Just. According to formulas and clear simple rules. At the first stage

we need to bring the given equation to standard form, i.e. to the view:

If the equation is already given to you in this form, you do not need to do the first stage. The most important thing is right

determine all coefficients but, b And c.

Formula for finding the roots of a quadratic equation.

The expression under the root sign is called discriminant . As you can see, to find x, we

use only a, b and c. Those. odds from quadratic equation. Just carefully insert

values a, b and c into this formula and count. Substitute with their signs!

For example, in the equation:

but =1; b = 3; c = -4.

Substitute the values ​​and write:

Example almost solved:

This is the answer.

The most common mistakes are confusion with the signs of values a, b And from. Rather, with substitution

negative values ​​into the formula for calculating the roots. Here the detailed formula saves

with specific numbers. If there are problems with calculations, do it!

Suppose we need to solve the following example:

Here a = -6; b = -5; c = -1

We paint everything in detail, carefully, without missing anything with all the signs and brackets:

Often quadratic equations look slightly different. For example, like this:

Now take note of the practical techniques that dramatically reduce the number of errors.

First reception. Don't be lazy before solving a quadratic equation bring it to standard form.

What does this mean?

Suppose, after any transformations, you get the following equation:

Do not rush to write the formula of the roots! You will almost certainly mix up the odds a, b and c.

Build the example correctly. First, x squared, then without a square, then a free member. Like this:

Get rid of the minus. How? We need to multiply the whole equation by -1. We get:

And now you can safely write down the formula for the roots, calculate the discriminant and complete the example.

Decide on your own. You should end up with roots 2 and -1.

Second reception. Check your roots! By Vieta's theorem.

To solve the given quadratic equations, i.e. if the coefficient

x2+bx+c=0,

thenx 1 x 2 =c

x1 +x2 =−b

For a complete quadratic equation in which a≠1:

x 2 +bx+c=0,

divide the whole equation by but:

→ →

where x 1 And x 2 - roots of the equation.

Reception third. If your equation has fractional coefficients, get rid of the fractions! Multiply

equation for a common denominator.

Output. Practical Tips:

1. Before solving, we bring the quadratic equation to the standard form, build it right.

2. If there is a negative coefficient in front of the x in the square, we eliminate it by multiplying everything

equations for -1.

3. If the coefficients are fractional, we eliminate the fractions by multiplying the entire equation by the corresponding

factor.

4. If x squared is pure, the coefficient for it is equal to one, the solution can be easily checked by

Having received a general idea of ​​\u200b\u200bequalities, and having become acquainted with one of their types - numerical equalities, you can start talking about another form of equality that is very important from a practical point of view - about equations. In this article, we will analyze what is the equation, and what is called the root of the equation. Here we give the corresponding definitions, and also give various examples of equations and their roots.

Page navigation.

What is an equation?

Purposeful familiarity with equations usually begins in math classes in grade 2. At this time the following equation definition:

Definition.

The equation is an equality containing an unknown number to be found.

Unknown numbers in equations are usually denoted using small Latin letters, for example, p, t, u, etc., but the letters x, y and z are most often used.

Thus, the equation is determined from the standpoint of the form of the notation. In other words, equality is an equation when it obeys the specified notation rules - it contains a letter whose value needs to be found.

Here are some examples of the first and most simple equations. Let's start with equations like x=8 , y=3 , etc. Equations that contain signs of arithmetic operations along with numbers and letters look a little more complicated, for example, x+2=3 , z−2=5 , 3 t=9 , 8:x=2 .

The variety of equations grows after acquaintance with - equations with brackets begin to appear, for example, 2 (x−1)=18 and x+3 (x+2 (x−2))=3 . An unknown letter can appear multiple times in an equation, for example, x+3+3 x−2−x=9 , and letters can be on the left side of the equation, on the right side, or both sides of the equation, for example, x (3+1)−4=8 , 7−3=z+1 or 3 x−4=2 (x+12) .

Further after studying natural numbers there is an acquaintance with whole, rational, real numbers, new mathematical objects are studied: degrees, roots, logarithms, etc., while more and more new types of equations appear that contain these things. Examples can be found in the article. main types of equations studied at school.

