With this math program, you can solve quadratic equation.
The program not only gives an answer to the problem, but also displays the solution process in two ways:
- using the discriminant
- using Vieta's theorem (if possible).
Moreover, the answer is displayed accurate, not approximate.
For example, for the equation \ (81x ^ 2-16x-1 = 0 \), the answer is displayed in this form:
This program can be useful for senior students of secondary schools in preparation for tests and exams, when checking knowledge before the exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it's too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with a detailed solution.
In this way, you can conduct your own teaching and / or the teaching of your younger siblings, while the level of education in the field of the problems being solved increases.
If you are not familiar with the rules for entering a square polynomial, we recommend that you familiarize yourself with them.
Rules for entering a square polynomial
Any Latin letter can be used as a variable.
For example: \ (x, y, z, a, b, c, o, p, q \) etc.
Numbers can be entered as whole or fractional numbers.
Moreover, fractional numbers can be entered not only in the form of a decimal, but also in the form of an ordinary fraction.
Rules for entering decimal fractions.
In decimal fractions, the fractional part from the whole can be separated by either a point or a comma.
For example, you can enter decimal fractions like this: 2.5x - 3.5x ^ 2
Rules for entering ordinary fractions.
Only an integer can be used as the numerator, denominator and whole part of a fraction.
The denominator cannot be negative.
When entering a numeric fraction, the numerator is separated from the denominator by a division sign: /
The whole part is separated from the fraction by an ampersand: &
Input: 3 & 1/3 - 5 & 6 / 5z + 1 / 7z ^ 2
Result: \ (3 \ frac (1) (3) - 5 \ frac (6) (5) z + \ frac (1) (7) z ^ 2 \)
When entering an expression brackets can be used... In this case, when solving a quadratic equation, the introduced expression is first simplified.
For example: 1/2 (y-1) (y + 1) - (5y-10 & 1/2)
Decide
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A bit of theory.
Quadratic equation and its roots. Incomplete quadratic equations
Each of the equations
\ (- x ^ 2 + 6x + 1,4 = 0, \ quad 8x ^ 2-7x = 0, \ quad x ^ 2- \ frac (4) (9) = 0 \)
has the form
\ (ax ^ 2 + bx + c = 0, \)
where x is a variable, a, b and c are numbers.
In the first equation a = -1, b = 6 and c = 1.4, in the second a = 8, b = -7 and c = 0, in the third a = 1, b = 0 and c = 4/9. Such equations are called quadratic equations.
Definition.
Quadratic equation is an equation of the form ax 2 + bx + c = 0, where x is a variable, a, b and c are some numbers, and \ (a \ neq 0 \).
The numbers a, b and c are the coefficients of the quadratic equation. The number a is called the first coefficient, the number b - the second coefficient, and the number c - the free term.
In each of the equations of the form ax 2 + bx + c = 0, where \ (a \ neq 0 \), the greatest power of the variable x is the square. Hence the name: quadratic equation.
Note that a quadratic equation is also called an equation of the second degree, since its left side is a polynomial of the second degree.
A quadratic equation in which the coefficient at x 2 is 1 is called reduced quadratic equation... For example, the reduced quadratic equations are the equations
\ (x ^ 2-11x + 30 = 0, \ quad x ^ 2-6x = 0, \ quad x ^ 2-8 = 0 \)
If in the quadratic equation ax 2 + bx + c = 0 at least one of the coefficients b or c is equal to zero, then such an equation is called incomplete quadratic equation... So, the equations -2x 2 + 7 = 0, 3x 2 -10x = 0, -4x 2 = 0 are incomplete quadratic equations. In the first of them b = 0, in the second c = 0, in the third b = 0 and c = 0.
Incomplete quadratic equations are of three types:
1) ax 2 + c = 0, where \ (c \ neq 0 \);
2) ax 2 + bx = 0, where \ (b \ neq 0 \);
3) ax 2 = 0.
Let's consider the solution of equations of each of these types.
To solve an incomplete quadratic equation of the form ax 2 + c = 0 for \ (c \ neq 0 \), transfer its free term to the right side and divide both sides of the equation by a:
\ (x ^ 2 = - \ frac (c) (a) \ Rightarrow x_ (1,2) = \ pm \ sqrt (- \ frac (c) (a)) \)
Since \ (c \ neq 0 \), then \ (- \ frac (c) (a) \ neq 0 \)
If \ (- \ frac (c) (a)> 0 \), then the equation has two roots.
