Some problems in mathematics require the ability to calculate the value of the square root. Such problems include solving second-order equations. In this article we will present an effective method for calculating square roots and use it when working with formulas for the roots of a quadratic equation.

What is a square root?

In mathematics, this concept corresponds to the symbol √. Historical data says that it was first used around the first half of the 16th century in Germany (the first German work on algebra by Christoph Rudolf). Scientists believe that the specified symbol is a transformed Latin letter r (radix means "root" in Latin).

The root of any number is equal to the value whose square corresponds to the radical expression. In the language of mathematics, this definition will look like this: √x = y, if y 2 = x.

The root of a positive number (x > 0) is also a positive number (y > 0), but if you take the root of a negative number (x< 0), то его результатом уже будет комплексное число, включающее мнимую единицу i.

Here are two simple examples:

√9 = 3, since 3 2 = 9; √(-9) = 3i, since i 2 = -1.

Heron's iterative formula for finding the values ​​of square roots

The above examples are very simple, and calculating the roots in them is not difficult. Difficulties begin to appear even when finding root values ​​for any value that cannot be represented as a square of a natural number, for example √10, √11, √12, √13, not to mention the fact that in practice it is necessary to find roots for non-integer numbers: for example √(12.15), √(8.5) and so on.

In all of the above cases, a special method for calculating the square root should be used. Currently, several such methods are known: for example, Taylor series expansion, column division and some others. Of all the known methods, perhaps the simplest and most effective is the use of Heron's iterative formula, which is also known as the Babylonian method of determining square roots (there is evidence that the ancient Babylonians used it in their practical calculations).

Let it be necessary to determine the value of √x. The formula for finding the square root is as follows:

a n+1 = 1/2(a n +x/a n), where lim n->∞ (a n) => x.

Let's decipher this mathematical notation. To calculate √x, you should take a certain number a 0 (it can be arbitrary, but to quickly obtain the result, you should choose it so that (a 0) 2 is as close as possible to x. Then substitute it into the indicated formula for calculating the square root and get a new the number a 1, which will already be closer to the desired value. After this, it is necessary to substitute a 1 into the expression and obtain a 2. This procedure should be repeated until the required accuracy is obtained.

An example of using Heron's iterative formula

The algorithm described above for obtaining the square root of a given number may sound quite complicated and confusing to many, but in reality everything turns out to be much simpler, since this formula converges very quickly (especially if a successful number a 0 is chosen).

Let's give a simple example: you need to calculate √11. Let's choose a 0 = 3, since 3 2 = 9, which is closer to 11 than 4 2 = 16. Substituting into the formula, we get:

a 1 = 1/2(3 + 11/3) = 3.333333;

a 2 = 1/2(3.33333 + 11/3.33333) = 3.316668;

a 3 = 1/2(3.316668 + 11/3.316668) = 3.31662.

There is no point in continuing the calculations, since we found that a 2 and a 3 begin to differ only in the 5th decimal place. Thus, it was enough to apply the formula only 2 times to calculate √11 with an accuracy of 0.0001.

Nowadays, calculators and computers are widely used to calculate roots, however, it is useful to remember the marked formula in order to be able to manually calculate their exact value.

Second order equations

Understanding what a square root is and the ability to calculate it is used in solving quadratic equations. These equations are called equalities with one unknown, the general form of which is shown in the figure below.

Here c, b and a represent some numbers, and a must not be equal to zero, and the values ​​of c and b can be completely arbitrary, including equal to zero.

Any values ​​of x that satisfy the equality indicated in the figure are called its roots (this concept should not be confused with the square root √). Since the equation under consideration is of the 2nd order (x 2), then there cannot be more than two roots for it. Let's look further in the article at how to find these roots.

