With this math program you can solve quadratic equation.
The program not only gives the answer to the problem, but also displays the solution process in two ways:
- using the discriminant
- using the Vieta theorem (if possible).
Moreover, the answer is displayed exact, not approximate.
For example, for the equation \(81x^2-16x-1=0\), the answer is displayed in this form:
This program can be useful for high school students in preparation for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it's too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with a detailed solution.
In this way, you can conduct your own training and/or the training of your younger brothers or sisters, while the level of education in the field of tasks to be solved is increased.
If you are not familiar with the rules for entering a square polynomial, we recommend that you familiarize yourself with them.
Rules for entering a square polynomial
Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q \) etc.
Numbers can be entered as integers or fractions.
Moreover, fractional numbers can be entered not only in the form of a decimal, but also in the form of an ordinary fraction.
Rules for entering decimal fractions.
In decimal fractions, the fractional part from the integer can be separated by either a dot or a comma.
For example, you can enter decimals like this: 2.5x - 3.5x^2
Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
The integer part is separated from the fraction by an ampersand: &
Input: 3&1/3 - 5&6/5z +1/7z^2
Result: \(3\frac(1)(3) - 5\frac(6)(5) z + \frac(1)(7)z^2 \)
When entering an expression you can use brackets. In this case, when solving a quadratic equation, the introduced expression is first simplified.
For example: 1/2(y-1)(y+1)-(5y-10&1/2)
Decide
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A bit of theory.
Quadratic equation and its roots. Incomplete quadratic equations
Each of the equations
\(-x^2+6x+1,4=0, \quad 8x^2-7x=0, \quad x^2-\frac(4)(9)=0 \)
has the form
\(ax^2+bx+c=0, \)
where x is a variable, a, b and c are numbers.
In the first equation a = -1, b = 6 and c = 1.4, in the second a = 8, b = -7 and c = 0, in the third a = 1, b = 0 and c = 4/9. Such equations are called quadratic equations.
Definition.
quadratic equation an equation of the form ax 2 +bx+c=0 is called, where x is a variable, a, b and c are some numbers, and \(a \neq 0 \).
The numbers a, b and c are the coefficients of the quadratic equation. The number a is called the first coefficient, the number b is the second coefficient and the number c is the intercept.
In each of the equations of the form ax 2 +bx+c=0, where \(a \neq 0 \), the largest power of the variable x is a square. Hence the name: quadratic equation.
Note that a quadratic equation is also called an equation of the second degree, since its left side is a polynomial of the second degree.
A quadratic equation in which the coefficient at x 2 is 1 is called reduced quadratic equation. For example, the given quadratic equations are the equations
\(x^2-11x+30=0, \quad x^2-6x=0, \quad x^2-8=0 \)
If in the quadratic equation ax 2 +bx+c=0 at least one of the coefficients b or c is equal to zero, then such an equation is called incomplete quadratic equation. So, the equations -2x 2 +7=0, 3x 2 -10x=0, -4x 2 =0 are incomplete quadratic equations. In the first of them b=0, in the second c=0, in the third b=0 and c=0.
Incomplete quadratic equations are of three types:
1) ax 2 +c=0, where \(c \neq 0 \);
2) ax 2 +bx=0, where \(b \neq 0 \);
3) ax2=0.
Consider the solution of equations of each of these types.
To solve an incomplete quadratic equation of the form ax 2 +c=0 for \(c \neq 0 \), its free term is transferred to the right side and both parts of the equation are divided by a:
\(x^2 = -\frac(c)(a) \Rightarrow x_(1,2) = \pm \sqrt( -\frac(c)(a)) \)
Since \(c \neq 0 \), then \(-\frac(c)(a) \neq 0 \)
If \(-\frac(c)(a)>0 \), then the equation has two roots.
If \(-\frac(c)(a) To solve an incomplete quadratic equation of the form ax 2 +bx=0 for \(b \neq 0 \) factorize its left side and obtain the equation
\(x(ax+b)=0 \Rightarrow \left\( \begin(array)(l) x=0 \\ ax+b=0 \end(array) \right. \Rightarrow \left\( \begin (array)(l) x=0 \\ x=-\frac(b)(a) \end(array) \right. \)
Hence, an incomplete quadratic equation of the form ax 2 +bx=0 for \(b \neq 0 \) always has two roots.
