How to solve quadratic equations. Solving complete quadratic equations

", 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”, “b” and “c” are given numbers.

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

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.

5x 2 − 14x + 17 = 0 −7x 2 − 13x + 8 = 0 −x 2 + x +

The equation	Odds
a = 5 b = −14 c = 17
a = −7 b = −13 c = 8

1

3

= 0

• a = −1
• b = 1
• c =

  1

  3

x 2 + 0.25x = 0

• a = 1
• b = 0.25
• c = 0

x 2 − 8 = 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.
Let us determine the coefficients “a”, “b” and “c” for this equation.
Let us determine the coefficients “a”, “b” and “c” for this equation.
Let us determine the coefficients “a”, “b” and “c” for this equation.

x 1;2 =

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 − 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 =

6

2

x = 3
Answer: x = 3

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

Continuing the topic “Solving Equations,” the material in this article will introduce you to quadratic equations.

Let's look at everything in detail: the essence and notation of a quadratic equation, define the accompanying terms, analyze the scheme for solving incomplete and complete equations, get acquainted with the formula of roots and the discriminant, establish connections between the roots and coefficients, and of course we will give a visual solution to practical examples.

Quadratic equation, its types

Definition 1

Quadratic equation is an equation written as a x 2 + b x + c = 0, Where x– variable, a , b and c– some numbers, while a is not zero.

Often, quadratic equations are also called equations of the second degree, since in essence a quadratic equation is an algebraic equation of the second degree.

Let's give an example to illustrate the given definition: 9 x 2 + 16 x + 2 = 0 ; 7, 5 x 2 + 3, 1 x + 0, 11 = 0, etc. These are quadratic equations.

Definition 2

Numbers a, b and c are the coefficients of the quadratic equation a x 2 + b x + c = 0, while the coefficient a is called the first, or senior, or coefficient at x 2, b - the second coefficient, or coefficient at x, A c called a free member.

For example, in the quadratic equation 6 x 2 − 2 x − 11 = 0 the leading coefficient is 6, the second coefficient is − 2 , and the free term is equal to − 11 . Let us pay attention to the fact that when the coefficients b and/or c are negative, then a short form of the form is used 6 x 2 − 2 x − 11 = 0, but not 6 x 2 + (− 2) x + (− 11) = 0.

Let us also clarify this aspect: if the coefficients a and/or b equal 1 or − 1 , then they may not take an explicit part in writing the quadratic equation, which is explained by the peculiarities of writing the indicated numerical coefficients. For example, in the quadratic equation y 2 − y + 7 = 0 the leading coefficient is 1, and the second coefficient is − 1 .

Reduced and unreduced quadratic equations

Based on the value of the first coefficient, quadratic equations are divided into reduced and unreduced.

Definition 3

Reduced quadratic equation is a quadratic equation where the leading coefficient is 1. For other values ​​of the leading coefficient, the quadratic equation is unreduced.

Let's give examples: quadratic equations x 2 − 4 · x + 3 = 0, x 2 − x − 4 5 = 0 are reduced, in each of which the leading coefficient is 1.

9 x 2 − x − 2 = 0- unreduced quadratic equation, where the first coefficient is different from 1 .

Any unreduced quadratic equation can be converted into a reduced equation by dividing both sides by the first coefficient (equivalent transformation). The transformed equation will have the same roots as the given unreduced equation or will also have no roots at all.

Consideration of a specific example will allow us to clearly demonstrate the transition from an unreduced quadratic equation to a reduced one.

Example 1

Given the equation 6 x 2 + 18 x − 7 = 0 . It is necessary to convert the original equation into the reduced form.

Solution

According to the above scheme, we divide both sides of the original equation by the leading coefficient 6. Then we get: (6 x 2 + 18 x − 7) : 3 = 0: 3, and this is the same as: (6 x 2) : 3 + (18 x) : 3 − 7: 3 = 0 and further: (6: 6) x 2 + (18: 6) x − 7: 6 = 0. From here: x 2 + 3 x - 1 1 6 = 0 . Thus, an equation equivalent to the given one is obtained.

Answer: x 2 + 3 x - 1 1 6 = 0 .

Complete and incomplete quadratic equations

Let's turn to the definition of a quadratic equation. In it we specified that a ≠ 0. A similar condition is necessary for the equation a x 2 + b x + c = 0 was precisely square, since at a = 0 it essentially transforms into a linear equation b x + c = 0.