In grade 7, along with letters, which mean some specific numbers, they begin to consider letters that can take various meanings, they are called variables (see the article). In this case, the word “variable” is introduced into the definition of the equation, and it becomes like this:

Definition.

Equation name an equality containing a variable whose value is to be found.

For example, the equation x+3=6 x+7 is an equation with variable x , and 3 z−1+z=0 is an equation with variable z .

In algebra lessons in the same 7th grade, there is a meeting with equations containing in their record not one, but two different unknown variables. They are called equations with two variables. In the future, the presence of three or more variables in the equation record is allowed.

Definition.

Equations with one, two, three, etc. variables- these are equations containing one, two, three, ... unknown variables in their record, respectively.

For example, the equation 3.2 x+0.5=1 is an equation with one variable x, in turn, an equation of the form x−y=3 is an equation with two variables x and y. And one more example: x 2 +(y−1) 2 +(z+0.5) 2 =27 . It is clear that such an equation is an equation with three unknown variables x, y and z.

What is the root of the equation?

The definition of the equation's root is directly related to the definition of the equation. We will carry out some reasoning that will help us understand what the root of the equation is.

Suppose we have an equation with one letter (variable). If instead of the letter included in the record of this equation, a certain number is substituted, then the equation will turn into a numerical equality. Moreover, the resulting equality can be both true and false. For example, if instead of the letter a in the equation a+1=5 we substitute the number 2 , then we get an incorrect numerical equality 2+1=5 . If we substitute the number 4 instead of a in this equation, then we get the correct equality 4+1=5.

In practice, in the overwhelming majority of cases, of interest are such values ​​of the variable, the substitution of which into the equation gives the correct equality, these values ​​are called roots or solutions. given equation.

Definition.

Root of the equation- this is the value of the letter (variable), when substituting which the equation turns into the correct numerical equality.

Note that the root of an equation with one variable is also called the solution of the equation. In other words, the solution to an equation and the root of the equation are the same thing.

Let us explain this definition with an example. To do this, we return to the equation written above a+1=5 . According to the voiced definition of the root of the equation, the number 4 is the root of this equation, since when substituting this number instead of the letter a, we get the correct equality 4+1=5, and the number 2 is not its root, since it corresponds to an incorrect equality of the form 2+1= five .

At this point, a number of natural questions arise: “Does any equation have a root, and how many roots does a given equation have”? We will answer them.

There are both equations with roots and equations without roots. For example, the equation x+1=5 has a root 4, and the equation 0 x=5 has no roots, since no matter what number we substitute into this equation instead of the variable x, we will get the wrong equality 0=5.

As for the number of roots of an equation, there are both equations that have some finite number of roots (one, two, three, etc.) and equations that have infinitely many roots. For example, the equation x−2=4 has a single root 6 , the roots of the equation x 2 =9 are two numbers −3 and 3 , the equation x (x−1) (x−2)=0 has three roots 0 , 1 and 2 , and the solution to the equation x=x is any number, that is, it has an infinite number of roots.

A few words should be said about the accepted notation of the roots of the equation. If the equation has no roots, then usually they write “the equation has no roots” or use the sign of the empty set ∅. If the equation has roots, then they are written separated by commas, or written as set elements in curly brackets. For example, if the roots of the equation are the numbers −1, 2 and 4, then write −1, 2, 4 or (−1, 2, 4) . It is also possible to write the roots of the equation in the form of simple equalities. For example, if the letter x enters the equation, and the roots of this equation are the numbers 3 and 5, then you can write x=3, x=5, and subscripts x 1 =3, x 2 =5 are often added to the variable, as if indicating numbers the roots of the equation. An infinite set of roots of an equation is usually written in the form, also, if possible, the notation of sets of natural numbers N, integers Z, real numbers R is used. For example, if the root of the equation with the variable x is any integer, then write , and if the roots of the equation with the variable y are any real number from 1 to 9 inclusive, then write .

