Two quadratic equations have a common root. Solution of quadratic equations, formula of roots, examples. How to solve quadratic equations

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

it is necessary to bring the given equation to the standard form, i.e. to the view:

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

determine all coefficients A, b And c.

Formula for finding the roots of a quadratic equation.

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

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

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

For example, in the equation:

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

Substitute the values and write:

Example almost solved:

This is the answer.

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

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

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

Suppose we need to solve the following example:

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

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

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

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

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

What does this mean?

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

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

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

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

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

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

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

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

x2+bx+c=0,

Thenx 1 x 2 =c

x1 +x2 =−b

For a complete quadratic equation in which a≠1:

x 2 +bx+c=0,

divide the whole equation by A:

→ →

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

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

equation for a common denominator.

Conclusion. Practical Tips:

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

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

equations for -1.

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

factor.

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

Kopyevskaya rural secondary school

10 Ways to Solve Quadratic Equations

Head: Patrikeeva Galina Anatolyevna,

mathematic teacher

s.Kopyevo, 2007

1. History of the development of quadratic equations

1.1 Quadratic equations in ancient Babylon

1.2 How Diophantus compiled and solved quadratic equations

1.3 Quadratic equations in India

1.4 Quadratic equations in al-Khwarizmi

1.5 Quadratic equations in Europe XIII - XVII centuries

1.6 About Vieta's theorem

2. Methods for solving quadratic equations

Conclusion

Literature

1. History of the development of quadratic equations

1.1 Quadratic equations in ancient Babylon

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

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

X 2 + X = ¾; X 2 - X = 14,5

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

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

1.2 How Diophantus compiled and solved quadratic equations.

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

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

Here, for example, is one of his tasks.

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

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

Hence the equation:

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

100 - x 2 = 96

x 2 - 4 = 0 (1)

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

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

y(20 - y) = 96,

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

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

1.3 Quadratic equations in India

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

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

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

In ancient India, public competitions in solving difficult problems were common. In one of the old Indian books, the following is said about such competitions: “As the sun outshines the stars with its brilliance, so a learned person will outshine the glory of another in public meetings, proposing and solving algebraic problems.” Tasks were often dressed in poetic form.

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

Task 13.

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

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

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

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

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

The equation corresponding to problem 13 is:

(x/8) 2 + 12 = x

Bhaskara writes under the guise of:

x 2 - 64x = -768

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

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

(x - 32) 2 = 256,

x - 32 = ± 16,

x 1 = 16, x 2 = 48.

1.4 Quadratic equations in al-Khorezmi

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

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

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

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

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

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

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

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

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

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

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

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

1.5 Quadratic equations in EuropeXIII - XVIIcenturies

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

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

x 2+bx= with,

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

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

1.6 About Vieta's theorem

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

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

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

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

x 1 = a, x 2 =b.

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

2. Methods for solving quadratic equations

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

The solution of equations in mathematics occupies a special place. This process is preceded by many hours of studying the theory, during which the student learns how to solve equations, determine their form and bring the skill to full automatism. However, the search for roots does not always make sense, since they may simply not exist. There are special methods for finding roots. In this article, we will analyze the main functions, their domains of definition, as well as cases where 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 such number x, the product of which 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 parts are presented 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 value . Which equation has no roots? Examples of linear equations are shown in the illustration below.

Basically, linear equations are solved by simply transferring the numerical part to one part and the contents of x to the other. It turns out an equation of the form mx \u003d n, where m and n are numbers, and x is an unknown. To find x, it is enough to divide both parts by m. Then x = n/m. Basically, linear equations have only one root, but there are cases when there are either infinitely many roots or none at all. When m = 0 and n = 0, the equation takes the form 0 * x = 0. Absolutely any number will be the solution to such an equation.

But what equation has no roots?

For m = 0 and n = 0, the equation has no roots from 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 \u003d 0 for a \u003d 0. The most common is the solution through the discriminant. The formula for finding the discriminant of a quadratic equation: D \u003d 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 on the graph of a function, which is a parabola. For a > 0, the branches are directed upwards, 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 top of the parabola and determine in which direction the branches are directed. You can determine the x-coordinate of the vertex by the formula: x 0 \u003d -b / 2a. In this case, the y-coordinate of the vertex is found by simply substituting the x0 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, hence the equation has no real roots.