Finding the roots of a quadratic equation (formula)

This method of solving the type of equalities under consideration is also called the universal method, or the discriminant method. It can be used for any quadratic equations. The formula for the discriminant and roots of the quadratic equation is as follows:

It shows that the roots depend on the value of each of the three coefficients of the equation. Moreover, the calculation of x 1 differs from the calculation of x 2 only by the sign in front of the square root. The radical expression, which is equal to b 2 - 4ac, is nothing more than the discriminant of the equality in question. The discriminant in the formula for the roots of a quadratic equation plays an important role because it determines the number and type of solutions. So, if it is equal to zero, then there will be only one solution, if it is positive, then the equation has two real roots, and finally, a negative discriminant leads to two complex roots x 1 and x 2.

Vieta's theorem or some properties of the roots of second-order equations

At the end of the 16th century, one of the founders of modern algebra, a Frenchman, studying second-order equations, was able to obtain the properties of its roots. Mathematically they can be written like this:

x 1 + x 2 = -b / a and x 1 * x 2 = c / a.

Both equalities can be easily obtained by anyone; to do this, you just need to perform the appropriate mathematical operations with the roots obtained through the formula with the discriminant.

The combination of these two expressions can rightfully be called the second formula for the roots of a quadratic equation, which makes it possible to guess its solutions without using a discriminant. Here it should be noted that although both expressions are always valid, it is convenient to use them to solve an equation only if it can be factorized.

The task of consolidating the acquired knowledge

Let's solve a mathematical problem in which we will demonstrate all the techniques discussed in the article. The conditions of the problem are as follows: you need to find two numbers for which the product is -13 and the sum is 4.

This condition immediately reminds us of Vieta’s theorem; using the formulas for the sum of square roots and their product, we write:

x 1 + x 2 = -b / a = 4;

x 1 * x 2 = c / a = -13.

If we assume that a = 1, then b = -4 and c = -13. These coefficients allow us to create a second-order equation:

x 2 - 4x - 13 = 0.

Let's use the formula with the discriminant and get the following roots:

x 1.2 = (4 ± √D)/2, D = 16 - 4 * 1 * (-13) = 68.

That is, the problem was reduced to finding the number √68. Note that 68 = 4 * 17, then, using the square root property, we get: √68 = 2√17.

Now let’s use the considered square root formula: a 0 = 4, then:

a 1 = 1/2(4 + 17/4) = 4.125;

a 2 = 1/2(4.125 + 17/4.125) = 4.1231.

There is no need to calculate a 3 since the values ​​found differ by only 0.02. Thus, √68 = 8.246. Substituting it into the formula for x 1,2, we get:

x 1 = (4 + 8.246)/2 = 6.123 and x 2 = (4 - 8.246)/2 = -2.123.

As we can see, the sum of the numbers found is really equal to 4, but if we find their product, then it will be equal to -12.999, which satisfies the conditions of the problem with an accuracy of 0.001.

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 a discriminant
- using Vieta's theorem (if possible).

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

This program can be useful for high school students in general education schools when preparing for tests and exams, when testing knowledge before the Unified State Exam, and for 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 your math or algebra homework done as quickly as possible? In this case, you can also use our programs with detailed solutions.

In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems increases.

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

Rules for entering a quadratic 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 whole or fractional numbers.
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 can be separated from the whole part by either a period or a comma.
For example, you can enter decimal fractions like this: 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: /
The whole part is separated from the fraction by the ampersand sign: &
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 parentheses. 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)

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A little theory.

Quadratic equation and its roots. Incomplete quadratic equations

Each of the equations
\(-x^2+6x+1.4=0, \quad 8x^2-7x=0, \quad x^2-\frac(4)(9)=0 \)
looks like
\(ax^2+bx+c=0, \)
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 is called an equation of the form ax 2 +bx+c=0, 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 free term.

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 of x 2 is equal to 1 is called given quadratic equation. For example, the quadratic equations given are the equations
\(x^2-11x+30=0, \quad x^2-6x=0, \quad x^2-8=0 \)

If in a 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. Thus, 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.

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

Let's consider solving equations of each of these types.

To solve an incomplete quadratic equation of the form ax 2 +c=0 for \(c \neq 0 \), move its free term to the right side and divide both sides of the equation 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 with \(b \neq 0 \) factor 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.

This means that 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 =0 is equivalent to the equation x 2 =0 and therefore has a single root 0.

