Solved quadratic equations with solution. Solving Quadratic Equations Using a Discriminant

Let's consider the problem. The base of the rectangle is 10 cm greater than its height, and its area is 24 cm². Find the height of the rectangle. Let X centimeters is the height of the rectangle, then its base is equal to ( X+10) cm. The area of ​​this rectangle is X(X+ 10) cm². According to the conditions of the problem X(X+ 10) = 24. Opening the brackets and moving the number 24 with the opposite sign into left side equations, we get: X² + 10 X-24 = 0. When solving this problem, an equation was obtained that is called quadratic.

A quadratic equation is an equation of the form

ax ²+ bx+c= 0

Where a, b, c- given numbers, and A≠ 0, and X- unknown.

Odds a, b, c quadratic equation usually called this: a— the first or highest coefficient, b- second coefficient, c- a free member. For example, in our problem, the leading coefficient is 1, the second coefficient is 10, and the free term is -24. Solving many problems in mathematics and physics comes down to solving quadratic equations.

Solving Quadratic Equations

Complete quadratic equations. The first step is to bring the given equation to standard form ax²+ bx+ c = 0. Let's return to our problem, in which the equation can be written as X(X+ 10) = 24 let’s bring it to standard form, open the brackets X² + 10 X- 24 = 0, we solve this equation using the formula for the roots of a general quadratic equation.

The expression under the root sign in this formula is called the discriminant D = b² - 4 ac

If D>0, then the quadratic equation has two different roots, which can be found using the formula for the roots of a quadratic equation.

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

If D<0, то квадратное уравнение не имеет действительных корней, т. е. не имеет решения.

Let's substitute the values ​​into our formula A= 1, b= 10, c= -24.

we get D>0, therefore we get two roots.

Let's consider an example where D=0, under this condition there should be one root.

25x² — 30 x+ 9 = 0

Consider an example where D<0, при этом условии решения не должно быть.

2x² + 3 x+ 4 = 0

The number under the root sign (discriminant) is negative; we write the answer as follows: the equation has no real roots.

Solving incomplete quadratic equations

Quadratic equation ax² + bx+ c= 0 is called incomplete if at least one of the coefficients b or c equal to zero. An incomplete quadratic equation is an equation of one of the following types:

ax² = 0,

ax² + c= 0, c≠ 0,

ax² + bx= 0, b≠ 0.

Let's look at a few examples and solve the equation

Dividing both sides of the equation by 5 gives the equation X² = 0, the answer will have one root X= 0.

Consider an equation of the form

3X² - 27 = 0

Dividing both sides by 3, we get the equation X² - 9 = 0, or it can be written X² = 9, the answer will have two roots X= 3 and X= -3.

Consider an equation of the form

2X² + 7 = 0

Dividing both sides by 2, we get the equation X² = -7/2. This equation has no real roots, since X² ≥ 0 for any real number X.

Consider an equation of the form

3X² + 5 X= 0

Factoring the left side of the equation, we get X(3X+ 5) = 0, the answer will have two roots X= 0, X=-5/3.

The most important thing when solving quadratic equations is to bring the quadratic equation to a standard form, memorize the formula for the roots of a general quadratic equation and not get confused in the signs.

Kopyevskaya rural secondary school

10 Ways to Solve Quadratic Equations

Head: Patrikeeva Galina Anatolyevna,

mathematic teacher

village Kopevo, 2007

1. History of the development of quadratic equations

1.1 Quadratic equations in Ancient Babylon

1.2 How Diophantus composed and solved quadratic equations

1.3 Quadratic equations in India

1.4 Quadratic equations by al-Khorezmi

1.5 Quadratic equations in Europe XIII - XVII centuries

1.6 About Vieta's theorem

2. Methods for solving quadratic equations

Conclusion

Literature

1. History of the development of quadratic equations

1.1 Quadratic equations in Ancient Babylon

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

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

X 2 + X = ¾; X 2 - X = 14,5

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

Despite 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 composed and solved quadratic equations.

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

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

Here, for example, is one of his tasks.