In the case when the coefficients b And c are equal to zero (which is possible, both individually and jointly), the quadratic equation is called incomplete.

Definition 4

Incomplete quadratic equation- such a quadratic equation a x 2 + b x + c = 0, where at least one of the coefficients b And c(or both) is zero.

Complete quadratic equation– a quadratic equation in which all numerical coefficients are not equal to zero.

Let's discuss why the types of quadratic equations are given exactly these names.

When b = 0, the quadratic equation takes the form a x 2 + 0 x + c = 0, which is the same as a x 2 + c = 0. At c = 0 quadratic equation written as a x 2 + b x + 0 = 0, which is equivalent a x 2 + b x = 0. At b = 0 And c = 0 the equation will take the form a x 2 = 0. The equations that we obtained differ from the complete quadratic equation in that their left-hand sides do not contain either a term with the variable x, or a free term, or both. Actually, this fact gave the name to this type of equation – incomplete.

For example, x 2 + 3 x + 4 = 0 and − 7 x 2 − 2 x + 1, 3 = 0 are complete quadratic equations; x 2 = 0, − 5 x 2 = 0; 11 x 2 + 2 = 0, − x 2 − 6 x = 0 – incomplete quadratic equations.

Solving incomplete quadratic equations

The definition given above makes it possible to distinguish the following types of incomplete quadratic equations:

• a x 2 = 0, this equation corresponds to the coefficients b = 0 and c = 0 ;
• a · x 2 + c = 0 at b = 0 ;
• a · x 2 + b · x = 0 at c = 0.

Let us consider sequentially the solution of each type of incomplete quadratic equation.

Solution of the equation a x 2 =0

As mentioned above, this equation corresponds to the coefficients b And c, equal to zero. The equation a x 2 = 0 can be converted into an equivalent equation x 2 = 0, which we get by dividing both sides of the original equation by the number a, not equal to zero. The obvious fact is that the root of the equation x 2 = 0 this is zero because 0 2 = 0 . This equation has no other roots, which can be explained by the properties of the degree: for any number p, not equal to zero, the inequality is true p 2 > 0, from which it follows that when p ≠ 0 equality p 2 = 0 will never be achieved.

Definition 5

Thus, for the incomplete quadratic equation a x 2 = 0 there is a unique root x = 0.

Example 2

For example, let’s solve an incomplete quadratic equation − 3 x 2 = 0. It is equivalent to the equation x 2 = 0, its only root is x = 0, then the original equation has a single root - zero.

Briefly, the solution is written as follows:

− 3 x 2 = 0, x 2 = 0, x = 0.

Solving the equation a x 2 + c = 0

Next in line is the solution of incomplete quadratic equations, where b = 0, c ≠ 0, that is, equations of the form a x 2 + c = 0. Let's transform this equation by moving a term from one side of the equation to the other, changing the sign to the opposite one and dividing both sides of the equation by a number that is not equal to zero:

• transfer c to the right hand side, which gives the equation a x 2 = − c;
• divide both sides of the equation by a, we end up with x = - c a .

Our transformations are equivalent; accordingly, the resulting equation is also equivalent to the original one, and this fact makes it possible to draw conclusions about the roots of the equation. From what the values ​​are a And c the value of the expression - c a depends: it can have a minus sign (for example, if a = 1 And c = 2, then - c a = - 2 1 = - 2) or a plus sign (for example, if a = − 2 And c = 6, then - c a = - 6 - 2 = 3); it is not zero because c ≠ 0. Let us dwell in more detail on situations when - c a< 0 и - c a > 0 .

In the case when - c a< 0 , уравнение x 2 = - c a не будет иметь корней. Утверждая это, мы опираемся на то, что квадратом любого числа является число неотрицательное. Из сказанного следует, что при - c a < 0 ни для какого числа p the equality p 2 = - c a cannot be true.

Everything is different when - c a > 0: remember the square root, and it will become obvious that the root of the equation x 2 = - c a will be the number - c a, since - c a 2 = - c a. It is not difficult to understand that the number - - c a is also the root of the equation x 2 = - c a: indeed, - - c a 2 = - c a.

The equation will have no other roots. We can demonstrate this using the method of contradiction. To begin with, let us define the notations for the roots found above as x 1 And − x 1. Let us assume that the equation x 2 = - c a also has a root x 2, which is different from the roots x 1 And − x 1. We know that by substituting into the equation x its roots, we transform the equation into a fair numerical equality.