For equations with two, three, and big amount variables, as a rule, do not use the term "root of the equation", in these cases they say "solution of the equation". What is called the solution of equations with several variables? Let us give an appropriate definition.

Definition.

Solving an equation with two, three, etc. variables call a pair, three, etc. values ​​of the variables, which turns this equation into a true numerical equality.

We will show explanatory examples. Consider an equation with two variables x+y=7 . We substitute the number 1 instead of x, and the number 2 instead of y, while we have the equality 1+2=7. Obviously, it is incorrect, therefore, the pair of values ​​x=1 , y=2 is not a solution to the written equation. If we take a pair of values ​​x=4 , y=3 , then after substitution into the equation we will come to the correct equality 4+3=7 , therefore, this pair of variable values ​​is, by definition, a solution to the equation x+y=7 .

Equations with multiple variables, like equations with one variable, may have no roots, may have a finite number of roots, or may have infinitely many roots.

Pairs, triples, fours, etc. variable values ​​are often written briefly, listing their values ​​separated by commas in parentheses. In this case, the written numbers in brackets correspond to the variables in alphabetical order. Let's clarify this point by returning to the previous equation x+y=7 . The solution to this equation x=4 , y=3 can be briefly written as (4, 3) .

The greatest attention in the school course of mathematics, algebra and the beginning of analysis is given to finding the roots of equations with one variable. We will analyze the rules of this process in great detail in the article. solution of equations.

Bibliography.

• Maths. 2 cells Proc. for general education institutions with adj. to an electron. carrier. At 2 o'clock, Part 1 / [M. I. Moro, M. A. Bantova, G. V. Beltyukova and others] - 3rd ed. - M.: Education, 2012. - 96 p.: ill. - (School of Russia). - ISBN 978-5-09-028297-0.
• Algebra: textbook for 7 cells. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 17th ed. - M. : Education, 2008. - 240 p. : ill. - ISBN 978-5-09-019315-3.
• Algebra: Grade 9: textbook. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.

Kopyevskaya rural secondary school

10 Ways to Solve Quadratic Equations

Head: Patrikeeva Galina Anatolyevna,

mathematic teacher

s.Kopyevo, 2007

1. History of the development of quadratic equations

1.1 Quadratic equations in ancient Babylon

1.2 How Diophantus compiled and solved quadratic equations

1.3 Quadratic equations in India

1.4 Quadratic equations in al-Khwarizmi

1.5 Quadratic equations in Europe XIII - XVII centuries

1.6 About Vieta's theorem

2. Methods for solving quadratic equations

Conclusion

Literature

1. History of the development of quadratic equations

1.1 Quadratic equations in ancient Babylon

The need to solve equations not only of the first, but also of the second degree in ancient times was caused by the need to solve problems related to finding the areas of land and earthworks military nature, as well as with the development of astronomy and mathematics itself. Quadratic equations were able to solve about 2000 BC. e. Babylonians.

Applying modern algebraic notation, we can say that in their cuneiform texts there are, in addition to incomplete ones, such, for example, complete quadratic equations:

X 2 + X = ¾; X 2 - X = 14,5

The rule for solving these equations, stated in the Babylonian texts, coincides essentially with the modern one, but it is not known how the Babylonians came to this rule. Almost all the cuneiform texts found so far give only problems with solutions stated in the form of recipes, with no indication of how they were found.

In spite of high level development of algebra in Babylon, in cuneiform texts there is no concept of a negative number and common methods solutions of quadratic equations.

1.2 How Diophantus compiled and solved quadratic equations.

Diophantus' Arithmetic does not contain a systematic exposition of algebra, but it contains a systematic series of problems, accompanied by explanations and solved by formulating equations of various degrees.

When compiling equations, Diophantus skillfully chooses unknowns to simplify the solution.

Here, for example, is one of his tasks.

Task 11."Find two numbers knowing that their sum is 20 and their product is 96"

Diophantus argues as follows: it follows from the condition of the problem that the desired numbers are not equal, since if they were equal, then their product would be equal not to 96, but to 100. Thus, one of them will be more than half of their sum, i.e. . 10+x, the other is smaller, i.e. 10's. The difference between them 2x .