3. Trigonometric equations

Trigonometric functions are considered on a trigonometric circle, but can also be represented in a Cartesian coordinate system. In this article, we will look at two basic 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 sinx function graph shown in the picture below.

We see that the function is symmetrical 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 it is greater than one in absolute value.

This is the simplest example of trigonometric equations. In fact, their solution can take many pages, at the end of which you realize that you used the wrong formula and you need to start all over again. Sometimes, even with the correct finding of the roots, you can forget to take into account the restrictions on the ODZ, due to which an extra root or interval appears in the answer, and the whole answer turns into an erroneous one. 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 combined with curly or square brackets. Curly braces denote the joint execution of all equations. That is, if at least one of the equations has no roots or contradicts the other, the whole system has no solution. Square brackets denote the word "or". This means that if at least one of the equations of the system has a solution, then the whole system has a solution.

The answer of system c is the totality 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 this complexity does not allow you to immediately say which equation has no roots.

In problem books and textbooks, there are different types of equations: those that have roots, and those that do not have them. First of all, if you can't find roots, don't think they don't exist at all. Perhaps you have made a mistake somewhere, then it is enough to carefully double-check your decision.

We have considered the most basic equations and their types. Now you can tell which equation has no roots. In most cases, this is not at all difficult to do. To achieve success in solving equations, only attention and concentration are required. 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 m = 0 and n = 0;
in a quadratic equation if the discriminant is 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.

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

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

Quadratic equation, its types

Definition 1

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

Often, quadratic equations are also called equations of the second degree, since in fact 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. 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 highest 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 the shorthand 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 senior coefficient is 1 and the second coefficient is − 1 .

Reduced and non-reduced quadratic equations

According to the value of the first coefficient, quadratic equations are divided into reduced and non-reduced.

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.

Here are some 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 its parts by the first coefficient (equivalent transformation). The transformed equation will have the same roots as the given non-reduced 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 parts 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 us 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 exactly square, since a = 0 it essentially transforms into a linear equation b x + c = 0.

In the case where 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 is a quadratic equation a x 2 + b x + c \u003d 0, where at least one of the coefficients b And c(or both) is zero.

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

Let's discuss why the types of quadratic equations are given precisely such names.

For 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 have obtained differ from the full quadratic equation in that their left-hand sides do not contain either a term with the variable x, or a free term, or both at once. Actually, this fact gave the name to this type of equations - incomplete.

For example, x 2 + 3 x + 4 = 0 and − 7 x 2 − 2 x + 1, 3 = 0 are complete quadratic equations; x 2 \u003d 0, − 5 x 2 \u003d 0; 11 x 2 + 2 = 0 , − x 2 − 6 x = 0 are 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, coefficients correspond to such an equation b = 0 and c = 0 ;
a x 2 + c \u003d 0 for b \u003d 0;
a x 2 + b x = 0 for c = 0 .

Consider successively the solution of each type of incomplete quadratic equation.

Solution of the equation a x 2 \u003d 0

As already mentioned above, such an equation corresponds to the coefficients b And c, equal to zero. The equation a x 2 = 0 can be converted into an equivalent equation x2 = 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 x2 = 0 is zero because 0 2 = 0 . This equation has no other roots, which is explained by the properties of the degree: for any number p , not equal to zero, the inequality is true p2 > 0, from which it follows that when p ≠ 0 equality p2 = 0 will never be reached.

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 x2 = 0, its only root is x=0, then the original equation has a single root - zero.

The solution is summarized as follows:

− 3 x 2 \u003d 0, x 2 \u003d 0, x \u003d 0.

Solution of the equation a x 2 + c \u003d 0

Next in line is the solution of incomplete quadratic equations, where b \u003d 0, c ≠ 0, that is, equations of the form a x 2 + c = 0. Let's transform this equation by transferring the term from one side of the equation to the other, changing the sign to the opposite and dividing both sides of the equation by a number that is not equal to zero:

endure c to the right side, which gives the equation a x 2 = − c;
divide both sides of the equation by a, we get as a result x = - c a .

