A quadratic equation is an equation of the form a*x^2 +b*x+c=0, where a,b,c are some arbitrary real (real) numbers, and x is a variable. And the number a=0.

The numbers a,b,c are called coefficients. The number a - is called the leading coefficient, the number b is the coefficient at x, and the number c is called the free member.

Solving quadratic equations

To solve a quadratic equation means to find all its roots, or to establish the fact that the quadratic equation has no roots. The root of the quadratic equation a * x ^ 2 + b * x + c \u003d 0 is any value of the variable x, such that the square trinomial a * x ^ 2 + b * x + c vanishes. Sometimes such a value of x is called the root of a square trinomial.

There are several ways to solve quadratic equations. Consider one of them - the most versatile. It can be used to solve any quadratic equation.

Formulas for solving quadratic equations

The formula for the roots of the quadratic equation is a*x^2 +b*x+c=0.

x=(-b±√D)/(2*a), where D =b^2-4*a*c.

This formula is obtained by solving the equation a * x ^ 2 + b * x + c \u003d 0 in general form, by highlighting the square of the binomial.

In the formula of the roots of a quadratic equation, the expression D (b^2-4*a*c) is called the discriminant of the quadratic equation a*x^2 +b*x+c=0. This name comes from the Latin language, translated "distinguisher". Depending on the value of the discriminant, the quadratic equation will have two or one root, or no roots at all.

If the discriminant is greater than zero, then the quadratic equation has two roots. (x=(-b±√D)/(2*a))

If the discriminant is zero, then the quadratic equation has one root. (x=(-b/(2*a))

If the discriminant is negative, then the quadratic equation has no roots.

General algorithm for solving a quadratic equation

Based on the foregoing, we formulate a general algorithm for solving the quadratic equation a*x^2 +b*x+c=0 using the formula:

1. Find the value of the discriminant using the formula D =b^2-4*a*c.

2. Depending on the value of the discriminant, calculate the roots using the formulas:

D<0, корней нет.

D=0, x=(-b/(2*a)

D>0, x=(-b+√D)/(2*a), x=(-b-√D)/(2*a)

This algorithm is universal and suitable for solving any quadratic equations. Complete and incomplete, cited and not cited.

Bibliographic description: Gasanov A. R., Kuramshin A. A., Elkov A. A., Shilnenkov N. V., Ulanov D. D., Shmeleva O. V. Methods for solving quadratic equations // Young scientist. - 2016. - No. 6.1. - S. 17-20..04.2019).



Our project is dedicated to the ways of solving quadratic equations. The purpose of the project: to learn how to solve quadratic equations in ways that are not included in the school curriculum. Task: find all possible ways to solve quadratic equations and learn how to use them yourself and introduce classmates to these methods.

What are "quadratic equations"?

Quadratic equation- equation of the form ax2 + bx + c = 0, where a, b, c- some numbers ( a ≠ 0), x- unknown.

The numbers a, b, c are called the coefficients of the quadratic equation.

• a is called the first coefficient;
• b is called the second coefficient;
• c - free member.

And who was the first to "invent" quadratic equations?

Some algebraic techniques for solving linear and quadratic equations were known as early as 4000 years ago in Ancient Babylon. The found ancient Babylonian clay tablets, dated somewhere between 1800 and 1600 BC, are the earliest evidence of the study of quadratic equations. The same tablets contain methods for solving certain types of quadratic equations.

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.

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.

Babylonian mathematicians from about the 4th century B.C. used the square complement method to solve equations with positive roots. Around 300 B.C. Euclid came up with a more general geometric solution method. The first mathematician who found solutions to an equation with negative roots in the form of an algebraic formula was an Indian scientist. Brahmagupta(India, 7th century AD).

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

ax2 + bx = c, a>0

In this equation, the coefficients can be negative. Brahmagupta's rule essentially coincides with ours.

In 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 in public meetings, proposing and solving algebraic problems.” Tasks were often dressed in poetic form.

In an algebraic treatise Al-Khwarizmi a classification of linear and quadratic equations is given. The author lists 6 types of equations, expressing them as follows:

1) “Squares are equal to roots”, i.e. ax2 = bx.

2) “Squares are equal to number”, i.e. ax2 = c.

3) "The roots are equal to the number", i.e. ax2 = c.

4) “Squares and numbers are equal to roots”, i.e. ax2 + c = bx.

5) “Squares and roots are equal to number”, i.e. ax2 + bx = c.

6) “Roots and numbers are equal to squares”, i.e. bx + c == ax2.