Formula for the roots of a quadratic equation

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

Let us solve the quadratic equation in general form and as a result we obtain the formula for the roots. This formula can then be used to solve any quadratic equation.

Let's solve the quadratic equation ax 2 +bx+c=0

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

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

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

Now, using the discriminant notation, 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 is 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, a quadratic equation can have two roots (for D > 0), one root (for D = 0) or have 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 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.

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\( \begin(array)(l) x_1+x_2=-p \\ x_1 \cdot x_2=q \end(array) \right. \)

In modern society, the ability to perform operations with equations containing a squared variable can be useful in many areas of activity and is widely used in practice in scientific and technical developments. Evidence of this can be found in the design of sea and river vessels, airplanes and rockets. Using such calculations, the trajectories of movement of a wide variety of bodies, including space objects, are determined. Examples with the solution of quadratic equations are used not only in economic forecasting, in the design and construction of buildings, but also in the most ordinary everyday circumstances. They may be needed on hiking trips, at sporting events, in stores when making purchases and in other very common situations.

Let's break the expression into its component factors

The degree of an equation is determined by the maximum value of the degree of the variable that the expression contains. If it is equal to 2, then such an equation is called quadratic.

If we speak in the language of formulas, then the indicated expressions, no matter how they look, can always be brought to the form when the left side of the expression consists of three terms. Among them: ax 2 (that is, a variable squared with its coefficient), bx (an unknown without a square with its coefficient) and c (a free component, that is, an ordinary number). All this on the right side is equal to 0. In the case when such a polynomial lacks one of its constituent terms, with the exception of ax 2, it is called an incomplete quadratic equation. Examples with the solution of such problems, the values ​​of the variables in which are easy to find, should be considered first.

If the expression looks like it has two terms on the right side, more precisely ax 2 and bx, the easiest way to find x is by putting the variable out of brackets. Now our equation will look like this: x(ax+b). Next, it becomes obvious that either x=0, or the problem comes down to finding a variable from the following expression: ax+b=0. This is dictated by one of the properties of multiplication. The rule states that the product of two factors results in 0 only if one of them is zero.

Example

x=0 or 8x - 3 = 0

As a result, we get two roots of the equation: 0 and 0.375.

Equations of this kind can describe the movement of bodies under the influence of gravity, which began to move from a certain point taken as the origin of coordinates. Here the mathematical notation takes the following form: y = v 0 t + gt 2 /2. By substituting the necessary values, equating the right side to 0 and finding possible unknowns, you can find out the time that passes from the moment the body rises to the moment it falls, as well as many other quantities. But we'll talk about this later.

Factoring an Expression

The rule described above makes it possible to solve these problems in more complex cases. Let's look at examples of solving quadratic equations of this type.

X 2 - 33x + 200 = 0

This quadratic trinomial is complete. First, let's transform the expression and factor it. There are two of them: (x-8) and (x-25) = 0. As a result, we have two roots 8 and 25.

Examples with solving quadratic equations in grade 9 allow this method to find a variable in expressions not only of the second, but even of the third and fourth orders.

For example: 2x 3 + 2x 2 - 18x - 18 = 0. When factoring the right side into factors with a variable, there are three of them, that is, (x+1), (x-3) and (x+3).

As a result, it becomes obvious that this equation has three roots: -3; -1; 3.

Square Root

Another case of an incomplete second-order equation is an expression represented in the language of letters in such a way that the right-hand side is constructed from the components ax 2 and c. Here, to obtain the value of the variable, the free term is transferred to the right side, and after that the square root is extracted from both sides of the equality. It should be noted that in this case there are usually two roots of the equation. The only exceptions can be equalities that do not contain a term with at all, where the variable is equal to zero, as well as variants of expressions when the right side turns out to be negative. In the latter case, there are no solutions at all, since the above actions cannot be performed with roots. Examples of solutions to quadratic equations of this type should be considered.

In this case, the roots of the equation will be the numbers -4 and 4.