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

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

Hence the equation:

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

100 - x 2 = 96

x 2 - 4 = 0 (1)

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

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

y(20 - y) = 96,

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

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

1.3 Quadratic Equations in India

Problems on quadratic equations are found already in the astronomical treatise “Aryabhattiam”, compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scientist, Brahmagupta (7th century), outlined a 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 A, can also be negative. Brahmagupta's rule is essentially the same as ours.

In ancient India, public competitions in solving difficult problems were common. One of the old Indian books says the following about such competitions: “As the sun eclipses the stars with its brilliance, so learned man will eclipse the glory of another people's assemblies, proposing and solving algebraic problems.” Problems were often presented in poetic form.

This is one of the problems of the famous Indian mathematician of the 12th century. Bhaskars.

Problem 13.

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

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

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

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

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

The equation corresponding to problem 13 is:

( x /8) 2 + 12 = x

Bhaskara writes under the guise:

x 2 - 64x = -768

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

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

(x - 32) 2 = 256,

x - 32 = ± 16,

x 1 = 16, x 2 = 48.

1.4 Quadratic equations in al - Khorezmi

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

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

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

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

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

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

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

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

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

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

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

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

1.5 Quadratic equations in Europe XIII - XVII bb

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

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

x 2 + bx = c,

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

Derivation of the formula for solving a quadratic equation in general view Viet has it, but Viet recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 16th century. In addition to positive ones, negative roots are also taken into account. Only in the 17th century. Thanks to the work of Girard, Descartes, Newton and other scientists, the method of solving quadratic equations takes on a modern form.

1.6 About Vieta's theorem

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

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

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

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

x 1 = a, x 2 = b .

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

2. Methods for solving quadratic equations

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

We continue to study the topic " solving equations" We have already become acquainted with linear equations and are moving on to getting acquainted with quadratic equations.

First, we will look at what a quadratic equation is, how it is written in general form, and give related definitions. After this, we will use examples to examine in detail how incomplete quadratic equations are solved. Let's move on to the solution complete equations, we will obtain the root formula, get acquainted with the discriminant of a quadratic equation, and consider solutions to typical examples. Finally, let's trace the connections between the roots and coefficients.

Page navigation.

What is a quadratic equation? Their types

First you need to clearly understand what a quadratic equation is. Therefore, it is logical to start a conversation about quadratic equations with the definition of a quadratic equation, as well as related definitions. After this, you can consider the main types of quadratic equations: reduced and unreduced, as well as complete and incomplete equations.

Definition and examples of quadratic equations

Definition.

Quadratic equation is an equation of the form a x 2 +b x+c=0, where x is a variable, a, b and c are some numbers, and a is non-zero.

Let's say right away that quadratic equations are often called equations of the second degree. This is due to the fact that the quadratic equation is algebraic equation second degree.

The stated definition allows us to give examples of quadratic equations. So 2 x 2 +6 x+1=0, 0.2 x 2 +2.5 x+0.03=0, etc. These are quadratic equations.

Definition.

Numbers a, b and c are called coefficients of the quadratic equation a·x 2 +b·x+c=0, and coefficient a is called the first, or the highest, or the coefficient of x 2, b is the second coefficient, or the coefficient of x, and c is the free term.

For example, let's take a quadratic equation of the form 5 x 2 −2 x −3=0, here the leading coefficient is 5, the second coefficient is equal to −2, and the free term is equal to −3. Note that when the coefficients b and/or c are negative, as in the example just given, then short form writing a quadratic equation of the form 5 x 2 −2 x−3=0, and not 5 x 2 +(−2) x+(−3)=0.

It is worth noting that when the coefficients a and/or b are equal to 1 or −1, they are usually not explicitly present in the quadratic equation, which is due to the peculiarities of writing such . For example, in the quadratic equation y 2 −y+3=0 the leading coefficient is one, and the coefficient of y is equal to −1.

Reduced and unreduced quadratic equations

Depending on the value of the leading coefficient, reduced and unreduced quadratic equations are distinguished. Let us give the corresponding definitions.

Definition.

A quadratic equation in which the leading coefficient is 1 is called given quadratic equation. Otherwise the quadratic equation is untouched.

