"The woman is created for a man, not a man for a woman" - such a postulate ...
5x (x - 4) \u003d 0
5 x \u003d 0 or x - 4 \u003d 0
x \u003d ± √ 25/4
Having learned to solve the equations of the first country, of course, I want to work with others, in particular, with the equations of the second degree, which are differently called square.
Square equations are equations of type ah ² + bx + c \u003d 0, where the variable is x, the numbers will be - a, b, С, where rather equals zero.
If one or another coefficient (C or B) is zero in the square equation, then this equation will refer to an incomplete square equation.
How to solve an incomplete square equation if students still knew how to solve only the first degree equations? Consider incomplete square equations of different species and simple ways to solve them.
a) if the coefficient C is equal to 0, and the coefficient B will not be zero, then AH ² + BX + 0 \u003d 0 is reduced to the equation of the form ah ² + bx \u003d 0.
To solve such an equation, you need to know the formula for solving an incomplete square equation, which is to decompose it to decompose it on multipliers and later to use the condition equality condition zero.
For example, 5x ² - 20x \u003d 0. Unchedule the left part of the factors' equation, while making a conventional mathematical operation: the transaction of a common factor for brackets
5x (x - 4) \u003d 0
We use the condition that the works are equal to zero.
5 x \u003d 0 or x - 4 \u003d 0
The answer will be: the first root - 0; The second root is 4.
b) if b \u003d 0, and the free term is not equal to zero, then the equation AH ² + 0x + C \u003d 0 is reduced to the equation of the form ah ² + C \u003d 0. solve the equations in two ways: a) decaying the polynomial equation in the left side of the multipliers ; b) Using the properties of an arithmetic square root. Such an equation is solved by one of the methods, for example:
x \u003d ± √ 25/4
x \u003d ± 5/2. The answer will be: the first root is 5/2; The second root is equal - 5/2.
c) If b is equal to 0 and C will be 0, then ah ² + 0 + 0 \u003d 0 is reduced to the equation of the form ah ² \u003d 0. In this equation X will be 0.
As you can see, incomplete square equations may have no more than two roots.
The formulas of the roots of the square equation. Cases of valid, multiple and complex roots are considered. Decomposition of square three-shred multipliers. Geometric interpretation. Examples of determining the roots and decomposition of multipliers.
Basic formulas
Consider a square equation:
(1)
.
Roots square equation (1) are determined by formulas:
;
.
These formulas can be combined like this:
.
When the roots of the square equation are known, the second degree polynomial can be represented as a work of the factors (decompose on multipliers):
.
Next, we believe that - the actual numbers.
Consider discriminant square equation:
.
If the discriminant is positive, then the square equation (1) has two different valid root:
;
.
Then the decomposition of the square three decreases on the factors has the form:
.
If the discriminant is zero, then the square equation (1) has two multiple (equal) valid root:
.
Factorization:
.
If the discriminant is negative, then the square equation (1) has two comprehensively conjugated root:
;
.
Here - the imaginary unit;
And - the actual and imaginary parts of the roots:
;
.
Then
.
Graphic interpretation
If you build a chart function
,
which is parabola, then the point of intersection of the graph with the axis will be roots of the equation
.
When, the schedule crosses the abscissa axis (axis) at two points.
When, the graph concerns the abscissa axis at one point.
When, the schedule does not intersect the abscissa axis.
Below are examples of such graphs.
Useful formulas associated with a square equation
(F.1) ;
(F.2) ;
(F.3) .
The output of the formula for the roots of the square equation
We carry out transformations and apply formulas (F.1) and (F.3):
,
Where
;
.
So, we got a formula for a polynomial of the second degree in the form:
.
From here it can be seen that the equation
performed at
and.
That is, the roots of the square equation are roots
.
Examples of determining the roots of the square equation
Example 1.
(1.1)
.
Decision
.
Comparing with our equation (1.1), we find the values \u200b\u200bof the coefficients:
.
We find discriminant:
.
Since the discriminant is positive, the equation has two valid root:
;
;
.
From here we get a decomposition of a square three-stakes on multipliers:
.
Schedule function y \u003d 2 x 2 + 7 x + 3 Crosses the abscissa axis at two points.
We construct a function schedule
.
The schedule of this function is parabola. She places the abscissa axis (axis) at two points:
and.
These points are roots of the initial equation (1.1).
Answer
;
;
.
Example 2.
Find the roots of the square equation:
(2.1)
.
Decision
We write the square equation in general form:
.
Comparing with the initial equation (2.1), we find the values \u200b\u200bof the coefficients:
.
We find discriminant:
.
Since the discriminant is zero, the equation has two multiple (equal) root:
;
.