Our transformations are equivalent, respectively, the resulting equation is also equivalent to the original one, and this fact makes it possible to draw a conclusion about the roots of the equation. From what are the values a And c depends on the value of the expression - c a: 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 equal to 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 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 \u003d - c a will be the number - c a, since - c a 2 \u003d - c a. It is easy 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 opposite method. First, let's set the notation of the roots found above as x 1 And − x 1. Let's assume that the equation x 2 = - c a also has a root x2, which is different from the roots x 1 And − x 1. We know that by substituting into the equation instead of x its roots, we transform the equation into a fair numerical equality.

For x 1 And − x 1 write: x 1 2 = - c a , and for x2- x 2 2 \u003d - c a. Based on the properties of numerical equalities, we subtract one true equality from another term by term, which will give us: x 1 2 − x 2 2 = 0. Use the properties of number operations 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 what has been said, it follows that x1 − x2 = 0 and/or x1 + x2 = 0, which is the same x2 = x1 and/or x 2 = − x 1. An obvious contradiction arose, because at first it was agreed that the root of the equation x2 differs from x 1 And − x 1. So, we have proved that the equation has no other roots than x = - c a and x = - - c a .

We 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 not have roots at - c a< 0 ;
will have two roots x = - c a and x = - - c a when - c a > 0 .

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

Example 3

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

Solution

We transfer the free term to the right side of the equation, then the equation will take the form 9 x 2 \u003d - 7.
We divide both sides of the resulting equation by 9 , we come to 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 not have roots.

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

Example 4

It is necessary to solve the equation − x2 + 36 = 0.

Solution

Let's move 36 to the right side: − x 2 = − 36.
Let's divide both parts into − 1 , we get x2 = 36. On the right side is a positive number, from which we can conclude that x = 36 or x = - 36 .
We extract the root and write the final result: an incomplete quadratic equation − x2 + 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 kind of incomplete quadratic equations, when c = 0. To find a solution to an incomplete quadratic equation a x 2 + b x = 0, we use the factorization method. Let us factorize the polynomial, which 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 the 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 consolidate the material with an example.

Example 5

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

Solution

Let's take out x outside the brackets and 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, we write the solution of 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 of the roots of a quadratic equation

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

Definition 8

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

Writing x \u003d - b ± D 2 a essentially means that x 1 \u003d - b + D 2 a, x 2 \u003d - b - D 2 a.

It will be useful to understand how the indicated formula was derived and how to apply it.

Derivation of the formula of the roots of a quadratic equation

Suppose we are faced with the task of solving a quadratic equation a x 2 + b x + c = 0. Let's carry out a number of equivalent transformations:

divide both sides of the equation by the number a, different from zero, we obtain the reduced quadratic equation: x 2 + b a x + c a \u003d 0;
select the full 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 \u003d 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 \u003d b 2 4 a 2 - c a \u003d b 2 4 a 2 - 4 a c 4 a 2 \u003d b 2 - 4 a c 4 a 2.

Thus, we have come to the equation x + b 2 a 2 = b 2 - 4 a c 4 a 2 , which is equivalent to the original equation a x 2 + b x + c = 0.

We discussed the solution of such equations in the previous paragraphs (the solution of 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:

for b 2 - 4 a c 4 a 2< 0 уравнение не имеет действительных решений;
for b 2 - 4 · a · c 4 · a 2 = 0, the equation has the form 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 correct one is: 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 the roots of the equation x + b 2 a 2 = b 2 - 4 a c 4 a 2 (and hence 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, (the 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 a name is given - the discriminant of a quadratic equation and the letter D is defined as its designation. Here you can write down the essence of the discriminant - by its value and sign, they conclude whether the quadratic equation will have real roots, and, if so, how many 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 the discriminant notation: x + b 2 · a 2 = D 4 · a 2 .