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

Forms for solving quadratic equations on the model of Al-Khwarizmi in Europe were first described in the "Book of the Abacus", written in 1202. Italian mathematician Leonard Fibonacci. The author independently developed some new algebraic examples of problem solving and was the first in Europe to approach the introduction of negative numbers.

This book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many tasks from this book were transferred to almost all European textbooks of the 14th-17th centuries. The general rule for solving quadratic equations reduced to a single canonical form x2 + bx = c with all possible combinations of signs and coefficients b, c, was formulated in Europe in 1544. M. Stiefel.

Vieta has a general derivation of the formula for solving a quadratic equation, but Vieta recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli 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 Girard, Descartes, Newton and other scientists, the way of solving quadratic equations takes on a modern form.

Consider several ways to solve quadratic equations.

Standard ways to solve quadratic equations from the school curriculum:

1. Factorization of the left side of the equation.
2. Full square selection method.
3. Solution of quadratic equations by formula.
4. Graphical solution of a quadratic equation.
5. Solution of equations using Vieta's theorem.

Let us dwell in more detail on the solution of reduced and non-reduced quadratic equations using the Vieta theorem.

Recall that to solve the above quadratic equations, it is enough to find two numbers such that the product of which is equal to the free term, and the sum is equal to the second coefficient with the opposite sign.

Example.x 2 -5x+6=0

You need to find numbers whose product is 6 and the sum is 5. These numbers will be 3 and 2.

Answer: x 1 =2,x 2 =3.

But you can use this method for equations with the first coefficient not equal to one.

Example.3x 2 +2x-5=0

We take the first coefficient and multiply it by the free term: x 2 +2x-15=0

The roots of this equation will be numbers whose product is - 15, and the sum is - 2. These numbers are 5 and 3. To find the roots of the original equation, we divide the obtained roots by the first coefficient.

Answer: x 1 =-5/3, x 2 =1

6. Solution of equations by the method of "transfer".

Consider the quadratic equation ax 2 + bx + c = 0, where a≠0.

Multiplying both its parts by a, we get the equation a 2 x 2 + abx + ac = 0.

Let ax = y, whence x = y/a; then we arrive at the equation y 2 + by + ac = 0, which is equivalent to the given one. We find its roots at 1 and at 2 using the Vieta theorem.

Finally we get x 1 = y 1 /a and x 2 = y 2 /a.

With this method, the coefficient a is multiplied by the free term, as if "transferred" to it, therefore it is called the "transfer" method. This method is used when it is easy to find the roots of an equation using Vieta's theorem and, most importantly, when the discriminant is an exact square.

Example.2x 2 - 11x + 15 = 0.

Let's "transfer" the coefficient 2 to the free term and making the replacement we get the equation y 2 - 11y + 30 = 0.

According to Vieta's inverse theorem

y 1 = 5, x 1 = 5/2, x 1 = 2.5; y 2 ​​= 6, x 2 = 6/2, x 2 = 3.

Answer: x 1 =2.5; X 2 = 3.

7. Properties of the coefficients of a quadratic equation.

Let the quadratic equation ax 2 + bx + c \u003d 0, a ≠ 0 be given.

1. If a + b + c \u003d 0 (i.e., the sum of the coefficients of the equation is zero), then x 1 \u003d 1.

2. If a - b + c \u003d 0, or b \u003d a + c, then x 1 \u003d - 1.

Example.345x 2 - 137x - 208 = 0.

Since a + b + c \u003d 0 (345 - 137 - 208 \u003d 0), then x 1 \u003d 1, x 2 \u003d -208/345.

Answer: x 1 =1; X 2 = -208/345 .

Example.132x 2 + 247x + 115 = 0

Because a-b + c \u003d 0 (132 - 247 + 115 \u003d 0), then x 1 \u003d - 1, x 2 \u003d - 115/132

Answer: x 1 = - 1; X 2 =- 115/132

There are other properties of the coefficients of a quadratic equation. but their usage is more complicated.

8. Solving quadratic equations using a nomogram.

Fig 1. Nomogram

This is an old and currently forgotten method for solving quadratic equations, placed on p. 83 of the collection: Bradis V.M. Four-digit mathematical tables. - M., Education, 1990.

Table XXII. Nomogram for Equation Solving z2 + pz + q = 0. This nomogram allows, without solving the quadratic equation, to determine the roots of the equation by its coefficients.

The curvilinear scale of the nomogram is built according to the formulas (Fig. 1):

Assuming OS = p, ED = q, OE = a(all in cm), from Fig. 1 similarity of triangles SAN and CDF we get the proportion

whence, after substitutions and simplifications, the equation follows z 2 + pz + q = 0, and the letter z means the label of any point on the curved scale.

