
called the first or highest coefficient,
· called the second or coefficient at ,
· called a free member.
Complete quadratic equation.
These quadratic equations have a full set of terms on the left. X squared c
coefficient A, x to the first power with coefficient b And free memberWith. IN all coefficients
must be different from zero.
Incomplete is a quadratic equation in which at least one of the coefficients, except
the leading term (either the second coefficient or the free term) is equal to zero.
Let's pretend that b= 0, - X to the first power will disappear. It turns out, for example:
2x 2 -6x=0,
And so on. And if both coefficients b And c are equal to zero, then everything is even simpler, For example:
2x 2 =0,
Note that x squared appears in all equations.
Why A can't be equal to zero? Then x squared will disappear and the equation will become linear .
And the solution is completely different...
Lesson summary
math teachers
MBOU Secondary School No. 2, Vorsma
Kiseleva Larisa Alekseevna
Topic: “Reduced quadratic equation. Vieta's theorem"
The purpose of the lesson: Introduction of the concept of a reduced quadratic equation, Vieta's theorem and its converse theorem.
Tasks:
Educational:
Introduce the concept of a reduced quadratic equation,
Derive the formula for the roots of the given quadratic equation,
Formulate and prove Vieta’s theorem,
Formulate and prove the theorem converse to Vieta’s theorem,
Teach students to solve the given quadratic equations using the theorem inverse to Vieta’s theorem.
Educational:
development of logical thinking, memory, attention, general educational skills, ability to compare and generalize;
Educational:
fostering hard work, mutual assistance, and mathematical culture.
Lesson type: lesson on introducing new material.
Equipment: algebra textbook ed. Alimova and others, notebook, handouts, presentation for the lesson.
Lesson plan.
Lesson stage
Content (goal) of the stage
Time (min)
Organizing time
Checking homework
Verification work
Analysis of work, answers to questions.
Learning new material
Formation of basic knowledge, formulation of rules, solving problems, analyzing results, answering student questions.
Mastering the studied material by applying it to solving problems by analogy under the supervision of a teacher.
Summing up the lesson
Assessing the knowledge of the students who responded. Testing knowledge and understanding of the wording of the rules using the frontal survey method.
Homework
Familiarize students with the content of the task and receive the necessary explanations.
Additional tasks
Multi-level tasks to ensure student development.
During the classes.
Organizing time. Setting the lesson goal. Creating favorable conditions for successful activities. Motivation for learning.
Checking homework. Frontal, individual testing and correction of students' knowledge and skills.
The equation
Number of roots
Teacher: How can you determine the number of its roots without solving a quadratic equation? (students' answers)
Verification work. Answers on questions.
Test text:
Option #1.
Solve the equations:
A) 
,
B) 
It has:
One root
Two different roots.
Option #2.
Solve the equations:
A) 
,
B) 
2.Find the value of parameter a for which the equation 
It has:
One root
Two different roots.
The test work is completed on separate sheets of paper and submitted to the teacher for checking.
After submitting the work, the solution is displayed on the screen.
Learning new material.
4.1. Francois Viet- French mathematician of the 16th century. He was a lawyer and later an adviser to the French kings Henry III and Henry II.
He once managed to decipher a very complex Spanish letter intercepted by the French. The Inquisition nearly burned him at the stake, accusing him of conspiring with the devil.
François Vieta is called the "father of modern letter algebra"
How are the roots of a quadratic trinomial and its coefficients p and q related to each other? The answer to this question is given by a theorem that bears the name of the “father of algebra,” the French mathematician F. Vieta, which we will study today.
The famous theorem was published in 1591.
4.2. Let us formulate the definition of the reduced quadratic equation.
Definition. Quadratic equation of the form 
is called reduced.
This means that the leading coefficient of the equation is equal to one.
Example. .
Any quadratic equation 
can be reduced to the form . To do this, you need to divide both sides of the equation by.
For example, the equation 7Х 2 – 12Х + 14 = 0 by dividing by 7 is reduced to the form
X 2 – 12/7X + 2 = 0
4.3. Derive formulas for the roots of the given quadratic equation.
a, b, c
a=1 , b=p , c=q
Solve the equation X 2 – 14X – 15 =0 (Student solves at the board)
Questions:
Name the coefficients p and q (-14, -15);
Write down the formula for the roots of the given quadratic equation;
Find the roots of this equation (X 1 = 15, X 2 = -1)
4.4. formulate and prove Vieta's theorem.
If and are the roots of the equation , then the formulas are valid, i.e. the sum of the roots of the reduced quadratic equation is equal to the second coefficient taken with the opposite sign, and the product of the roots is equal to the free term.
After this, the teacher proves the theorem. Then, together with the students, he draws a conclusion.
Example. 
. p =-5,q =6.
So numbers and are numbers
positive. You need to find two positive numbers whose product is
equals 6, and the sum equals 5. =2, =3 are the roots of the equation.
4.5. Application of Vieta's theorem .
With its help you can:
Find the sum and product of the roots of a quadratic equation without solving it,
Knowing one of the roots, find another,
Determine the signs of the roots of the equation,
Find the roots of an equation without solving it.
4.6. Let us formulate the theorem inverse to Vieta’s theorem.
If the numbers p, q, and are such that they satisfy the relations, then , are the roots of the quadratic equation .
The proof of the theorem converse to Vieta's theorem is taken home for strong students to study independently.
4.7. Consider the solution to problem 5 on textbook page 125.
Reinforcing the material learned
№ 450 (1)
№ 451 (1, 3, 5) - orally
№ 452 (oral)
№ 455 (1,3)
№ 456 (1, 3)
Summing up the lesson.
Answer the questions:
State Vieta's theorem.