Let's recap the conclusions:

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 \u003d - b 2 a + D 4 a 2 or x \u003d - b 2 a - D 4 a 2. Based on the properties of radicals, these roots can be written as: x \u003d - b 2 a + D 2 a or - b 2 a - D 2 a. And when we open the modules and reduce the fractions to a common denominator, we get: x \u003d - b + D 2 a, x \u003d - b - D 2 a.

So, the result of our reasoning was the derivation of the formula for the roots of the 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, when the discriminant is greater than zero, to determine both real roots. When the discriminant is zero, applying both formulas will give the same root as the only solution to the quadratic equation. In the case when the discriminant is negative, trying to use the quadratic root formula, we will be faced with the need to extract the square root of a negative number, which will take us beyond 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 basically this is done when it is necessary to find complex roots.

In the bulk of cases, the search is usually meant not for complex, but for real roots of a quadratic equation. Then it is optimal, before using the formulas for the roots of the quadratic equation, first to 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 value of the discriminant;
at D< 0 сделать вывод об отсутствии у квадратного уравнения действительных корней;
for D = 0 find the only root of the equation by the formula x = - b 2 · a ;
for D > 0, determine two real roots of the quadratic equation by 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 .

Consider examples.

Examples of solving quadratic equations

We present the solution of examples for various values of the discriminant.

Example 6

It is necessary to find the roots of the equation x 2 + 2 x - 6 = 0.

Solution

We write the numerical coefficients of the quadratic equation: a \u003d 1, b \u003d 2 and c = − 6. Next, we act according to the algorithm, i.e. Let's start calculating the discriminant, for which we 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 got D > 0, which means that the original equation will have two real roots.
To find them, we use the root formula x \u003d - b ± D 2 · a and, substituting the appropriate values, we get: x \u003d - 2 ± 28 2 · 1. We simplify the resulting expression by taking the factor out of the sign of the root, followed by reduction of 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

It is necessary 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

It is necessary to solve the equation 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 computed 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 by performing operations with complex numbers:

x \u003d - 6 ± - 4 2 5,

x \u003d - 6 + 2 i 10 or x \u003d - 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: - 3 5 + 1 5 i , - 3 5 - 1 5 i .

In the school curriculum, as a standard, there is no requirement to look for complex roots, therefore, if the discriminant is defined as negative during the decision, the answer is immediately recorded 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 you to find solutions to quadratic equations with an even coefficient at x (or with a coefficient of the form 2 a n, for example, 2 3 or 14 ln 5 = 2 7 ln 5). Let us show how this formula is derived.

Suppose we are faced with the task of finding a solution to the quadratic equation a · x 2 + 2 · n · x + c = 0. We act 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 \u003d - 2 n ± D 2 a, x \u003d - 2 n ± 4 n 2 - a c 2 a, x \u003d - 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 considered quadratic equation with the second coefficient 2 n will take the form:

x \u003d - n ± D 1 a, where D 1 \u003d 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 that the sign of D 1 can also serve as an indicator of the presence or absence of the 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;
for D 1 = 0, determine the only root of the equation by 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

The second coefficient of the given equation can be represented 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. We define them by the corresponding formula of the roots:

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 perform 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 .

Simplification of 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 \u003d 0 is clearly more convenient for solving than 1200 x 2 - 400 x - 700 \u003d 0.

More often, the simplification of the form of a quadratic equation is performed by multiplying or dividing its both parts 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 of its parts by 100.

Such a transformation is possible when the coefficients of the quadratic equation are not relatively prime numbers. Then, usually, both parts of the equation are divided 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's define the gcd of the absolute values of its coefficients: gcd (12 , 42 , 48) = gcd(gcd (12 , 42) , 48) = gcd (6 , 48) = 6 . Let's divide both parts of the original quadratic equation by 6 and get the equivalent quadratic equation 2 · x 2 − 7 · x + 8 = 0 .

By multiplying both sides of the quadratic equation, fractional coefficients are usually eliminated. In this case, 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 \u003d 0 is multiplied with LCM (6, 3, 1) \u003d 6, then it will be written in a simpler form x 2 + 4 x - 18 = 0 .

