Solving equations in mathematics occupies a special place. This process is preceded by many hours of studying theory, during which the student learns how to solve equations, determine their type, and brings the skill to complete automation. However, searching for roots does not always make sense, since they may simply not exist. There are special techniques for finding roots. In this article we will analyze the main functions, their domains of definition, as well as cases when their roots are missing.

Which equation has no roots?

An equation has no roots if there are no real arguments x for which the equation is identically true. For a non-specialist, this formulation, like most mathematical theorems and formulas, looks very vague and abstract, but this is in theory. In practice, everything becomes extremely simple. For example: the equation 0 * x = -53 has no solution, since there is no number x whose product with zero would give something other than zero.

Now we will look at the most basic types of equations.

1. Linear equation

An equation is called linear if its right and left sides are represented as linear functions: ax + b = cx + d or in generalized form kx + b = 0. Where a, b, c, d are known numbers, and x is an unknown quantity . Which equation has no roots? Examples of linear equations are presented in the illustration below.

Basically, linear equations are solved by simply transferring the number part to one part and the contents of x to another. The result is an equation of the form mx = n, where m and n are numbers, and x is an unknown. To find x, just divide both sides by m. Then x = n/m. Most linear equations have only one root, but there are cases when there are either infinitely many roots or no roots at all. When m = 0 and n = 0, the equation takes the form 0 * x = 0. The solution to such an equation will be absolutely any number.

However, what equation has no roots?

For m = 0 and n = 0, the equation has no roots in the set of real numbers. 0 * x = -1; 0 * x = 200 - these equations have no roots.

2. Quadratic equation

A quadratic equation is an equation of the form ax 2 + bx + c = 0 for a = 0. The most common solution is through the discriminant. The formula for finding the discriminant of a quadratic equation is: D = b 2 - 4 * a * c. Next there are two roots x 1.2 = (-b ± √D) / 2 * a.

For D > 0 the equation has two roots, for D = 0 it has one root. But what quadratic equation has no roots? The easiest way to observe the number of roots of a quadratic equation is by graphing the function, which is a parabola. For a > 0 the branches are directed upward, for a< 0 ветви опущены вниз. Если дискриминант отрицателен, такое квадратное уравнение не имеет корней на множестве действительных чисел.

You can also visually determine the number of roots without calculating the discriminant. To do this, you need to find the vertex of the parabola and determine in which direction the branches are directed. The x coordinate of the vertex can be determined using the formula: x 0 = -b / 2a. In this case, the y coordinate of the vertex is found by simply substituting the x 0 value into the original equation.

The quadratic equation x 2 - 8x + 72 = 0 has no roots, since it has a negative discriminant D = (-8) 2 - 4 * 1 * 72 = -224. This means that the parabola does not touch the x-axis and the function never takes the value 0, therefore the equation has no real roots.

3. Trigonometric equations

Trigonometric functions are considered on the trigonometric circle, but can also be represented in Cartesian system coordinates In this article we will look at two main trigonometric functions and their equations: sinx and cosx. Since these functions form a trigonometric circle with radius 1, |sinx| and |cosx| cannot be greater than 1. So, which sinx equation has no roots? Consider the graph of the sinx function shown in the picture below.

We see that the function is symmetric and has a repetition period of 2pi. Based on this, we can say that the maximum value of this function can be 1, and the minimum -1. For example, the expression cosx = 5 will not have roots, since its absolute value is greater than one.

This is the simplest example of trigonometric equations. In fact, solving them can take many pages, at the end of which you realize that you used the wrong formula and need to start all over again. Sometimes, even if you find the roots correctly, you may forget to take into account the restrictions on OD, which is why an extra root or interval appears in the answer, and the entire answer turns into an error. Therefore, strictly follow all the restrictions, because not all roots fit into the scope of the task.

4. Systems of equations

A system of equations is a set of equations joined by curly or square brackets. The curly brackets indicate that all equations are run together. That is, if at least one of the equations does not have roots or contradicts another, the entire system has no solution. Square brackets indicate the word "or". This means that if at least one of the equations of the system has a solution, then the entire system has a solution.

The answer of the system c is the set of all the roots of the individual equations. And systems with curly braces have only common roots. Systems of equations can include completely different functions, so such complexity does not allow us to immediately say which equation does not have roots.