Hence the equation:

(10 + x)(10 - x) = 96

100 - x 2 = 96

x 2 - 4 = 0 (1)

From here x = 2. One of the desired numbers is 12 , other 8 . Solution x = -2 for Diophantus does not exist, since Greek mathematics knew only positive numbers.

If we solve this problem by choosing one of the desired numbers as the unknown, then we will come to the solution of the equation

y(20 - y) = 96,

y 2 - 20y + 96 = 0. (2)

It is clear that Diophantus simplifies the solution by choosing the half-difference of the desired numbers as the unknown; he manages to reduce the problem to solving an incomplete quadratic equation (1).

1.3 Quadratic equations in India

Problems for quadratic equations are already found in the astronomical tract "Aryabhattam", compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scholar, Brahmagupta (7th century), expounded general rule solutions of quadratic equations reduced to a single canonical form:

ah 2+ b x = c, a > 0. (1)

In equation (1), the coefficients, except for but, can also be negative. Brahmagupta's rule essentially coincides with ours.

IN ancient india public competitions in solving difficult problems were common. In one of the old Indian books, the following is said about such competitions: “As the sun outshines the stars with its brilliance, so scientist man eclipse the glory of another in public meetings, proposing and solving algebraic problems. Tasks were often dressed in poetic form.

Here is one of the problems of the famous Indian mathematician of the XII century. Bhaskara.

Task 13.

“A frisky flock of monkeys And twelve in vines ...

Having eaten power, had fun. They began to jump, hanging ...

Part eight of them in a square How many monkeys were there,

Having fun in the meadow. You tell me, in this flock?

Bhaskara's solution indicates that he knew about the two-valuedness of the roots of quadratic equations (Fig. 3).

The equation corresponding to problem 13 is:

( x /8) 2 + 12 = x

Bhaskara writes under the guise of:

x 2 - 64x = -768

and, to complete the left side of this equation to a square, he adds to both sides 32 2 , getting then:

x 2 - 64x + 32 2 = -768 + 1024,

(x - 32) 2 = 256,

x - 32 = ± 16,

x 1 = 16, x 2 = 48.

1.4 Quadratic equations in al-Khorezmi

Al-Khorezmi's algebraic treatise gives a classification of linear and quadratic equations. The author lists 6 types of equations, expressing them as follows:

1) "Squares are equal to roots", i.e. ax 2 + c = b X.

2) "Squares are equal to number", i.e. ax 2 = s.

3) "The roots are equal to the number", i.e. ah = s.

4) "Squares and numbers are equal to roots", i.e. ax 2 + c = b X.

5) "Squares and roots are equal to the number", i.e. ah 2+ bx = s.

6) "Roots and numbers are equal to squares", i.e. bx + c \u003d ax 2.

For al-Khorezmi, who avoided the use of negative numbers, the terms of each of these equations are summands, not subtrahends. In this case, equations that do not have positive solutions are obviously not taken into account. The author outlines ways to solve the indicated equations, using the techniques of al-jabr and al-muqabala. His decisions, of course, do not completely coincide with ours. Not to mention the fact 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 before the 17th century, does not take into account the zero solution, probably because it does not matter in specific practical problems. When solving the complete quadratic equations of al - Khorezmi on partial numerical examples sets out the decision rules, and then the geometric proofs.

Task 14.“The square and the number 21 are equal to 10 roots. Find the root" (assuming the root of the equation x 2 + 21 = 10x).

The author's solution goes something like this: divide the number of roots in half, you get 5, multiply 5 by itself, subtract 21 from the product, 4 remains. Take the root of 4, you get 2. Subtract 2 from 5, you get 3, this will be the desired root. Or add 2 to 5, which will give 7, this is also a root.

Treatise al - Khorezmi is the first book that has come down to us, in which the classification of quadratic equations is systematically stated and formulas for their solution are given.

