Solution of complete quadratic equations using the discriminant. Derivation of the formula for the roots of a quadratic equation. How an incomplete quadratic equation is solved

", That is, equations of the first degree. In this lesson we will analyze what is called a quadratic equation and how to solve it.

What is called a quadratic equation

Important!

The degree of the equation is determined by the largest degree in which the unknown stands.

If the maximum power in which the unknown stands is "2", then you have a quadratic equation.

Examples of quadratic equations

• 5x 2 - 14x + 17 = 0
• −x 2 + x +

  1

  3

  = 0
• x 2 + 0.25x = 0
• x 2 - 8 = 0

Important! The general view of the quadratic equation looks like this:

A x 2 + b x + c = 0

"A", "b" and "c" are given numbers.

• "A" - the first or most significant coefficient;
• “B” is the second coefficient;
• "C" is a free member.

To find "a", "b" and "c" you need to compare your equation with the general form of the quadratic equation "ax 2 + bx + c = 0".

Let's practice defining the coefficients "a", "b" and "c" in quadratic equations.

5x 2 - 14x + 17 = 0 −7x 2 - 13x + 8 = 0 −x 2 + x +

The equation	Odds
a = 5 b = −14 c = 17
a = −7 b = −13 c = 8

1

3

= 0

• a = −1
• b = 1
• c =

  1

  3

x 2 + 0.25x = 0

• a = 1
• b = 0.25
• c = 0

x 2 - 8 = 0

• a = 1
• b = 0
• c = −8

How to solve quadratic equations

Unlike linear equations for solving quadratic equations special formula for finding roots.

Remember!

To solve a quadratic equation you need:

• bring the quadratic equation to the general form "ax 2 + bx + c = 0". That is, only "0" should remain on the right side;
• use formula for roots:

Let's take an example of how to use a formula to find the roots of a quadratic equation. Let's solve the quadratic equation.

X 2 - 3x - 4 = 0

The equation "x 2 - 3x - 4 = 0" has already been reduced to the general form "ax 2 + bx + c = 0" and does not require additional simplifications. To solve it, we just need to apply the formula for finding the roots of a quadratic equation.

Let's define the coefficients "a", "b" and "c" for this equation.

x 1; 2 =
x 1; 2 =
x 1; 2 =
x 1; 2 =

With its help, any quadratic equation is solved.

In the formula "x 1; 2 =" the radical expression is often replaced
"B 2 - 4ac" with the letter "D" and is called the discriminant. The concept of a discriminant is discussed in more detail in the lesson "What is a discriminant".

Consider another example of a quadratic equation.

x 2 + 9 + x = 7x

It is rather difficult to determine the coefficients "a", "b" and "c" in this form. Let's first bring the equation to the general form "ax 2 + bx + c = 0".

X 2 + 9 + x = 7x
x 2 + 9 + x - 7x = 0
x 2 + 9 - 6x = 0
x 2 - 6x + 9 = 0

Now you can use the root formula.

X 1; 2 =
x 1; 2 =
x 1; 2 =
x 1; 2 =
x =

6

2

x = 3
Answer: x = 3

There are times when there are no roots in quadratic equations. This situation occurs when a negative number is found under the root in the formula.

This topic may seem difficult at first due to the many not very simple formulas... Not only do the quadratic equations themselves have long records, but also the roots are found through the discriminant. There are three new formulas in total. It's not easy to remember. This succeeds only after frequent decision such equations. Then all the formulas will be remembered by themselves.

General view of the quadratic equation

Here, their explicit recording is proposed, when the highest degree is recorded first, and then in descending order. There are often situations when the terms are out of order. Then it is better to rewrite the equation in decreasing order of the degree of the variable.

Let us introduce the notation. They are presented in the table below.

If we accept these designations, all quadratic equations are reduced to the following record.

Moreover, the coefficient a ≠ 0. Let this formula be denoted by number one.

When the equation is given, it is not clear how many roots there will be in the answer. Because one of three options is always possible:

• there will be two roots in the solution;
• the answer is one number;
• the equation will have no roots at all.