Rice. 2 Solving a quadratic equation using a nomogram

Examples.

1) For the equation z 2 - 9z + 8 = 0 the nomogram gives the roots z 1 = 8.0 and z 2 = 1.0

Answer: 8.0; 1.0.

2) Solve the equation using the nomogram

2z 2 - 9z + 2 = 0.

Divide the coefficients of this equation by 2, we get the equation z 2 - 4.5z + 1 = 0.

The nomogram gives the roots z 1 = 4 and z 2 = 0.5.

Answer: 4; 0.5.

9. Geometric method for solving quadratic equations.

Example.X 2 + 10x = 39.

In the original, this problem is formulated as follows: "The square and ten roots are equal to 39."

Consider a square with side x, rectangles are built on its sides so that the other side of each of them is 2.5, therefore, the area of ​​​​each is 2.5x. The resulting figure is then supplemented to a new square ABCD, completing four equal squares in the corners, the side of each of them is 2.5, and the area is 6.25

Rice. 3 Graphical way to solve the equation x 2 + 10x = 39

The area S of square ABCD can be represented as the sum of the areas: the original square x 2, four rectangles (4 ∙ 2.5x = 10x) and four attached squares (6.25 ∙ 4 = 25), i.e. S \u003d x 2 + 10x \u003d 25. Replacing x 2 + 10x with the number 39, we get that S \u003d 39 + 25 \u003d 64, which implies that the side of the square ABCD, i.e. segment AB \u003d 8. For the desired side x of the original square, we get

10. Solution of equations using Bezout's theorem.

Bezout's theorem. The remainder after dividing the polynomial P(x) by the binomial x - α is equal to P(α) (that is, the value of P(x) at x = α).

If the number α is the root of the polynomial P(x), then this polynomial is divisible by x -α without remainder.

Example.x²-4x+3=0

Р(x)= x²-4x+3, α: ±1,±3, α=1, 1-4+3=0. Divide P(x) by (x-1): (x²-4x+3)/(x-1)=x-3

x²-4x+3=(x-1)(x-3), (x-1)(x-3)=0

x-1=0; x=1, or x-3=0, x=3; Answer: x1 =2, x2 =3.

Conclusion: The ability to quickly and rationally solve quadratic equations is simply necessary for solving more complex equations, for example, fractional rational equations, equations of higher powers, biquadratic equations, and in high school trigonometric, exponential and logarithmic equations. Having studied all the methods found for solving quadratic equations, we can advise classmates, in addition to standard methods, to solve by the transfer method (6) and solve equations by the property of coefficients (7), since they are more accessible for understanding.

Literature:

1. Bradis V.M. Four-digit mathematical tables. - M., Education, 1990.
2. Algebra grade 8: textbook for grade 8. general education institutions Makarychev Yu. N., Mindyuk N. G., Neshkov K. I., Suvorova S. B. ed. S. A. Telyakovsky 15th ed., revised. - M.: Enlightenment, 2015
3. https://en.wikipedia.org/wiki/%D0%9A%D0%B2%D0%B0%D0%B4%D1%80%D0%B0%D1%82%D0%BD%D0%BE%D0 %B5_%D1%83%D1%80%D0%B0%D0%B2%D0%BD%D0%B5%D0%BD%D0%B8%D0%B5
4. Glazer G.I. History of mathematics at school. A guide for teachers. / Ed. V.N. Younger. - M.: Enlightenment, 1964.

1. Find the discriminant D according to the formula D= -4ac.

2.If D<0, то квадратное уравнение не имеет корней.

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

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

Now let's start solving our equation 3 -10x+3=0,

where =3, b=-10 and c=3.

Finding the discriminant:

D= -4*3*3=64

Since D>0, then this equation has two roots. We find them:

; .

Thus, the roots of the polynomial f(x)=3 -10+3 will be the numbers 3 and .

Horner's scheme

Horner's scheme(or Horner's rule, Horner's method) - an algorithm for calculating the value of a polynomial, written as a sum of polynomials (monomials), for a given value of a variable . She, in turn, helps us find out whether the number is the root of a given polynomial or not.

First, consider how the polynomial is divided f(x) into a binomial g(x).

This can be written as follows: f(x):g(x)=n(x), where f(x)- dividend, g(x)- divisor a n(x)- private.

But in the case when f(x) not divisible by g(x) there is a general notation of the expression

Here, the degree r(x)< deg s(x), в таком случае можно сказать, что делится на с остатком .

Consider dividing a polynomial by a binomial. Let

,

We get

Where r is a number because the degree of r must be less than the degree of (x-c).