If \ (- \ frac (c) (a) To solve an incomplete quadratic equation of the form ax 2 + bx = 0 with \ (b \ neq 0 \) factor its left side into factors and obtain the equation
\ (x (ax + b) = 0 \ Rightarrow \ left \ (\ begin (array) (l) x = 0 \\ ax + b = 0 \ end (array) \ right. \ Rightarrow \ left \ (\ begin (array) (l) x = 0 \\ x = - \ frac (b) (a) \ end (array) \ right. \)
This means that an incomplete quadratic equation of the form ax 2 + bx = 0 for \ (b \ neq 0 \) always has two roots.
An incomplete quadratic equation of the form ax 2 = 0 is equivalent to the equation x 2 = 0 and therefore has a unique root 0.
The formula for the roots of a quadratic equation
Let us now consider how quadratic equations are solved in which both the coefficients of the unknowns and the free term are nonzero.
Let's solve the quadratic equation in general form and as a result we get the formula for the roots. Then this formula can be applied to solve any quadratic equation.
Solve the quadratic equation ax 2 + bx + c = 0
Dividing both of its parts by a, we obtain the equivalent reduced quadratic equation
\ (x ^ 2 + \ frac (b) (a) x + \ frac (c) (a) = 0 \)
We transform this equation by selecting the square of the binomial:
\ (x ^ 2 + 2x \ cdot \ frac (b) (2a) + \ left (\ frac (b) (2a) \ right) ^ 2- \ left (\ frac (b) (2a) \ right) ^ 2 + \ frac (c) (a) = 0 \ Rightarrow \)
The radical expression is called the discriminant of the quadratic equation ax 2 + bx + c = 0 (Latin "discriminant" is a discriminator). It is designated by the letter D, i.e.
\ (D = b ^ 2-4ac \)
Now, using the notation of the discriminant, we rewrite the formula for the roots of the quadratic equation:
\ (x_ (1,2) = \ frac (-b \ pm \ sqrt (D)) (2a) \), where \ (D = b ^ 2-4ac \)
It's obvious that:
1) If D> 0, then the quadratic equation has two roots.
2) If D = 0, then the quadratic equation has one root \ (x = - \ frac (b) (2a) \).
3) If D Thus, depending on the value of the discriminant, the quadratic equation can have two roots (for D> 0), one root (for D = 0) or not have roots (for D When solving a quadratic equation using this formula, it is advisable to proceed as follows way:
1) calculate the discriminant and compare it with zero;
2) if the discriminant is positive or equal to zero, then use the root formula, if the discriminant is negative, then write down that there are no roots.
Vieta's theorem
The given quadratic equation ax 2 -7x + 10 = 0 has roots 2 and 5. The sum of the roots is 7, and the product is 10. We see that the sum of the roots is equal to the second coefficient taken with the opposite sign, and the product of the roots is equal to the free term. Any given quadratic equation with roots possesses this property.
The sum of the roots of the given quadratic equation is equal to the second coefficient, taken with the opposite sign, and the product of the roots is equal to the free term.
Those. Vieta's theorem states that the roots x 1 and x 2 of the reduced quadratic equation x 2 + px + q = 0 have the property:
\ (\ left \ (\ begin (array) (l) x_1 + x_2 = -p \\ x_1 \ cdot x_2 = q \ end (array) \ right. \)
In modern society, the ability to perform actions with equations containing a variable squared can be useful in many areas of activity and is widely used in practice in scientific and technical developments. This is evidenced by the design of sea and river vessels, airplanes and missiles. With the help of such calculations, the trajectories of movement of a wide variety of bodies, including space objects, are determined. Examples with the solution of quadratic equations are used not only in economic forecasting, in the design and construction of buildings, but also in the most ordinary everyday circumstances. They may be needed on camping trips, at sports events, in stores when shopping, and in other very common situations.
Let's break the expression into its constituent factors
The degree of an equation is determined by the maximum value of the degree of the variable that the given expression contains. If it is equal to 2, then such an equation is called square.
If we use the language of formulas, then these expressions, no matter how they look, can always be reduced 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 (an unknown without a square with its coefficient) and c (a free component, that is, an ordinary number). All this on the right side equals 0. In the case when a similar polynomial is missing one of its constituent terms, with the exception of ax 2, it is called an incomplete quadratic equation. Examples with the solution of such problems, the value of variables in which is easy to find, should be considered first.