According to this definition, quadratic equations x 2 −3·x+1=0, x 2 −x−2/3=0, etc. – given, in each of them the first coefficient is equal to one. A 5 x 2 −x−1=0, etc. - unreduced quadratic equations, their leading coefficients are different from 1.

From any unreduced quadratic equation, by dividing both sides by the leading coefficient, you can go to the reduced one. This action is an equivalent transformation, that is, the reduced quadratic equation obtained in this way has the same roots as the original unreduced quadratic equation, or, like it, has no roots.

Let us look at an example of how the transition from an unreduced quadratic equation to a reduced one is performed.

Example.

From the equation 3 x 2 +12 x−7=0, go to the corresponding reduced quadratic equation.

Solution.

We just need to divide both sides of the original equation by the leading coefficient 3, it is non-zero, so we can perform this action. We have (3 x 2 +12 x−7):3=0:3, which is the same, (3 x 2):3+(12 x):3−7:3=0, and then (3:3) x 2 +(12:3) x−7:3=0, from where . This is how we obtained the reduced quadratic equation, which is equivalent to the original one.

Answer:

Complete and incomplete quadratic equations

The definition of a quadratic equation contains the condition a≠0. This condition is necessary so that the equation a x 2 + b x + c = 0 is quadratic, since when a = 0 it actually becomes a linear equation of the form b x + c = 0.

As for the coefficients b and c, they can be equal to zero, both individually and together. In these cases, the quadratic equation is called incomplete.

Definition.

The quadratic equation a x 2 +b x+c=0 is called incomplete, if at least one of the coefficients b, c is equal to zero.

In its turn

Definition.

Complete quadratic equation is an equation in which all coefficients are different from zero.

Such names were not given by chance. This will become clear from the following discussions.

If the coefficient b is zero, then the quadratic equation takes the form a·x 2 +0·x+c=0, and it is equivalent to the equation a·x 2 +c=0. If c=0, that is, the quadratic equation has the form a·x 2 +b·x+0=0, then it can be rewritten as a·x 2 +b·x=0. And with b=0 and c=0 we get the quadratic equation a·x 2 =0. The resulting equations differ from the complete quadratic equation in that their left-hand sides do not contain either a term with the variable x, or a free term, or both. Hence their name - incomplete quadratic equations.

So the equations x 2 +x+1=0 and −2 x 2 −5 x+0.2=0 are examples of complete quadratic equations, and x 2 =0, −2 x 2 =0, 5 x 2 +3=0 , −x 2 −5 x=0 are incomplete quadratic equations.

Solving incomplete quadratic equations

From the information in the previous paragraph it follows that there is three types of incomplete quadratic equations:

• a·x 2 =0, the coefficients b=0 and c=0 correspond to it;
• a x 2 +c=0 when b=0 ;
• and a·x 2 +b·x=0 when c=0.

Let us examine in order how incomplete quadratic equations of each of these types are solved.

a x 2 =0

Let's start with solving incomplete quadratic equations in which the coefficients b and c are equal to zero, that is, with equations of the form a x 2 =0. The equation a·x 2 =0 is equivalent to the equation x 2 =0, which is obtained from the original by dividing both parts by a non-zero number a. Obviously, the root of the equation x 2 =0 is zero, since 0 2 =0. This equation has no other roots, which is explained by the fact that for any non-zero number p the inequality p 2 >0 holds, which means that for p≠0 the equality p 2 =0 is never achieved.

So, the incomplete quadratic equation a·x 2 =0 has a single root x=0.

As an example, we give the solution to the incomplete quadratic equation −4 x 2 =0. It is equivalent to the equation x 2 =0, its only root is x=0, therefore, the original equation has a single root zero.

A short solution in this case can be written as follows:
−4 x 2 =0 ,
x 2 =0,
x=0 .

a x 2 +c=0

Now let's look at how incomplete quadratic equations are solved in which the coefficient b is zero and c≠0, that is, equations of the form a x 2 +c=0. We know that moving a term from one side of the equation to the other with the opposite sign, as well as dividing both sides of the equation by a non-zero number, gives an equivalent equation. Therefore, we can carry out the following equivalent transformations of the incomplete quadratic equation a x 2 +c=0:

• move c to the right side, which gives the equation a x 2 =−c,
• and divide both sides by a, we get .