Then the decomposition of three decisions on multipliers has the form:
.
Function graph y \u003d x 2 - 4 x + 4 Requests the abscissa axis at one point.
We construct a function schedule
.
The schedule of this function is parabola. It concerns the abscissa axis (axis) at one point:
.
This point is the root of the initial equation (2.1). Since this root enters the expansion of multipliers twice:
,
That such root is called multiple. That is, it is believed that there are two equal root:
.
Answer
;
.
Example 3.
Find the roots of the square equation:
(3.1)
.
Decision
We write the square equation in general form:
(1)
.
We rewrite the initial equation (3.1):
.
Compare C (1), we find the values \u200b\u200bof the coefficients:
.
We find discriminant:
.
Discriminant is negative. Therefore, there are no valid roots.
You can find complex roots:
;
;
.
Then
.
The function graph does not cross the abscissa axis. There are no valid roots.
We construct a function schedule
.
The schedule of this function is parabola. It does not intersect the abscissa axis (axis). Therefore, there are no valid roots.
Answer
There are no valid roots. Roings are integrated:
;
;
.
The conversion of a complete square equation in incomplete looks like that (for the case \\ (B \u003d 0 \\)):
For cases when \\ (C \u003d 0 \\) or when both coefficients are zero - everything is similar.
Note that there is no speech about the equality of zero \\ (a \\), it can not be zero, since in this case turn into:
Decision of incomplete square equations.
First of all, it is necessary to understand that the incomplete square equation is still, therefore it can be solved as well as the usual square (through). To do this, simply add the missing component of the equation with a zero coefficient.
Example
: Find the roots of the equation \\ (3x ^ 2-27 \u003d 0 \\)
Decision
:
|
We have an incomplete square equation with a coefficient \\ (b \u003d 0 \\). That is, we can write the equation in the following form: |
||
|
\\ (3x ^ 2 + 0 \\ Cdot X-27 \u003d 0 \\) |
In fact, here is the same equation as at the beginning, but now it can be solved as an ordinary square. First we write the coefficients. |
|
|
\\ (a \u003d 3; \\) \\ (b \u003d 0; \\) \\ (c \u003d -27; \\) |
Calculate discriminant by formula \\ (d \u003d b ^ 2-4ac \\) |
|
|
\\ (D \u003d 0 ^ 2-4 \\ cdot3 \\ cdot (-27) \u003d \\) |
Find the roots of the equation by formulas |
|
|
\\ (x_ (1) \u003d \\) \\ (\\ FRAC (-0+ \\ SQRT (324)) (2 \\ CDOT3) \\)\\ (\u003d \\) \\ (\\ FRAC (18) (6) \\) \\ (\u003d 3 \\) \\ (x_ (2) \u003d \\) \\ (\\ FRAC (-0- \\ SQRT (324)) (2 \\ CDOT3) \\)\\ (\u003d \\) \\ (\\ FRAC (-18) (6) \\) \\ (\u003d - 3 \\) |
|
Record the answer |
Answer : \\ (x_ (1) \u003d 3 \\); \\ (x_ (2) \u003d - 3 \\)
Example
: Find the roots of the equation \\ (- x ^ 2 + x \u003d 0 \\)
Decision
:
|
Again, an incomplete square equation, but now zero is equal to the coefficient \\ (C \\). Record equation as complete. |
||
We will work by S. square equations. These are very popular equations! In the most general form, the square equation looks like this:
For example:
Here but =1; b. = 3; c. = -4
Here but =2; b. = -0,5; c. = 2,2
Here but =-3; b. = 6; c. = -18
Well, you understood ...
How to solve square equations? If you have a square equation in such a form, then everything is simple. Remember the magic word discriminant . A rare high school student did not hear the word! The phrase "decide through the discriminant" will instill confidence and encourages. Because it is not necessary to wait for the tricks from the discriminant! It is simple and trouble-free in circulation. So, the formula for finding the roots of the square equation looks like this:
Expression under the sign of the root - and there is the same discriminant. As you can see, to find the ICA, we use only a, b and with. Those. The coefficients of the square equation. Just neatly substitute the values a, b and with In this formula and we consider. Substitute with your signs! For example, for the first equation but =1; b. = 3; c. \u003d -4. Here and write:
An example is practically solved:
That's all.
What cases are possible when using this formula? Total three cases.
1. Discriminant positive. This means that it is possible to extract the root. Good root is extracted, or bad - the question is different. It is important that it is extracted in principle. Then your square 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 it plays a role in inequalities, there we are more detailed to study the question.
3. The discriminant is negative. Of the negative number, the square root is not removed. Well, okay. This means that there are no solutions.