If the expression looks in such a way that there are two terms on the right side of the expression, more precisely ax 2 and bx, it is easiest to find x by placing the variable outside the brackets. Now our equation will look like this: x (ax + b). Further, it becomes obvious that either x = 0, or the problem is reduced to finding a variable from the following expression: ax + b = 0. This is dictated by one of the properties of multiplication. The rule is that the product of two factors results in 0 only if one of them is equal to zero.
Example
x = 0 or 8x - 3 = 0
As a result, we get 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 side to 0 and finding possible unknowns, you can find out the time elapsing from the moment the body rises to the moment it falls, as well as many other quantities. But we'll talk about this later.
Factoring an Expression
The rule described above makes it possible to solve these problems in more complex cases. Let's consider examples with the solution of quadratic equations of this type.
X 2 - 33x + 200 = 0
This square trinomial is complete. First, let's transform 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 the solution of quadratic equations in grade 9 allow 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 factoring the right side into factors with a variable, there are three of them, that is, (x + 1), (x-3) and (x + 3).
As a result, it becomes obvious that this equation has three roots: -3; -1; 3.
Extraction of the square root
Another case of an incomplete second-order equation is an expression represented in the language of letters in such a way that the right-hand side is constructed from the components ax 2 and c. Here, to obtain the value of the variable, the free term is transferred to the right side, and then the square root is extracted from both sides of the equality. It should be noted that in this case, there are usually two roots of the equation. The only exceptions are equalities that do not contain the term c at all, where the variable is equal to zero, as well as variants of expressions when the right-hand side turns out to be negative. In the latter case, there are no solutions at all, since the above actions cannot be performed with roots. Examples of solutions to quadratic equations of this type should be considered.
In this case, the roots of the equation will be the numbers -4 and 4.
Calculation of the area of the land plot
The need for this kind of calculations appeared in ancient times, because the development of mathematics in many respects in those distant times was due to the need to determine with the greatest accuracy the areas and perimeters of land plots.
Examples with the solution of quadratic equations, compiled on the basis of problems of this kind, should be considered by us.
So, let's say there is a rectangular piece of land, the length of which is 16 meters longer than the width. Find the length, width and perimeter of the site if it is known that its area is 612 m 2.
Getting down to business, let's first draw up the necessary equation. Let's 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 done in the same way. Why? Although the left side of it still contains two factors, the product does not equal 0 at all, so other methods apply 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 that we have received an expression in the form corresponding to the previously indicated standard, where a = 1, b = 16, c = -612.
This can be an example of solving quadratic equations through the discriminant. Here the necessary calculations are made according to the scheme: D = b 2 - 4ac. This auxiliary quantity not only makes it possible to find the required quantities in the second-order equation, it determines the number of possible options. If D> 0, there are two of them; for D = 0 there is one root. If D<0, никаких шансов для решения у уравнения вообще не имеется.
About roots and their formula
In our case, the discriminant is: 256 - 4 (-612) = 2704. This indicates that our problem has an answer. If you know, k, the solution of 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 dimensions of the land plot cannot be measured in negative values, which means x (that is, the width of the plot) is 18 m.From here we calculate the length: 18 + 16 = 34, and the perimeter 2 (34+ 18) = 104 (m 2).
Examples and tasks
We continue to study quadratic equations. Examples and a detailed solution to 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, make a transformation, that is, we get the form of the equation, which is usually called standard, and equate it to zero.
15x 2 + 20x + 5 - 12x 2 - 27x - 1 = 0
Adding similar ones, we define the discriminant: D = 49 - 48 = 1. This means that our equation will have two roots. Let's calculate them according to the above formula, which means that the first of them will be equal to 4/3, and the second 1.
2) Now we will reveal the riddles of a different kind.
Let us find out if there are any roots here at all x 2 - 4x + 5 = 1? To obtain an exhaustive answer, let us bring the polynomial to the appropriate familiar form and calculate the discriminant. In this example, the solution of the quadratic equation is not necessary, because the essence of the problem is not at all in this. In this case, D = 16 - 20 = -4, which means that there really are no roots.
Vieta's theorem
It is convenient to solve quadratic equations using the above formulas and the 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 get the values of variables in this case. Example: solving quadratic equations by Vieta's theorem. She is named after a man who lived in 16th century France and made a brilliant career thanks to his mathematical talent and connections at court. His portrait can be seen in the article.
The pattern noticed by the famous Frenchman was as follows. He proved that the roots of the equation in the sum are numerically equal to -p = b / a, and their product corresponds to q = c / a.
Now let's look at specific tasks.