Finally, we note that almost always get rid of the minus at the first coefficient of the quadratic equation, changing the signs of each term of the equation, which is achieved by multiplying (or dividing) both parts by − 1. For example, from the quadratic equation - 2 x 2 - 3 x + 7 \u003d 0, you can go to its simplified version 2 x 2 + 3 x - 7 \u003d 0.

Relationship between roots and coefficients

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

The most famous and applicable are the formulas of the Vieta theorem:

x 1 + x 2 \u003d - b a and x 2 \u003d 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 the form of the quadratic equation 3 · x 2 − 7 · x + 22 \u003d 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 relationships 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.

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

We remind you that the complete quadratic equation is an equation of the form:

Solving full quadratic equations is a bit more complicated (just a little bit) than those given.

Remember, any quadratic equation can be solved using the discriminant!

Even incomplete.

The rest of the methods will help you do it faster, but if you have problems with quadratic equations, first master the solution using the discriminant.

1. Solving quadratic equations using the discriminant.

Solving quadratic equations in this way is very simple, the main thing is to remember the sequence of actions and a couple of formulas.

If, then the equation has 2 roots. Pay special attention to step 2.

The discriminant D tells us the number of roots of the equation.

If, then the formula at the step will be reduced to. Thus, the equation will have only a root.
If, then we will not be able to extract the root of the discriminant at the step. This indicates that the equation has no roots.

Let us turn to the geometric meaning of the quadratic equation.

The graph of the function is a parabola:

Let's go back to our equations and look at a few examples.

Example 9

Solve the Equation

Step 1 skip.

Step 2

Finding the discriminant:

So the equation has two roots.

Step 3

Answer:

Example 10

Solve the Equation

The equation is in standard form, so Step 1 skip.

Step 2

Finding the discriminant:

So the equation has one root.

Answer:

Example 11

Solve the Equation

The equation is in standard form, so Step 1 skip.

Step 2

Finding the discriminant:

This means that we will not be able to extract the root from the discriminant. There are no roots of the equation.

Now we know how to write down such answers correctly.

Answer: no roots

2. Solving quadratic equations using the Vieta theorem

If you remember, then there is such a type of equations that are called reduced (when the coefficient a is equal to):

Such equations are very easy to solve using Vieta's theorem:

The sum of the roots given quadratic equation is equal, and the product of the roots is equal.

You just need to choose a pair of numbers whose product is equal to the free term of the equation, and the sum is equal to the second coefficient, taken with the opposite sign.

Example 12

Solve the Equation

This equation is suitable for solution using Vieta's theorem, because .

The sum of the roots of the equation is, i.e. we get the first equation:

And the product is:

Let's create and solve the system:

And. The sum is;
And. The sum is;
And. The amount is equal.

and are the solution of the system:

Answer: ; .

Example 13

Solve the Equation

Answer:

Example 14

Solve the Equation

The equation is reduced, which means:

Answer:

QUADRATIC EQUATIONS. AVERAGE LEVEL

What is a quadratic equation?

In other words, a quadratic equation is an equation of the form, where - unknown, - some numbers, moreover.

The number is called the highest or first coefficient quadratic equation, - second coefficient, A - free member.

Because if, the equation will immediately become linear, because will disappear.

In this case, and can be equal to zero. In this chair equation is called incomplete.

If all the terms are in place, that is, the equation - complete.

Methods for solving incomplete quadratic equations

To begin with, we will analyze the methods for solving incomplete quadratic equations - they are simpler.

The following types of equations can be distinguished:

I. , in this equation the coefficient and the free term are equal.

II. , in this equation the coefficient is equal.

III. , in this equation the free term is equal to.

Now consider the solution of each of these subtypes.

Obviously, this equation always has only one root:

A number squared cannot be negative, because when multiplying two negative or two positive numbers, the result will always be a positive number. That's why:

if, then the equation has no solutions;

if we have two roots

These formulas do not need to be memorized. The main thing to remember is that it cannot be less.

Examples of solving quadratic equations

Example 15

Answer:

Never forget about roots with a negative sign!

Example 16

The square of a number cannot be negative, which means that the equation

no roots.