1.5 Quadratic equations in Europe XIII - XVII centuries

Formulas for solving quadratic equations on the model of al - Khorezmi in Europe were first set forth in the "Book of the Abacus", written in 1202 by the Italian mathematician Leonardo Fibonacci. This voluminous work, which reflects the influence of mathematics, both the countries of Islam and Ancient Greece, differs in both completeness and clarity of presentation. The author independently developed some new algebraic examples problem solving and was the first in Europe to approach the introduction of negative numbers. His book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many tasks from the "Book of the Abacus" passed into almost all European textbooks of the 16th - 17th centuries. and partly XVIII.

The general rule for solving quadratic equations reduced to a single canonical form:

x 2+ bx = with,

for all possible combinations of signs of the coefficients b , from was formulated in Europe only in 1544 by M. Stiefel.

Derivation of the formula for solving a quadratic equation in general view Viet has, but Viet recognized only positive roots. The Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 16th century. Take into account, in addition to positive, and negative roots. Only in the XVII century. Thanks to the work of Girard, Descartes, Newton and other scientists, the way to solve quadratic equations takes on a modern look.

1.6 About Vieta's theorem

The theorem expressing the relationship between the coefficients of a quadratic equation and its roots, bearing the name of Vieta, was formulated by him for the first time in 1591 as follows: “If B + D multiplied by A - A 2 , equals BD, then A equals IN and equal D ».

To understand Vieta, one must remember that BUT, like any vowel, meant for him the unknown (our X), the vowels IN, D- coefficients for the unknown. In the language of modern algebra, Vieta's formulation above means: if

(a + b )x - x 2 = ab ,

x 2 - (a + b )x + a b = 0,

x 1 = a, x 2 = b .

Expressing the relationship between the roots and coefficients of equations by general formulas written using symbols, Viet established uniformity in the methods of solving equations. However, the symbolism of Vieta is still far from modern look. He did not recognize negative numbers, and therefore, when solving equations, he considered only cases where all roots are positive.

2. Methods for solving quadratic equations

Quadratic equations are the foundation on which the majestic edifice of algebra rests. Quadratic equations are widely used in solving trigonometric, exponential, logarithmic, irrational and transcendental equations and inequalities. We all know how to solve quadratic equations from school (grade 8) until graduation.

Type equation

Expression D= b 2 - 4ac called discriminant quadratic equation. IfD = 0, then the equation has one real root; if D> 0, then the equation has two real roots.
In case when D = 0 , it is sometimes said that a quadratic equation has two identical roots.
Using the notation D= b 2 - 4ac, formula (2) can be rewritten as

If b= 2 k, then formula (2) takes the form:

where k= b / 2 .
The last formula is especially convenient when b / 2 is an integer, i.e. coefficient b - even number.
Example 1: solve the equation 2 x 2 - 5 x + 2 = 0 . Here a=2, b=-5, c=2. We have D= b 2 - 4ac = (-5) 2- 4*2*2 = 9 . Because D > 0 , then the equation has two roots. Let's find them by the formula (2)

so x 1 =(5 + 3) / 4 = 2,x 2 =(5 - 3) / 4 = 1 / 2 ,
i.e x 1 = 2 And x 2 = 1 / 2 - roots given equation.
Example 2: solve the equation 2 x 2 - 3 x + 5 = 0 . Here a=2, b=-3, c=5. Finding the discriminant D= b 2 - 4ac = (-3) 2- 4*2*5 = -31 . Because D 0 , then the equation has no real roots.

Incomplete quadratic equations. If in a quadratic equation ax 2 +bx+c =0 second factor b or free member c zero, then the quadratic equation is called incomplete. Incomplete equations are distinguished because to find their roots, you can not use the formula for the roots of a quadratic equation - it is easier to solve the equation by factoring its left side into factors.
Example 1: solve the equation 2 x 2 - 5 x = 0 .
We have x(2 x - 5) = 0 . So either x = 0 , or 2 x - 5 = 0 , i.e x = 2.5 . So the equation has two roots: 0 And 2.5
Example 2: solve the equation 3 x 2 - 27 = 0 .
We have 3 x 2 = 27 . Therefore, the roots of this equation are 3 And -3 .