Everything is very simple. And what do you think it is impossible to make a mistake? Well, yes, how ...
The most common mistakes - confusion with signs of values a, b and with. Rather, not with their signs (where is there confused?), But with the substitution of negative values \u200b\u200bin the formula for calculating the roots. Here is a detailed entry of the formula with specific numbers. If there are problems with computing, do so!
Suppose you need to solve this one:
Here a \u003d -6; b \u003d -5; C \u003d -1.
Suppose you know that you rarely have answers from the first time.
Well, do not be lazy. Write an excess line will take seconds 30. And the number of errors sharply cut. Here we write in detail, with all brackets and signs:
It seems incredibly difficult, so carefully paint. But it only seems. Try. Well, or choose. What is better, fast, or right? Also, I'll kick you. After a while, there will disappear so carefully to paint everything. Itself will be right. Especially if you apply practical techniques, which are described just below. This evil example with a bunch of minuses will be solved easily and without errors!
So, how to solve square equations Through the discriminant we remembered. Or learned that it is also good. Know how to define correctly a, b and with. Knowledge carefully substitute them in the root formula and carefully count the result. You understood that the key word is here - carefully?
However, the square equations look slightly different. For example, like this:
it incomplete square equations . They can also be solved through the discriminant. It is only necessary to correctly imagine what is equal to a, b and with.
Corrected? In the first example a \u003d 1; b \u003d4; but c.? There is no one at all! Well, yes, right. In mathematics, this means that c \u003d 0. ! That's all. We substitute in the zero formula instead c, And everything will turn out. Similarly, with the second example. Only zero here do not from, but b. !
But incomplete square equations can be solved much easier. Without any discriminant. Consider the first incomplete equation. What can be done there in the left side? You can make the IS for brackets! Let's bring out.
And what from this? And the fact that the work is zero then, and only when some of the multipliers equals zero! Do not believe? Well, come up with two non-zero numbers, which will give zero with multiply!
Does not work? That's something ...
Consequently, you can confidently write: x \u003d 0., or x \u003d 4.
Everything. This will be the roots of our equation. Both are suitable. When substituting any of them into the original equation, we obtain a faithful identity 0 \u003d 0. As you can see, the solution is much simpler than through the discriminant.
The second equation can also be solved simply. We carry 9 to the right side. We get:
It remains the root to extract out of 9, and that's it. It turns out:
Also two roots . x \u003d +3 and x \u003d -3.
So all incomplete square equations are solved. Either by means of making a bracket, or by simply transferring the number to the right, followed by the extraction of the root.
It is extremely difficult to confuse these techniques. Simply because in the first case you will have to extract the root from XCA, which is somehow it is not clear, and in the second case, it is nothing for brackets ...
And now take note of practical techniques that dramatically reduce the number of errors. The most that because of the inattention. ... for which then it happens hurt and hurt ...
Reception First. Do not be lazy before solving the square equation to bring it to the standard form. What does this mean?
Suppose, after all transformations, you received such an equation:
Do not rush to write the root formula! Almost probably, you confuse the coefficients a, b and s. Build an example correctly. First, X is in the square, then without a square, then a free dick. Like this:
And do not rush again! The minus in front of the ix in the square can be healthy to upset you. Forget it easy ... Get rid of a minus. How? Yes, as taught in the previous topic! It is necessary to multiply the entire equation on -1. We get:
But now you can safely record the formula for the roots, consider the discriminant and the example. Dore yourself. You must have roots 2 and -1.
Reception second. Check the roots! On the Vieta Theorem. Do not scare, I will explain everything! Check last thing the equation. Those. That we recorded the roots formula. If (as in this example) coefficient a \u003d 1., Check the roots easily. Enough to multiply them. There should be a free member, i.e. In our case -2. Note, not 2, A -2! Free dick with your sign
. If it did not work, it means somewhere they have accumulated. Look for an error. If it happened - it is necessary to fold the roots. Last and final check. Must happen the coefficient b. from opposite
sign. In our case -1 + 2 \u003d +1. And coefficient b.which is in front of the ix, equal to -1. So, everything is right!
It is a pity that it is so simple for examples, where X is clean, with a coefficient a \u003d 1. But at least check in such equations! There will be less errors.
Taking third. If there are fractional coefficients in your equation, - get rid of fractions! Multiply the equation on the general denominator as described in the previous section. When working with fractions of the error, for some reason and climb ...
By the way, I promised an evil example with a bunch of minuses to simplify. You are welcome! Here it is.
In order not to be confused in the minuses, the equation on -1 is dominant. We get:
That's all! Decide - one pleasure!
So, summarize the topic.
