Equation - what is it? Definition of the term, examples

In the course of school mathematics, the child first hears the term "equation". What is this, let's try to figure it out together. In this article, we will consider the types and methods of solution.

Mathematics. Equations

To begin with, we offer to understand the very concept, what is it? As many textbooks of mathematics say, the equation is some expressions between which there is necessarily an equal sign. In these expressions there are letters, the so-called variables, the meaning of which must be found.

What is a variable? It is an attribute of a system that changes its meaning. A clear example of variables are:

• air temperature;
• child's height;
• Weight and so on.

In mathematics they are denoted by letters, for example, x, a, b, c ... Usually the task in mathematics sounds like this: find the value of the equation. This means that you need to find the value of these variables.

Varieties

The equation (what is it, we disassembled in the previous paragraph) can be of the following form:

• Linear;
• Square;
• Cubic;
• Algebraic;
• Transcendental.

For more detailed acquaintance with all kinds, we will consider each separately.

The linear equation

This is the first kind that schoolchildren get to know. They are solved fairly quickly and simply. So, the linear equation, what is it? This is an expression of the form: ax = c. So it is not especially clear, therefore we will result some examples: 2х = 26; 5x = 40; 1,2x = 6.

Let us analyze examples of equations. For this we need to collect all known data from one side, and unknowns in the other: x = 26/2; X = 40/5; X = 6 / 1.2. Here we used the elementary rules of mathematics: a * c = e, from this c = e / a; A = e / c. In order to complete the solution of the equation, we perform one action (in our case, division) x = 13; X = 8; X = 5. These were examples of multiplication, now look at subtraction and addition: x + 3 = 9; 10x-5 = 15. We transfer the known data to one side: x = 9-3; X = 20/10. We perform the last action: x = 6; X = 2.

Also, variants of linear equations are possible, where more than one variable is used: 2x-2y = 4. In order to solve, it is necessary to add 2y to each part, we obtain 2x-2y + 2y = 4-2y, as we have seen, on the left-hand side of the sign -2y and + 2y cancel, while we have: 2x = 4 -2y. The last step divides each part into two, we get the answer: X is equal to two minus the game.

Problems with equations are encountered even on Ahmess papyri. Here is one of the tasks: the number and the fourth part give a total of 15. To solve it, we write the following equation: x plus one fourth x is fifteen. We see one more example of a linear equation, as a result of the solution, we get the answer: x = 12. But this problem can be solved in another way, namely the Egyptian or, as it is called in another way, the method of assumption. In papyrus, the following solution is used: take four and a fourth part of it, that is, one. In total, they give five, now fifteen must be divided into a sum, we get three, the last action three multiplied by four. We get the answer: 12. Why do we divide by fifteen into five in a decision? So we know how many times fifteen, that is, the result that we need to get less than five. This was the way to solve problems in the Middle Ages, he was called the method of falsehood.

Quadratic equations

In addition to the examples considered earlier, there are others. Which ones? A quadratic equation, what is? They have the form ax ² + bx + c = 0. To solve them, you need to familiarize yourself with certain concepts and rules.

First, we need to find the discriminant by the formula: b ² -4ac. There are three options for the outcome of the solution:

• The discriminant is greater than zero;
• Less than zero;
• Is equal to zero.

In the first variant, we can get a response from two roots, which are found by the formula: -b + -cres from the discriminant divided by twice the first coefficient, that is 2a.

In the second case, the equation does not have roots. In the third case, the root is found by the formula: -b / 2a.

Consider an example of a quadratic equation for a more detailed acquaintance: three x-squares minus fourteen x minus five equals zero. To begin with, as we wrote earlier, we are looking for a discriminant, in our case it is equal to 256. Note that the number obtained is greater than zero, hence we must get a response consisting of two roots. We substitute the received discriminant into the formula for finding the roots. As a result, we have: X is equal to five and minus one third.

Special cases in quadratic equations

These are examples in which some values are zero (a, b or c), and possibly several.

For example, let's take the following equation, which is square: two x in the square equals zero, here we see that b and c are zero. Let's try to solve it, for this we divide both parts of the equation into two, we have: x ² = 0. As a result, we get x = 0.