Let's multiply s(x) on and get

Thus, when dividing by a binomial, it is possible to determine the coefficients of the quotient from the obtained formulas. This method of determining the coefficients is called Horner's scheme.

				...
	+			...
c				...	r

Now let's look at a few examples of the application of Horner's scheme.

Example. Perform polynomial division f(x)= on the x+3.

Solution. At the beginning it is necessary to write x+3) as ( x-(-3)), since exactly -3 will participate in the scheme itself. In the top line we will write the coefficients, in the bottom line - the result of the actions.

f(x)=(x-2)(1)+16.

Finding roots according to Horner's scheme. Root types

According to Horner's scheme, one can find integer roots of a polynomial f(x). Let's look at this with an example.

Example. Find all integer roots of a polynomial f(x)= , using the Horner scheme.

Solution. The coefficients of this polynomial are integers. The coefficient before the highest degree (in our case before) is equal to one. Therefore, we will look for the integer roots of the polynomial among the divisors of the free term (we have 15), these are numbers:

Let's start with the number 1.

Table #1

			-21	-20
	+			-18	-38
			-18	-38

From the resulting table it can be seen that for =1 the polynomial of the polynomial f(x)= , we got the remainder r=192, not 0, which means that the unit is not a root. Therefore, we continue the check at =-1. To do this, we will not create a new table, but continue in the old one, and cross out the data that is no longer necessary.

Table number 2

			-21	-20
	+			-18	-38
			-18	-38
	+	-1	-1		-2	-69	-45
-1			-22

As we can see from the table, the last cell turned out to be zero, which means that r=0. Consequently? the number -1 is the root of this polynomial. Dividing our polynomial polynomial f(x)= on ()=x+1 we got a polynomial

f(x)=(x+1)(),

the coefficients for which we took from the third line of table No. 2.

We can also make the equivalent notation

(x+1)(). Tag him (1)

Now it is necessary to continue the search for integer roots, but only now we will already look for the roots of the polynomial. We will look for these roots among the free term of the polynomial, the number 45.

Let's check the number -1 again.

Table #3

			-21	-20
	+			-18	-38
			-18	-38
	+	-1	-1		-2	-69	-45
-1			-22
	+	-1			-24	-45
-1			-22

Thus, the number -1 is the root of the polynomial, it can be written as

Taking into account equality (2), we can write equality (1) in the following form

Now we are looking for roots for the polynomial, again among the divisors of the free term. Let's check the number -1 again.

Table No. 4

			-21	-20
	+			-18	-38
			-18	-38
	+	-1	-1		-2	-69	-45
-1			-22
	+	-1			-24	-45
-1			-22
	+	-1			-45
-1		-1	-21

According to the table, we see that the number -1 is the root of the polynomial.

Given (3*), we can rewrite equality (2*) as:

Now we will look for the root for . Again we look at the divisors of the free term. Let's start checking again with the number -1.

Table number 5

			-21	-20
	+			-18	-38
			-18	-38
	+	-1	-1		-2	-69	-45
-1			-22
	+	-1			-24	-45
-1			-22
	+	-1			-45
-1		-1	-21
	+	-1
-1		-2	-19

We got a remainder that is not equal to zero, which means that the number -1 is not a root for the polynomial. Let's check the next number 1.

Table No. 6

			-21	-20
	+			-18	-38
			-18	-38
	+	-1	-1		-2	-69	-45
-1			-22
	+	-1			-24	-45
-1			-22
	+	-1			-45
-1		-1	-21
	+	-1
-1		-2	-19
	+			-21
			-21

And we see that again it does not fit, the remainder is r(x) = 24. We take a new number.

Let's check the number 3.

			-21	-20
	+			-18	-38
			-18	-38
	+	-1	-1		-2	-69	-45
-1			-22
	+	-1			-24	-45
-1			-22
	+	-1			-45
-1		-1	-21
	+	-1
-1		-2	-19
	+			-21
			-21
	+			-45
			-15

Table number 7

r(x)= 0, this means that the number 3 is the root of the polynomial, we can write this polynomial as:

=(x-3)( )

Given the resulting expression, we can write equality (5) as follows:

(x-3)( ) (6)

Let's check now for the polynomial

			-21	-20
	+			-18	-38
			-18	-38
	+	-1	-1		-2	-69	-45
-1			-22
	+	-1			-24	-45
-1			-22
	+	-1			-45
-1		-1	-21
	+	-1
-1		-2	-19
	+			-21
			-21
	+			-45
			-15
	+

Table No. 8

Based on the table, we see that the number 3 is the root of the polynomial . Now let's write the following:

We write the equality (5*), taking into account the resulting expression, as follows:

(x-3)()= = .