The resulting equation allows us to draw conclusions about its roots. Depending on the values ​​of a and c, the value of the expression can be negative (for example, if a=1 and c=2, then ) or positive (for example, if a=−2 and c=6, then ), it is not equal to zero , since by condition c≠0. Let's look at the cases separately.

If , then the equation has no roots. This statement follows from the fact that the square of any number is a non-negative number. It follows from this that when , then for any number p the equality cannot be true.

If , then the situation with the roots of the equation is different. In this case, if we remember about , then the root of the equation immediately becomes obvious; it is the number, since . It’s easy to guess that the number is also the root of the equation, indeed, . This equation has no other roots, which can be shown, for example, by contradiction. Let's do it.

Let us denote the roots of the equation just announced as x 1 and −x 1 . Suppose that the equation has one more root x 2, different from the indicated roots x 1 and −x 1. It is known that substituting its roots into an equation instead of x turns the equation into a correct numerical equality. For x 1 and −x 1 we have , and for x 2 we have . The properties of numerical equalities allow us to perform term-by-term subtraction of correct numerical equalities, so subtracting the corresponding parts of the equalities gives x 1 2 −x 2 2 =0. The properties of operations with numbers allow us to rewrite the resulting equality as (x 1 −x 2)·(x 1 +x 2)=0. We know that the product of two numbers is equal to zero if and only if at least one of them is equal to zero. Therefore, from the resulting equality it follows that x 1 −x 2 =0 and/or x 1 +x 2 =0, which is the same, x 2 =x 1 and/or x 2 =−x 1. So we came to a contradiction, since at the beginning we said that the root of the equation x 2 is different from x 1 and −x 1. This proves that the equation has no roots other than and .

Let us summarize the information in this paragraph. The incomplete quadratic equation a x 2 +c=0 is equivalent to the equation that

• has no roots if ,
• has two roots and , if .

Let's consider examples of solving incomplete quadratic equations of the form a·x 2 +c=0.

Let's start with the quadratic equation 9 x 2 +7=0. After moving the free term to the right side of the equation, it will take the form 9 x 2 =−7. Dividing both sides of the resulting equation by 9, we arrive at . Since the right side has a negative number, this equation has no roots, therefore, the original incomplete quadratic equation 9 x 2 +7 = 0 has no roots.

Let's solve another incomplete quadratic equation −x 2 +9=0. We move the nine to the right side: −x 2 =−9. Now we divide both sides by −1, we get x 2 =9. On the right side is positive number, from which we conclude that or . Then we write down the final answer: the incomplete quadratic equation −x 2 +9=0 has two roots x=3 or x=−3.

a x 2 +b x=0

It remains to deal with the solution of the last type of incomplete quadratic equations for c=0. Incomplete quadratic equations of the form a x 2 + b x = 0 allows you to solve factorization method. Obviously, we can, located on the left side of the equation, for which it is enough to take the common factor x out of brackets. This allows us to move from the original incomplete quadratic equation to an equivalent equation of the form x·(a·x+b)=0. And this equation is equivalent to a set of two equations x=0 and a·x+b=0, the latter of which is linear and has a root x=−b/a.

So, the incomplete quadratic equation a·x 2 +b·x=0 has two roots x=0 and x=−b/a.

To consolidate the material, we will analyze the solution to a specific example.

Example.

Solve the equation.

Solution.

Taking x out of brackets gives the equation . It is equivalent to two equations x=0 and . Solving what we got linear equation: , and performing the division mixed number on common fraction, we find . Therefore, the roots of the original equation are x=0 and .

After gaining the necessary practice, solutions to such equations can be written briefly:

Answer:

x=0 , .

Discriminant, formula for the roots of a quadratic equation

To solve quadratic equations, there is a root formula. Let's write it down formula for the roots of a quadratic equation: , Where D=b 2 −4 a c- so-called discriminant of a quadratic equation. The entry essentially means that .