Another case is 16x ² -9 = 0. Here, only b = 0. We solve the equation, we transfer the free coefficient to the right-hand side: 16x2 = 9, now we divide each part into sixteen: x ² = nine sixteenth. Since we have x in the square, the root of 9/16 can be either negative or positive. The answer is written as follows: X is equal to plus / minus three fourths.

A variant of the answer is also possible, as the root equation does not. Let's look at an example: 5x2 + 80 = 0, here b = 0. To solve the free term, throw it to the right side, after these actions we get: 5x2 = -80, now each part is divided into five: x2 = minus sixteen. If any number is squared, then we do not get a negative value. Therefore our answer is: the root equation does not.

Decomposition of a trinomial

The task of quadratic equations can also sound in another way: decompose the square trinomial into multipliers. This can be done using the following formula: a (x-x ₁ ) (x-x ₂ ). For this, as in another variant of the task, it is necessary to find the discriminant.

Consider the following example: 3x ² -14x-5, decompose the trinomial into multipliers. We find the discriminant, using the formula already known to us, it is obtained equal to 256. We note at once that 256 is greater than zero, hence, the equation will have two roots. We find them, as in the previous paragraph, we have: x = five and minus one third. We use the formula for expanding a trinomial into multipliers: 3 (x-5) (x + 1/3). In the second bracket, we received the equal sign, because the formula contains a minus sign, and the root is also negative, using the elementary knowledge of mathematics, in the sum we have a plus sign. For simplicity, we multiply the first and third term of the equation to get rid of the fraction: (x-5) (x + 1).

Equations that reduce to a quadratic

In this paragraph we will learn to solve more complicated equations. Let's start right from the example:

(X ² - 2x) ² - 2 (x ² - 2x) - 3 = 0. We can notice the repeating elements: (x ² - 2x), for the solution it is convenient to replace it with another variable, and then solve the usual quadratic equation, immediately Note that in this task we will get four roots, this should not frighten you. We denote the repetition of the variable a. We get: a ² -2a-3 = 0. Our next step is to find the discriminant of the new equation. We get 16, we find two roots: minus one and three. We recall that we made a substitution, we substitute these values, in the end we have the equations: x ² - 2x = -1; X ² - 2x = 3. We solve them in the first answer: x is equal to one, in the second: x is equal to minus one and three. We write down the answer as follows: plus / minus one and three. As a rule, the answer is written in ascending order.

Cubic equations

Let's consider one more possible variant. We shall discuss cubic equations. They have the form: ax ³ + bx ² + cx + d = 0. Examples of equations we will consider below, but for the beginning a little theory. They can have three roots, so there is a formula for finding the discriminant for a cubic equation.

Let's consider an example: 3х ³ + 4х ² + 2х = 0. How to solve it? To do this, we just put x in parentheses: x (3x2 + 4x + 2) = 0. All we have to do is calculate the roots of the equation in parentheses. The discriminant of the quadratic equation in brackets is less than zero, on this basis, the expression has a root: x = 0.

Algebra. Equations

We proceed to the next form. Now we briefly consider algebraic equations. One of the tasks sounds as follows: by grouping the method into 3x ⁴ + 2x ³ + 8x ² + 2x + 5 multipliers. The most convenient way will be the following grouping: (3x4 + 3x2) + (2x3 + 2x) + (5x2 +5). Note that 8x2 from the first expression was represented as the sum of 3x2 and 5x2. Now we take out from each bracket the common factor 3x2 (x2 + 1) + 2x (x2 + 1) +5 (x2 + 1). We see that we have a common multiplier: x in a square plus one, we take it out of brackets: (x2 + 1) (3x2 + 2x + 5). Further decomposition is impossible, since both equations have a negative discriminant.

Transcendental equations

We propose to deal with the following type. These are equations that contain transcendental functions, namely, logarithmic, trigonometric or exponential. Examples: 6sin ² x + tgx-1 = 0, x + 5lgx = 3 and so on. How they are solved you will learn from the course of trigonometry.

Function

The final step is to consider the concept of a function equation. Unlike previous versions, this type is not solved, and on it a graph is built. To do this, the equation should be well analyzed, find all the necessary points for the construction, calculate the minimum and maximum points.