An incomplete quadratic equation of the form ax 2 \u003d 0 is equivalent to the equation x 2 \u003d 0 and therefore has a single root 0.
The formula for the roots of a quadratic equation
Let us now consider how quadratic equations are solved in which both coefficients of the unknowns and the free term are nonzero.
We solve the quadratic equation in general form and as a result we obtain the formula of the roots. Then this formula can be applied to solve any quadratic equation.
Solve the quadratic equation ax 2 +bx+c=0
Dividing both its parts by a, we obtain the equivalent reduced quadratic equation
\(x^2+\frac(b)(a)x +\frac(c)(a)=0 \)
We transform this equation by highlighting the square of the binomial:
\(x^2+2x \cdot \frac(b)(2a)+\left(\frac(b)(2a)\right)^2- \left(\frac(b)(2a)\right)^ 2 + \frac(c)(a) = 0 \Rightarrow \)
The root expression is called discriminant of a quadratic equation ax 2 +bx+c=0 (“discriminant” in Latin - distinguisher). It is denoted by the letter D, i.e.
\(D = b^2-4ac\)
Now, using the notation of the discriminant, we rewrite the formula for the roots of the quadratic equation:
\(x_(1,2) = \frac( -b \pm \sqrt(D) )(2a) \), where \(D= b^2-4ac \)
It's obvious that:
1) If D>0, then the quadratic equation has two roots.
2) If D=0, then the quadratic equation has one root \(x=-\frac(b)(2a)\).
3) If D Thus, depending on the value of the discriminant, the quadratic equation can have two roots (for D > 0), one root (for D = 0) or no roots (for D When solving a quadratic equation using this formula, it is advisable to do the following way:
1) calculate the discriminant and compare it with zero;
2) if the discriminant is positive or equal to zero, then use the root formula, if the discriminant is negative, then write down that there are no roots.
Vieta's theorem
The given quadratic equation ax 2 -7x+10=0 has roots 2 and 5. The sum of the roots is 7, and the product is 10. We see that the sum of the roots is equal to the second coefficient, taken with the opposite sign, and the product of the roots is equal to the free term. Any reduced quadratic equation that has roots has this property.
The sum of the roots of the given quadratic equation is equal to the second coefficient, taken with the opposite sign, and the product of the roots is equal to the free term.
Those. Vieta's theorem states that the roots x 1 and x 2 of the reduced quadratic equation x 2 +px+q=0 have the property:
\(\left\( \begin(array)(l) x_1+x_2=-p \\ x_1 \cdot x_2=q \end(array) \right. \)
I hope that after studying this article, you will learn how to find the roots of a complete quadratic equation.
With the help of the discriminant, only complete quadratic equations are solved; to solve incomplete quadratic equations, other methods are used, which you will find in the article "Solving incomplete quadratic equations".
What quadratic equations are called complete? This is equations of the form ax 2 + b x + c = 0, where the coefficients a, b and c are not equal to zero. So, to solve the complete quadratic equation, you need to calculate the discriminant D.
D \u003d b 2 - 4ac.
Depending on what value the discriminant has, we will write down the answer.
If the discriminant is a negative number (D< 0),то корней нет.
If the discriminant is zero, then x \u003d (-b) / 2a. When the discriminant is a positive number (D > 0),
then x 1 = (-b - √D)/2a, and x 2 = (-b + √D)/2a.
For example. solve the equation x 2– 4x + 4= 0.
D \u003d 4 2 - 4 4 \u003d 0
x = (- (-4))/2 = 2
Answer: 2.
Solve Equation 2 x 2 + x + 3 = 0.
D \u003d 1 2 - 4 2 3 \u003d - 23
Answer: no roots.
Solve Equation 2 x 2 + 5x - 7 = 0.
D \u003d 5 2 - 4 2 (-7) \u003d 81
x 1 \u003d (-5 - √81) / (2 2) \u003d (-5 - 9) / 4 \u003d - 3.5
x 2 \u003d (-5 + √81) / (2 2) \u003d (-5 + 9) / 4 \u003d 1
Answer: - 3.5; one.
So let's imagine the solution of complete quadratic equations by the scheme in Figure 1.
