In modern society, the ability to perform actions with equations containing a squared variable can be useful in many fields of activity and is widely used in practice in scientific and technical developments. Evidence of this can serve as the design of sea and river vessels, aircraft and missiles. With the help of such calculations, the trajectories of the movement of various bodies, including space objects, are determined. Examples with the solution of quadratic equations find application not only in economic forecasting, in the design and construction of buildings, but also in the most ordinary everyday circumstances. They can be used in hiking, in sports, in shops when shopping and in other very common situations.
Divide the expression into component factors
The degree of the equation is determined by the maximum degree of the variable that the given expression contains. If it is equal to 2, then such an equation is called square.
If expressed in the language of formulas, then these expressions, no matter how they look, can always be brought to the form when the left side of the expression consists of three terms. Among them: ax 2 (that is, a variable squared with its coefficient), bx (unknown without a square with its coefficient) and c (free component, that is, an ordinary number). All this on the right-hand side is equal to 0. In the case when such a polynomial lacks one of its component terms, with the exception of ax 2 , it is called an incomplete quadratic equation. Examples with the solution of such problems, the value of the variables in which it is not difficult to find, should be considered first.
If the expression looks so that the terms in the expression on the right side are two, more precisely ax 2 and bx, it is easiest to find x by putting the variable out of brackets. Now our equation will look like this: x (ax + b). Then it becomes obvious that either x = 0, or the task is reduced to finding a variable from the following expression: ax + b = 0. The indicated is dictated by one of the properties of multiplication. The rule says that the product of two factors yields 0 only if one of them is equal to zero.
Example
8x 2 - 3x = 0
x (8x - 3) = 0
Next, we act according to the rule just described.
x = 0 or 8x - 3 = 0
As a result, we obtain two roots of the equation: 0 and 0.375.
Equations of this kind can describe the movement of bodies under the action of gravity, which began to move from a certain point taken as the origin. Here, the mathematical notation takes the following form: y = v 0 t + gt 2/2. Substituting the necessary values, equating the right-hand side 0 and finding possible unknowns, you can find out the time elapsing from the moment the body rises until it falls, as well as many other values. But we will talk about this later.
Factorization of an expression
The rule described above makes it possible to solve these problems in more complex cases. Consider examples of solving quadratic equations of this type.
X 2 - 33x + 200 = 0
This square trinomial is complete. First, convert the expression and factor it. There are two of them: (x-8) and (x-25) = 0. As a result, we have two roots 8 and 25.
Examples with solving quadratic equations in the 9th grade allow using this method to find a variable in expressions not only of the second, but even of the third and fourth orders.
For example: 2x 3 + 2x 2 - 18x - 18 = 0. When factorizing the right side of the variable, they get three, that is (x + 1), (x-3) and (x + 3).
As a result, it becomes obvious that this equation has three roots: -3; -1; 3.
Square root extraction
Another case of an incomplete second-order equation is an expression in the language of letters represented in such a way that the right-hand side is constructed from components ax 2 and c. Here, to obtain the value of the variable, the free term is transferred to the right side, and after that the square root is extracted from both sides of the equality. It should be noted that in this case, too, there are usually two roots of the equation. An exception may be only equalities that do not contain the term c, where the variable is equal to zero, and also variants of expressions when the right-hand side turns out to be negative. In the latter case, solutions do not exist at all, since the above actions cannot be performed with roots. Examples of solutions of quadratic equations of this type must be considered.
3x 2 - 48 = 0
3x 2 = 48
In this case, the roots of the equation will be the numbers -4 and 4.
Calculation of land
The need for such calculations appeared in ancient times, because the development of mathematics was largely due to the need to determine the areas and perimeters of land plots with the greatest accuracy.
Examples with the solution of quadratic equations based on problems of this kind should be considered by us.
So, let's say there is a rectangular piece of land whose length is 16 meters more than the width. The length, width and perimeter of the site should be found if it is known that its area is 612 m 2 .
Getting down to business, we first draw up the necessary equation. Denote by x the width of the section, then its length will be (x + 16). It follows from what has been written that the area is determined by the expression x (x + 16), which, according to the condition of our problem, is 612. This means that x (x + 16) = 612.
The solution of complete quadratic equations, and this expression is just that, cannot be carried out in the same way. Why? Although the left side of it still contains two factors, the product of them is not at all 0, so other methods are used here.
Discriminant
First of all, we will make the necessary transformations, then the appearance of this expression will look like this: x 2 + 16x - 612 = 0. This means we got the expression in the form corresponding to the standard indicated earlier, where a = 1, b = 16, c = -612.
This can be an example of solving quadratic equations through discriminant. Here, the necessary calculations are made according to the scheme: D = b 2 - 4ac. This auxiliary value not only makes it possible to find the desired values in the second-order equation, it determines the number of possible options. In case D> 0, there are two of them; for D = 0 there is one root. If D <0, the equation does not have any chances to solve.
About the roots and their formula
In our case, the discriminant is: 256 - 4 (-612) = 2704. This suggests that the answer to our problem exists. If you know, for example, the discriminant , the solution of the quadratic equations must be continued using the formula below. It allows you to calculate the roots.