Calculation of land area

The need for this kind of calculations appeared in ancient times, because the development of mathematics in those distant times was largely determined by the need to determine with the greatest accuracy the areas and perimeters of land plots.

We should also consider examples of solving quadratic equations based on problems of this kind.

So, let's say there is a rectangular plot of land, the length of which is 16 meters greater than the width. You should find the length, width and perimeter of the site if you know that its area is 612 m2.

To get started, let's first create the necessary equation. Let us denote by x the width of the area, then its length will be (x+16). From what has been written it follows that the area is determined by the expression x(x+16), which, according to the conditions of our problem, is 612. This means that x(x+16) = 612.

Solving complete quadratic equations, and this expression is exactly that, cannot be done in the same way. Why? Although the left side still contains two factors, their product does not equal 0 at all, so different methods are used here.

Discriminant

First of all, we will make the necessary transformations, then the appearance of this expression will look like this: x 2 + 16x - 612 = 0. This means that we have received the expression in a form corresponding to the previously specified standard, where a=1, b=16, c= -612.

This could be an example of solving quadratic equations using a discriminant. Here the necessary calculations are made according to the scheme: D = b 2 - 4ac. This auxiliary quantity not only makes it possible to find the required quantities in a second-order equation, it determines the number of possible options. If D>0, there are two of them; for D=0 there is one root. In case D<0, никаких шансов для решения у уравнения вообще не имеется.

About roots and their formula

In our case, the discriminant is equal to: 256 - 4(-612) = 2704. This suggests that our problem has an answer. If you know k, the solution of quadratic equations must be continued using the formula below. It allows you to calculate the roots.

This means that in the presented case: x 1 =18, x 2 =-34. The second option in this dilemma cannot be a solution, because the dimensions of the land plot cannot be measured in negative quantities, which means x (that is, the width of the plot) is 18 m. From here we calculate the length: 18+16=34, and the perimeter 2(34+ 18)=104(m2).

Examples and tasks

We continue our study of quadratic equations. Examples and detailed solutions of several of them will be given below.

1) 15x 2 + 20x + 5 = 12x 2 + 27x + 1

Let’s move everything to the left side of the equality, make a transformation, that is, we’ll get the type of equation that is usually called standard, and equate it to zero.

15x 2 + 20x + 5 - 12x 2 - 27x - 1 = 0

Adding similar ones, we determine the discriminant: D = 49 - 48 = 1. This means our equation will have two roots. Let's calculate them according to the above formula, which means that the first of them will be equal to 4/3, and the second to 1.

2) Now let's solve mysteries of a different kind.

Let's find out if there are any roots here x 2 - 4x + 5 = 1? To obtain a comprehensive answer, let’s reduce the polynomial to the corresponding usual form and calculate the discriminant. In the above example, it is not necessary to solve the quadratic equation, because this is not the essence of the problem at all. In this case, D = 16 - 20 = -4, which means there really are no roots.

Vieta's theorem

It is convenient to solve quadratic equations using the above formulas and the discriminant, when the square root is taken from the value of the latter. But this does not always happen. However, there are many ways to obtain the values ​​of variables in this case. Example: solving quadratic equations using Vieta's theorem. She is named after someone who lived in 16th-century France and made a brilliant career thanks to his mathematical talent and connections at court. His portrait can be seen in the article.

The pattern that the famous Frenchman noticed was as follows. He proved that the roots of the equation add up numerically to -p=b/a, and their product corresponds to q=c/a.

Now let's look at specific tasks.

3x 2 + 21x - 54 = 0

For simplicity, let's transform the expression:

x 2 + 7x - 18 = 0

Let's use Vieta's theorem, this will give us the following: the sum of the roots is -7, and their product is -18. From here we get that the roots of the equation are the numbers -9 and 2. After checking, we will make sure that these variable values ​​really fit into the expression.

Parabola graph and equation

The concepts of quadratic function and quadratic equations are closely related. Examples of this have already been given earlier. Now let's look at some mathematical riddles in a little more detail. Any equation of the described type can be represented visually. Such a relationship, drawn as a graph, is called a parabola. Its various types are presented in the figure below.