3x 2 + 21x - 54 = 0
For simplicity, let's transform the expression:
x 2 + 7x - 18 = 0
We will use Vieta's theorem, this will give us the following: the sum of the roots is -7, and their product is -18. From this we get that the roots of the equation are the numbers -9 and 2. Having made a 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 earlier. Now let's look at some of the math riddles in a little more detail. Any equation of the described type can be visualized. Such a relationship, drawn in the form of a graph, is called a parabola. Its various types are shown in the figure below.
Any parabola has a vertex, that is, a point from which its branches emerge. If a> 0, they go high to infinity, and when a<0, они рисуются вниз. Простейшим примером подобной зависимости является функция y = x 2 . В данном случае в уравнении x 2 =0 неизвестное может принимать только одно значение, то есть х=0, а значит существует только один корень. Это неудивительно, ведь здесь D=0, потому что a=1, b=0, c=0. Выходит формула корней (точнее одного корня) квадратного уравнения запишется так: x = -b/2a.
Visual representations of functions help to 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 coordinates of the vertex can be found by the just given formula x 0 = -b / 2a. And, substituting the resulting value into the original equation of the function, you can find out y 0, that is, the second coordinate of the vertex of the parabola, belonging to the ordinate axis.
The intersection of the branches of the parabola with the abscissa axis
There are a lot of examples with the solution of quadratic equations, but there are also general patterns. Let's 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 координата у 0 должна быть положительна. Для указанных вариантов D>0. Otherwise, D<0. А когда D=0, вершина параболы расположена непосредственно на оси 0х.
The roots can also be determined from the parabola graph. 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 points of intersection with the 0x axis, it is easier to build a graph.
From the history
With the help of equations containing a variable squared, in the old days they not only made mathematical calculations and determined the areas of geometric shapes. The ancients needed such calculations for grandiose discoveries in the field of physics and astronomy, as well as for making astrological forecasts.
As modern scientists assume, the inhabitants of Babylon were among the first to solve quadratic equations. It happened four centuries before our era. Of course, their calculations were fundamentally different from those currently accepted and turned out to be much more primitive. For example, the Mesopotamian mathematicians had no idea about the existence of negative numbers. They were also unfamiliar with other subtleties from those that any schoolchild of our time knows.
Perhaps even earlier than the scientists of Babylon, the sage from India Baudhayama took up the solution of quadratic equations. It happened about eight centuries before the advent of the era of Christ. True, the equations of the second order, the methods of solving which he gave, were the simplest. In addition to him, Chinese mathematicians were also interested in similar questions in the old days. In Europe, quadratic equations began to be solved only at the beginning of the 13th century, but later they were used in their works by such great scientists as Newton, Descartes and many others.
COMPLEX NUMBERS XI
§ 253. Extraction of square roots from negative numbers.
Solving quadratic equations with negative discriminants
As we know,
i 2 = - 1.
At the same time
(- i ) 2 = (- 1 i ) 2 = (- 1) 2 i 2 = -1.
Thus, there are at least two values for the square root of - 1, namely i and - i ... But maybe there are some other complex numbers whose squares are equal to - 1?
To clarify this question, suppose that the square of a complex number a + bi equals - 1. Then
(a + bi ) 2 = - 1,
a 2 + 2abi - b 2 = - 1
Two complex numbers are equal if and only if their real parts and coefficients at imaginary parts are equal. That's why
| { |
a
2 - b
2 = - 1 |
According to the second equation of system (1), at least one of the numbers a and b should be zero. If b = 0, then from the first equation we obtain a 2 = - 1. Number a valid and therefore a 2 > 0. Non-negative number a 2 cannot be equal to a negative number - 1. Therefore, the equality b = 0 in this case is impossible. It remains to admit that a = 0, but then from the first equation of the system we obtain: - b 2 = - 1, b = ± 1.
Therefore, complex numbers with squares equal to -1 are only the numbers i and - i This is conventionally written as:
√-1 = ± i .
By similar reasoning, students can make sure that there are exactly two numbers whose squares are equal to a negative number - a ... These numbers are √ a i and -√ a i ... This is conventionally written as follows:
√- a = ± √ a i .
Under √ a here the arithmetic, that is, positive, root is meant. For example, √4 = 2, √9 = .3; therefore
√-4 = + 2i , √-9 = ± 3 i
If earlier, when considering quadratic equations with negative discriminants, we said that such equations have no roots, now it is no longer possible to say so. Quadratic equations with negative discriminants have complex roots. These roots are obtained according to the formulas known to us. For example, let the equation be given x 2 + 2NS + 5 = 0; then
NS 1,2 = - 1 ± √1 -5 = - 1 ± √-4 = - 1 ± 2 i .