And until the decision is not brought to the end, it is difficult to understand which of the options will fall out in a particular case.

Types of records of quadratic equations

Tasks may contain their different records. They will not always look like a general quadratic equation. Sometimes it will lack some terms. What was written above is a complete equation. If you remove the second or third term in it, you get something different. These records are also called quadratic equations, only incomplete.

Moreover, only the terms in which the coefficients "b" and "c" can disappear. The number "a" cannot be zero under any circumstances. Because in this case, the formula turns into a linear equation. Formulas for an incomplete form of equations will be as follows:

So, there are only two types, besides the complete ones, there are also incomplete quadratic equations. Let the first formula be number two and the second number three.

Discriminant and dependence of the number of roots on its value

You need to know this number in order to calculate the roots of the equation. It can always be calculated, no matter what the formula for the quadratic equation. In order to calculate the discriminant, you need to use the equality written below, which will have the number four.

After substituting the values ​​of the coefficients into this formula, you can get numbers with different signs... If the answer is yes, then the answer to the equation will be two different roots. With a negative number, the roots of the quadratic equation will be absent. If it is equal to zero, the answer will be one.

How is a complete quadratic equation solved?

In fact, consideration of this issue has already begun. Because first you need to find the discriminant. After it has been found that there are roots of the quadratic equation, and their number is known, you need to use the formulas for the variables. If there are two roots, then you need to apply the following formula.

Since it contains the “±” sign, there will be two values. The square root expression is the discriminant. Therefore, the formula can be rewritten in a different way.

Formula number five. The same record shows that if the discriminant is zero, then both roots will take the same values.

If the solution of quadratic equations has not yet been worked out, then it is better to write down the values ​​of all coefficients before applying the discriminant and variable formulas. Later this moment will not cause difficulties. But at the very beginning, there is confusion.

How is an incomplete quadratic equation solved?

Everything is much simpler here. There is even no need for additional formulas. And you will not need those that have already been recorded for the discriminant and the unknown.

First, consider the incomplete equation number two. In this equality, it is supposed to take the unknown quantity out of the bracket and solve the linear equation, which remains in the brackets. The answer will have two roots. The first one is necessarily equal to zero, because there is a factor consisting of the variable itself. The second is obtained when solving a linear equation.

Incomplete equation number three is solved by transferring the number from the left side of the equation to the right. Then you need to divide by the factor in front of the unknown. All that remains is to extract the square root and remember to write it down twice with opposite signs.

The following are some actions to help you learn how to solve all kinds of equalities that turn into quadratic equations. They will help the student to avoid careless mistakes. These flaws are the reason bad grades while studying the extensive topic "Quadratic equations (grade 8)". Subsequently, these actions will not need to be constantly performed. Because a stable skill will appear.

• First, you need to write the equation in standard form. That is, first the term with the highest degree of the variable, and then - without the degree and the last - just a number.

• If a minus appears in front of the coefficient "a", then it can complicate the work for a beginner to study quadratic equations. It is better to get rid of it. For this purpose, all equality must be multiplied by "-1". This means that all the terms will change their sign to the opposite.

• In the same way, it is recommended to get rid of fractions. Simply multiply the equation by the appropriate factor to cancel out the denominators.

Examples of

It is required to solve the following quadratic equations:

x 2 - 7x = 0;

15 - 2x - x 2 = 0;

x 2 + 8 + 3x = 0;

12x + x 2 + 36 = 0;

(x + 1) 2 + x + 1 = (x + 1) (x + 2).

The first equation: x 2 - 7x = 0. It is incomplete, therefore it is solved as described for the formula number two.

After leaving the brackets, it turns out: x (x - 7) = 0.

The first root takes on the value: x 1 = 0. The second will be found from linear equation: x - 7 = 0. It is easy to see that x 2 = 7.

Second equation: 5x 2 + 30 = 0. Again incomplete. Only it is solved as described for the third formula.

After transferring 30 to the right side of the equality: 5x 2 = 30. Now you need to divide by 5. It turns out: x 2 = 6. The answers will be the numbers: x 1 = √6, x 2 = - √6.