Vieta's theorem. If the given quadratic equation x 2 +px+ q =0 has real roots, then their sum is equal to - p, and the product is q, i.e

x 1 + x 2 \u003d -p,
x 1 x 2 = q

(the sum of the roots of the given 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).

With this math program you can solve quadratic equation.

The program not only gives the answer to the problem, but also displays the solution process in two ways:
- using the discriminant
- using the Vieta theorem (if possible).

Moreover, the answer is displayed exact, not approximate.
For example, for the equation \(81x^2-16x-1=0\), the answer is displayed in this form:

$$ x_1 = \frac(8+\sqrt(145))(81), \quad x_2 = \frac(8-\sqrt(145))(81) $$ instead of this: \(x_1 = 0.247; \quad x_2 = -0.05 \)

This program May be useful for high school students general education schools in preparation for control work and exams, when testing knowledge before the exam, parents to control the solution of many problems in mathematics and algebra. Or maybe it's too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as soon as possible? homework math or algebra? In this case, you can also use our programs with a detailed solution.

In this way, you can conduct your own training and/or the training of your younger brothers or sisters, while the level of education in the field of tasks to be solved is increased.

If you are not familiar with the rules for entering a square polynomial, we recommend that you familiarize yourself with them.

Rules for entering a square polynomial

Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q \) etc.

Numbers can be entered as integers or fractions.
Moreover, fractional numbers can be entered not only in the form of a decimal, but also in the form of an ordinary fraction.

Rules for entering decimal fractions.
In decimal fractions, the fractional part from the integer can be separated by either a dot or a comma.
For example, you can enter decimals so: 2.5x - 3.5x^2

Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.

The denominator cannot be negative.

When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
whole part separated from the fraction by an ampersand: &
Input: 3&1/3 - 5&6/5z +1/7z^2
Result: \(3\frac(1)(3) - 5\frac(6)(5) z + \frac(1)(7)z^2 \)

When entering an expression you can use brackets. In this case, when solving a quadratic equation, the introduced expression is first simplified.
For example: 1/2(y-1)(y+1)-(5y-10&1/2)

=0

Solve

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A bit of theory.

Quadratic equation and its roots. Incomplete quadratic equations

Each of the equations
\(-x^2+6x+1,4=0, \quad 8x^2-7x=0, \quad x^2-\frac(4)(9)=0 \)
has the form
\(ax^2+bx+c=0, \)
where x is a variable, a, b and c are numbers.
In the first equation a = -1, b = 6 and c = 1.4, in the second a = 8, b = -7 and c = 0, in the third a = 1, b = 0 and c = 4/9. Such equations are called quadratic equations.

Definition.
quadratic equation an equation of the form ax 2 +bx+c=0 is called, where x is a variable, a, b and c are some numbers, and \(a \neq 0 \).

The numbers a, b and c are the coefficients of the quadratic equation. The number a is called the first coefficient, the number b is the second coefficient and the number c is the intercept.

In each of the equations of the form ax 2 +bx+c=0, where \(a \neq 0 \), the largest power of the variable x is a square. Hence the name: quadratic equation.

Note that a quadratic equation is also called an equation of the second degree, since its left side is a polynomial of the second degree.

A quadratic equation in which the coefficient at x 2 is 1 is called reduced quadratic equation. For example, the given quadratic equations are the equations
\(x^2-11x+30=0, \quad x^2-6x=0, \quad x^2-8=0 \)

If in the quadratic equation ax 2 +bx+c=0 at least one of the coefficients b or c is equal to zero, then such an equation is called incomplete quadratic equation. So, the equations -2x 2 +7=0, 3x 2 -10x=0, -4x 2 =0 are incomplete quadratic equations. In the first of them b=0, in the second c=0, in the third b=0 and c=0.

Incomplete quadratic equations are of three types:
1) ax 2 +c=0, where \(c \neq 0 \);
2) ax 2 +bx=0, where \(b \neq 0 \);
3) ax2=0.

Consider the solution of equations of each of these types.