So, this equation has two roots: NS 1 = - 1 +2i , NS 2 = - 1 - 2i ... These roots are mutually conjugate. It is interesting to note that their sum is - 2, and the product is 5, so Vieta's theorem holds.
Exercises
2022. (Us tn about.) Solve the equations:
a) x 2 = - 16; b) x 2 = - 2; at 3 x 2 = - 5.
2023. Find all complex numbers whose squares are equal:
a) i ; b) 1/2 - √ 3/2 i ;
2024. Solve quadratic equations:
a) x 2 - 2x + 2 = 0; b) 4 x 2 + 4x + 5 = 0; v) x 2 - 14x + 74 = 0.
Solve systems of equations (No. 2025, 2026):
| { |
x + y
= 6 |
| { |
2x -
3y
= 1 |
2027. Prove that the roots of a quadratic equation with real coefficients and negative discriminant are mutually conjugate.
2028. Prove that Vieta's theorem is true for any quadratic equations, and not only for equations with non-negative discriminant.
2029. Make a quadratic equation with real coefficients, the roots of which are:
a) NS 1 = 5 - i , NS 2 = 5 + i ; b) NS 1 = 3i , NS 2 = - 3i .
2030. Make a quadratic equation with real coefficients, one of the roots of which is (3 - i ) (2i - 4).
2031. Write a quadratic equation with real coefficients, one of the roots of which is equal to 32 -
i
1- 3i
.
Let's work with quadratic equations... These are very popular equations! In its most general form, the quadratic equation looks like this:
For example:
Here a =1; b = 3; c = -4
Here a =2; b = -0,5; c = 2,2
Here a =-3; b = 6; c = -18
Well, you get the idea ...
How to solve quadratic equations? If you have a quadratic equation in this form, then everything is already simple. Remembering the magic word discriminant ... A rare high school student has not heard this word! The phrase “deciding through the discriminant” is reassuring and reassuring. Because there is no need to wait for dirty tricks from the discriminant! It is simple and trouble-free to use. So, the formula for finding the roots of a quadratic equation looks like this:
The expression under the root sign is the same discriminant... As you can see, to find x, we use only a, b and c... Those. coefficients from the quadratic equation. Just carefully substitute the values a, b and c into this formula and count. Substitute with your signs! For example, for the first equation a =1; b = 3; c= -4. So we write down:
The example is practically solved:
That's all.
What cases are possible when using this formula? There are only three cases.
1. The discriminant is positive. This means you can extract the root from it. Good root is extracted, or bad - another question. It is important what is extracted in principle. Then your quadratic equation has two roots. Two different solutions.
2. The discriminant is zero. Then you have one solution. Strictly speaking, this is not one root, but two identical... But this plays a role in inequalities, there we will study the issue in more detail.
3. The discriminant is negative. No square root is taken from a negative number. Well, okay. This means that there are no solutions.
Everything is very simple. And what, you think, is impossible to be mistaken? Well, yes, how ...
The most common mistakes are confusion with meaning signs. a, b and c... Rather, not with their signs (where to get confused?), But with the substitution of negative values in the formula for calculating the roots. Here, a detailed notation of the formula with specific numbers saves. If there are computational problems, do so!
Suppose you need to solve this example:
Here a = -6; b = -5; c = -1
Let's say you know that you rarely get answers the first time.
Well, don't be lazy. It will take 30 seconds to write an extra line. And the number of errors will sharply decrease... So we write in detail, with all the brackets and signs:
It seems incredibly difficult to paint so carefully. But it only seems to be. Try it. Well, or choose. Which is better, fast, or right? Besides, I will make you happy. After a while, there will be no need to paint everything so carefully. It will work out right by itself. Especially if you use the practical techniques described below. This evil example with a bunch of drawbacks can be solved easily and without errors!
So, how to solve quadratic equations we remembered through the discriminant. Or have learned, which is also not bad. Know how to correctly identify a, b and c... You know how attentively substitute them in the root formula and attentively read the result. You get the idea that the key word here is attentively?
However, quadratic equations often look slightly different. For example, like this:
it incomplete quadratic equations ... They can also be solved through the discriminant. You just need to figure out correctly what they are equal to a, b and c.
Have you figured it out? In the first example a = 1; b = -4; a c? He's not there at all! Well, yes, that's right. In mathematics, this means that c = 0 ! That's all. Substitute zero in the formula instead of c, and we will succeed. The same is with the second example. Only zero we have here not with, a b !
