How to solve the equations correctly. Solution of square equations. When the roots are infinitely a lot

After we studied the concept of equalities, namely, one of their species - numerical equality, one can go to another important form - equations. As part of this material, we explain what equation and its root, we formulate the main definitions and give various examples of equations and finding their roots.

The concept of equation

Usually the concept of the equation is studied at the very beginning school Course Algebras. Then it is determined like this:

Definition 1.

Equation It is called equality with an unknown number to find.

It is customary to designate unknown to small Latin letters, for example, t, r, m, etc., but the X, Y, Z is most often used. In other words, the equation determines the form of its record, that is, equality will be equation only when it is shown to a certain type - it should be the letter, the value that needs to be found.

Let us give a few examples of the simplest equations. These may be equalities of the form x \u003d 5, y \u003d 6, etc., as well as those that include arithmetic action, for example, x + 7 \u003d 38, z - 4 \u003d 2, 8 · t \u003d 4, 6: X \u003d 3.

After the concept of brackets is studied, the concept of equations with brackets appears. These include 7 · (x - 1) \u003d 19, x + 6 · (x + 6 · (x - 8)) \u003d 3, etc. The letter that needs to be found may occur more than once, but a few, like For example, in equation X + 2 + 4 · X - 2 - x \u003d 10. Also unknown can be located not only to the left, but also to the right or in both parts at the same time, for example, x · (8 + 1) - 7 \u003d 8, 3 - 3 \u003d z + 3 or 8 · x - 9 \u003d 2 · (x + 17).

Next, after students get acquainted with the concept of integers, valid, rational, natural numbers, as well as logarithms, roots and degrees, new equations appear, including all these objects. Examples of such expressions, we devoted a separate article.

In the program for grade 7, for the first time, the concept of variables arises. These are the letters that can take different values \u200b\u200b(see the article on numerical, alphabone expressions and expressions with variables). Based on this concept, we can give a new definition of equation:

Definition 2.

The equation - It is equality that includes a variable whose value must be calculated.

That is, for example, the expression X + 3 \u003d 6 · X + 7 is an equation with a variable x, a 3 · y - 1 + y \u003d 0 - equation from the variable y.

In one equation, there may be not one variable, but two or more. They are called respectively equations with two, three variables, etc. We write down the definition:

Definition 3.

Equations with two (three, four or more) variables are called equations that include the appropriate number of unknown.

For example, the equality of the form 3, 7 · x + 0, 6 \u003d 1 is an equation with one variable x, and x - z \u003d 5 is an equation with two variables x and z. An example of an equation with three variables may be an expression x 2 + (y - 6) 2 + (Z + 0, 6) 2 \u003d 26.

Root of the equation

When we talk about the equation, immediately arises the need to decide on the concept of its root. Let's try to explain what it means.

Example 1.

We are given a certain equation that includes one variable. If we substitute instead of an unknown letter, the equation will become a numerical equality - faithful or incorrect. So, if we replace the letter 2 in equation A + 1 \u003d 5, then the equality becomes incorrect, and if 4, then the faithful equality is 4 + 1 \u003d 5.

We are more interested in exactly the meanings with which the variable will turn into faithful equality. They are called roots or solutions. We write down.

Definition 4.

The root of the equation Call such a value of a variable that draws this equation to true equality.

The root can also be called a solution, or vice versa - both of these concepts mean the same thing.

Example 2.

Take an example to explain this definition. Above, we led the equation a + 1 \u003d 5. According to the definition, the root in this case will be 4, because in the substitution instead of the letter it gives the right numerical equality, and the deuce will not be a solution, since it is responsible for the incorrect equality 2 + 1 \u003d 5.

How many roots may have one equation? Is any equation have a root? Reply to these questions.

Equations that do not have a single root also exist. An example may be 0 · X \u003d 5. We can substitute in it infinitely a lot different numbersBut none of them will turn it into faithful equality, since multiplication by 0 always gives 0.

There are also equations having several roots. They can have both finite, and infinitely a large number of roots.

Example 3.