It is useful to know how the root formula was derived and how it is used in finding the roots of quadratic equations. Let's figure this out.

Derivation of the formula for the roots of a quadratic equation

Let us need to solve the quadratic equation a·x 2 +b·x+c=0. Let's perform some equivalent transformations:

• We can divide both sides of this equation by a non-zero number a, resulting in the following quadratic equation.
• Now let's highlight perfect square on its left side: . After this, the equation will take the form .
• At this stage, it is possible to transfer the last two terms to the right side with the opposite sign, we have .
• And let’s also transform the expression on the right side: .

As a result, we arrive at an equation that is equivalent to the original quadratic equation a·x 2 +b·x+c=0.

We have already solved equations similar in form in the previous paragraphs, when we examined. This allows us to draw the following conclusions regarding the roots of the equation:

• if , then the equation does not have valid solutions;
• if , then the equation has the form , therefore, , from which its only root is visible;
• if , then or , which is the same as or , that is, the equation has two roots.

Thus, the presence or absence of roots of the equation, and therefore the original quadratic equation, depends on the sign of the expression on the right side. In turn, the sign of this expression is determined by the sign of the numerator, since the denominator 4·a 2 is always positive, that is, by the sign of the expression b 2 −4·a·c. This expression b 2 −4 a c was called discriminant of a quadratic equation and designated by the letter D. From here the essence of the discriminant is clear - based on its value and sign, they conclude whether the quadratic equation has real roots, and if so, what is their number - one or two.

Let's return to the equation and rewrite it using the discriminant notation: . And we draw conclusions:

• if D<0 , то это уравнение не имеет действительных корней;
• if D=0, then this equation has a single root;
• finally, if D>0, then the equation has two roots or, which can be rewritten in the form or, and after expanding and bringing the fractions to a common denominator we obtain.

So we derived the formulas for the roots of the quadratic equation, they look like , where the discriminant D is calculated by the formula D=b 2 −4·a·c.

With their help, with a positive discriminant, you can calculate both real roots of a quadratic equation. When the discriminant is equal to zero, both formulas give the same value of the root, corresponding to the only solution quadratic equation. And with a negative discriminant, when trying to use the formula for the roots of a quadratic equation, we are faced with extracting the square root of a negative number, which takes us beyond the scope and school curriculum. With a negative discriminant, the quadratic equation has no real roots, but has a pair complex conjugate roots, which can be found using the same root formulas we obtained.

Algorithm for solving quadratic equations using root formulas

In practice, when solving quadratic equations, you can immediately use the root formula to calculate their values. But this is more related to finding complex roots.

However, in school course Algebra usually deals not with complex, but with real roots of a quadratic equation. In this case, it is advisable, before using the formulas for the roots of a quadratic equation, to first find the discriminant, make sure that it is non-negative (otherwise, we can conclude that the equation does not have real roots), and only then calculate the values ​​of the roots.

The above reasoning allows us to write algorithm for solving a quadratic equation. To solve the quadratic equation a x 2 +b x+c=0, you need to:

• using the discriminant formula D=b 2 −4·a·c, calculate its value;
• conclude that a quadratic equation has no real roots if the discriminant is negative;
• calculate the only root of the equation using the formula if D=0;
• find two real roots of a quadratic equation using the root formula if the discriminant is positive.

Here we just note that if the discriminant is equal to zero, you can also use the formula; it will give the same value as .

You can move on to examples of using the algorithm for solving quadratic equations.

Examples of solving quadratic equations

Let's consider solutions to three quadratic equations with a positive, negative and zero discriminant. Having dealt with their solution, by analogy it will be possible to solve any other quadratic equation. Let's begin.

Example.

Find the roots of the equation x 2 +2·x−6=0.

Solution.

In this case, we have the following coefficients of the quadratic equation: a=1, b=2 and c=−6. According to the algorithm, you first need to calculate the discriminant; to do this, we substitute the indicated a, b and c into the discriminant formula, we have D=b 2 −4·a·c=2 2 −4·1·(−6)=4+24=28. Since 28>0, that is, the discriminant is greater than zero, the quadratic equation has two real roots. Let's find them using the root formula, we get , here you can simplify the resulting expressions by doing moving the multiplier beyond the root sign followed by reduction of the fraction:

Answer:

Let's move on to the next typical example.