But incomplete quadratic equations can be solved much easier. Without any discriminant. Consider the first incomplete equation. What can you do there on the left side? You can put the x out of the parentheses! Let's take it out.
And what of it? And the fact that the product is equal to zero if and only if any of the factors is equal to zero! Don't believe me? Well, then think of two non-zero numbers that, when multiplied, will give zero!
Does not work? That's it ...
Therefore, we can confidently write: x = 0, or x = 4
Everything. These will be the roots of our equation. Both fit. When substituting any of them into the original equation, we get the correct identity 0 = 0. As you can see, the solution is much simpler than through the discriminant.
The second equation can also be solved simply. Move 9 to the right side. We get:
It remains to extract the root from 9, and that's it. It will turn out:
Also two roots ... x = +3 and x = -3.
This is how all incomplete quadratic equations are solved. Either by placing the x in parentheses, or by simply moving the number to the right and then extracting the root.
It is extremely difficult to confuse these techniques. Simply because in the first case you will have to extract the root from the x, which is somehow incomprehensible, and in the second case there is nothing to put out of the brackets ...
For now, take note of the best practices that will drastically reduce errors. The very ones that are due to inattention. ... For which then it hurts and insults ...
First reception... Do not be lazy to bring it to the standard form before solving the quadratic equation. What does this mean?
Let's say, after some transformations, you got the following equation:
Don't rush to write the root formula! You will almost certainly mix up the odds. a, b and c. Build the example correctly. First, the X is squared, then without the square, then the free term. Like this:
And again, do not rush! The minus in front of the x in the square can make you really sad. It's easy to forget it ... Get rid of the minus. How? Yes, as taught in the previous topic! You have to multiply the whole equation by -1. We get:
But now you can safely write down the formula for the roots, calculate the discriminant and complete the example. Do it yourself. You should have roots 2 and -1.
Reception second. Check the roots! By Vieta's theorem. Do not be alarmed, I will explain everything! Checking last thing the equation. Those. the one by which we wrote down the formula for the roots. If (as in this example) the coefficient a = 1, checking the roots is easy. It is enough to multiply them. You should get a free member, i.e. in our case, -2. Pay attention, not 2, but -2! Free member with my sign
... If it didn’t work, then it’s already screwed up somewhere. Look for the error. If it works out, you need to fold the roots. The last and final check. You should get a coefficient b with opposite
familiar. In our case, -1 + 2 = +1. And the coefficient b which is before the x is -1. So, everything is correct!
It is a pity that this is so simple only for examples where the x squared is pure, with a coefficient a = 1. But at least in such equations, check! There will be fewer mistakes.
Reception third... If you have fractional coefficients in your equation, get rid of fractions! Multiply the equation by the common denominator as described in the previous section. When working with fractions, for some reason, errors tend to pop in ...
By the way, I promised to simplify the evil example with a bunch of cons. Please! Here it is.
In order not to get confused in the minuses, we multiply the equation by -1. We get:
That's all! It's a pleasure to decide!
So, to summarize the topic.
Practical advice:
1. Before solving, we bring the quadratic equation to the standard form, build it right.
2. If there is a negative coefficient in front of the x in the square, we eliminate it by multiplying the entire equation by -1.
3. If the coefficients are fractional, we eliminate the fractions by multiplying the entire equation by the appropriate factor.
4. If x squared is pure, the coefficient at it is equal to one, the solution can be easily verified by Vieta's theorem. Do it!
Fractional equations. ODZ.
We continue to master the equations. We already know how to work with linear and quadratic equations. The last look remains - fractional equations... Or they are also called much more solidly - fractional rational equations... This is the same.
Fractional equations.
As the name implies, fractions are always present in these equations. But not just fractions, but fractions that have unknown in denominator... At least one. For example:
Let me remind you that if the denominators contain only the numbers, these are linear equations.
How to solve fractional equations? First of all, get rid of fractions! After that, the equation, most often, turns into linear or quadratic. And then we know what to do ... In some cases, it can turn into an identity, such as 5 = 5, or an incorrect expression, such as 7 = 2. But this rarely happens. I will mention this below.
But how to get rid of fractions !? Very simple. Applying all the same identical transformations.
We need to multiply the whole equation by the same expression. So that all denominators are reduced! Everything will become easier at once. Let me explain with an example. Suppose we need to solve the equation:
How did you teach in the lower grades? We transfer everything in one direction, bring to a common denominator, etc. Forget it like a bad dream! This should be done when you add or subtract fractional expressions. Or working with inequalities. And in the equations, we immediately multiply both sides by an expression that will give us the opportunity to reduce all denominators (i.e., in essence, by a common denominator). And what is this expression?