Why is Vieta's theorem needed?
State the converse of Vieta's theorem.
Homework.
§29 (before task 6), No. 450(2,4,6); 455(2.4); 456(2,4,6).
Additional tasks.
Level A.
Find the sum and product of the roots of the equation:
2. Using the inverse theorem of Vieta’s theorem, create a quadratic equation whose roots are 2 and 5.
Level B.
1.Find the sum and product of the roots of the equation:
2. Using the theorem inverse to Vieta’s theorem, create a quadratic equation whose roots are equal to and .
Level C.
1. Analyze the proof of the theorem converse to Vieta’s theorem
2. Solve the equation and check using the inverse theorem of Vieta’s theorem:
Lesson outline diagram
Stages of work
Contents of the stage
Organizing time, including:
setting a goal that must be achieved by students at this stage of the lesson (what must be done by students in order for their further work in the lesson to be effective)
description of methods for organizing students’ work at the initial stage of the lesson, setting students up for learning activities, the subject and topic of the lesson (taking into account the real characteristics of the class with which the teacher works)
The program requirements for the mathematical preparation of students on this topic is to introduce the concept of a reduced quadratic equation, Vieta's theorem and its inverse theorem (from the program for general education institutions).
8th grade students are children of adolescence, which is characterized by instability of attention. The best way to organize attention is to organize learning activities in such a way that students have neither the time, nor the desire, nor the opportunity to be distracted for a long time.
Based on the above, the purpose of the lesson is to solve the following problems:
a) educational: introduction of the concept of a reduced quadratic equation, Vieta’s theorem and its converse theorem.
b) developing: development of logical thinking, memory, attention, general educational skills, abilities to compare and generalize;
c) educational: fostering hard work, mutual assistance, and mathematical culture.
In order for students to perceive the lesson as a logically complete, holistic, time-limited segment of the educational process, it begins with setting the rationale for the tasks and ends with summing up the results and setting tasks for the next lessons.
Survey of students on homework assignments, including:
determining the goals that the teacher sets for students at this stage of the lesson (what result should be achieved by students);
determining the goals and objectives that the teacher wants to achieve at this stage of the lesson;
description of methods that contribute to solving set goals and objectives;
description of the criteria for achieving the goals and objectives of this stage of the lesson;
determining possible actions of the teacher if he or the students fail to achieve their goals;
a description of methods for organizing joint activities of students, taking into account the characteristics of the class with which the teacher works;
description of methods of motivating (stimulating) students' learning activity during the survey;
description of methods and criteria for assessing student responses during the survey.
At the first stage, a frontal, individual check and correction of students’ knowledge and skills takes place. In this case, the solution of quadratic equations is repeated and the determination of the number of roots by its discriminant is consolidated. The transition is made to the definition of the reduced quadratic equation.
At the second stage, equations of two types are considered. To ensure that students do not get tired of monotonous work, various forms of work and task options are used, and tasks of a higher level (with a parameter) are included.
Students' oral work alternates with written work, which consists of justifying the choice of method for solving a quadratic equation and analyzing the solution to the equation
One of the methods of pedagogical support is the use of information technologies as visual aids, which help students of different levels of preparedness to easily learn the material, so certain moments of the lesson are conducted using a presentation (showing solutions to independent work, questions, homework)
Studying new educational material. This stage involves:
presentation of the main provisions of the new educational material that must be mastered by students;
description of forms and methods of presentation (presentation) of new educational material;
description of the main forms and methods of organizing individual and group activities of students, taking into account the characteristics of the class in which the teacher works;
a description of the criteria for determining the level of attention and interest of students in the educational material presented by the teacher;
description of methods of motivating (stimulating) students' educational activity during the development of new educational material
The definition of the reduced quadratic equation is given. The teacher, together with the students, deduces the formulas for the roots of the given quadratic equation, the students realize the importance of the educational material of the lesson. Analysis of the formulation and proof of Vieta’s theorem also occurs jointly with students
Such work is also a consolidation of the study of new material.
Methods:
visual;
practical;
verbal;
partial-search
Reinforcing educational material, suggesting:
setting a specific educational goal for students (what result should be achieved by students at this stage of the lesson);
determining the goals and objectives that the teacher sets for himself at this stage of the lesson;
description of the forms and methods of achieving set goals during the consolidation of new educational material, taking into account the individual characteristics of the students with whom the teacher works.
a description of the criteria to determine the degree to which students have mastered new educational material;
a description of possible ways and methods of responding to situations when the teacher determines that some students have not mastered the new educational material.
Reinforcement of educational material occurs when answering questions and working with the textbook:
Analysis of problem No. 5 on page 125;
Solution of exercises
№ 450 (1), 451 (1, 3, 5) – orally, 452 (orally);
455 (1,3); 456 (1, 3)
Throughout the lesson, students are highly active; the teacher has the opportunity to interview all students in the class, and some even more than once.
The lesson is summarized in the form of a frontal survey of students on the following questions:
What equations are called reduced?
Can an ordinary quadratic equation be reduced?
Write down the formula for the roots of the given quadratic equation
State Vieta's theorem.
What is the sum and product of the roots of the equation:
Homework assignment, including:
setting independent work goals for students (what students should do while completing homework);
determining the goals that the teacher wants to achieve by assigning homework;
defining and explaining to students the criteria for successfully completing homework.
In homework, students are expected to work within their capabilities. Strong students work independently and at the end of the work have the opportunity to check the correctness of their solutions by checking them with the solutions written on the board at the beginning of the next lesson. Other students can get advice from their classmates or teacher. Weak students work based on examples and use solutions to equations discussed in class. Thus, conditions are created for working at various levels of complexity.
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