So, in equation x - 2 \u003d 4 there is only one root - six, in x 2 \u003d 9 two roots - three and minus three, in x · (x - 1) · (x - 2) \u003d 0 three roots - zero, One and two, in the equation x \u003d x the roots are infinitely a lot.

Now explain how to record the roots of the equation correctly. If they are not, then we are writing: "The root equation does not have." You can also indicate the sign of an empty set ∅. If the roots are, then we write them through the comma or indicate as elements of the set, enclosing in curly brackets. So, if some equation has three roots - 2, 1 and 5, then we write - 2, 1, 5 or (- 2, 1, 5).

It is allowed to record the roots in the form of simple equalities. So, unknown in the equation is indicated by the letter Y, and the roots are 2 and 7, then we write y \u003d 2 and y \u003d 7. Sometimes lower indexes are added to the letters, for example, x 1 \u003d 3, x 2 \u003d 5. So we point to the roots numbers. If the solutions at the equation are infinitely a lot, then we record an answer as a numerical gap or use generally accepted designations: the set of natural numbers is denoted by n, integers - z, valid - R. Let's say if we need to record that the solution of the equation is any integer, then we write that x ∈ Z, and if any actual from one to nine, then y ∈ 1, 9.

When the equation is two, three roots or more, then, as a rule, they are not talking about roots, but about solutions of the equation. We formulate the definition of solving an equation with several variables.

Definition 5.

The solution of the equation with two, three and more variables is two, three and more values \u200b\u200bof the variables that draw this equation to the correct numerical equality.

Let us explain the definition on the examples.

Example 4.

Suppose we have an expression X + Y \u003d 7, which is an equation with two variables. Substitute instead of the first unit, and instead of the second two. We will have incorrect equality, it means that this pair of values \u200b\u200bwill not be solved by this equation. If we take a couple of 3 and 4, the equality will be true, it means that we have found a solution.

Such equations may also not have roots or have an infinite number. If we need to write two, three, four or more values, we write them through the comma in parentheses. That is, in the example above the answer will look like (3, 4).

In practice, most often have to deal with equations containing one variable. The algorithm of their solutions will consider in detail in the article on solving equations.

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In the course of mathematics grade 7 for the first time meet with equations with two variablesBut they are studied only in the context of systems of equations with two unknowns. That is why, from the field of view, a number of tasks fall out, in which some conditions that are limited to the coefficients of the equation are introduced. In addition, methods of solving problems of type "solve equation in natural or integers" are ignored, although eME Materials and on entrance exams The tasks of this kind are increasingly and more often.

What equation will be called an equation with two variables?

For example, equations 5x + 2y \u003d 10, x 2 + y 2 \u003d 20 or xy \u003d 12 are equations with two variables.

Consider equation 2x - y \u003d 1. It refers to the right equality at x \u003d 2 and y \u003d 3, therefore, this pair of variable values \u200b\u200bis the solution of the equation under consideration.

Thus, the solution of any equation with two variables is a plurality of ordered pairs (x; y), values \u200b\u200bof variables, which this equation is drawn to the right numerical equality.

The equation with two unknowns can:

but) have one solution. For example, the equation x 2 + 5y 2 \u003d 0 has a single solution (0; 0);

b) have several solutions. For example, (5 - | x |) 2 + (| y | - 2) 2 \u003d 0 has 4 solutions: (5; 2), (-5; 2), (5; -2), (-5; - 2);

in) do not have solutions. For example, the equation x 2 + y 2 + 1 \u003d 0 does not have solutions;

d) have infinitely a lot of solutions. For example, X + Y \u003d 3. The solutions of this equation will be the number, the sum of which is 3. The set of solutions of this equation can be written in the form (k; 3 - k), where k is any valid number.

The main methods of solving the equations with two variables are methods based on the decomposition of expressions on multipliers, the release of a complete square, the use of the properties of a square equation, limited expressions, evaluation methods. The equation, as a rule, is converted to the form from which you can get the system to find unknown.

Factorization

Example 1.

Solve equation: xy - 2 \u003d 2x - y.

Decision.