The third equation: 15 - 2x - x 2 = 0. Hereinafter, the solution of quadratic equations will begin with rewriting them into standard view: - x 2 - 2x + 15 = 0. Now it's time to use the second useful advice and multiply everything by minus one. It turns out x 2 + 2x - 15 = 0. According to the fourth formula, you need to calculate the discriminant: D = 2 2 - 4 * (- 15) = 4 + 60 = 64. It is a positive number. From what was said above, it turns out that the equation has two roots. They need to be calculated using the fifth formula. It turns out that x = (-2 ± √64) / 2 = (-2 ± 8) / 2. Then x 1 = 3, x 2 = - 5.

The fourth equation x 2 + 8 + 3x = 0 is transformed into this: x 2 + 3x + 8 = 0. Its discriminant is equal to this value: -23. Since this number is negative, the answer to this task will be the following entry: "There are no roots."

The fifth equation 12x + x 2 + 36 = 0 should be rewritten as follows: x 2 + 12x + 36 = 0. After applying the formula for the discriminant, the number zero is obtained. This means that it will have one root, namely: x = -12 / (2 * 1) = -6.

The sixth equation (x + 1) 2 + x + 1 = (x + 1) (x + 2) requires transformations, which consist in the fact that you need to bring similar terms, before opening the brackets. In place of the first, there will be such an expression: x 2 + 2x + 1. After the equality, this record will appear: x 2 + 3x + 2. After such terms are counted, the equation will take the form: x 2 - x = 0. It turned into incomplete ... Something similar to it has already been considered a little higher. The roots of this will be the numbers 0 and 1.

An incomplete quadratic equation differs from classical (complete) equations in that its factors or intercept are equal to zero. The graph of such functions are parabolas. Depending on their general appearance, they are divided into 3 groups. The principles of solving for all types of equations are the same.

There is nothing difficult in determining the type of an incomplete polynomial. It is best to consider the main differences with illustrative examples:

1. If b = 0, then the equation is ax 2 + c = 0.
2. If c = 0, then the expression ax 2 + bx = 0 should be solved.
3. If b = 0 and c = 0, then the polynomial becomes an equality of the type ax 2 = 0.

The latter case is more of a theoretical possibility and never occurs in knowledge testing tasks, since the only valid value of the variable x in the expression is zero. In the future, methods and examples of solving incomplete quadratic equations of 1) and 2) types will be considered.

General algorithm for finding variables and examples with a solution

Regardless of the type of equation, the solution algorithm boils down to the following steps:

1. Bring the expression to a form that is convenient for finding roots.
2. Perform calculations.
3. Record your answer.

The easiest way to solve incomplete equations is by factoring the left side and leaving zero on the right. Thus, the formula for an incomplete quadratic equation for finding the roots is reduced to calculating the value of x for each of the factors.

You can only learn how to solve it in practice, so consider specific example finding the roots of an incomplete equation:

As you can see, in this case b = 0. Factor the left side and get the expression:

4 (x - 0.5) ⋅ (x + 0.5) = 0.

Obviously, the product is zero when at least one of the factors is zero. The values ​​of the variable x1 = 0.5 and (or) x2 = -0.5 meet these requirements.

In order to easily and quickly cope with the task of decomposition square trinomial by factors, you should remember the following formula:

If there is no free term in the expression, the task is greatly simplified. It will be enough just to find and take out of the brackets common denominator... For clarity, consider an example of how to solve incomplete quadratic equations of the form ax2 + bx = 0.

Let's take the variable x out of the parentheses and get the following expression:

x ⋅ (x + 3) = 0.

Guided by logic, we come to the conclusion that x1 = 0, and x2 = -3.

Traditional solution and incomplete quadratic equations

What will happen if you apply the discriminant formula and try to find the roots of the polynomial, with the coefficients equal to zero? Let's take an example from the collection typical assignments for the exam in mathematics in 2017, we will solve it using standard formulas and the factorization method.

7x 2 - 3x = 0.

Let's calculate the value of the discriminant: D = (-3) 2 - 4 ⋅ (-7) ⋅ 0 = 9. It turns out that the polynomial has two roots:

Now, let's solve the equation by factoring and compare the results.