Example.

Solve the quadratic equation −4 x 2 +28 x−49=0 .

Solution.

We start by finding the discriminant: D=28 2 −4·(−4)·(−49)=784−784=0. Therefore, this quadratic equation has a single root, which we find as , that is,

Answer:

x=3.5.

It remains to consider solving quadratic equations with a negative discriminant.

Example.

Solve the equation 5·y 2 +6·y+2=0.

Solution.

Here are the coefficients of the quadratic equation: a=5, b=6 and c=2. We substitute these values ​​into the discriminant formula, we have D=b 2 −4·a·c=6 2 −4·5·2=36−40=−4. The discriminant is negative, therefore, this quadratic equation has no real roots.

If you need to indicate complex roots, then we apply the well-known formula for the roots of a quadratic equation, and perform operations with complex numbers:

Answer:

there are no real roots, complex roots are: .

Let us note once again that if the discriminant of a quadratic equation is negative, then in school they usually immediately write down an answer in which they indicate that there are no real roots, and complex roots are not found.

Root formula for even second coefficients

The formula for the roots of a quadratic equation, where D=b 2 −4·a·c allows you to obtain a formula of a more compact form, allowing you to solve quadratic equations with an even coefficient for x (or simply with a coefficient having the form 2·n, for example, or 14· ln5=2·7·ln5 ). Let's get her out.

Let's say we need to solve a quadratic equation of the form a x 2 +2 n x+c=0. Let's find its roots using the formula we know. To do this, we calculate the discriminant D=(2 n) 2 −4 a c=4 n 2 −4 a c=4 (n 2 −a c), and then we use the root formula:

Let us denote the expression n 2 −a c as D 1 (sometimes it is denoted D "). Then the formula for the roots of the quadratic equation under consideration with the second coefficient 2 n will take the form , where D 1 =n 2 −a·c.

It is easy to see that D=4·D 1, or D 1 =D/4. In other words, D 1 is the fourth part of the discriminant. It is clear that the sign of D 1 is the same as the sign of D . That is, the sign D 1 is also an indicator of the presence or absence of roots of a quadratic equation.

So, to solve a quadratic equation with a second coefficient 2·n, you need

• Calculate D 1 =n 2 −a·c ;
• If D 1<0 , то сделать вывод, что действительных корней нет;
• If D 1 =0, then calculate the only root of the equation using the formula;
• If D 1 >0, then find two real roots using the formula.

Let's consider solving the example using the root formula obtained in this paragraph.

Example.

Solve the quadratic equation 5 x 2 −6 x −32=0 .

Solution.

The second coefficient of this equation can be represented as 2·(−3) . That is, you can rewrite the original quadratic equation in the form 5 x 2 +2 (−3) x−32=0, here a=5, n=−3 and c=−32, and calculate the fourth part of the discriminant: D 1 =n 2 −a·c=(−3) 2 −5·(−32)=9+160=169. Since its value is positive, the equation has two real roots. Let's find them using the appropriate root formula:

Note that it was possible to use the usual formula for the roots of a quadratic equation, but in this case more computational work would have to be performed.

Answer:

Simplifying the form of quadratic equations

Sometimes, before starting to calculate the roots of a quadratic equation using formulas, it doesn’t hurt to ask the question: “Is it possible to simplify the form of this equation?” Agree that in terms of calculations it will be easier to solve the quadratic equation 11 x 2 −4 x−6=0 than 1100 x 2 −400 x−600=0.

Typically, simplifying the form of a quadratic equation is achieved by multiplying or dividing both sides by a certain number. For example, in the previous paragraph it was possible to simplify the equation 1100 x 2 −400 x −600=0 by dividing both sides by 100.