On the left side, to cancel the denominator, multiply by x + 2... And on the right, multiplication by 2 is required. Hence, the equation must be multiplied by 2 (x + 2)... We multiply:
This is the usual multiplication of fractions, but I will write it in detail:
Please note that I am not expanding the parenthesis yet. (x + 2)! So, in its entirety, I write it:
On the left side, it is reduced entirely (x + 2), and in the right 2. Which is required! After reduction, we get linear the equation:
And everyone will solve this equation! x = 2.
Let's solve one more example, a little more complicated:
If we remember that 3 = 3/1, and 2x = 2x / 1, you can write:
And again, we get rid of what we don't really like - fractions.
We see that to cancel the denominator with x, you need to multiply the fraction by (x - 2)... A few are not a hindrance to us. Well, we multiply. The whole left side and the whole right side:
Again brackets (x - 2) I do not disclose. I work with the parenthesis as a whole, as if it were one number! This should always be done, otherwise nothing will be reduced.
With a feeling of deep satisfaction, we cut (x - 2) and we get the equation without any fractions, in a ruler!
And now we open the brackets:
We give similar ones, transfer everything to the left side and get:
Classical quadratic equation. But the minus ahead is not good. You can always get rid of it, by multiplying or dividing by -1. But if you look closely at the example, you will notice that it is best to divide this equation by -2! In one fell swoop, the minus will disappear, and the odds will become prettier! Divide by -2. On the left side - term by term, and on the right - simply divide zero by -2, zero and get:
We solve through the discriminant and check by Vieta's theorem. We get x = 1 and x = 3... Two roots.
As you can see, in the first case, the equation after the transformation became linear, but here it is quadratic. It so happens that after getting rid of fractions, all xes are reduced. Remains something like 5 = 5. It means that x can be anything... Whatever it is, it will still shrink. And you get the honest truth, 5 = 5. But, after getting rid of fractions, it may turn out to be completely untrue, like 2 = 7. This means that no solutions! With any x, it turns out to be untrue.
Realized the main solution fractional equations? It is simple and logical. We change the original expression so that whatever we don't like disappears. Or interferes. In this case, these are fractions. We will do the same with all sorts of complex examples with logarithms, sines and other horrors. We always we will get rid of all this.
However, we need to change the original expression in the direction we need. according to the rules, yes ... Mastering which is preparation for the exam in mathematics. So we master it.
Now we will learn how to bypass one of the main ambushes on the exam! But first, let's see if you get into it, or not?
Let's look at a simple example:
The matter is already familiar, we multiply both parts by (x - 2), we get:
I remind you, with brackets (x - 2) we work as with one whole expression!
Here I no longer wrote 1 in the denominators, it is undignified ... And I did not draw brackets in the denominators, except for x - 2 there is nothing, you don't have to draw. We shorten:
We open the brackets, move everything to the left, give similar ones:
We solve, we check, we get two roots. x = 2 and x = 3... Fine.
Suppose the task says to write down the root, or their sum, if there are more than one root. What are we going to write?
If you decide the answer is 5, you were ambushed... And the task will not be counted for you. They worked in vain ... Correct answer 3.
What's the matter?! And you try to make a check. Substitute the values of the unknown into original example. And if at x = 3 everything will grow together wonderfully with us, we get 9 = 9, then with x = 2 division by zero! What can not be done categorically. Means x = 2 is not a solution, and is not taken into account in the answer. This is the so-called extraneous or extra root. We just drop it. The final root is one. x = 3.
How so ?! - I hear outraged exclamations. We were taught that an equation can be multiplied by an expression! This is an identical transformation!
Yes, identical. With a small condition - the expression by which we multiply (divide) - nonzero... A x - 2 at x = 2 is equal to zero! So everything is fair.
And now what i can do?! Don't multiply by expression? Do you need to check every time? Again it is not clear!
Calmly! Don't panic!
In this difficult situation, three magic letters will save us. I know what you are thinking. Right! it ODZ ... Range of Allowed Values.
I hope, after studying this article, you will learn how to find the roots of a complete quadratic equation.
Using the discriminant, only complete quadratic equations are solved; other methods are used to solve incomplete quadratic equations, which you will find in the article "Solving incomplete quadratic equations".