Any parabola has a vertex, that is, a point from which its branches emerge. If a>0, they go high to infinity, and when a<0, они рисуются вниз. Простейшим примером подобной зависимости является функция y = x 2 . В данном случае в уравнении x 2 =0 неизвестное может принимать только одно значение, то есть х=0, а значит существует только один корень. Это неудивительно, ведь здесь D=0, потому что a=1, b=0, c=0. Выходит формула корней (точнее одного корня) квадратного уравнения запишется так: x = -b/2a.

Visual representations of functions help solve any equations, including quadratic ones. This method is called graphical. And the value of the x variable is the abscissa coordinate at the points where the graph line intersects with 0x. The coordinates of the vertex can be found using the formula just given x 0 = -b/2a. And by substituting the resulting value into the original equation of the function, you can find out y 0, that is, the second coordinate of the vertex of the parabola, which belongs to the ordinate axis.

The intersection of the branches of a parabola with the abscissa axis

There are a lot of examples of solving quadratic equations, but there are also general patterns. Let's look at them. It is clear that the intersection of the graph with the 0x axis for a>0 is possible only if 0 takes negative values. And for a<0 координата у 0 должна быть положительна. Для указанных вариантов D>0. Otherwise D<0. А когда D=0, вершина параболы расположена непосредственно на оси 0х.

From the graph of the parabola you can also determine the roots. The opposite is also true. That is, if it is not easy to obtain a visual representation of a quadratic function, you can equate the right side of the expression to 0 and solve the resulting equation. And knowing the points of intersection with the 0x axis, it is easier to construct a graph.

From history

Using equations containing a squared variable, in the old days they not only made mathematical calculations and determined the areas of geometric figures. The ancients needed such calculations for grand discoveries in the fields of physics and astronomy, as well as for making astrological forecasts.

As modern scientists suggest, the inhabitants of Babylon were among the first to solve quadratic equations. This happened four centuries before our era. Of course, their calculations were radically different from those currently accepted and turned out to be much more primitive. For example, Mesopotamian mathematicians had no idea about the existence of negative numbers. They were also unfamiliar with other subtleties that any modern schoolchild knows.

Perhaps even earlier than the scientists of Babylon, the sage from India Baudhayama began solving quadratic equations. This happened about eight centuries before the era of Christ. True, the second-order equations, the methods for solving which he gave, were the simplest. Besides him, Chinese mathematicians were also interested in similar questions in the old days. In Europe, quadratic equations began to be solved only at the beginning of the 13th century, but later they were used in their works by such great scientists as Newton, Descartes and many others.

Yakupova M.I. 1

Smirnova Yu.V. 1

1 Municipal budgetary educational institution secondary school No. 11

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History of quadratic equations

Babylon

The need to solve equations not only of the first degree, but also of the second, in ancient times was caused by the need to solve problems related to finding the areas of land plots, with the development of astronomy and mathematics itself. Quadratic equations could be solved around 2000 BC. e. Babylonians. The rules for solving these equations set out in the Babylonian texts are essentially the same as modern ones, but these texts lack the concept of a negative number and general methods for solving quadratic equations.

Ancient Greece

In Ancient Greece, scientists such as Diophantus, Euclid and Heron also worked on solving quadratic equations. Diophantus Diophantus of Alexandria is an ancient Greek mathematician who presumably lived in the 3rd century AD. The main work of Diophantus is “Arithmetic” in 13 books. Euclid. Euclid is an ancient Greek mathematician, the author of the first theoretical treatise on mathematics that has come down to us, Heron. Heron - Greek mathematician and engineer first in Greece in the 1st century AD. gives a purely algebraic way to solve a quadratic equation

India

Problems on quadratic equations are found already in the astronomical treatise “Aryabhattiam”, compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scientist, Brahmagupta (VII century), outlined the general rule for solving quadratic equations reduced to a single canonical form: ax2 + bx = c, a> 0. (1) In equation (1) the coefficients can 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.

This is one of the problems of the famous Indian mathematician of the 12th century. Bhaskars.