A similar transformation is carried out with quadratic equations, the coefficients of which are not . In this case, we usually divide both sides of the equation by absolute values its coefficients. For example, let's take the quadratic equation 12 x 2 −42 x+48=0. absolute values ​​of its coefficients: GCD(12, 42, 48)= GCD(GCD(12, 42), 48)= GCD(6, 48)=6. Dividing both sides of the original quadratic equation by 6, we arrive at the equivalent quadratic equation 2 x 2 −7 x+8=0.

And multiplying both sides of a quadratic equation is usually done to get rid of fractional coefficients. In this case, multiplication is carried out by the denominators of its coefficients. For example, if both sides of the quadratic equation are multiplied by LCM(6, 3, 1)=6, then it will take the simpler form x 2 +4·x−18=0.

In conclusion of this point, we note that they almost always get rid of the minus at the highest coefficient of a quadratic equation by changing the signs of all terms, which corresponds to multiplying (or dividing) both sides by −1. For example, usually one moves from the quadratic equation −2 x 2 −3 x+7=0 to the solution 2 x 2 +3 x−7=0 .

Relationship between roots and coefficients of a quadratic equation

The formula for the roots of a quadratic equation expresses the roots of the equation through its coefficients. Based on the root formula, you can obtain other relationships between roots and coefficients.

The most well-known and applicable formulas from Vieta’s theorem are of the form and . In particular, for the given quadratic equation, the sum of the roots is equal to the second coefficient with the opposite sign, and the product of the roots is equal to the free term. For example, by looking at the form of the quadratic equation 3 x 2 −7 x + 22 = 0, we can immediately say that the sum of its roots is equal to 7/3, and the product of the roots is equal to 22/3.

Using the already written formulas, you can obtain a number of other connections between the roots and coefficients of the quadratic equation. For example, you can express the sum of the squares of the roots of a quadratic equation through its coefficients: .

Bibliography.

• Algebra: textbook for 8th grade. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
• Mordkovich A. G. Algebra. 8th grade. At 2 p.m. Part 1. Textbook for students educational institutions/ A. G. Mordkovich. - 11th ed., erased. - M.: Mnemosyne, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.

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 a discriminant
- using Vieta's theorem (if possible).

Moreover, the answer is displayed as exact, not approximate.
For example, for the equation \(81x^2-16x-1=0\) the answer is displayed in the following form:

$$ x_1 = \frac(8+\sqrt(145))(81), \quad x_2 = \frac(8-\sqrt(145))(81) $$ and not like this: \(x_1 = 0.247; \quad x_2 = -0.05\)

This program may be useful for high school students secondary schools 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 detailed solutions.

This way you can conduct your own training and/or training of yours. younger brothers or sisters, while the level of education in the field of problems being solved increases.

If you are not familiar with the rules for entering a quadratic polynomial, we recommend that you familiarize yourself with them.

Rules for entering a quadratic 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 whole or fractional numbers.
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 can be separated from the whole part by either a period 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 numerical fraction The numerator is separated from the denominator by a division sign: /
Whole part 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 parentheses. 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)

=0

Decide

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A little theory.

Quadratic equation and its roots. Incomplete quadratic equations

Each of the equations
\(-x^2+6x+1.4=0, \quad 8x^2-7x=0, \quad x^2-\frac(4)(9)=0 \)
looks like
\(ax^2+bx+c=0, \)
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 is called an equation of the form ax 2 +bx+c=0, 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 free term.

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 of x 2 is equal to 1 is called given quadratic equation. For example, the quadratic equations given are the equations
\(x^2-11x+30=0, \quad x^2-6x=0, \quad x^2-8=0 \)

If in a 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. Thus, 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.

There are three types of incomplete quadratic equations:
1) ax 2 +c=0, where \(c \neq 0 \);
2) ax 2 +bx=0, where \(b \neq 0 \);
3) ax 2 =0.

Let's consider solving equations of each of these types.

To solve an incomplete quadratic equation of the form ax 2 +c=0 for \(c \neq 0 \), move its free term to the right side and divide both sides of the equation 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 with \(b \neq 0 \) factor 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\( \begin (array)(l) x=0 \\ x=-\frac(b)(a) \end(array) \right. \)

This means that 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 =0 is equivalent to the equation x 2 =0 and therefore has a single root 0.