For x 1 And − x 1 we write: x 1 2 = - c a , and for x 2- x 2 2 = - c a . Based on the properties of numerical equalities, we subtract one correct equality term by term from another, which will give us: x 1 2 − x 2 2 = 0. We use the properties of operations with numbers to rewrite the last equality as (x 1 − x 2) · (x 1 + x 2) = 0. It is known that the product of two numbers is zero if and only if at least one of the numbers is zero. From the above it follows that x 1 − x 2 = 0 and/or x 1 + x 2 = 0, which is the same x 2 = x 1 and/or x 2 = − x 1. An obvious contradiction arose, because at first it was agreed that the root of the equation x 2 differs from x 1 And − x 1. So, we have proven that the equation has no roots other than x = - c a and x = - - c a.

Let us summarize all the arguments above.

Definition 6

Incomplete quadratic equation a x 2 + c = 0 is equivalent to the equation x 2 = - c a, which:

• will have no roots at - c a< 0 ;
• will have two roots x = - c a and x = - - c a for - c a > 0.

Let us give examples of solving the equations a x 2 + c = 0.

Example 3

Given a quadratic equation 9 x 2 + 7 = 0. It is necessary to find a solution.

Solution

Let's move the free term to the right side of the equation, then the equation will take the form 9 x 2 = − 7.
Let us divide both sides of the resulting equation by 9 , we arrive at x 2 = - 7 9 . On the right side we see a number with a minus sign, which means: the given equation has no roots. Then the original incomplete quadratic equation 9 x 2 + 7 = 0 will have no roots.

Answer: the equation 9 x 2 + 7 = 0 has no roots.

Example 4

The equation needs to be solved − x 2 + 36 = 0.

Solution

Let's move 36 to the right side: − x 2 = − 36.
Let's divide both parts by − 1 , we get x 2 = 36. On the right side there is a positive number, from which we can conclude that x = 36 or x = - 36 .
Let's extract the root and write down the final result: incomplete quadratic equation − x 2 + 36 = 0 has two roots x = 6 or x = − 6.

Answer: x = 6 or x = − 6.

Solution of the equation a x 2 +b x=0

Let us analyze the third type of incomplete quadratic equations, when c = 0. To find a solution to an incomplete quadratic equation a x 2 + b x = 0, we will use the factorization method. Let's factorize the polynomial that is on the left side of the equation, taking the common factor out of brackets x. This step will make it possible to transform the original incomplete quadratic equation into its equivalent x (a x + b) = 0. And this equation, in turn, is equivalent to a set of equations x = 0 And a x + b = 0. The equation a x + b = 0 linear, and its root: x = − b a.

Definition 7

Thus, the incomplete quadratic equation a x 2 + b x = 0 will have two roots x = 0 And x = − b a.

Let's reinforce the material with an example.

Example 5

It is necessary to find a solution to the equation 2 3 · x 2 - 2 2 7 · x = 0.

Solution

We'll take it out x outside the brackets we get the equation x · 2 3 · x - 2 2 7 = 0 . This equation is equivalent to the equations x = 0 and 2 3 x - 2 2 7 = 0. Now you should solve the resulting linear equation: 2 3 · x = 2 2 7, x = 2 2 7 2 3.

Briefly write the solution to the equation as follows:

2 3 x 2 - 2 2 7 x = 0 x 2 3 x - 2 2 7 = 0

x = 0 or 2 3 x - 2 2 7 = 0

x = 0 or x = 3 3 7

Answer: x = 0, x = 3 3 7.

Discriminant, formula for the roots of a quadratic equation

To find solutions to quadratic equations, there is a root formula:

Definition 8

x = - b ± D 2 · a, where D = b 2 − 4 a c– the so-called discriminant of a quadratic equation.

Writing x = - b ± D 2 · a essentially means that x 1 = - b + D 2 · a, x 2 = - b - D 2 · a.

It would be useful to understand how this formula was derived and how to apply it.

Derivation of the formula for the roots of a quadratic equation

Let us be faced with the task of solving a quadratic equation a x 2 + b x + c = 0. Let us carry out a number of equivalent transformations:

• divide both sides of the equation by a number a, different from zero, we obtain the following quadratic equation: x 2 + b a · x + c a = 0 ;
• Let's select the complete square on the left side of the resulting equation:
  x 2 + b a · x + c a = x 2 + 2 · b 2 · a · x + b 2 · a 2 - b 2 · a 2 + c a = = x + b 2 · a 2 - b 2 · a 2 + c a
  After this, the equation will take the form: x + b 2 · a 2 - b 2 · a 2 + c a = 0;
• Now it is possible to transfer the last two terms to the right side, changing the sign to the opposite, after which we get: x + b 2 · a 2 = b 2 · a 2 - c a ;
• Finally, we transform the expression written on the right side of the last equality:
  b 2 · a 2 - c a = b 2 4 · a 2 - c a = b 2 4 · a 2 - 4 · a · c 4 · a 2 = b 2 - 4 · a · c 4 · a 2 .