These formulas can be used to solve any complete quadratic equation. You just need to be careful to the equation was written as a polynomial of standard form
a x 2 + bx + c, otherwise you can make a mistake. For example, in writing the equation x + 3 + 2x 2 = 0, you can mistakenly decide that
a = 1, b = 3 and c = 2. Then
D \u003d 3 2 - 4 1 2 \u003d 1 and then the equation has two roots. And this is not true. (See example 2 solution above).
Therefore, if the equation is not written as a polynomial of the standard form, first the complete quadratic equation must be written as a polynomial of the standard form (the monomial with the largest exponent should be in the first place, that is a x 2 , then with less – bx, and then the free term with.
When solving the above quadratic equation and the quadratic equation with an even coefficient for the second term, other formulas can also be used. Let's get acquainted with these formulas. If in the full quadratic equation with the second term the coefficient is even (b = 2k), then the equation can be solved using the formulas shown in the diagram of Figure 2.
A complete quadratic equation is called reduced if the coefficient at x 2 equals unity and the equation takes the form x 2 + px + q = 0. Such an equation can be given to solve, or is obtained by dividing all the coefficients of the equation by the coefficient a standing at x 2 .
Figure 3 shows a diagram of the solution of the reduced square 
equations. Consider the example of the application of the formulas discussed in this article.
Example. solve the equation
3x 2 + 6x - 6 = 0.
Let's solve this equation using the formulas shown in Figure 1.
D \u003d 6 2 - 4 3 (- 6) \u003d 36 + 72 \u003d 108
√D = √108 = √(36 3) = 6√3
x 1 \u003d (-6 - 6 √ 3) / (2 3) \u003d (6 (-1- √ (3))) / 6 \u003d -1 - √ 3
x 2 \u003d (-6 + 6 √ 3) / (2 3) \u003d (6 (-1 + √ (3))) / 6 \u003d -1 + √ 3
Answer: -1 - √3; –1 + √3
You can see that the coefficient at x in this equation is an even number, that is, b \u003d 6 or b \u003d 2k, whence k \u003d 3. Then let's try to solve the equation using the formulas shown in the figure diagram D 1 \u003d 3 2 - 3 (- 6 ) = 9 + 18 = 27
√(D 1) = √27 = √(9 3) = 3√3
x 1 \u003d (-3 - 3√3) / 3 \u003d (3 (-1 - √ (3))) / 3 \u003d - 1 - √3
x 2 \u003d (-3 + 3√3) / 3 \u003d (3 (-1 + √ (3))) / 3 \u003d - 1 + √3
Answer: -1 - √3; –1 + √3. Noticing that all the coefficients in this quadratic equation are divisible by 3 and dividing, we get the reduced quadratic equation x 2 + 2x - 2 = 0 We solve this equation using the formulas for the reduced quadratic 
equations figure 3.
D 2 \u003d 2 2 - 4 (- 2) \u003d 4 + 8 \u003d 12
√(D 2) = √12 = √(4 3) = 2√3
x 1 \u003d (-2 - 2√3) / 2 \u003d (2 (-1 - √ (3))) / 2 \u003d - 1 - √3
x 2 \u003d (-2 + 2 √ 3) / 2 \u003d (2 (-1 + √ (3))) / 2 \u003d - 1 + √ 3
Answer: -1 - √3; –1 + √3.
As you can see, when solving this equation using different formulas, we got the same answer. Therefore, having well mastered the formulas shown in the diagram of Figure 1, you can always solve any complete quadratic equation.
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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:
;
;
.
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 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 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.
I hope that after studying this article, you will learn how to find the roots of a complete quadratic equation.
With the help of the discriminant, only complete quadratic equations are solved; to solve incomplete quadratic equations, other methods are used, which you will find in the article "Solving incomplete quadratic equations".
What quadratic equations are called complete? This is equations of the form ax 2 + b x + c = 0, where the coefficients a, b and c are not equal to zero. So, to solve the complete quadratic equation, you need to calculate the discriminant D.
D \u003d b 2 - 4ac.
Depending on what value the discriminant has, we will write down the answer.
If the discriminant is a negative number (D< 0),то корней нет.
If the discriminant is zero, then x \u003d (-b) / 2a. When the discriminant is a positive number (D > 0),
then x 1 = (-b - √D)/2a, and x 2 = (-b + √D)/2a.