X ⋅ (7x + 3) = 0,

2) 7x + 3 = 0,
7x = -3,
x = -.

As you can see, both methods give the same result, but solving the equation by the second method turned out to be much easier and faster.

Vieta's theorem

But what to do with the beloved Vieta's theorem? Can I apply this method with an incomplete trinomial? Let's try to understand the aspects of reducing incomplete equations to classic look ax2 + bx + c = 0.

In fact, it is possible to apply Vieta's theorem in this case. It is only necessary to bring the expression to a general form, replacing the missing members with zero.

For example, with b = 0 and a = 1, in order to eliminate the likelihood of confusion, the task should be written in the form: ax2 + 0 + c = 0. Then the ratio of the sum and product of the roots and factors of the polynomial can be expressed as follows:

Theoretical calculations help to get acquainted with the essence of the issue, and always require practicing the skill in solving specific problems. Let's turn again to the reference book of typical tasks for the exam and find a suitable example:

Let us write the expression in a form convenient for the application of Vieta's theorem:

x 2 + 0 - 16 = 0.

The next step is to create a system of conditions:

Obviously, the roots of a square polynomial will be x 1 = 4 and x 2 = -4.

Now, let's practice bringing the equation to a general form. Take the following example: 1/4 × x 2 - 1 = 0

In order to apply Vieta's theorem to an expression, it is necessary to get rid of the fraction. Multiply the left and right sides by 4, and look at the result: x2– 4 = 0. The resulting equality is ready to be solved by Vieta's theorem, but it is much easier and faster to get the answer simply by transferring c = 4 to the right side of the equation: x2 = 4.

Summing up, it should be said that the best way solving incomplete equations is factorization, is the simplest and quick method... If you encounter difficulties in the process of finding roots, you can turn to the traditional method of finding roots through the discriminant.

Kopyevskaya rural secondary school

10 ways to solve quadratic equations

Head: Galina Anatolyevna Patrikeyeva,

mathematic teacher

the village of Kopyevo, 2007

1. The 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 from 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. The 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 antiquity was caused by the need to solve problems associated with finding areas land plots and with earthworks military character, as well as with the development of astronomy and mathematics itself. They were able to solve quadratic equations around 2000 BC. e. Babylonians.

Applying the 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, coincides essentially with the modern one, but it is not known how the Babylonians got to this rule. Almost all the cuneiform texts found so far only give problems with solutions set out in the form of recipes, without instructions as to how they were found.

In spite of high level the development of algebra in Babylon, in cuneiform texts there is no concept of a negative number and general methods solutions of quadratic equations.

1.2 How Diophantus compiled and solved quadratic equations.

In the "Arithmetic" of Diophantus there is no systematic presentation of algebra, but it contains a systematized series of problems, accompanied by explanations and solved by drawing up equations of different degrees.

When drawing up equations, Diophantus skillfully chooses 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 the product is 96"

Diophantus argues as follows: from the condition of the problem it follows that the sought numbers are not equal, since if they were equal, then their product would equal not 96, but 100. Thus, one of them will be more than half of their sum, i.e. ... 10 + x, the other is less, i.e. 10 - x... 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 12 , other 8 ... Solution x = -2 for Diophantus does not exist, since Greek mathematics knew only positive numbers.

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

y (20 - y) = 96,

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

It is clear that, choosing the half-difference of the sought 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 for quadratic equations are already encountered in the astronomical tract "Aryabhattiam", compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scholar, Brahmagupta (VII century), outlined general rule solutions of quadratic equations reduced to a single canonical form:

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

In equation (1), the coefficients, except a, can be negative. The Brahmagupta rule is essentially the same as ours.

V Ancient India public competition in solving difficult problems was widespread. One of the ancient Indian books says the following about such competitions: “As the sun eclipses the stars with its brilliance, so a learned man will eclipse the glory of another in popular assemblies proposing and solving algebraic problems ”. The tasks were often clothed in poetic form.

Here is one of the tasks of the famous Indian mathematician of the XII century. Bhaskaras.

Problem 13.