Formula for the roots of a quadratic equation

Let us now consider how to solve quadratic equations in which both the coefficients of the unknowns and the free term are nonzero.

Let us solve the quadratic equation in general form and as a result we obtain the formula for the roots. This formula can then be used to solve any quadratic equation.

Solve the quadratic equation ax 2 +bx+c=0

Dividing both sides by a, we obtain the equivalent reduced quadratic equation
\(x^2+\frac(b)(a)x +\frac(c)(a)=0 \)

Let's transform this equation by selecting the square of the binomial:
\(x^2+2x \cdot \frac(b)(2a)+\left(\frac(b)(2a)\right)^2- \left(\frac(b)(2a)\right)^ 2 + \frac(c)(a) = 0 \Rightarrow \)

\(x^2+2x \cdot \frac(b)(2a)+\left(\frac(b)(2a)\right)^2 = \left(\frac(b)(2a)\right)^ 2 - \frac(c)(a) \Rightarrow \) \(\left(x+\frac(b)(2a)\right)^2 = \frac(b^2)(4a^2) - \frac( c)(a) \Rightarrow \left(x+\frac(b)(2a)\right)^2 = \frac(b^2-4ac)(4a^2) \Rightarrow \) \(x+\frac(b )(2a) = \pm \sqrt( \frac(b^2-4ac)(4a^2) ) \Rightarrow x = -\frac(b)(2a) + \frac( \pm \sqrt(b^2 -4ac) )(2a) \Rightarrow \) \(x = \frac( -b \pm \sqrt(b^2-4ac) )(2a) \)

The radical expression is called discriminant of a quadratic equation ax 2 +bx+c=0 (“discriminant” in Latin - discriminator). It is designated by the letter D, i.e.
\(D = b^2-4ac\)

Now, using the discriminant notation, 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, a quadratic equation can have two roots (for D > 0), one root (for D = 0) or have 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 above 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\( \begin(array)(l) x_1+x_2=-p \\ x_1 \cdot x_2=q \end(array) \right. \)

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

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 solutions, note that all quadratic equations can be divided into three classes:

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

This is an important difference between quadratic equations and linear ones, 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.

You need to know this formula 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 believe. 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. 5x 2 + 3x + 7 = 0;
3. x 2 − 6x + 9 = 0.

Let's write out 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 a similar way:
a = 5; b = 3; c = 7;
D = 3 2 − 4 5 7 = 9 − 140 = −131.

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

The discriminant is zero - the root will be one.

Please note that coefficients have been written down for each equation. Yes, it’s long, yes, it’s tedious, but you won’t mix up the odds and make stupid mistakes. Choose for yourself: speed or quality.

By the way, if you get the hang of it, after a while you won’t need to write down 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 that much.

Roots of a quadratic equation

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

Basic formula for the roots of a quadratic equation

When D = 0, you can use any of these formulas - you will 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. x 2 + 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 can count, there will be no problems. Most often, errors occur when substituting negative coefficients into the formula. Here again, the technique described above will help: look at the formula literally, write down each step - and very soon you will get rid of errors.

Incomplete quadratic equations

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

1. x 2 + 9x = 0;
2. x 2 − 16 = 0.

It is easy to notice that these equations are missing one of the terms. Such quadratic equations are even easier to solve than standard ones: they don’t even require calculating 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 = c = 0. In this case, the equation takes the form ax 2 = 0. Obviously, such an equation has a single root: x = 0.

Let's consider the remaining cases. Let b = 0, then we obtain an incomplete quadratic equation of the form ax 2 + c = 0. Let us transform it a little:

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

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

As you can see, a 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 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 it is negative, there will be no roots at all.

Now let's look at 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 factor the polynomial:

Taking the common factor out of brackets

The product is zero when at least one of the factors is zero. This is where the roots come from. In conclusion, let’s look at a few of these equations:

Task. Solve quadratic equations:

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

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

5x 2 + 30 = 0 ⇒ 5x 2 = −30 ⇒ x 2 = −6. There are no roots, because a 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 = −1.5.