Find the root for the binomial among the divisors of the free term.

Let's take the number 5

Table No. 9

			-21	-20
	+			-18	-38
			-18	-38
	+	-1	-1		-2	-69	-45
-1			-22
	+	-1			-24	-45
-1			-22
	+	-1			-45
-1		-1	-21
	+	-1
-1		-2	-19
	+			-21
			-21
	+			-45
			-15
	+
	+	-5
-5

r(x)=0, so 5 is the root of the binomial.

Thus, we can write

The solution to this example will be table number 8.

As can be seen from the table, the numbers -1; 3; 5 are the roots of the polynomial.

Now let's go directly to types of roots.

1 is the root of the third degree, since the bracket (x + 1) is in the third degree;

3- root of the second degree, bracket (x-3) in the second degree;

5 is the root of the first degree or, in other words, simple.

Quadratic equations often appear in a number of problems in mathematics and physics, so every student should be able to solve them. This article discusses in detail the main methods for solving quadratic equations, and also provides examples of their use.

What equation is called quadratic

First of all, we will answer the question of this paragraph in order to better understand what will be discussed in the article. So, the quadratic equation has the following general form: c + b * x + a * x 2 \u003d 0, where a, b, c are some numbers, which are called coefficients. Here a≠0 is a mandatory condition, otherwise the indicated equation degenerates into a linear one. The remaining coefficients (b, c) can take absolutely any values, including zero. So, expressions like a*x 2 =0, where b=0 and c=0 or c+a*x 2 =0, where b=0, or b*x+a*x 2 =0, where c=0 - these are also quadratic equations, which are called incomplete, since in them either the linear coefficient b is equal to zero, or the free term c is zero, or they both vanish.

An equation in which a \u003d 1 is called reduced, that is, it has the form: x 2 + c / a + (b / a) * x \u003d 0.

The solution of a quadratic equation is to find such values ​​of x that satisfy its equality. These values ​​are called roots. Since the equation under consideration is an expression of the second degree, this means that the maximum number of its roots cannot exceed two.

What methods for solving square equations exist

In general, there are 4 solution methods. Their names are listed below:

1. Factorization.
2. Complement to the square.
3. Using a well-known formula (through the discriminant).
4. The solution is geometric.

As is clear from the above list, the first three methods are algebraic, so they are used more often than the last one, which involves plotting a function graph.

There is another way to solve square equations using the Vieta theorem. It could be included 5th in the list above, however, this is not done, since Vieta's theorem is a simple consequence of the 3rd method.

Method number 1. Factorization

There is a beautiful name for this method in the mathematics of quadratic equations: factorization. The essence of this method is as follows: it is necessary to present the quadratic equation as a product of two terms (expressions), which must equal zero. After such a representation, one can use the product property, which will be equal to zero only when one or more (all) of its members are zero.

Now consider the sequence of specific actions that need to be performed in order to find the roots of the equation:

1. Transfer all members to one part of the expression (for example, to the left) so that only 0 remains in its other part (right).
2. Express the sum of the terms in one part of the equation as a product of two linear equations.
3. Equate each of the linear expressions to zero and solve them.

As you can see, the factorization algorithm is quite simple, however, most students have difficulties during the implementation of the 2nd point, so we will explain it in more detail.

To guess which 2 linear expressions, when multiplied by each other, will give the desired quadratic equation, you need to remember two simple rules:

• The linear coefficients of two linear expressions, when multiplied by each other, should give the first coefficient of the quadratic equation, that is, the number a.
• The free terms of linear expressions, when they are multiplied, must give the number c of the desired equation.

After all the numbers of factors have been selected, they should be multiplied, and if they give the desired equation, then go to step 3 in the above algorithm, otherwise the factors should be changed, but this should be done so that the above rules are always fulfilled.

An example of a factorization solution

We will show clearly how to compose an algorithm for solving a quadratic equation and find unknown roots. Let an arbitrary expression be given, for example, 2*x-5+5*x 2 -2*x 2 = x 2 +2+x 2 +1. Let's move on to its solution, observing the sequence of points from 1 to 3, which are set out in the previous paragraph of the article.

Point 1. Let's move all the terms to the left side and build them in the classical sequence for a quadratic equation. We have the following equality: 2*x+(-8)+x 2 =0.

Point 2. We break it into a product of linear equations. Since a=1, and c=-8, then we will select, for example, such a product (x-2)*(x+4). It satisfies the rules for finding the expected factors set out in the paragraph above. If we open the brackets, we get: -8+2*x+x 2 , that is, we get exactly the same expression as on the left side of the equation. This means that we correctly guessed the multipliers, and we can proceed to the 3rd step of the algorithm.