Grouping terms in order to decompose the factors:

(XY + Y) - (2x + 2) \u003d 0. From each bracket, we will summarize:

y (x + 1) - 2 (x + 1) \u003d 0;

(x + 1) (y - 2) \u003d 0. We have:

y \u003d 2, X is any valid number or x \u003d -1, y - any valid number.

In this way, the answer is all pairs of the form (x; 2), X € R and (-1; y), y € R.

Equality zero non-negative numbers

Example 2.

Solve equation: 9x 2 + 4Y 2 + 13 \u003d 12 (x + y).

Decision.

We group:

(9x 2 - 12x + 4) + (4Y 2 - 12Y + 9) \u003d 0. Now each bracket can be collapsed by the square formula.

(3x - 2) 2 + (2y - 3) 2 \u003d 0.

The sum of two non-negative expressions is zero, only if 3x is 2 \u003d 0 and 2y - 3 \u003d 0.

So, x \u003d 2/3 and y \u003d 3/2.

Answer: (2/3; 3/2).

Evaluation method

Example 3.

Solve equation: (x 2 + 2x + 2) (y 2 - 4Y + 6) \u003d 2.

Decision.

In each bracket, highlight the full square:

((x + 1) 2 + 1) ((y - 2) 2 + 2) \u003d 2. Establish the value of the expressions in brackets.

(x + 1) 2 + 1 ≥ 1 and (y - 2) 2 + 2 ≥ 2, then the left part of the equation is always at least 2. Equality is possible if:

(x + 1) 2 + 1 \u003d 1 and (y - 2) 2 + 2 \u003d 2, which means x \u003d -1, y \u003d 2.

Answer: (-1; 2).

We will get acquainted with another method of solving equations with two variables of the second degree. This method is that the equation is considered as square relative to any variable.

Example 4.

Solve equation: x 2 - 6x + y - 4√y + 13 \u003d 0.

Decision.

I solve equation as a square relative to X. We find discriminant:

D \u003d 36 - 4 (y - 4√y + 13) \u003d -4y + 16√y - 16 \u003d -4 (√Y - 2) 2. The equation will have a solution only at d \u003d 0, i.e., if Y \u003d 4. We substitute the value of y to the original equation and find that x \u003d 3.

Answer: (3; 4).

Often in equations with two unknowns indicate restrictions on variables.

Example 5.

Solve equation in integers: x 2 + 5y 2 \u003d 20x + 2.

Decision.

I rewrite the equation in the form x 2 \u003d -5y 2 + 20x + 2. The right side of the resulting equation during division by 5 gives in the residue 2. Therefore, x 2 is not divided by 5. But the square of the number that does not divide on 5 gives in the residancy 1 or 4. Thus, equality is impossible and there are no solutions.

Answer: No roots.

Example 6.

Solve equation: (x 2 - 4 | x | + 5) (Y 2 + 6Y + 12) \u003d 3.

Decision.

We highlight full squares in each bracket:

((| x | - 2) 2 + 1) ((y + 3) 2 + 3) \u003d 3. The left part of the equation is always greater than or equal to 3. Equality is possible provided | x | - 2 \u003d 0 and y + 3 \u003d 0. Thus, x \u003d ± 2, y \u003d -3.

Answer: (2; -3) and (-2; -3).

Example 7.

For each pair of integer negative numbers (x; y), satisfying the equation
x 2 - 2XY + 2Y 2 + 4Y \u003d 33, calculate the amount (x + y). In response, specify the smallest summary.

Decision.

We highlight full squares:

(x 2 - 2XY + Y 2) + (Y 2 + 4Y + 4) \u003d 37;

(x - y) 2 + (y + 2) 2 \u003d 37. Since X and Y are integers, then their squares are also integers. The sum of the squares of two integers equal to 37, we obtain, if we fold 1 + 36. Consequently:

(x - y) 2 \u003d 36 and (y + 2) 2 \u003d 1

(x - y) 2 \u003d 1 and (y + 2) 2 \u003d 36.

Solving these systems and considering that X and Y are negative, find solutions: (-7; -1), (-9; -3), (-7; -8), (-9; -8).