To briefly write that the problem has no solutions, we use the empty set icon.

Answer:

Example 17

So, this equation has two roots: and.

Answer:

Let's take the common factor out of brackets:

The product is equal to zero if at least one of the factors is equal to zero. This means that the equation has a solution when:

So, this quadratic equation has two roots: and.

Example:

Solve the equation.

Solution:

We factorize the left side of the equation and find the roots:

Answer:

Methods for solving complete quadratic equations

1. Discriminant

Solving quadratic equations in this way is easy, the main thing is to remember the sequence of actions and a couple of formulas. Remember, any quadratic equation can be solved using the discriminant! Even incomplete.

Did you notice the root of the discriminant in the root formula?

But the discriminant can be negative.

What to do?

We need to pay special attention to step 2. The discriminant tells us the number of roots of the equation.

If, then the equation has a root:
If, then the equation has the same root, but in fact, one root:
Such roots are called double roots.
If, then the root of the discriminant is not extracted. This indicates that the equation has no roots.

Why are there different numbers of roots?

Let us turn to the geometric meaning of the quadratic equation. The graph of the function is a parabola:

In a particular case, which is a quadratic equation, .

And this means that the roots of the quadratic equation are the points of intersection with the x-axis (axis).

The parabola may not cross the axis at all, or it may intersect it at one (when the top of the parabola lies on the axis) or two points.

In addition, the coefficient is responsible for the direction of the branches of the parabola. If, then the branches of the parabola are directed upwards, and if - then downwards.

4 examples of solving quadratic equations

Example 18

Answer:

Example 19

Answer: .

Example 20

Answer:

Example 21

This means there are no solutions.

Answer: .

2. Vieta's theorem

Using Vieta's theorem is very easy.

All you need is pick up such a pair of numbers, the product of which is equal to the free term of the equation, and the sum is equal to the second coefficient, taken with the opposite sign.

It is important to remember that Vieta's theorem can only be applied to given quadratic equations ().

Let's look at a few examples:

Example 22

Solve the equation.

Solution:

This equation is suitable for solution using Vieta's theorem, because . Other coefficients: ; .

The sum of the roots of the equation is:

And the product is:

Let's select such pairs of numbers, the product of which is equal, and check if their sum is equal:

And. The sum is;
And. The sum is;
And. The amount is equal.

and are the solution of the system:

Thus, and are the roots of our equation.

Answer: ; .

Example 23

Solution:

We select such pairs of numbers that give in the product, and then check whether their sum is equal:

and: give in total.

and: give in total. To get it, you just need to change the signs of the alleged roots: and, after all, the product.

Answer:

Example 24

Solution:

The free term of the equation is negative, and hence the product of the roots is a negative number. This is possible only if one of the roots is negative and the other is positive. So the sum of the roots is differences of their modules.

We select such pairs of numbers that give in the product, and the difference of which is equal to:

and: their difference is - not suitable;

and: - not suitable;

and: - suitable. It remains only to remember that one of the roots is negative. Since their sum must be equal, then the root, which is smaller in absolute value, must be negative: . We check:

Answer:

Example 25

Solve the equation.

Solution:

The equation is reduced, which means:

The free term is negative, and hence the product of the roots is negative. And this is possible only when one root of the equation is negative and the other is positive.

We select such pairs of numbers whose product is equal, and then determine which roots should have a negative sign:

Obviously, only roots and are suitable for the first condition:

Answer:

Example 26

Solve the equation.

Solution:

The equation is reduced, which means:

The sum of the roots is negative, which means that at least one of the roots is negative. But since their product is positive, it means both roots are minus.

We select such pairs of numbers, the product of which is equal to:

Obviously, the roots are the numbers and.

Answer:

Agree, it is very convenient - to invent roots orally, instead of counting this nasty discriminant.

Try to use Vieta's theorem as often as possible!

But the Vieta theorem is needed in order to facilitate and speed up finding the roots.

To make it profitable for you to use it, you must bring the actions to automatism. And for this, solve five more examples.

But don't cheat: you can't use the discriminant! Only Vieta's theorem!