To solve an incomplete quadratic equation of the form ax 2 +c=0 for \(c \neq 0 \), its free term is transferred to the right side and both parts of the equation are divided by a:
\(x^2 = -\frac(c)(a) \Rightarrow x_(1,2) = \pm \sqrt( -\frac(c)(a)) \)

Since \(c \neq 0 \), then \(-\frac(c)(a) \neq 0 \)

If \(-\frac(c)(a)>0 \), then the equation has two roots.

If \(-\frac(c)(a) To solve an incomplete quadratic equation of the form ax 2 +bx=0 for \(b \neq 0 \) factorize its left side and obtain the equation
\(x(ax+b)=0 \Rightarrow \left\( \begin(array)(l) x=0 \\ ax+b=0 \end(array) \right. \Rightarrow \left\( \begin (array)(l) x=0 \\ x=-\frac(b)(a) \end(array) \right. \)

Hence, an incomplete quadratic equation of the form ax 2 +bx=0 for \(b \neq 0 \) always has two roots.

An incomplete quadratic equation of the form ax 2 \u003d 0 is equivalent to the equation x 2 \u003d 0 and therefore has a single root 0.

The formula for the roots of a quadratic equation

Let us now consider how quadratic equations are solved in which both coefficients of the unknowns and the free term are nonzero.

We solve the quadratic equation in general form and as a result we obtain the formula of the roots. Then this formula can be applied to solve any quadratic equation.

Solve the quadratic equation ax 2 +bx+c=0

Dividing both its parts by a, we obtain the equivalent reduced quadratic equation
\(x^2+\frac(b)(a)x +\frac(c)(a)=0 \)

We transform this equation by highlighting the square of the binomial:
\(x^2+2x \cdot \frac(b)(2a)+\left(\frac(b)(2a)\right)^2- \left(\frac(b)(2a)\right)^ 2 + \frac(c)(a) = 0 \Rightarrow \)

\(x^2+2x \cdot \frac(b)(2a)+\left(\frac(b)(2a)\right)^2 = \left(\frac(b)(2a)\right)^ 2 - \frac(c)(a) \Rightarrow \) \(\left(x+\frac(b)(2a)\right)^2 = \frac(b^2)(4a^2) - \frac( c)(a) \Rightarrow \left(x+\frac(b)(2a)\right)^2 = \frac(b^2-4ac)(4a^2) \Rightarrow \) \(x+\frac(b )(2a) = \pm \sqrt( \frac(b^2-4ac)(4a^2) ) \Rightarrow x = -\frac(b)(2a) + \frac( \pm \sqrt(b^2 -4ac) )(2a) \Rightarrow \) \(x = \frac( -b \pm \sqrt(b^2-4ac) )(2a) \)

The root expression is called discriminant of a quadratic equation ax 2 +bx+c=0 (“discriminant” in Latin - distinguisher). It is denoted by the letter D, i.e.
\(D = b^2-4ac\)

Now, using the notation of the discriminant, we rewrite the formula for the roots of the quadratic equation:
\(x_(1,2) = \frac( -b \pm \sqrt(D) )(2a) \), where \(D= b^2-4ac \)

It's obvious that:
1) If D>0, then the quadratic equation has two roots.
2) If D=0, then the quadratic equation has one root \(x=-\frac(b)(2a)\).
3) If D Thus, depending on the value of the discriminant, the quadratic equation can have two roots (for D > 0), one root (for D = 0) or no roots (for D When solving a quadratic equation using this formula, it is advisable to do the following way:
1) calculate the discriminant and compare it with zero;
2) if the discriminant is positive or equal to zero, then use the root formula, if the discriminant is negative, then write down that there are no roots.

Vieta's theorem

The given quadratic equation ax 2 -7x+10=0 has roots 2 and 5. The sum of the roots is 7, and the product is 10. We see that the sum of the roots is equal to the second coefficient, taken with the opposite sign, and the product of the roots is equal to the free term. Any reduced quadratic equation that has roots has this property.

The sum of the roots of the given 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.

Those. Vieta's theorem states that the roots x 1 and x 2 of the reduced quadratic equation x 2 +px+q=0 have the property:
\(\left\( \begin(array)(l) x_1+x_2=-p \\ x_1 \cdot x_2=q \end(array) \right. \)