“A flock of frisky monkeys

And twelve along the vines, having eaten to my heart’s content, had fun

They began to jump, hanging

Part eight of them squared

How many monkeys were there?

I was having fun in the clearing

Tell me, in this pack?

Bhaskara's solution indicates that the author knew that the roots of quadratic equations are two-valued. Bhaskar writes the equation corresponding to the problem as x2 - 64x = - 768 and, in order to complete the left side of this equation to a square, adds 322 to both sides, then obtaining: x2 - b4x + 322 = -768 + 1024, (x - 32)2 = 256, x - 32= ±16, x1 = 16, x2 = 48.

Quadratic equations in 17th century Europe

Formulas for solving quadratic equations along the lines of Al-Khorezmi in Europe were first set forth in the Book of Abacus, written in 1202 by the Italian mathematician Leonardo Fibonacci. This voluminous work, which reflects the influence of mathematics, both from the countries of Islam and from ancient Greece, is distinguished by both completeness and clarity of presentation. The author independently developed some new algebraic examples of solving problems 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 problems from the Book of Abacus were used in almost all European textbooks of the 16th - 17th centuries. and partly XVIII. 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 were 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 work of Girard, Descartes, Newton and other scientists, the method of solving quadratic equations takes on a modern form.

Definition of a quadratic equation

An equation of the form ax 2 + bx + c = 0, where a, b, c are numbers, is called quadratic.

Quadratic equation coefficients

Numbers a, b, c are the coefficients of the quadratic equation. a is the first coefficient (before x²), a ≠ 0; b is the second coefficient (before x); c is the free term (without x).

Which of these equations are not quadratic??

1. 4x² + 4x + 1 = 0;2. 5x - 7 = 0;3. - x² - 5x - 1 = 0;4. 2/x² + 3x + 4 = 0;5. ¼ x² - 6x + 1 = 0;6. 2x² = 0;

7. 4x² + 1 = 0;8. x² - 1/x = 0;9. 2x² - x = 0;10. x² -16 = 0;11. 7x² + 5x = 0;12. -8x²= 0;13. 5x³ +6x -8= 0.

Types of quadratic equations

Reduced is a quadratic equation in which the leading coefficient is equal to one. Such an equation can be obtained by dividing the entire expression by the leading coefficient a:

x 2 + px + q =0, p = b/a, q = c/a

A quadratic equation is called complete if all its coefficients are nonzero.

A quadratic equation is called incomplete in which at least one of the coefficients, except the leading one (either the second coefficient or the free term), is equal to zero.

Methods for solving quadratic equations

Method I General formula for calculating roots

To find the roots of a quadratic equation ax 2 + b + c = 0 In general, you should use the algorithm below:

Calculate the value of the discriminant of a quadratic equation: this is the expression for it D= b 2 - 4ac

Derivation of the formula:

Note: It is obvious that the formula for a root of multiplicity 2 is a special case of the general formula, obtained by substituting the equality D=0 into it, and the conclusion about the absence of real roots at D0, and (displaystyle (sqrt (-1))=i) = i.

The presented method is universal, but it is far from the only one. Solving a single equation can be approached in a variety of ways, with preferences usually depending on the solver. In addition, often for this purpose some of the methods turn out to be much more elegant, simple, and less labor-intensive than the standard one.

II method. Roots of a quadratic equation with an even coefficient b III method. Solving incomplete quadratic equations

IV method. Using partial ratios of coefficients

There are special cases of quadratic equations in which the coefficients are in relationships with each other, making them much easier to solve.

Roots of a quadratic equation in which the sum of the leading coefficient and the free term is equal to the second coefficient

If in a quadratic equation ax 2 + bx + c = 0 the sum of the first coefficient and the free term is equal to the second coefficient: a+b=c, then its roots are -1 and the number opposite to the ratio of the free term to the leading coefficient ( -c/a).

Hence, before solving any quadratic equation, you should check the possibility of applying this theorem to it: compare the sum of the leading coefficient and the free term with the second coefficient.