Answer: -17.

It is not necessary to despair if you have difficulty solving equations with two unknowns. A little practice, and you can cope with any equations.

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

The square equation is the equation of the form AX 2 + BX + C \u003d 0, where the coefficients A, B and C are arbitrary numbers, and a ≠ 0.

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

1. Do not have roots;
2. Have exactly one root;
3. Have two different roots.

This is an important difference square equations From linear where the root always exists and is unique. How to determine how many roots have an equation? For this there is a wonderful thing - discriminant.

Discriminant

Let the square equation AX 2 + BX + C \u003d 0. Then the discriminant is just the number d \u003d b 2 - 4ac.

This formula must be known by heart. Where she takes - now it does not matter. Other It is important: the discriminant sign can be determined how many roots has a square equation. Namely:

1. If D< 0, корней нет;
2. If D \u003d 0, there is exactly one root;
3. If D\u003e 0, there will be two roots.

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

A task. How many roots are square equations:

1. x 2 - 8x + 12 \u003d 0;
2. 5x 2 + 3x + 7 \u003d 0;
3. x 2 - 6x + 9 \u003d 0.

We repel the coefficients for the first equation and find the discriminant:
a \u003d 1, b \u003d -8, c \u003d 12;
D \u003d (-8) 2 - 4 · 1 · 12 \u003d 64 - 48 \u003d 16

So, the discriminant is positive, so the equation has two different roots. Similarly, disassemble the second equation:
a \u003d 5; b \u003d 3; C \u003d 7;
D \u003d 3 2 - 4 · 5 · 7 \u003d 9 - 140 \u003d -131.

The discriminant is negative, no roots. The last equation remains:
a \u003d 1; b \u003d -6; c \u003d 9;
D \u003d (-6) 2 - 4 · 1 · 9 \u003d 36 - 36 \u003d 0.

The discriminant is zero - the root will be one.

Please note that for each equation the coefficients were discharged. Yes, it's a long time, yes, it's a tedious - but you will not confuse the coefficients and do not allow stupid mistakes. Choose yourself: speed or quality.

By the way, if you "fill the hand", after a while no longer need to write all the coefficients. Such operations you will be performed in your head. Most people begin to do so somewhere after 50-70 solved equations - in general, not so much.

Roots square equation

We now turn, actually, to the decision. If discriminant D\u003e 0, roots can be found by formulas:

The basic formula of the roots of the square equation

When D \u003d 0, you can use any of these formulas - it will be the same number that will be the answer. Finally, if D< 0, корней нет — ничего считать не надо.

1. x 2 - 2x - 3 \u003d 0;
2. 15 - 2x - x 2 \u003d 0;
3. x 2 + 12x + 36 \u003d 0.

First equation:
x 2 - 2x - 3 \u003d 0 ⇒ a \u003d 1; b \u003d -2; c \u003d -3;
D \u003d (-2) 2 - 4 · 1 · (-3) \u003d 16.

D\u003e 0 ⇒ The equation has two roots. Find them:

Second equation:
15 - 2x - x 2 \u003d 0 ⇒ a \u003d -1; b \u003d -2; C \u003d 15;
D \u003d (-2) 2 - 4 · (-1) · 15 \u003d 64.

D\u003e 0 ⇒ The equation again has two roots. We find them

\\ [\\ begin (Align) & ((x) _ (1)) \u003d \\ FRAC (2+ \\ SQRT (64)) (2 \\ Cdot \\ left (-1 \\ right)) \u003d - 5; \\\\ & ((x) _ (2)) \u003d \\ FRAC (2- \\ SQRT (64)) (2 \\ Cdot \\ left (-1 \\ right)) \u003d 3. \\\\ \\ END (ALIGN) \\]

Finally, the third equation:
x 2 + 12x + 36 \u003d 0 ⇒ a \u003d 1; b \u003d 12; C \u003d 36;
D \u003d 12 2 - 4 · 1 · 36 \u003d 0.