Found in problem books and textbooks different types equations: those that have roots and those that do not. First of all, if you can’t find the roots, don’t think that they are not there at all. Perhaps you made a mistake somewhere, then you just need to carefully double-check your decision.

We looked at the most basic equations and their types. Now you can tell which equation has no roots. In most cases this is not difficult to do. Achieving success in solving equations requires only attention and concentration. Practice more, it will help you navigate the material much better and faster.

So, the equation has no roots if:

• in the linear equation mx = n the value is m = 0 and n = 0;
• in a quadratic equation, if the discriminant less than zero;
• in a trigonometric equation of the form cosx = m / sinx = n, if |m| > 0, |n| > 0;
• in a system of equations with curly brackets, if at least one equation has no roots, and with square brackets, if all equations have no roots.

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 recording of the quadratic equation, define the associated terms, analyze the scheme for solving incomplete and complete equations, let's get acquainted with the formula of roots and discriminant, establish connections between roots and coefficients, and of course we will give a visual solution to practical examples.

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Quadratic equation, its types

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 use short form records like 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 concrete 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 the quadratic equation is 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 single 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 V right 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 open 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 formula for the root of a quadratic equation, we will be faced with the need to extract Square root from a negative number, which will take us beyond the 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 a solution to the examples for different meanings 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 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 become written in more in simple 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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Quadratic equation - easy to solve! *Hereinafter referred to as “KU”. Friends, it would seem that there could be nothing simpler in mathematics than solving such an equation. But something told me that many people have problems with him. I decided to see how many on-demand impressions Yandex gives out per month. Here's what happened, look:

What does it mean? This means that about 70,000 people a month are looking for this information, what does this summer have to do with it, and what will happen among school year— there will be twice as many requests. This is not surprising, because those guys and girls who graduated from school a long time ago and are preparing for the Unified State Exam are looking for this information, and schoolchildren also strive to refresh their memory.

Despite the fact that there are a lot of sites that tell you how to solve this equation, I decided to also contribute and publish the material. Firstly, I would like visitors to come to my site based on this request; secondly, in other articles, when the topic of “KU” comes up, I will provide a link to this article; thirdly, I’ll tell you a little more about his solution than is usually stated on other sites. Let's get started! The content of the article:

A quadratic equation is an equation of the form:

where coefficients a,band c are arbitrary numbers, with a≠0.

In the school course, the material is given in the following form– the equations are divided into three classes:

1. They have two roots.

2. *Have only one root.

3. They have no roots. It is worth especially noting here that they do not have real roots

How are roots calculated? Just!

We calculate the discriminant. Underneath this “terrible” word lies a very simple formula:

The root formulas are as follows:

*You need to know these formulas by heart.

You can immediately write down and solve:

Example:

1. If D > 0, then the equation has two roots.

2. If D = 0, then the equation has one root.

3. If D< 0, то уравнение не имеет действительных корней.

Let's look at the equation:

In this regard, when the discriminant is equal to zero, the school course says that one root is obtained, here it is equal to nine. Everything is correct, it is so, but...

This idea is somewhat incorrect. In fact, there are two roots. Yes, yes, don’t be surprised, you get two equal roots, and to be mathematically precise, then the answer should write two roots:

x 1 = 3 x 2 = 3

But this is so - a small digression. At school you can write it down and say that there is one root.

Now the next example:

As we know, the root of a negative number cannot be taken, so there is no solution in this case.

That's the whole decision process.

Quadratic function.

This shows what the solution looks like geometrically. This is extremely important to understand (in the future, in one of the articles we will analyze in detail the solution to the quadratic inequality).

This is a function of the form:

where x and y are variables

a, b, c – given numbers, with a ≠ 0

The graph is a parabola:

That is, it turns out that by solving a quadratic equation with “y” equal to zero, we find the points of intersection of the parabola with the x axis. There can be two of these points (the discriminant is positive), one (the discriminant is zero) and none (the discriminant is negative). Details about the quadratic function You can view article by Inna Feldman.

Let's look at examples:

Example 1: Solve 2x 2 +8 x–192=0

a=2 b=8 c= –192

D=b 2 –4ac = 8 2 –4∙2∙(–192) = 64+1536 = 1600

Answer: x 1 = 8 x 2 = –12

*It was possible to immediately divide the left and right sides of the equation by 2, that is, simplify it. The calculations will be easier.