This means that in the presented case: x 1 = 18, x 2 = -34. The second option in this dilemma cannot be a solution, because the size of the land cannot be measured in negative values, so x (that is, the width of the plot) is 18 m. From here, we calculate the length: 18 + 16 = 34, and perimeter 2 (34+ 18) = 104 (m 2 ).
Examples and tasks
We continue the study of quadratic equations. Examples and a detailed solution of several of them will be given below.
1) 15x 2 + 20x + 5 = 12x 2 + 27x + 1
We transfer everything to the left side of the equality, do the transformation, that is, we get the form of the equation, which is usually called the standard, and equate it to zero.
15x 2 + 20x + 5 - 12x 2 - 27x - 1 = 0
Putting these together, we define the discriminant: D = 49 - 48 = 1. Therefore, our equation will have two roots. We calculate them according to the above formula, which means that the first of them will be 4/3, and the second 1.
2) Now we will reveal riddles of another kind.
Find out if there are any roots x 2 - 4x + 5 = 1 here? To get an exhaustive answer, we bring the polynomial to the corresponding familiar form and calculate the discriminants. In this example, the solution of the quadratic equation is not necessary, because the essence of the problem is not at all this. In this case, D = 16 - 20 = -4, which means that there really are no roots.
Vieta Theorem
Quadratic equations are conveniently solved through the above formulas and discriminant, when the square root is extracted from the value of the latter. But this is not always the case. However, there are many ways to obtain the values of variables in this case. Example: solutions of quadratic equations by the Vieta theorem. It is named after Francois Viet, who lived in France in the 16th century and made a brilliant career thanks to his mathematical talent and connections at court. His portrait can be seen in the article.
The pattern observed by the illustrious Frenchman was as follows. He proved that the roots of the equation are numerically equal to -p = b / a, and their product corresponds to q = c / a.
Now consider specific tasks.
3x 2 + 21x - 54 = 0
For simplicity, we transform the expression:
x 2 + 7x - 18 = 0
Using the Vieta theorem, this will give us the following: the sum of the roots is -7, and their product is -18. From here we get that the roots of the equation are the numbers -9 and 2. Having done the check, we will make sure that these values of the variables really fit into the expression.
Parabola graph and equation
The concepts of a quadratic function and quadratic equations are closely related. Examples of this have already been given previously. Now let's look at some mathematical puzzles in a bit more detail. Any equation of the described type can be visualized. A similar dependence drawn in a graph is called a parabola. Its various types are presented in the figure below.
Any parabola has a vertex, that is, a point from which its branches come out. If a> 0, they go high to infinity, and when a <0, they are drawn down. The simplest example of such a dependence is the function y = x 2 . In this case, in the equation x 2 = 0, the unknown can take only one value, that is, x = 0, which means that there is only one root. This is not surprising, because here D = 0, because a = 1, b = 0, c = 0. It turns out the root formula (more precisely, one root) of the quadratic equation is written as follows: x = -b / 2a.
Visual images of functions help solve any equations, including quadratic ones. This method is called graphical. And the value of the variable x is the abscissa coordinate at the points where the graph line intersects with 0x. The vertex coordinates can be found by the formula just given x 0 = -b / 2a. And, substituting the obtained value in the initial equation of the function, you can find y 0 , that is, the second coordinate of the vertex of the parabola, which belongs to the ordinate axis.
The intersection of the branches of the parabola with the abscissa axis
There are a lot of examples with solving quadratic equations, but there are general patterns. Consider them. It is clear that the intersection of the graph with the 0x axis for a> 0 is possible only if y 0 takes negative values. And for a <0, the coordinate at 0 must be positive. For the indicated options, D> 0. Otherwise, D <0. And when D = 0, the vertex of the parabola is located directly on the axis 0x.
According to the schedule of the parabola, you can also determine the roots. The converse is also true. That is, if it is not easy to get a visual image of a quadratic function, you can equate the right side of the expression to 0 and solve the resulting equation. And knowing the intersection points with the 0x axis, it is easier to plot.
From the history
Using equations containing a squared variable, in the old days not only did mathematical calculations and determined the area of geometric shapes. The ancients needed such calculations for grandiose discoveries in the fields of physics and astronomy, as well as for making astrological forecasts.
As modern scientists suggest, the inhabitants of Babylon were among the first to solve the quadratic equations. It happened four centuries before the advent of our era. Of course, their calculations were fundamentally different from those currently accepted and turned out to be much more primitive. For example, Mesopotamian mathematicians had no idea about the existence of negative numbers. They were also unfamiliar with other subtleties of those that any schoolboy of our time knows.
Perhaps even earlier than the scientists of Babylon, the sage of India, Baudhayama, took up the solution of quadratic equations. It happened about eight centuries before the advent of Christ. True, second-order equations, the methods of solution of which he cited, were the simplest. In addition to him, Chinese mathematicians were interested in such questions in antiquity. In Europe, quadratic equations began to be solved only at the beginning of the XIII century, but later such great scientists as Newton, Descartes and many others used them in their works.