D \u003d 0 ⇒ The equation has one root. You can use any formula. For example, the first:

As can be seen from examples, everything is very simple. If you know the formula and be able to consider, there will be no problems. Most often, errors occur during substitution in the formula of negative coefficients. Here, again, the reception described above will help: look at the formula literally, paint every step - and very soon get rid of errors.

Incomplete square equations

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

1. x 2 + 9x \u003d 0;
2. x 2 - 16 \u003d 0.

It is easy to see that in these equations there is no one of the terms. Such square equations are even easier than standard: they do not even need to consider discriminant. So, we introduce a new concept:

The AX 2 + BX + C \u003d 0 equation is called an incomplete square equation if B \u003d 0 or C \u003d 0, i.e. The coefficient with a variable x or the free element is zero.

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

Consider the remaining cases. Let b \u003d 0 be 0, then we obtain an incomplete square equation of the form AX 2 + C \u003d 0. We convert it a little:

Since arithmetic square root There is only non-negative number, the latter equality makes sense exclusively at (-C / a) ≥ 0. Conclusion:

1. If in an incomplete square equation of the form AX 2 + C \u003d 0, inequality (-C / a) is performed ≥ 0, there will be two roots. The formula is given above;
2. If (-c / a)< 0, корней нет.

As you can see, the discriminant did not need - in incomplete square equations there are no complex computing. In fact, even it is not necessary to remember the inequality (-c / a) ≥ 0. It is enough to express the value of X 2 and see what stands on the other side of the equality sign. If there is a positive number - the roots will be two. If negative - the roots will not be at all.

Now we will understand with the equations of the form AX 2 + BX \u003d 0, in which the free element is zero. Everything is simple here: the roots will always be two. It is enough to decompose a polynomial to multipliers:

Multiplier for bracket

The work is zero, when at least one of the multipliers is zero. From here there are roots. In conclusion, we will analyze several such equations:

A task. Square square equations:

1. x 2 - 7x \u003d 0;
2. 5x 2 + 30 \u003d 0;
3. 4x 2 - 9 \u003d 0.

x 2 - 7x \u003d 0 ⇒ x · (x - 7) \u003d 0 ⇒ x 1 \u003d 0; x 2 \u003d - (- 7) / 1 \u003d 7.

5x 2 + 30 \u003d 0 ⇒ 5x 2 \u003d -30 ⇒ x 2 \u003d -6. No roots, because Square cannot be equal to a negative number.

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

How to learn to solve simple and complex equations

Dear Parents!

Without basic mathematical preparation, the formulation of the formation of a modern person is impossible. In school, mathematics serves as a reference subject for many adjacent disciplines. In the post-school life, the real need becomes continuing educationWhat requires basic communal training, including mathematical.

IN primary school not only knowledge is laid on the main topics, but also develops logical thinking, imagination and spatial representations, and also formed interest in this subject.

Observing the principle of continuity, we will focus on the most important topic, namely, the "relationship of the components of actions in solving composite equations".

With this lesson, you can easily learn to solve complicated equations. In the lesson, you will get acquainted in detail with step-by-step instructions for solving complicated equations.

Many parents put a question in a dead end - how to make children learn to solve simple and complex equations. If the equations are simple - this is still half of the trouble, but they are both complex - for example integral. By the way, for information, there are also such equations over the solution of which the best minds of our planet are fighting and for the solution of which are given very weighty cash premiums. For example, if you rememberPerelman and the monetary premium unclaimed them in the amount of several million.

However, back to begin with simple mathematical equations and repeat the types of equations and the names of the components. Small warm-up:

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Workout

Find an excess number in each column:

2) What word is missing in every column?

3) Connect the words from the first column with words from 2 columns.

"Equation" "Equality"

4) How do you explain what "equality" is?

5) and "equation"? Is it equality? What is special in it?

summer sum

reduced difference

subdued piece

factorequality

dividend

the equation

Conclusion: The equation is equality with a variable, the value of which must be found.

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I offer each group to write an equation on a sheet with a felt-tip pen: (on the board)

1 group - with an unknown alignment;

2 group - with unknown reduced;

3 group - with unknown subtracted;

4 group - with an unknown divider;

5 group - with unknown divisible;

6 group - with an unknown multiplier.