Thus, we arrive at the equation x + b 2 · a 2 = b 2 - 4 · a · c 4 · a 2 , equivalent to the original equation a x 2 + b x + c = 0.

We examined the solution of such equations in the previous paragraphs (solving incomplete quadratic equations). The experience already gained makes it possible to draw a conclusion regarding the roots of the equation x + b 2 · a 2 = b 2 - 4 · a · c 4 · a 2:

• with b 2 - 4 a c 4 a 2< 0 уравнение не имеет действительных решений;
• when b 2 - 4 · a · c 4 · a 2 = 0 the equation is x + b 2 · a 2 = 0, then x + b 2 · a = 0.

From here the only root x = - b 2 · a is obvious;

• for b 2 - 4 · a · c 4 · a 2 > 0, the following will be true: x + b 2 · a = b 2 - 4 · a · c 4 · a 2 or x = b 2 · a - b 2 - 4 · a · c 4 · a 2 , which is the same as x + - b 2 · a = b 2 - 4 · a · c 4 · a 2 or x = - b 2 · a - b 2 - 4 · a · c 4 · a 2 , i.e. the equation has two roots.

It is possible to conclude that the presence or absence of roots of the equation x + b 2 · a 2 = b 2 - 4 · a · c 4 · a 2 (and therefore the original equation) depends on the sign of the expression b 2 - 4 · a · c 4 · a 2 written on the right side. And the sign of this expression is given by the sign of the numerator, (denominator 4 a 2 will always be positive), that is, the sign of the expression b 2 − 4 a c. This expression b 2 − 4 a c the name is given - the discriminant of the quadratic equation and the letter D is defined as its designation. Here you can write down the essence of the discriminant - based on its value and sign, they can conclude whether the quadratic equation will have real roots, and, if so, what is the number of roots - one or two.

Let's return to the equation x + b 2 · a 2 = b 2 - 4 · a · c 4 · a 2 . Let's rewrite it using discriminant notation: x + b 2 · a 2 = D 4 · a 2 .

Let us formulate our conclusions again:

Definition 9

• at D< 0 the equation has no real roots;
• at D=0 the equation has a single root x = - b 2 · a ;
• at D > 0 the equation has two roots: x = - b 2 · a + D 4 · a 2 or x = - b 2 · a - D 4 · a 2. Based on the properties of radicals, these roots can be written in the form: x = - b 2 · a + D 2 · a or - b 2 · a - D 2 · a. And, when we expand the modules and bring the fractions to a common denominator, we get: x = - b + D 2 · a, x = - b - D 2 · a.

So, the result of our reasoning was the derivation of the formula for the roots of a quadratic equation:

x = - b + D 2 a, x = - b - D 2 a, discriminant D calculated by the formula D = b 2 − 4 a c.

These formulas make it possible to determine both real roots when the discriminant is greater than zero. When the discriminant is zero, applying both formulas will give the same root as the only solution to the quadratic equation. In the case where the discriminant is negative, if we try to use the quadratic root formula, we will be faced with the need to take the square root of a negative number, which will take us beyond the scope of real numbers. With a negative discriminant, the quadratic equation will not have real roots, but a pair of complex conjugate roots is possible, determined by the same root formulas we obtained.

Algorithm for solving quadratic equations using root formulas

It is possible to solve a quadratic equation by immediately using the root formula, but this is generally done when it is necessary to find complex roots.

In the majority of cases, it usually means searching not for complex, but for real roots of a quadratic equation. Then it is optimal, before using the formulas for the roots of a quadratic equation, to first determine the discriminant and make sure that it is not negative (otherwise we will conclude that the equation has no real roots), and then proceed to calculate the value of the roots.

The reasoning above makes it possible to formulate an algorithm for solving a quadratic equation.

Definition 10

To solve a quadratic equation a x 2 + b x + c = 0, necessary:

• according to the formula D = b 2 − 4 a c find the discriminant value;
• at D< 0 сделать вывод об отсутствии у квадратного уравнения действительных корней;
• for D = 0, find the only root of the equation using the formula x = - b 2 · a ;
• for D > 0, determine two real roots of the quadratic equation using the formula x = - b ± D 2 · a.