5 examples of Vieta's theorem for self-study

Example 27

Task 1. ((x)^(2))-8x+12=0

According to Vieta's theorem:

As usual, we start the selection with the product:

Not suitable because the amount;

: the amount is what you need.

Answer: ; .

Example 28

Task 2.

And again, our favorite Vieta theorem: the sum should work out, but the product is equal.

But since it should be not, but, we change the signs of the roots: and (in total).

Answer: ; .

Example 29

Task 3.

Hmm... Where is it?

It is necessary to transfer all the terms into one part:

The sum of the roots is equal to the product.

Yes, stop! The equation is not given.

But Vieta's theorem is applicable only in the given equations.

So first you need to bring the equation.

If you can’t bring it up, drop this idea and solve it in another way (for example, through the discriminant).

Let me remind you that to bring a quadratic equation means to make the leading coefficient equal to:

Then the sum of the roots is equal, and the product.

It's easier to pick up here: after all - a prime number (sorry for the tautology).

Answer: ; .

Example 30

Task 4.

The free term is negative.

What's so special about it?

And the fact that the roots will be of different signs.

And now, during the selection, we check not the sum of the roots, but the difference between their modules: this difference is equal, but the product.

So, the roots are equal and, but one of them is with a minus.

Vieta's theorem tells us that the sum of the roots is equal to the second coefficient with the opposite sign, that is.

This means that the smaller root will have a minus: and, since.

Answer: ; .

Example 31

Task 5.

What needs to be done first?

That's right, give the equation:

Again: we select the factors of the number, and their difference should be equal to:

The roots are equal and, but one of them is minus. Which? Their sum must be equal, which means that with a minus there will be a larger root.

Answer: ; .

Summarize

Vieta's theorem is used only in the given quadratic equations.
Using the Vieta theorem, you can find the roots by selection, orally.
If the equation is not given or no suitable pair of factors of the free term was found, then there are no integer roots, and you need to solve it in another way (for example, through the discriminant).

3. Full square selection method

If all the terms containing the unknown are represented as terms from the formulas of abbreviated multiplication - the square of the sum or difference - then after the change of variables, the equation can be represented as an incomplete quadratic equation of the type.

For example:

Example 32

Solve the equation: .

Solution:

Answer:

Example 33

Solve the equation: .

Solution:

Answer:

In general, the transformation will look like this:

This implies: .

Doesn't it remind you of anything?

It's the discriminant! That's exactly how the discriminant formula was obtained.

QUADRATIC EQUATIONS. BRIEFLY ABOUT THE MAIN

Quadratic equation is an equation of the form, where is the unknown, are the coefficients of the quadratic equation, is the free term.

Complete quadratic equation- an equation in which the coefficients are not equal to zero.

Reduced quadratic equation- an equation in which the coefficient, that is: .

Incomplete quadratic equation- an equation in which the coefficient and or free term c are equal to zero:

if the coefficient, the equation has the form: ,
if a free term, the equation has the form: ,
if and, the equation has the form: .

1. Algorithm for solving incomplete quadratic equations

1.1. An incomplete quadratic equation of the form, where, :

1) Express the unknown: ,

2) Check the sign of the expression:

if, then the equation has no solutions,
if, then the equation has two roots.

1.2. An incomplete quadratic equation of the form, where, :

1) Let's take the common factor out of brackets: ,

2) The product is equal to zero if at least one of the factors is equal to zero. Therefore, the equation has two roots:

1.3. An incomplete quadratic equation of the form, where:

This equation always has only one root: .

2. Algorithm for solving complete quadratic equations of the form where

2.1. Solution using the discriminant

1) Let's bring the equation to the standard form: ,

2) Calculate the discriminant using the formula: , which indicates the number of roots of the equation:

3) Find the roots of the equation:

if, then the equation has a root, which are found by the formula:
if, then the equation has a root, which is found by the formula:
if, then the equation has no roots.

2.2. Solution using Vieta's theorem

The sum of the roots of the reduced quadratic equation (an equation of the form, where) is equal, and the product of the roots is equal, i.e. , A.

2.3. Full square solution