Roots of a quadratic equation whose sum of all coefficients is zero

If in a quadratic equation the sum of all its coefficients is zero, then the roots of such an equation are 1 and the ratio of the free term to the leading coefficient ( c/a).

Hence, before solving an equation using standard methods, you should check the applicability of this theorem to it: add up all the coefficients of this equation and see if this sum is not equal to zero.

V method. Factoring a quadratic trinomial into linear factors

If the trinomial is of the form (displaystyle ax^(2)+bx+c(anot =0))ax 2 + bx + c(a ≠ 0) can somehow be represented as a product of linear factors (displaystyle (kx+m)(lx+n)=0)(kx + m)(lx + n), then we can find the roots of the equation ax 2 + bx + c = 0- they will be -m/k and n/l, indeed, after all (displaystyle (kx+m)(lx+n)=0Longleftrightarrow kx+m=0cup lx+n=0)(kx + m)(lx + n) = 0 kx + mUlx + n, and having solved the indicated linear equations, we obtain the above. Note that a quadratic trinomial does not always decompose into linear factors with real coefficients: this is possible if the corresponding equation has real roots.

Let's consider some special cases

Using the squared sum (difference) formula

If the quadratic trinomial has the form (displaystyle (ax)^(2)+2abx+b^(2))ax 2 + 2abx + b 2 , then by applying the above formula to it, we can factor it into linear factors and, therefore, find roots:

(ax) 2 + 2abx + b 2 = (ax + b) 2

Isolating the full square of the sum (difference)

The above formula is also used using a method called “selecting the full square of the sum (difference).” In relation to the above quadratic equation with the previously introduced notation, this means the following:

Note: If you notice, this formula coincides with the one proposed in the section “Roots of the reduced quadratic equation”, which, in turn, can be obtained from the general formula (1) by substituting the equality a=1. This fact is not just a coincidence: using the described method, albeit with some additional reasoning, one can derive a general formula and also prove the properties of the discriminant.

VI method. Using the direct and inverse Vieta theorem

Vieta's direct theorem (see below in the section of the same name) and its inverse theorem allow you to solve the above quadratic equations orally, without resorting to rather cumbersome calculations using formula (1).

According to the converse theorem, every pair of numbers (number) (displaystyle x_(1),x_(2))x 1, x 2, being a solution to the system of equations below, are the roots of the equation

In the general case, that is, for an unreduced quadratic equation ax 2 + bx + c = 0

x 1 + x 2 = -b/a, x 1 * x 2 = c/a

A direct theorem will help you find numbers that satisfy these equations orally. With its help, you can determine the signs of the roots without knowing the roots themselves. To do this, you should follow the rule:

1) if the free term is negative, then the roots have different signs, and the largest in absolute value of the roots has a sign opposite to the sign of the second coefficient of the equation;

2) if the free term is positive, then both roots have the same sign, and this is the sign opposite to the sign of the second coefficient.

VII method. Transfer method

The so-called “transfer” method allows you to reduce the solution of unreduced and irreducible equations to the form of reduced equations with integer coefficients by dividing them by the leading coefficient to the solution of reduced equations with integer coefficients. It is as follows:

Next, the equation is solved orally in the manner described above, then they return to the original variable and find the roots of the equations (displaystyle y_(1)=ax_(1)) y 1 =ax 1 And y 2 =ax 2 .(displaystyle y_(2)=ax_(2))

Geometric meaning

The graph of a quadratic function is a parabola. The solutions (roots) of a quadratic equation are the abscissas of the points of intersection of the parabola with the abscissa axis. If the parabola described by a quadratic function does not intersect the x-axis, the equation has no real roots. If a parabola intersects the x-axis at one point (at the vertex of the parabola), the equation has one real root (the equation is also said to have two coinciding roots). If the parabola intersects the x-axis at two points, the equation has two real roots (see image on the right.)

If coefficient (displaystyle a) a positive, the branches of the parabola are directed upward and vice versa. If the coefficient (displaystyle b) bpositive (if positive (displaystyle a) a, if negative, vice versa), then the vertex of the parabola lies in the left half-plane and vice versa.