Example 2: Decide x 2–22 x+121 = 0

a=1 b=–22 c=121

D = b 2 –4ac =(–22) 2 –4∙1∙121 = 484–484 = 0

We found that x 1 = 11 and x 2 = 11

It is permissible to write x = 11 in the answer.

Answer: x = 11

Example 3: Decide x 2 –8x+72 = 0

a=1 b= –8 c=72

D = b 2 –4ac =(–8) 2 –4∙1∙72 = 64–288 = –224

The discriminant is negative, there is no solution in real numbers.

Answer: no solution

The discriminant is negative. There is a solution!

Here we will talk about solving the equation in the case when a negative discriminant is obtained. Do you know anything about complex numbers? I will not go into detail here about why and where they arose and what their specific role and necessity in mathematics is; this is a topic for a large separate article.

The concept of a complex number.

A little theory.

A complex number z is a number of the form

z = a + bi

where a and b are real numbers, i is the so-called imaginary unit.

a+bi – this is a SINGLE NUMBER, not an addition.

The imaginary unit is equal to the root of minus one:

Now consider the equation:

We get two conjugate roots.

Incomplete quadratic equation.

Let's consider special cases, this is when the coefficient “b” or “c” is equal to zero (or both are equal to zero). They can be solved easily without any discriminatory issues.

Case 1. Coefficient b = 0.

The equation becomes:

Let's transform:

Example:

4x 2 –16 = 0 => 4x 2 =16 => x 2 = 4 => x 1 = 2 x 2 = –2

Case 2. Coefficient c = 0.

The equation becomes:

Let's transform and factorize:

*The product is equal to zero when at least one of the factors is equal to zero.

Example:

9x 2 –45x = 0 => 9x (x–5) =0 => x = 0 or x–5 =0

x 1 = 0 x 2 = 5

Case 3. Coefficients b = 0 and c = 0.

Here it is clear that the solution to the equation will always be x = 0.

Useful properties and patterns of coefficients.

There are properties that allow you to solve equations with large coefficients.

Ax 2 + bx+ c=0 equality holds

a + b+ c = 0, That

- if for the coefficients of the equation Ax 2 + bx+ c=0 equality holds

a+ s =b, That

These properties help solve a certain type of equation.

Example 1: 5001 x 2 –4995 x – 6=0

The sum of the odds is 5001+( – 4995)+(– 6) = 0, which means

Example 2: 2501 x 2 +2507 x+6=0

Equality holds a+ s =b, Means

Regularities of coefficients.

1. If in the equation ax 2 + bx + c = 0 the coefficient “b” is equal to (a 2 +1), and the coefficient “c” is numerically equal to the coefficient “a”, then its roots are equal

ax 2 + (a 2 +1)∙x+ a= 0 = > x 1 = –a x 2 = –1/a.

Example. Consider the equation 6x 2 + 37x + 6 = 0.

x 1 = –6 x 2 = –1/6.

2. If in the equation ax 2 – bx + c = 0 the coefficient “b” is equal to (a 2 +1), and the coefficient “c” is numerically equal to the coefficient “a”, then its roots are equal

ax 2 – (a 2 +1)∙x+ a= 0 = > x 1 = a x 2 = 1/a.

Example. Consider the equation 15x 2 –226x +15 = 0.

x 1 = 15 x 2 = 1/15.

3. If in Eq. ax 2 + bx – c = 0 coefficient “b” is equal to (a 2 – 1), and coefficient “c” is numerically equal to the coefficient “a”, then its roots are equal

ax 2 + (a 2 –1)∙x – a= 0 = > x 1 = – a x 2 = 1/a.

Example. Consider the equation 17x 2 +288x – 17 = 0.

x 1 = – 17 x 2 = 1/17.

4. If in the equation ax 2 – bx – c = 0 the coefficient “b” is equal to (a 2 – 1), and the coefficient c is numerically equal to the coefficient “a”, then its roots are equal

ax 2 – (a 2 –1)∙x – a= 0 = > x 1 = a x 2 = – 1/a.

Example. Consider the equation 10x 2 – 99x –10 = 0.

x 1 = 10 x 2 = – 1/10

Vieta's theorem.

Vieta's theorem is named after the famous French mathematician Francois Vieta. Using Vieta's theorem, we can express the sum and product of the roots of an arbitrary KU in terms of its coefficients.