Item 3. We equate each factor to zero, we get: x=-4 and x=2.

If there are any doubts about the result obtained, it is recommended to check by substituting the found roots into the original equation. In this case, we have: 2*2+2 2 -8=0 and 2*(-4)+(-4) 2 -8=0. Roots found correctly.

Thus, by the factorization method, we found that the given equation has two different roots: 2 and -4.

Method #2. Complement to the full square

In the algebra of square equations, the multiplier method cannot always be used, since in the case of fractional values ​​of the coefficients of the quadratic equation, difficulties arise in the implementation of paragraph 2 of the algorithm.

The full square method, in turn, is universal and can be applied to quadratic equations of any type. Its essence is to perform the following operations:

1. The terms of the equation containing the coefficients a and b must be transferred to one part of the equality, and the free term c to the other.
2. Further, the parts of the equality (right and left) should be divided by the coefficient a, that is, the equation should be presented in the reduced form (a=1).
3. The sum of terms with coefficients a and b is represented as a square of a linear equation. Since a \u003d 1, then the linear coefficient will be equal to 1, as for the free term of the linear equation, then it should be equal to half the linear coefficient of the reduced quadratic equation. After the square of the linear expression has been drawn up, it is necessary to add the corresponding number to the right side of the equality, where the free term is located, which is obtained by opening the square.
4. Take the square root with the signs "+" and "-" and solve the linear equation already obtained.

The described algorithm may at first glance be perceived as rather complicated, however, in practice it is easier to implement than the factorization method.

An example of a solution using the full square's complement

We give an example of a quadratic equation for training its solution by the method described in the previous paragraph. Let the quadratic equation -10 - 6*x+5*x 2 = 0 be given. We begin to solve it, following the algorithm described above.

Point 1. We use the transfer method when solving square equations, we get: - 6 * x + 5 * x 2 = 10.

Point 2. The reduced form of this equation is obtained by dividing by the number 5 of each of its members (if the equalities are both parts divided or multiplied by the same number, then the equality will be preserved). As a result of the transformations, we get: x 2 - 6/5 * x = 2.

Item 3. Half of the coefficient - 6/5 is equal to -6/10 = -3/5, we use this number to make a full square, we get: (-3/5 + x) 2 . We expand it and the resulting free term should be subtracted from the left side of the equality in order to satisfy the original form of the quadratic equation, which is equivalent to adding it to the right side. As a result, we get: (-3/5+x) 2 = 59/25.

Point 4. We calculate the square root with positive and negative signs and find the roots: x = 3/5±√59/5 = (3±√59)/5. The two found roots have the following values: x 1 = (√59+3)/5 and x 1 = (3-√59)/5.

Since the calculations performed are related to the roots, there is a high probability of making a mistake. Therefore, it is recommended to check the correctness of the roots x 2 and x 1 . We get for x 1: 5*((3+√59)/5) 2 -6*(3+√59)/5 - 10 = (9+59+6*√59)/5 - 18/5 - 6 *√59/5-10 = 68/5-68/5 = 0. Now substitute x 2: 5*((3-√59)/5) 2 -6*(3-√59)/5 - 10 = (9+59-6*√59)/5 - 18/5 + 6*√59/5-10 = 68/5-68/5 = 0.

Thus, we have shown that the found roots of the equation are true.

Method number 3. Application of the well-known formula

This method of solving quadratic equations is perhaps the simplest, since it consists in substituting the coefficients into a known formula. To use it, you do not need to think about compiling solution algorithms, it is enough to remember only one formula. It is shown in the figure above.

In this formula, the root expression (b 2 -4*a*c) is called the discriminant (D). From its value depends on what roots are obtained. 3 cases are possible:

• D>0, then the root two equation has real and different ones.
• D=0, then one root is obtained, which can be calculated from the expression x = -b / (a ​​* 2).
• D<0, тогда получается два различных мнимых корня, которые представляются в виде комплексных чисел. Например, число 3-5*i является комплексным, при этом мнимая единица i удовлетворяет свойству: i 2 =-1.

An example of a solution through the calculation of the discriminant

Here is an example of a quadratic equation to practice using the above formula. Find the roots for -3*x 2 -6+3*x+4*x = 0. First, calculate the value of the discriminant, we get: D = b 2 -4*a*c = 7 2 -4*(-3)* (-6) = -23.