1 group x + 8 \u003d 15

2 Group X - 8 \u003d 7

3 Group 48 - x \u003d 36

4 Group 540: x \u003d 9

5 Group x: 15 \u003d 9

6 group x * 10 \u003d 360

One of the group should mathematical language To read their equation and comment on their solution, i.e., speak the operation with the known components of action (algorithm).

Conclusion: We can solve the simple equations of all kinds of algorithm, read and record alphabetic expressions.

I propose to solve the problem in which a new type of equation appears.

X + 2kg 5kg and 3 kg

What size is the drawing?

Make the equation on this figure:

Choose a suitable equation for the obtained equation:

x + a \u003d in A: x \u003d in

x: a \u003d in x * a \u003d in

x - a \u003d in a - x \u003d in

Conclusion: I got acquainted with the solution of equations, in one of the parts of which contains a numerical expression, the value of which must be found and obtaining a simple equation.

________________________________________________________________________

Consider another option of the equation, the solution of which is reduced to solving the chain of simple equations. Here is one of the introduction of composite equations.

a + in * c (x - y): 3 2 * d + (m - n)

Are the record equations?

Why?

What do such actions call?

Read them, calling the last action:

Not. This is not the equation, since the equation should be a sign "\u003d".

Expressions

a + in * C - the sum of the number A and the works of numbers in and c;

(x - y): 3 - the private difference between the numbers x and y;

2 * d + (m - n) - the sum of the double number D and the difference between the numbers M and N.

I offer everyone to write off the proposal on the mathematical language:

The product of the difference in numbers x and 4 and numbers 3 is equal to 15.

Write down the proposal on the mathematical language: the product of the difference in numbers x and 4 and numbers 3 is equal to 15

(x - 4) * 3 \u003d 15

Conclusion: The problem that has arisen motivates the formulation of the lesson: learn to solve equations in which an unknown component is an expression. Such equations are composite equations.

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And maybe we will help you already studied types of equations? (algorithms)

Which of the known equations does our equation look like? X * a \u003d in

VERY IMPORTANT QUESTION : What is the expression on the left side - the amount, difference, work or private?

(x - 4) * 3 \u003d 15 (works)

Why? (Because Last Action - Multiplication)

Output: Such equations have not yet been considered. But you can decide if an expression x - 4. To impose a card (y - igrek), and the equation will be able to solve, using a simple algorithm for finding an unknown component.

When solving composite equations, it is necessary to select action at an automated level at each step, commenting, calling the components of the action.

Find Last Action

Select unknown component

Apply Rule

Simplify part

The root of the equation is found?

↓ Yes

Make check

(y - 5) * 4 = 28 y - 5. = 28: 4
y - 5 \u003d 7
y \u003d 5 +7
y \u003d 12.
(12 - 5) * 4 = 28
28 \u003d 28 (and)

Output: In classes with different preparation, this work can be organized in different ways. In more trained classes, even for primary consolidation, expressions can be used in which not two, and three or more action, but their solution requires more Steps with each step simply simplifying the equation, until it turns out a simple equation. And every time you can observe how an unknown component of action is changing.

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Conclusion:

When it comes to something very simple, understandable, we often say: "It's clear how twice two are four!".

But before you think that twice two - four, people had to learn a lot, many thousands of years.

Many rules from school textbooks of arithmetic and geometry were known to the ancient Greeks two more than a thousand years ago.

Everywhere where you need to consider something, measure, compare, without mathematics can not do.

It is difficult to imagine how people would live if they were not able to count, measure, compare. This teaches mathematics.

Today you plunged into school lifeWe visited the role of students and I suggest you, dear parents, evaluate your skills on the scale:

My skills

Date and evaluation

Components of action.

Drawing up an equation with an unknown component.

Reading and writing expressions.

Find the root of the equation in the simple equation.

Find the root of the equation, in one of the parts of which contains a numerical expression.

Find the root equations in which an unknown component of action is an expression.