For example. solve the equation x 2– 4x + 4= 0.
D \u003d 4 2 - 4 4 \u003d 0
x = (- (-4))/2 = 2
Answer: 2.
Solve Equation 2 x 2 + x + 3 = 0.
D \u003d 1 2 - 4 2 3 \u003d - 23
Answer: no roots.
Solve Equation 2 x 2 + 5x - 7 = 0.
D \u003d 5 2 - 4 2 (-7) \u003d 81
x 1 \u003d (-5 - √81) / (2 2) \u003d (-5 - 9) / 4 \u003d - 3.5
x 2 \u003d (-5 + √81) / (2 2) \u003d (-5 + 9) / 4 \u003d 1
Answer: - 3.5; one.
So let's imagine the solution of complete quadratic equations by the scheme in Figure 1.
These formulas can be used to solve any complete quadratic equation. You just need to be careful to the equation was written as a polynomial of standard form
a x 2 + bx + c, otherwise you can make a mistake. For example, in writing the equation x + 3 + 2x 2 = 0, you can mistakenly decide that
a = 1, b = 3 and c = 2. Then
D \u003d 3 2 - 4 1 2 \u003d 1 and then the equation has two roots. And this is not true. (See example 2 solution above).
Therefore, if the equation is not written as a polynomial of the standard form, first the complete quadratic equation must be written as a polynomial of the standard form (the monomial with the largest exponent should be in the first place, that is a x 2 , then with less – bx, and then the free term with.
When solving the above quadratic equation and the quadratic equation with an even coefficient for the second term, other formulas can also be used. Let's get acquainted with these formulas. If in the full quadratic equation with the second term the coefficient is even (b = 2k), then the equation can be solved using the formulas shown in the diagram of Figure 2.
A complete quadratic equation is called reduced if the coefficient at x 2 equals unity and the equation takes the form x 2 + px + q = 0. Such an equation can be given to solve, or is obtained by dividing all the coefficients of the equation by the coefficient a standing at x 2 .
Figure 3 shows a diagram of the solution of the reduced square equations. Consider the example of the application of the formulas discussed in this article.
Example. solve the equation
3x 2 + 6x - 6 = 0.
Let's solve this equation using the formulas shown in Figure 1.
D \u003d 6 2 - 4 3 (- 6) \u003d 36 + 72 \u003d 108
√D = √108 = √(36 3) = 6√3
x 1 \u003d (-6 - 6 √ 3) / (2 3) \u003d (6 (-1- √ (3))) / 6 \u003d -1 - √ 3
x 2 \u003d (-6 + 6 √ 3) / (2 3) \u003d (6 (-1 + √ (3))) / 6 \u003d -1 + √ 3
Answer: -1 - √3; –1 + √3
You can see that the coefficient at x in this equation is an even number, that is, b \u003d 6 or b \u003d 2k, whence k \u003d 3. Then let's try to solve the equation using the formulas shown in the figure diagram D 1 \u003d 3 2 - 3 (- 6 ) = 9 + 18 = 27
√(D 1) = √27 = √(9 3) = 3√3
x 1 \u003d (-3 - 3√3) / 3 \u003d (3 (-1 - √ (3))) / 3 \u003d - 1 - √3
x 2 \u003d (-3 + 3√3) / 3 \u003d (3 (-1 + √ (3))) / 3 \u003d - 1 + √3
Answer: -1 - √3; –1 + √3. Noticing that all the coefficients in this quadratic equation are divisible by 3 and dividing, we get the reduced quadratic equation x 2 + 2x - 2 = 0 We solve this equation using the formulas for the reduced quadratic equations figure 3.
D 2 \u003d 2 2 - 4 (- 2) \u003d 4 + 8 \u003d 12
√(D 2) = √12 = √(4 3) = 2√3
x 1 \u003d (-2 - 2√3) / 2 \u003d (2 (-1 - √ (3))) / 2 \u003d - 1 - √3
x 2 \u003d (-2 + 2 √ 3) / 2 \u003d (2 (-1 + √ (3))) / 2 \u003d - 1 + √ 3
Answer: -1 - √3; –1 + √3.
As you can see, when solving this equation using different formulas, we got the same answer. Therefore, having well mastered the formulas shown in the diagram of Figure 1, you can always solve any complete quadratic equation.
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