In the course of school mathematics, the child first hears the term "equation". What is it, let's try to figure it out together. In this article, we consider types and solutions.
Mathematics. Equations
To begin with, we offer to deal with the very concept of what it is? As many mathematics textbooks say, an equation is some expression between which there is always an equal sign. In these expressions there are letters, the so-called variables, the value of which must be found.
What is a variable? This is an attribute of a system that changes its meaning. A good 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 is as follows: 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 examined in the previous paragraph) can be of the following form:
- linear
- square;
- cubic;
- algebraic;
- transcendental.
For a more detailed acquaintance with all species, we will consider each separately.
Linear equation
This is the first view that schoolchildren get to know. They are solved quite quickly and easily. So, a linear equation, what is? This is an expression of the form: ah = s. So itβs not very clear, so here are a few examples: 2x = 26; 5x = 40; 1,2x = 6.
Let us examine examples of equations. To do this, we need to collect all the known data on the one hand, and the unknown on the other: x = 26/2; x = 40/5; x = 6 / 1.2. Here the elementary rules of mathematics were used: a * c = e, from this c = e / a; a = e / s. 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 let's look at subtraction and addition: x + 3 = 9; 10x-5 = 15. We transfer the known data in one direction: x = 9-3; x = 20/10. We perform the last action: x = 6; x = 2.
Variants of linear equations are also 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 get 2x-2y + 2y = 4-2u, as we noticed, on the left side of the equal sign -2u and + 2u are reduced, while we have: 2x = 4 -2y. The last step is to divide each part into two, we get the answer: x is two minus the game.
Equation problems are found even on Ahmes papyrus. 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 equals fifteen. We see another 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. The following solution is used in papyrus: take four and a fourth of it, that is, one. In total, they give five, now fifteen must be divided by the sum, we get three, with the last action we multiply three by four. We get the answer: 12. Why do we divide fifteen into five in the decision? So we find out how many times fifteen, that is, the result that we need to get is less than five. In this way, problems were solved in the Middle Ages, he began to be called the method of false position.
Quadratic equations
In addition to the examples discussed earlier, there are others. Which ones? The quadratic equation, what is? They have the form ax 2 + bx + c = 0. To solve them, you need to familiarize yourself with some concepts and rules.
First, you need to find the discriminant by the formula: b 2 -4ac. There are three options for the outcome of the decision:
- discriminant is greater than zero;
- less than zero;
- equal to zero.
In the first option, we can get an answer from two roots, which are found by the formula: -b + -Root from the discriminant divided by the doubled first coefficient, that is, 2a.
In the second case, the equation has no 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 squared minus fourteen x minus five is zero. To begin with, as was written earlier, we are looking for a discriminant, in our case it is 256. Note that the resulting number is greater than zero, therefore, we must get an answer consisting of two roots. Substitute the resulting discriminant in the formula for finding the roots. As a result, we have: X equals five and minus one third.
Special cases in quadratic equations
These are examples in which some values ββare zero (a, b or c), and possibly several.
For example, take the following equation, which is quadratic: two x squared equals zero, here we see that b and c are equal to 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 16x2 -9 = 0. Here, only b = 0. We solve the equation, transfer the free coefficient to the right side: 16x 2 = 9, now we divide each part by sixteen: x 2 = nine sixteenths. Since we have x squared, the root of 9/16 can be both negative and positive. The answer is written as follows: x equals plus / minus three fourths.
A possible answer is also possible, as the equation of roots has no roots at all. Let's look at an example: 5x 2 + 80 = 0, here b = 0. To solve the free term, throw it to the right side, after these actions we get: 5x 2 = -80, now we divide each part by five: x 2 = minus sixteen. If any number is squared, then we will not get a negative value. Therefore, our answer is: the equation has no roots.
Trinomial Decomposition
The quadratic assignment may sound differently: factor the quadratic trinomial. This can be done using the following formula: a (x-x 1 ) (x-x 2 ). For this, as in another version of the task, it is necessary to find the discriminant.
Consider the following example: 3x 2 -14x-5, factor the trinomial into factors. We find the discriminant, using the formula already known to us, it turns out to be 256. We immediately note that 256 is greater than zero, therefore, 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 the trinomial into factors: 3 (x-5) (x + 1/3). In the second bracket we received the equal sign, because the minus sign is in the formula, and the root is also negative, using basic knowledge of mathematics, in total we have a plus sign. To simplify, we multiply the first and third terms of the equation to get rid of the fraction: (x-5) (x + 1).
Quadratic equations
In this section, we will learn how to solve more complex equations. Let's start right away with an example:
(x 2 - 2x) 2 - 2 (x 2 - 2x) - 3 = 0. We can notice the repeating elements: (x 2 - 2x), for our solution it is convenient to replace it with another variable, and then solve the usual quadratic equation, immediately we note that in this task we get four roots, this should not scare you. 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 did the replacement, 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. Typically, the response is recorded in ascending order.
Cubic equations
Consider another possible option. We will talk about cubic equations. They have the form: ax 3 + bx 2 + cx + d = 0. We will consider examples of equations below, and for a start a little theory. They can have three roots, there is also a formula for finding the discriminant for the cubic equation.
Consider an example: 3x 3 + 4x 2 + 2x = 0. How to solve it? To do this, we simply put x out of the brackets: x (3x 2 + 4x + 2) = 0. All that remains for us to do is to calculate the roots of the equation in brackets. The discriminant of the quadratic equation in brackets is less than zero, on the basis of this, the expression has a root: x = 0.
Algebra. Equations
We pass to the following view. We will now briefly review algebraic equations. One of the tasks is as follows: by grouping, factor 3x4 + 2x3 + 8x2 + 2x + 5 into factors. The most convenient way would be the following grouping: (3x4 + 3x2) + (2x3 + 2x) + (5x2 +5). Note that we presented 8x2 from the first expression as the sum of 3x2 and 5x2. Now we take out from each bracket the common factor 3x 2 (x2 + 1) + 2x (x 2 +1) +5 (x 2 +1). We see that we have a common factor: x squared plus one, put it outside the brackets: (x 2 +1) (3x 2 + 2x + 5). Further decomposition is impossible, since both equations have a negative discriminant.
Transcendental equations
We offer 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 will consider the concept of the equation of function. Unlike previous options, this type is not solved, and a schedule is built on it. For this, the equation should be well analyzed, find all the necessary points for construction, calculate the minimum and maximum points.