Application of quadratic equations in life

The quadratic equation is widely used. It is used in many calculations, structures, sports, and also around us.

Let us consider and give some examples of the application of the quadratic equation.

Sport. High jumps: during the jumper's run-up, calculations related to the parabola are used to get the most accurate shot at the take-off bar and fly high.

Also, similar calculations are needed in throwing. The flight range of an object depends on the quadratic equation.

Astronomy. The trajectory of the planets can be found using a quadratic equation.

Airplane flight. Airplane takeoff is the main component of flight. Here we take the calculation for low resistance and acceleration of takeoff.

Quadratic equations are also used in various economic disciplines, in programs for processing audio, video, vector and raster graphics.

Conclusion

As a result of the work done, it turned out that quadratic equations attracted scientists back in ancient times; they had already encountered them when solving some problems and tried to solve them. Looking at different ways to solve quadratic equations, I came to the conclusion that not all of them are simple. In my opinion, the best way to solve quadratic equations is to solve them using formulas. The formulas are easy to remember, this method is universal. The hypothesis that equations are widely used in life and mathematics was confirmed. After studying the topic, I learned many interesting facts about quadratic equations, their use, application, types, solutions. And I will be happy to continue studying them. I hope this will help me do well in my exams.

List of used literature

Site materials:

Wikipedia

Open lesson.rf

Handbook of Elementary Mathematics Vygodsky M. Ya.

", that is, equations of the first degree. In this lesson we will look at what is called a quadratic equation and how to solve it.

What is a quadratic equation?

Important!

The degree of an equation is determined by the highest degree to which the unknown stands.

If the maximum power in which the unknown is “2”, then you have a quadratic equation.

Examples of quadratic equations

• 5x 2 − 14x + 17 = 0
• −x 2 + x +

  1

  3

  = 0
• x 2 + 0.25x = 0
• x 2 − 8 = 0

Important! The general form of a quadratic equation looks like this:

A x 2 + b x + c = 0

• “a” is the first or highest coefficient;
• “b” is the second coefficient;
• “c” is a free term.

To find “a”, “b” and “c” you need to compare your equation with the general form of the quadratic equation “ax 2 + bx + c = 0”.

Let's practice determining the coefficients "a", "b" and "c" in quadratic equations.

• a = −1
• b = 1
• c =

  1

  3

• a = 1
• b = 0.25
• c = 0

• a = 1
• b = 0
• c = −8

How to Solve Quadratic Equations

Unlike linear equations, a special method is used to solve quadratic equations. formula for finding roots.

Remember!

To solve a quadratic equation you need:

• bring the quadratic equation to the general form “ax 2 + bx + c = 0”. That is, only “0” should remain on the right side;
• use formula for roots:

Let's look at an example of how to use the formula to find the roots of a quadratic equation. Let's solve a quadratic equation.

X 2 − 3x − 4 = 0

The equation “x 2 − 3x − 4 = 0” has already been reduced to the general form “ax 2 + bx + c = 0” and does not require additional simplifications. To solve it, we just need to apply formula for finding the roots of a quadratic equation.

Let us determine the coefficients “a”, “b” and “c” for this equation.

It can be used to solve any quadratic equation.

In the formula “x 1;2 = ” the radical expression is often replaced
“b 2 − 4ac” for the letter “D” and is called discriminant. The concept of a discriminant is discussed in more detail in the lesson “What is a discriminant”.

Let's look at another example of a quadratic equation.

x 2 + 9 + x = 7x

In this form, it is quite difficult to determine the coefficients “a”, “b” and “c”. Let's first reduce the equation to the general form “ax 2 + bx + c = 0”.

X 2 + 9 + x = 7x
x 2 + 9 + x − 7x = 0
x 2 + 9 − 6x = 0
x 2 − 6x + 9 = 0

Now you can use the formula for the roots.

X 1;2 =
x 1;2 =
x 1;2 =
x 1;2 =
x =

There are times when quadratic equations have no roots. This situation occurs when the formula contains a negative number under the root.