How to solve the discriminant of a quadratic equation. Quadratic equations

The discriminant, as well as quadratic equations, begin to be studied in the algebra course in grade 8. You can solve a quadratic equation through the discriminant and using the Vieta theorem. The methodology for studying quadratic equations, as well as the discriminant formula, is rather unsuccessfully instilled in schoolchildren, like much in real education. Therefore, school years pass, education in grades 9-11 replaces "higher education" and everyone is again looking for - "How to solve a quadratic equation?", "How to find the roots of an equation?", "How to find the discriminant?" and...

Discriminant Formula

The discriminant D of the quadratic equation a*x^2+bx+c=0 is D=b^2–4*a*c.
The roots (solutions) of the quadratic equation depend on the sign of the discriminant (D):
D>0 - the equation has 2 different real roots;
D=0 - the equation has 1 root (2 coinciding roots):
D<0 – не имеет действительных корней (в школьной теории). В ВУЗах изучают комплексные числа и уже на множестве комплексных чисел уравнение с отрицательным дискриминантом имеет два комплексных корня.
The formula for calculating the discriminant is quite simple, so many sites offer an online discriminant calculator. We have not figured out this kind of scripts yet, so who knows how to implement this, please write to the mail This email address is being protected from spambots. You must have JavaScript enabled to view. .

General formula for finding the roots of a quadratic equation:

The roots of the equation are found by the formula
If the coefficient of the variable in the square is paired, then it is advisable to calculate not the discriminant, but its fourth part
In such cases, the roots of the equation are found by the formula

The second way to find roots is Vieta's Theorem.

The theorem is formulated not only for quadratic equations, but also for polynomials. You can read this on Wikipedia or other electronic resources. However, to simplify, consider that part of it that concerns the reduced quadratic equations, that is, equations of the form (a=1)
The essence of the Vieta formulas is that the sum of the roots of the equation is equal to the coefficient of the variable, taken with the opposite sign. The product of the roots of the equation is equal to the free term. The formulas of Vieta's theorem have a notation.
The derivation of the Vieta formula is quite simple. Let's write the quadratic equation in terms of prime factors
As you can see, everything ingenious is simple at the same time. It is effective to use the Vieta formula when the difference in the modulus of the roots or the difference in the modulus of the roots is 1, 2. For example, the following equations, according to the Vieta theorem, have roots




Up to 4 equation analysis should look like this. The product of the roots of the equation is 6, so the roots can be the values ​​(1, 6) and (2, 3) or pairs with the opposite sign. The sum of the roots is 7 (the coefficient of the variable with the opposite sign). From here we conclude that the solutions of the quadratic equation are equal to x=2; x=3.
It is easier to select the roots of the equation among the divisors of the free term, correcting their sign in order to fulfill the Vieta formulas. At the beginning, this seems difficult to do, but with practice on a number of quadratic equations, this technique will be more efficient than calculating the discriminant and finding the roots of the quadratic equation in the classical way.
As you can see, the school theory of studying the discriminant and ways to find solutions to the equation is devoid of practical meaning - "Why do schoolchildren need a quadratic equation?", "What is the physical meaning of the discriminant?".

Let's try to figure it out what does the discriminant describe?

In the course of algebra, they study functions, schemes for studying functions and plotting functions. Of all the functions, an important place is occupied by a parabola, the equation of which can be written in the form
So the physical meaning of the quadratic equation is the zeros of the parabola, that is, the points of intersection of the graph of the function with the abscissa axis Ox
I ask you to remember the properties of parabolas that are described below. The time will come to take exams, tests, or entrance exams and you will be grateful for the reference material. The sign of the variable in the square corresponds to whether the branches of the parabola on the graph will go up (a>0),

or a parabola with branches down (a<0) .

The vertex of the parabola lies midway between the roots

The physical meaning of the discriminant:

If the discriminant is greater than zero (D>0), the parabola has two points of intersection with the Ox axis.
If the discriminant is equal to zero (D=0), then the parabola at the top touches the x-axis.
And the last case, when the discriminant is less than zero (D<0) – график параболы принадлежит плоскости над осью абсцисс (ветки параболы вверх), или график полностью под осью абсцисс (ветки параболы опущены вниз).

Incomplete quadratic equations

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 methods of solving, 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:

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 many.

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 - x2 = 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:

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.

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 formulating 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 be equal not to 96, but to 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 . Decision 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+ b x = 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 = b X.

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 = b X.

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 sets out 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 Europe XIII - XVII centuries

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, then A equals AT and equal D ».

To understand Vieta, one must remember that BUT, like any vowel, meant for him the unknown (our X), the vowels AT, 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 + a b = 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.

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 otherwise it will no longer be a quadratic equation. Quadratic equations either have no roots, or have exactly one root, or two different roots. The first step is to look for the discriminant. Formula: D = b^2 − 4ac. 1. If D< 0, корней нет; 2. Если D = 0, есть ровно один корень; 3. Если D >0, there will be two roots. The first option is clear, there are no roots. If the discriminant D > 0, the roots can be found as follows: x12 = (-b +- √D) / 2a. As for the second option, when D = 0, the upper formula can be used.