Note that when the discriminant is zero, you can use the formula x = - b ± D 2 · a, it will give the same result as the formula x = - b 2 · a.

Let's look at examples.

Examples of solving quadratic equations

Let us give solutions to examples for different values ​​of the discriminant.

Example 6

We need to find the roots of the equation x 2 + 2 x − 6 = 0.

Solution

Let's write down the numerical coefficients of the quadratic equation: a = 1, b = 2 and c = − 6. Next we proceed according to the algorithm, i.e. Let's start calculating the discriminant, for which we will substitute the coefficients a, b And c into the discriminant formula: D = b 2 − 4 · a · c = 2 2 − 4 · 1 · (− 6) = 4 + 24 = 28 .

So we get D > 0, which means that the original equation will have two real roots.
To find them, we use the root formula x = - b ± D 2 · a and, substituting the corresponding values, we get: x = - 2 ± 28 2 · 1. Let us simplify the resulting expression by taking the factor out of the root sign and then reducing the fraction:

x = - 2 ± 2 7 2

x = - 2 + 2 7 2 or x = - 2 - 2 7 2

x = - 1 + 7 or x = - 1 - 7

Answer: x = - 1 + 7 ​​​​​​, x = - 1 - 7 .

Example 7

Need to solve a quadratic equation − 4 x 2 + 28 x − 49 = 0.

Solution

Let's define the discriminant: D = 28 2 − 4 · (− 4) · (− 49) = 784 − 784 = 0. With this value of the discriminant, the original equation will have only one root, determined by the formula x = - b 2 · a.

x = - 28 2 (- 4) x = 3.5

Answer: x = 3.5.

Example 8

The equation needs to be solved 5 y 2 + 6 y + 2 = 0

Solution

The numerical coefficients of this equation will be: a = 5, b = 6 and c = 2. We use these values ​​to find the discriminant: D = b 2 − 4 · a · c = 6 2 − 4 · 5 · 2 = 36 − 40 = − 4 . The calculated discriminant is negative, so the original quadratic equation has no real roots.

In the case when the task is to indicate complex roots, we apply the root formula, performing actions with complex numbers:

x = - 6 ± - 4 2 5,

x = - 6 + 2 i 10 or x = - 6 - 2 i 10,

x = - 3 5 + 1 5 · i or x = - 3 5 - 1 5 · i.

Answer: there are no real roots; the complex roots are as follows: - 3 5 + 1 5 · i, - 3 5 - 1 5 · i.

In the school curriculum, there is no standard requirement to look for complex roots, therefore, if during the solution the discriminant is determined to be negative, the answer is immediately written down that there are no real roots.

Root formula for even second coefficients

The root formula x = - b ± D 2 · a (D = b 2 − 4 · a · c) makes it possible to obtain another formula, more compact, allowing one to find solutions to quadratic equations with an even coefficient for x (or with a coefficient of the form 2 · n, for example, 2 3 or 14 ln 5 = 2 7 ln 5). Let us show how this formula is derived.

Let us be faced with the task of finding a solution to the quadratic equation a · x 2 + 2 · n · x + c = 0 . We proceed according to the algorithm: we determine the discriminant D = (2 n) 2 − 4 a c = 4 n 2 − 4 a c = 4 (n 2 − a c), and then use the root formula:

x = - 2 · n ± D 2 · a , x = - 2 · n ± 4 · n 2 - a · c 2 · a , x = - 2 · n ± 2 n 2 - a · c 2 · a , x = - n ± n 2 - a · c a .

Let the expression n 2 − a · c be denoted as D 1 (sometimes it is denoted D "). Then the formula for the roots of the quadratic equation under consideration with the second coefficient 2 · n will take the form:

x = - n ± D 1 a, where D 1 = n 2 − a · c.

It is easy to see that D = 4 · D 1, or D 1 = D 4. In other words, D 1 is a quarter of the discriminant. Obviously, the sign of D 1 is the same as the sign of D, which means the sign of D 1 can also serve as an indicator of the presence or absence of roots of a quadratic equation.

Definition 11

Thus, to find a solution to a quadratic equation with a second coefficient of 2 n, it is necessary:

• find D 1 = n 2 − a · c ;
• at D 1< 0 сделать вывод, что действительных корней нет;
• when D 1 = 0, determine the only root of the equation using the formula x = - n a;
• for D 1 > 0, determine two real roots using the formula x = - n ± D 1 a.