Since received D<0, значит, корни рассматриваемого уравнения являются числами комплексными. Найдем их, подставив найденное значение D в приведенную в предыдущем пункте формулу (она также представлена на фото выше). Получим: x = 7/6±√(-23)/(-6) = (7±i*√23)/6.

Method number 4. Using the Graph of a Function

It is also called the graphical method for solving quadratic equations. It should be said that it is used, as a rule, not for a quantitative, but for a qualitative analysis of the equation under consideration.

The essence of the method is to plot a quadratic function y = f(x), which is a parabola. Then, it is necessary to determine at what points the abscissa (X) axis of the parabola intersects, they will be the roots of the corresponding equation.

To tell whether a parabola will intersect the x-axis, it is enough to know the position of its minimum (maximum) and the direction of its branches (they can either increase or decrease). There are two properties of this curve to remember:

• If a>0 - the parabolas of the branch are directed upwards, vice versa, if a<0, то они идут вниз.
• The coordinate of the minimum (maximum) of the parabola is always x = -b/(2*a).

For example, it is necessary to determine whether the equation -4*x+5*x 2 +10 = 0 has roots. The corresponding parabola will be directed upwards, since a=5>0. Its extremum has coordinates: x=4/10=2/5, y=-4*2/5+5*(2/5) 2 +10 = 9.2. Since the minimum of the curve lies above the x-axis (y=9.2), it does not intersect the latter for any values ​​of x. That is, the given equation has no real roots.

Vieta's theorem

As noted above, this theorem is a consequence of method No. 3, which is based on the application of a formula with a discriminant. The essence of the Vieta theorem is that it allows you to connect the coefficients of the equation and its roots into equality. We obtain the corresponding equalities.

Let's use the formula for calculating the roots through the discriminant. Let's add two roots, we get: x 1 + x 2 \u003d -b / a. Now we multiply the roots by each other: x 1 * x 2, after a series of simplifications, we get the number c / a.

Thus, to solve the quadratic equations by the Vieta theorem, you can use the two equalities obtained. If all three coefficients of an equation are known, then the roots can be found by solving the appropriate system of these two equations.

An example of using Vieta's theorem

It is necessary to draw up a quadratic equation if it is known that it has the form x 2 + c \u003d -b * x and its roots are 3 and -4.

Since in the equation under consideration a \u003d 1, then the Vieta formulas will look like: x 2 + x 1 \u003d -b and x 2 * x 1 \u003d c. Substituting the known values ​​of the roots, we get: b = 1 and c = -12. As a result, the restored quadratic equation will look like: x 2 -12 = -1*x. You can substitute the value of the roots into it and make sure that the equality holds.

The reverse application of the Vieta theorem, that is, the calculation of the roots according to the known form of the equation, allows you to quickly (intuitively) find solutions for small integers a, b and c.

Quadratic equations are studied in grade 8, so there is nothing complicated here. The ability to solve them is essential.

A quadratic equation is an equation of the form ax 2 + bx + c = 0, where the coefficients a , b and c are arbitrary numbers, and a ≠ 0.

Before studying specific solution methods, we note that all quadratic equations can be divided into three classes:

1. Have no roots;
2. They have exactly one root;
3. They have two different roots.

This is an important difference between quadratic and linear equations, where the root always exists and is unique. How to determine how many roots an equation has? There is a wonderful thing for this - discriminant.

Discriminant

Let the quadratic equation ax 2 + bx + c = 0 be given. Then the discriminant is simply the number D = b 2 − 4ac .

This formula must be known by heart. Where it comes from is not important now. Another thing is important: by the sign of the discriminant, you can determine how many roots a quadratic equation has. Namely:

1. If D< 0, корней нет;
2. If D = 0, there is exactly one root;
3. If D > 0, there will be two roots.

Please note: the discriminant indicates the number of roots, and not at all their signs, as for some reason many people think. Take a look at the examples and you will understand everything yourself:

A task. How many roots do quadratic equations have:

1. x 2 - 8x + 12 = 0;
2. 5x2 + 3x + 7 = 0;
3. x 2 − 6x + 9 = 0.

We write the coefficients for the first equation and find the discriminant:
a = 1, b = −8, c = 12;
D = (−8) 2 − 4 1 12 = 64 − 48 = 16

So, the discriminant is positive, so the equation has two different roots. We analyze the second equation in the same way:
a = 5; b = 3; c = 7;
D \u003d 3 2 - 4 5 7 \u003d 9 - 140 \u003d -131.

The discriminant is negative, there are no roots. The last equation remains:
a = 1; b = -6; c = 9;
D = (−6) 2 − 4 1 9 = 36 − 36 = 0.