Quadratic equations are beginning to be studied in the school curriculum in the course of mathematics. But, unfortunately, not everyone understands and knows how to correctly solve a quadratic equation and calculate its roots. First, let's understand what a quadratic equation is.

What is a quadratic equation

The term quadratic equation is usually understood as an algebraic equation of a general form. This equation has the following form: ax2 + bx + c = 0, while a, b and c are some definite numbers, x is unknown. These three numbers are usually called the coefficients of the quadratic equation:

• a - first coefficient;
• b - second coefficient;
• c is the third coefficient.

How to find the roots of a quadratic equation

In order to calculate what the roots of a quadratic equation will equal, it is necessary to find the discriminant of the equation. The discriminant of a quadratic equation is an expression that equals and is calculated by the formula b2 - 4ac. If the discriminant is greater than zero, the root is calculated by the formula: x \u003d -b + - the root of the discriminant divided by 2 a.

Consider the example of the equation 5x squared - 8x +3 = 0

The discriminant is eight squared, minus four times five times three, i.e. = 64 - 4*5*3 = 64-60 = 4

x1 \u003d 8 + - root of four divided by two times five \u003d 8 + 2/10 \u003d 1

x2 = 8-2/10 = 6/10 = 3/5 = 0.6

Accordingly, the roots of this quadratic equation will be 1 and 0.6.

Formulas for the roots of a quadratic equation. The cases of real, multiple and complex roots are considered. Factorization of a square trinomial. Geometric interpretation. Examples of determining roots and factorization.

Basic Formulas

Consider the quadratic equation:
(1) .
The roots of a quadratic equation(1) are determined by the formulas:
; .
These formulas can be combined like this:
.
When the roots of the quadratic equation are known, then the polynomial of the second degree can be represented as a product of factors (factored):
.

Further, we assume that are real numbers.
Consider discriminant of a quadratic equation:
.
If the discriminant is positive, then the quadratic equation (1) has two different real roots:
; .
Then the factorization of the square trinomial has the form:
.
If the discriminant is zero, then the quadratic equation (1) has two multiple (equal) real roots:
.
Factorization:
.
If the discriminant is negative, then the quadratic equation (1) has two complex conjugate roots:
;
.
Here is the imaginary unit, ;
and are the real and imaginary parts of the roots:
; .
Then

.

Graphic interpretation

If we graph the function
,
which is a parabola, then the points of intersection of the graph with the axis will be the roots of the equation
.
When , the graph intersects the abscissa axis (axis) at two points.
When , the graph touches the x-axis at one point.
When , the graph does not cross the x-axis.

Below are examples of such graphs.

Useful Formulas Related to Quadratic Equation

(f.1) ;
(f.2) ;
(f.3) .

Derivation of the formula for the roots of a quadratic equation

We perform transformations and apply formulas (f.1) and (f.3):




,
where
; .

So, we got the formula for the polynomial of the second degree in the form:
.
From this it can be seen that the equation

performed at
and .
That is, and are the roots of the quadratic equation
.

Examples of determining the roots of a quadratic equation

Example 1

(1.1) .

Decision

.
Comparing with our equation (1.1), we find the values ​​of the coefficients:
.
Finding the discriminant:
.
Since the discriminant is positive, the equation has two real roots:
;
;
.

From here we obtain the decomposition of the square trinomial into factors:

.

Graph of the function y = 2 x 2 + 7 x + 3 crosses the x-axis at two points.

Let's plot the function
.
The graph of this function is a parabola. It crosses the x-axis (axis) at two points:
and .
These points are the roots of the original equation (1.1).

Answer

;
;
.

Example 2

Find the roots of a quadratic equation:
(2.1) .

Decision

We write the quadratic equation in general form:
.
Comparing with the original equation (2.1), we find the values ​​of the coefficients:
.
Finding the discriminant:
.
Since the discriminant is zero, the equation has two multiple (equal) roots:
;
.

Then the factorization of the trinomial has the form:
.

Graph of the function y = x 2 - 4 x + 4 touches the x-axis at one point.

Let's plot the function
.
The graph of this function is a parabola. It touches the x-axis (axis) at one point:
.
This point is the root of the original equation (2.1). Since this root is factored twice:
,
then such a root is called a multiple. That is, they consider that there are two equal roots:
.

Answer

;
.

Example 3

Find the roots of a quadratic equation:
(3.1) .

Decision

We write the quadratic equation in general form:
(1) .
Let us rewrite the original equation (3.1):
.
Comparing with (1), we find the values ​​of the coefficients:
.
Finding the discriminant:
.
The discriminant is negative, . Therefore, there are no real roots.

You can find complex roots:
;
;
.

Then


.

The graph of the function does not cross the x-axis. There are no real roots.

Let's plot the function
.
The graph of this function is a parabola. It does not cross the abscissa (axis). Therefore, there are no real roots.

Answer

There are no real roots. Complex roots:
;
;
.