What quadratic equations are called complete? it equations of the form ax 2 + b x + c = 0, where the coefficients a, b and c are not equal to zero. So, to solve the full quadratic equation, you need to calculate the discriminant D.
D = b 2 - 4ac.
Depending on what value the discriminant has, we will write down the answer.
If the discriminant is negative (D< 0),то корней нет.
If the discriminant is zero, then x = (-b) / 2a. When the discriminant is a positive number (D> 0),
then x 1 = (-b - √D) / 2a, and x 2 = (-b + √D) / 2a.
For example. Solve the equation x 2- 4x + 4 = 0.
D = 4 2 - 4 4 = 0
x = (- (-4)) / 2 = 2
Answer: 2.
Solve Equation 2 x 2 + x + 3 = 0.
D = 1 2 - 4 2 3 = - 23
Answer: no roots.
Solve Equation 2 x 2 + 5x - 7 = 0.
D = 5 2 - 4 · 2 · (–7) = 81
x 1 = (-5 - √81) / (2 2) = (-5 - 9) / 4 = - 3.5
x 2 = (-5 + √81) / (2 2) = (-5 + 9) / 4 = 1
Answer: - 3.5; 1.
So we will present the solution of complete quadratic equations by the scheme in Figure 1.
These formulas can be used to solve any complete quadratic equation. You just need to be careful to ensure that the equation was written as a standard polynomial
a x 2 + bx + c, otherwise, you can make a mistake. For example, in writing the equation x + 3 + 2x 2 = 0, you can erroneously decide that
a = 1, b = 3 and c = 2. Then
D = 3 2 - 4 · 1 · 2 = 1 and then the equation has two roots. And this is not true. (See solution to Example 2 above).
Therefore, if the equation is not written as a polynomial of the standard form, first the complete quadratic equation must be written as a polynomial of the standard form (in the first place should be the monomial with the largest exponent, that is a x 2 , then with less – bx and then a free member with.
When solving a reduced quadratic equation and a quadratic equation with an even coefficient at the second term, you can use other formulas. Let's get to know these formulas as well. If in the full quadratic equation for the second term the coefficient is even (b = 2k), then the equation can be solved using the formulas shown in the diagram in Figure 2.
A complete quadratic equation is called reduced if the coefficient at x 2 is equal to one and the equation takes the form x 2 + px + q = 0... Such an equation can be given for the solution, or it is obtained by dividing all the coefficients of the equation by the coefficient a standing at x 2 .
Figure 3 shows a scheme for solving the reduced square 
equations. Let's look at an example of the application of the formulas discussed in this article.
Example. Solve the equation
3x 2 + 6x - 6 = 0.
Let's solve this equation using the formulas shown in the diagram in Figure 1.
D = 6 2 - 4 3 (- 6) = 36 + 72 = 108
√D = √108 = √ (363) = 6√3
x 1 = (-6 - 6√3) / (2 3) = (6 (-1- √ (3))) / 6 = –1 - √3
x 2 = (-6 + 6√3) / (2 3) = (6 (-1+ √ (3))) / 6 = –1 + √3
Answer: -1 - √3; –1 + √3
You can notice that the coefficient at x in this equation is an even number, that is, b = 6 or b = 2k, whence k = 3. Then we will try to solve the equation using the formulas shown in the diagram in the figure D 1 = 3 2 - 3 · (- 6 ) = 9 + 18 = 27
√ (D 1) = √27 = √ (9 3) = 3√3
x 1 = (-3 - 3√3) / 3 = (3 (-1 - √ (3))) / 3 = - 1 - √3
x 2 = (-3 + 3√3) / 3 = (3 (-1 + √ (3))) / 3 = - 1 + √3
Answer: -1 - √3; –1 + √3... Noticing that all the coefficients in this quadratic equation are divided by 3 and performing division, we obtain the reduced quadratic equation x 2 + 2x - 2 = 0 Solve this equation using the formulas for the reduced quadratic 
Equations Figure 3.
D 2 = 2 2 - 4 (- 2) = 4 + 8 = 12
√ (D 2) = √12 = √ (4 3) = 2√3
x 1 = (-2 - 2√3) / 2 = (2 (-1 - √ (3))) / 2 = - 1 - √3
x 2 = (-2 + 2√3) / 2 = (2 (-1+ √ (3))) / 2 = - 1 + √3
Answer: -1 - √3; –1 + √3.
As you can see, when solving this equation using different formulas, we received the same answer. Therefore, having mastered the formulas shown in the diagram of Figure 1 well, you can always solve any complete quadratic equation.
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