The discriminant is equal to zero - the root will be one.

Note that coefficients have been written out for each equation. Yes, it's long, yes, it's tedious - but you won't mix up the odds and don't make stupid mistakes. Choose for yourself: speed or quality.

By the way, if you “fill your hand”, after a while you will no longer need to write out all the coefficients. You will perform such operations in your head. Most people start doing this somewhere after 50-70 solved equations - in general, not so much.

The roots of a quadratic equation

Now let's move on to the solution. If the discriminant D > 0, the roots can be found using the formulas:

The basic formula for the roots of a quadratic equation

When D = 0, you can use any of these formulas - you get the same number, which will be the answer. Finally, if D< 0, корней нет — ничего считать не надо.

1. x 2 - 2x - 3 = 0;
2. 15 - 2x - x 2 = 0;
3. x2 + 12x + 36 = 0.

First equation:
x 2 - 2x - 3 = 0 ⇒ a = 1; b = −2; c = -3;
D = (−2) 2 − 4 1 (−3) = 16.

D > 0 ⇒ the equation has two roots. Let's find them:

Second equation:
15 − 2x − x 2 = 0 ⇒ a = −1; b = −2; c = 15;
D = (−2) 2 − 4 (−1) 15 = 64.

D > 0 ⇒ the equation again has two roots. Let's find them

\[\begin(align) & ((x)_(1))=\frac(2+\sqrt(64))(2\cdot \left(-1 \right))=-5; \\ & ((x)_(2))=\frac(2-\sqrt(64))(2\cdot \left(-1 \right))=3. \\ \end(align)\]

Finally, the third equation:
x 2 + 12x + 36 = 0 ⇒ a = 1; b = 12; c = 36;
D = 12 2 − 4 1 36 = 0.

D = 0 ⇒ the equation has one root. Any formula can be used. For example, the first one:

As you can see from the examples, everything is very simple. If you know the formulas and be able to count, there will be no problems. Most often, errors occur when negative coefficients are substituted into the formula. Here, again, the technique described above will help: look at the formula literally, paint each step - and get rid of mistakes very soon.

Incomplete quadratic equations

It happens that the quadratic equation is somewhat different from what is given in the definition. For example:

1. x2 + 9x = 0;
2. x2 − 16 = 0.

It is easy to see that one of the terms is missing in these equations. Such quadratic equations are even easier to solve than standard ones: they do not even need to calculate the discriminant. So let's introduce a new concept:

The equation ax 2 + bx + c = 0 is called an incomplete quadratic equation if b = 0 or c = 0, i.e. the coefficient of the variable x or the free element is equal to zero.

Of course, a very difficult case is possible when both of these coefficients are equal to zero: b \u003d c \u003d 0. In this case, the equation takes the form ax 2 \u003d 0. Obviously, such an equation has a single root: x \u003d 0.

Let's consider other cases. Let b \u003d 0, then we get an incomplete quadratic equation of the form ax 2 + c \u003d 0. Let's slightly transform it:

Since the arithmetic square root exists only from a non-negative number, the last equality only makes sense when (−c / a ) ≥ 0. Conclusion:

1. If an incomplete quadratic equation of the form ax 2 + c = 0 satisfies the inequality (−c / a ) ≥ 0, there will be two roots. The formula is given above;
2. If (−c / a )< 0, корней нет.

As you can see, the discriminant was not required - there are no complex calculations at all in incomplete quadratic equations. In fact, it is not even necessary to remember the inequality (−c / a ) ≥ 0. It is enough to express the value of x 2 and see what is on the other side of the equal sign. If there is a positive number, there will be two roots. If negative, there will be no roots at all.

Now let's deal with equations of the form ax 2 + bx = 0, in which the free element is equal to zero. Everything is simple here: there will always be two roots. It is enough to factorize the polynomial:

Taking the common factor out of the bracket

The product is equal to zero when at least one of the factors is equal to zero. This is where the roots come from. In conclusion, we will analyze several of these equations:

A task. Solve quadratic equations:

1. x2 − 7x = 0;
2. 5x2 + 30 = 0;
3. 4x2 − 9 = 0.

x 2 − 7x = 0 ⇒ x (x − 7) = 0 ⇒ x 1 = 0; x2 = −(−7)/1 = 7.

5x2 + 30 = 0 ⇒ 5x2 = -30 ⇒ x2 = -6. There are no roots, because the square cannot be equal to a negative number.

4x 2 − 9 = 0 ⇒ 4x 2 = 9 ⇒ x 2 = 9/4 ⇒ x 1 = 3/2 = 1.5; x 2 \u003d -1.5.