Example 9

It is necessary to solve the quadratic equation 5 x 2 − 6 x − 32 = 0.

Solution

We can represent the second coefficient of the given equation as 2 · (− 3) . Then we rewrite the given quadratic equation as 5 x 2 + 2 (− 3) x − 32 = 0, where a = 5, n = − 3 and c = − 32.

Let's calculate the fourth part of the discriminant: D 1 = n 2 − a · c = (− 3) 2 − 5 · (− 32) = 9 + 160 = 169. The resulting value is positive, which means that the equation has two real roots. Let us determine them using the corresponding root formula:

x = - n ± D 1 a, x = - - 3 ± 169 5, x = 3 ± 13 5,

x = 3 + 13 5 or x = 3 - 13 5

x = 3 1 5 or x = - 2

It would be possible to carry out calculations using the usual formula for the roots of a quadratic equation, but in this case the solution would be more cumbersome.

Answer: x = 3 1 5 or x = - 2 .

Simplifying the form of quadratic equations

Sometimes it is possible to optimize the form of the original equation, which will simplify the process of calculating the roots.

For example, the quadratic equation 12 x 2 − 4 x − 7 = 0 is clearly more convenient to solve than 1200 x 2 − 400 x − 700 = 0.

More often, simplification of the form of a quadratic equation is carried out by multiplying or dividing its both sides by a certain number. For example, above we showed a simplified representation of the equation 1200 x 2 − 400 x − 700 = 0, obtained by dividing both sides by 100.

Such a transformation is possible when the coefficients of the quadratic equation are not coprime numbers. Then we usually divide both sides of the equation by the greatest common divisor of the absolute values ​​of its coefficients.

As an example, we use the quadratic equation 12 x 2 − 42 x + 48 = 0. Let us determine the GCD of the absolute values ​​of its coefficients: GCD (12, 42, 48) = GCD(GCD (12, 42), 48) = GCD (6, 48) = 6. Let us divide both sides of the original quadratic equation by 6 and obtain the equivalent quadratic equation 2 x 2 − 7 x + 8 = 0.

By multiplying both sides of a quadratic equation, you usually get rid of fractional coefficients. In this case, they multiply by the least common multiple of the denominators of its coefficients. For example, if each part of the quadratic equation 1 6 x 2 + 2 3 x - 3 = 0 is multiplied with LCM (6, 3, 1) = 6, then it will be written in a simpler form x 2 + 4 x − 18 = 0 .

Finally, we note that we almost always get rid of the minus at the first coefficient of a quadratic equation by changing the signs of each term of the equation, which is achieved by multiplying (or dividing) both sides by − 1. For example, from the quadratic equation − 2 x 2 − 3 x + 7 = 0, you can go to its simplified version 2 x 2 + 3 x − 7 = 0.

Relationship between roots and coefficients

The formula for the roots of quadratic equations, already known to us, x = - b ± D 2 · a, expresses the roots of the equation through its numerical coefficients. Based on this formula, we have the opportunity to specify other dependencies between the roots and coefficients.

The most famous and applicable formulas are Vieta’s theorem:

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

In particular, for the given quadratic equation, the sum of the roots is the second coefficient with the opposite sign, and the product of the roots is equal to the free term. For example, by looking at the form of the quadratic equation 3 x 2 − 7 x + 22 = 0, it is possible to immediately determine that the sum of its roots is 7 3 and the product of the roots is 22 3.

You can also find a number of other connections between the roots and coefficients of a quadratic equation. For example, the sum of the squares of the roots of a quadratic equation can be expressed in terms of coefficients:

x 1 2 + x 2 2 = (x 1 + x 2) 2 - 2 x 1 x 2 = - b a 2 - 2 c a = b 2 a 2 - 2 c a = b 2 - 2 a c a 2.

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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 modeled after 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 its 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

Name	General form of the equation	Feature (what are the coefficients)	Examples of equations
	ax 2 + bx + c = 0	a, b, c - numbers other than 0	1/3x 2 + 5x - 1 = 0
Incomplete
			x 2 - 1/5x = 0
Given	x 2 + bx + c = 0		x 2 - 3x + 5 = 0

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.

Method II. 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 a given 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 the 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 achieve the clearest possible impact on the take-off bar and high flight.

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.

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...

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, aircraft and missiles. 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 is 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 who lived in the 16th century in 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 the 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.