"Frisky flock of monkeys And twelve over the vines ...

After eating the power, having fun. They began to jump, hanging ...

There are eighth part of them in a square How many monkeys were there,

I was amusing myself in the clearing. You tell me, in this pack? "

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

Equation corresponding to problem 13:

( x /8) 2 + 12 = x

Bhaskara writes under the guise:

x 2 - 64x = -768

and, to complete the left side of this equation to a 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 for al - Khorezmi

In the algebraic treatise 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 a number", i.e. ax 2 = c.

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

4) "Squares and numbers are equal to roots", ie ax 2 + c = b X.

5) "Squares and roots are equal to a number", 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 use negative numbers, the terms of each of these equations are addends, not subtracted. In this case, equations that do not have positive solutions are certainly not taken into account. The author outlines the ways of solving these equations, using the techniques of al - jabr and al - muqabal. His decision, of course, does not completely coincide with ours. Apart from 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 up to 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, using particular numerical examples, sets out the rules for solving, and then geometric proofs.

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

The author's solution reads something like this: divide the number of roots in half, you get 5, multiply 5 by itself, subtract 21 from the product, there will be 4. Extract 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 gives 7, this is also a root.

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

1.5 Quadratic equations in Europe XIII - Xvii cc

Formulas for solving quadratic equations on the model of al - Khorezmi in Europe were first presented in the "Book of 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, differs in both completeness and clarity of presentation. The author independently developed some new algebraic examples of solving problems and was the first in Europe to approach the introduction of negative numbers. His book contributed to the dissemination of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many problems from the "Book of the Abacus" were transferred to 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 = s,

with all possible combinations of odds 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 is in Viet, however Viet recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 16th century. Consider, in addition to positive, and negative roots. Only in the 17th century. Thanks to the work of Girard, Descartes, Newton and other scientists, the method for solving quadratic equations takes on a modern form.

1.6 About Vieta's theorem

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

To understand Vieta, one should remember that A, like any vowel, meant for him the unknown (our X), the vowels V, D- coefficients for the unknown. In the language of modern algebra, Vieta's above formulation 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 the coefficients of the equations general formulas written with symbols, Viet established uniformity in the methods of solving equations. However, Vieta's symbolism is still far from modern look... He did not recognize negative numbers and therefore, when solving equations, he considered only cases when all roots are positive.

2. Methods for solving quadratic equations

Quadratic equations are the foundation on which the magnificent 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.

Quadratic equations are studied in grade 8, so there is nothing difficult here. The ability to solve them is absolutely 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 for solving, we note that all quadratic equations can be conditionally divided into three classes:

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

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

Discriminant

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

You need to know this formula by heart. Where it comes from - it doesn't matter 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 believe. Take a look at the examples - and you yourself will understand everything:

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 us write down 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 remains:
a = 1; b = −6; c = 9;
D = (−6) 2 - 4 1 9 = 36 - 36 = 0.

The discriminant is zero - there will be one root.

Note that coefficients have been written for each equation. Yes, it’s long, yes, it’s boring - but you won’t mix up the coefficients 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 equations are solved - in general, not that much.

Quadratic Roots

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

Basic formula for the roots of a quadratic equation

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

1. x 2 - 2x - 3 = 0;
2. 15 - 2x - x 2 = 0;
3. 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 has two roots again. 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 substituting negative coefficients in the formula. Here, again, the technique described above will help: look at the formula literally, describe each step - and very soon you will get rid of mistakes.

Incomplete quadratic equations

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

1. x 2 + 9x = 0;
2. x 2 - 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. coefficient at variable x or 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 rest of the cases. Let b = 0, then we get an incomplete quadratic equation of the form ax 2 + c = 0. Let's transform it a little:

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

1. If the inequality (−c / a) ≥ 0 holds in an incomplete quadratic equation of the form ax 2 + c = 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 - in incomplete quadratic equations there are no complicated calculations at all. 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 stands 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 factor out the polynomial:

Bracketing a common factor

The product is equal to zero when at least one of the factors is equal to zero. From here are the roots. In conclusion, we will analyze several such 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, tk. 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.