45 = 1∙45 45 = 3∙15 45 = 5∙9.

In total, the number 14 gives only 5 and 9. These are roots. With a certain skill, using the presented theorem, you can solve many quadratic equations orally immediately.

Vieta's theorem, in addition. It is convenient in that after solving a quadratic equation in the usual way (through a discriminant), the resulting roots can be checked. I recommend doing this always.

TRANSPORTATION METHOD

With this method, the coefficient “a” is multiplied by the free term, as if “thrown” to it, which is why it is called "transfer" method. This method is used when the roots of the equation can be easily found using Vieta's theorem and, most importantly, when the discriminant is an exact square.

If A± b+c≠ 0, then the transfer technique is used, for example:

2X 2 – 11x+ 5 = 0 (1) => X 2 – 11x+ 10 = 0 (2)

Using Vieta's theorem in equation (2), it is easy to determine that x 1 = 10 x 2 = 1

The resulting roots of the equation must be divided by 2 (since the two were “thrown” from x 2), we get

x 1 = 5 x 2 = 0.5.

What is the rationale? Look what's happening.

The discriminants of equations (1) and (2) are equal:

If you look at the roots of the equations, you only get different denominators, and the result depends precisely on the coefficient of x 2:

The second (modified) one has roots that are 2 times larger.

Therefore, we divide the result by 2.

*If we roll three, we will divide the result by 3, etc.

Answer: x 1 = 5 x 2 = 0.5

Sq. ur-ie and Unified State Examination.

I’ll tell you briefly about its importance - YOU MUST BE ABLE TO DECIDE quickly and without thinking, you need to know the formulas of roots and discriminants by heart. Many problems included in the Unified State Examination tasks boil down to solving a quadratic equation (geometric ones included).

Something worth noting!

1. The form of writing an equation can be “implicit”. For example, the following entry is possible:

15+ 9x 2 - 45x = 0 or 15x+42+9x 2 - 45x=0 or 15 -5x+10x 2 = 0.

You need to bring it to a standard form (so as not to get confused when solving).

2. Remember that x is an unknown quantity and it can be denoted by any other letter - t, q, p, h and others.

Video tutorial 2: Solving Quadratic Equations

Lecture: Quadratic equations

The equation- this is a kind of equality in the expressions of which there is a variable.

Solve the equation- means finding a number instead of a variable that will bring it into correct equality.

An equation may have one solution, several, or none at all.

To solve any equation, it should be simplified as much as possible to the form:

Linear: a*x = b;

Square: a*x 2 + b*x + c = 0.

That is, any equations must be converted to standard form before solving.

Any equation can be solved in two ways: analytical and graphical.

On the graph, the solution to the equation is considered to be the points at which the graph intersects the OX axis.

Quadratic equations

a*x 2 + b*x + c = 0.

Wherein a, b, c are coefficients of the equation that differ from zero. A "X"- root of the equation. It is believed that a quadratic equation has two roots or may not have a solution at all. The resulting roots may be the same.

"A"- the coefficient that stands before the squared root.

"b"- stands before the unknown in the first degree.

"With" is the free term of the equation.

If, for example, we have an equation of the form:

2x 2 -5x+3=0

In it, “2” is the coefficient of the leading term of the equation, “-5” is the second coefficient, and “3” is the free term.

Solving a quadratic equation

There are a huge variety of ways to solve a quadratic equation. However, in the school mathematics course, the solution is studied using Vieta’s theorem, as well as using the discriminant.

Discriminant solution:

When solving using this method, it is necessary to calculate the discriminant using the formula:

If during calculations you find that the discriminant is less than zero, this means that this equation has no solutions.

If the discriminant is zero, then the equation has two identical solutions. In this case, the polynomial can be collapsed using the abbreviated multiplication formula into the square of the sum or difference. Then solve it as a linear equation. Or use the formula:

If the discriminant is greater than zero, then you must use the following method:

Vieta's theorem

If the equation is given, that is, the coefficient of the leading term is equal to one, then you can use Vieta's theorem.

So let's assume the equation is:

The roots of the equation are found as follows:

Incomplete quadratic equation

There are several options for obtaining an incomplete quadratic equation, the form of which depends on the presence of coefficients.

1. If the second and third coefficients are zero (b = 0, c = 0), then the quadratic equation will look like:

This equation will have a unique solution. The equality will be true only if the solution to the equation is zero.

Kopyevskaya rural secondary school

10 Ways to Solve Quadratic Equations

Head: Patrikeeva Galina Anatolyevna,

mathematic teacher

village Kopevo, 2007

1. History of the development of quadratic equations

1.1 Quadratic equations in Ancient Babylon

1.2 How Diophantus composed and solved quadratic equations

1.3 Quadratic equations in India

1.4 Quadratic equations by al-Khorezmi

1.5 Quadratic equations in Europe XIII - XVII centuries

1.6 About Vieta's theorem

2. Methods for solving quadratic equations

Conclusion

Literature

1. History of the development of quadratic equations

1.1 Quadratic equations in Ancient Babylon

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

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

X 2 + X = ¾; X 2 - X = 14,5

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

Despite high level development of algebra in Babylon, the cuneiform texts lack the concept of a negative number and general methods for solving quadratic equations.

1.2 How Diophantus composed and solved quadratic equations.

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

When composing equations, Diophantus skillfully selects unknowns to simplify the solution.

Here, for example, is one of his tasks.

Problem 11.“Find two numbers, knowing that their sum is 20 and their product is 96”

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

Hence the equation:

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

100 - x 2 = 96

x 2 - 4 = 0 (1)

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

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

y(20 - y) = 96,

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

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

1.3 Quadratic Equations in 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 (7th century), outlined general rule solutions of quadratic equations reduced to a single canonical form:

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

In equation (1), the coefficients, except A, can also be negative. Brahmagupta's rule is essentially the same as ours.

In ancient India, public competitions in solving difficult problems were common. One of the old Indian books says the following about such competitions: “As the sun eclipses the stars with its brilliance, so learned man eclipse the glory of another in popular 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.

Problem 13.

“A flock of frisky monkeys, and twelve along the vines...

The authorities, having eaten, had fun. They started jumping, hanging...

There are them in the square, part eight. How many monkeys were there?

I was having fun in the clearing. Tell me, in this pack?

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

The equation corresponding to problem 13 is:

( x /8) 2 + 12 = x

Bhaskara writes under the guise:

x 2 - 64x = -768

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

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

(x - 32) 2 = 256,

x - 32 = ± 16,

x 1 = 16, x 2 = 48.

1.4 Quadratic equations in al - Khorezmi

In the algebraic treatise of al-Khorezmi, a classification of linear and quadratic equations is given. The author counts 6 types of equations, expressing them as follows:

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

2) “Squares are equal to numbers”, i.e. ax 2 = c.

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

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

5) “Squares and roots are equal to numbers”, i.e. ah 2 + bx = s.

6) “Roots and numbers are equal to squares,” i.e. bx + c = ax 2 .

For al-Khorezmi, who avoided the use of negative numbers, the terms of each of these equations are addends and not subtractables. In this case, equations that do not have positive solutions are obviously not taken into account. The author sets out methods for solving these equations using the techniques of al-jabr and al-muqabala. His decisions, of course, do not completely coincide with ours. Not to mention that it is purely rhetorical, it should be noted, for example, that when solving an incomplete quadratic equation of the first type

al-Khorezmi, like all mathematicians before the 17th century, does not take into account the zero solution, probably because in specific practical problems it does not matter. When solving complete quadratic equations, al-Khorezmi sets out the rules for solving them using particular numerical examples, and then geometric proofs.

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

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

The treatise of al-Khorezmi is the first book that has come down to us, which systematically sets out the classification of quadratic equations and gives formulas for their solution.

1.5 Quadratic equations in Europe XIII - XVII bb

Formulas for solving quadratic equations along the lines of al-Khwarizmi 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 Islamic countries and 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 general rule for solving quadratic equations reduced to a single canonical form:

x 2 + bx = c,

for all possible combinations of coefficient signs b , With was formulated in Europe only in 1544 by M. Stiefel.

Derivation of the formula for solving a quadratic equation in general view Viet has it, but Viet 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.

1.6 About Vieta's theorem

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

To understand Vieta, we should remember that A, like any vowel letter, meant the unknown (our X), vowels IN, D- coefficients for the unknown. In the language of modern algebra, the above Vieta formulation means: if there is

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

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

x 1 = a, x 2 = b .

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

2. Methods for solving quadratic equations

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