Practical tips:
1. Before solving, we give a square equation to the standard form, build it right.
2. If a negative coefficient is worth a negative coefficient before X, eliminate its multiplication of the entire equation on -1.
3. If fractional coefficients are eliminating the fraction by multiplying the entire equation to the corresponding multiplier.
4. If X is in the square - clean, the coefficient is equal to one, the solution can be easily checked by the Vieta theorem. Do it!
Fractional equations. Odd
We continue to explore the equations. We are already aware of how to work with linear equations and square. Last view remained - fractional equations. Or they are also called much more solid - fractional rational equations. This is the same.
Fractional equations.
As clearly from the name, the fractions are necessarily present in these equations. But not just a fraction, and the fraraty who have unknown in denominator. At least in one. For example:
Let me remind you if in the denominators only numbersThese are linear equations.
How to decide fractional equations? First of all - get rid of fractions! After that, the equation is most often turning into linear or square. And then we know what to do ... In some cases it can turn into identity, type 5 \u003d 5 or an incorrect expression, type 7 \u003d 2. But it rarely happens. Below I'm talking about it.
But how to get rid of fractions!? Very simple. Applying all the same identity conversions.
We need to multiply all the equation for the same expression. So that all the denominators are quiet! Everything will be easier immediately. I explain on the example. Let us need to solve the equation:
How did you learn in junior grades? We carry everything in one direction, lead to a common denominator, etc. Forget how a terrible dream! So you need to do when you fold or deduct fractional expressions. Or work with inequalities. And in the equations, we immediately multiply both parts on the expression that will give us the opportunity to reduce all the denominators (that is, in essence, on the general denominator). And what is this expression?
In the left part to reduce the denominator, multiplication is required to x + 2. . And in the right required multiplication by 2. So, the equation must be multiplied by 2 (x + 2). Multiply:
This is the usual multiplication of fractions, but I will write in detail:
Note, I still do not reveal the bracket (x + 2)! So, I will write entirely:
In the left side is reduced entirely (x + 2), and in right 2. What was required! After cutting, we get linear the equation:
And this equation will already decide anyone! x \u003d 2..
I decide another example, a little more complicated:
If you remember that 3 \u003d 3/1, and 2x \u003d 2x /1, you can write:
And again we get rid of what we do not really like - from fractions.
We see that to reduce the denominator with the XA, you must multiply the fraction on (x - 2). And units we do not interfere. Well, multiply. All Left part I. all Right part:
Above brackets (x - 2) I do not reveal. I work with a bracket as a whole, as if it is one number! So you should always do, otherwise nothing will be reduced.
With a sense of deep satisfaction reducing (x - 2) And we get the equation without any fractions, in Lineshek!
But now we already reveal the brackets:
We give these things, we transfer everything to the left and we get:
Classic square equation. But minus ahead is not good. You can always get rid of it, multiplying or dividing on -1. But if you look at the example, you can see that it is best to divide this equation to -2! One smear and minus will disappear, and the coefficients prettier will become! Delim on -2. In the left side - soil, and in the right - just zero divide on -2, zero and get:
We decide through the discriminant and check by the Vieta theorem. Receive x \u003d 1 and x \u003d 3. Two roots.
As we see, in the first case, the equation after the transformation has become linear, and here is square. It happens that after getting rid of fractions, all Xers are reduced. Something remains, such as 5 \u003d 5. It means that x can be any. Whatever it is, it will still be reduced. And it turns out pure truth, 5 \u003d 5. But, after getting rid of fractions, it may turn out to be completely untrue, type 2 \u003d 7. And this means that no solutions! With any IQA, it turns out not true.
Realized the main way to solve fractional equations? It is simple and logical. We change the original expression so that everything that we do not like is disappeared. Or interferes. IN this case It is a fraction. Similarly, we will come with all sorts of complex examples with logarithms, sinus and other horrors. we always We will get rid of all this.
However, to change the original expression in the direction you need according to the rulesYes ... the development of which is the preparation for the exam in mathematics. So we master.
Now we will learn to bypass one of main ambigu on the exam! But for a start, let's see, do you get into it, or not?
We will analyze a simple example:
The case is already familiar, we multiply both parts on (x - 2)We get:
I remind you with brackets (x - 2) We work like with one, solid expression!
Here I no longer wrote a one in the denominators, no collapse ... and brackets did not draw in the denominators, there are except x - 2. No, you can not draw. Redfish:
We reveal brackets, transfer everything to the left, give these things:
We decide whether we check, we get two roots. x \u003d 2. and x \u003d 3.. Excellent.
Suppose in the task says to write the root, or their sum, if the roots are more than one. What will we write?
If you decide that the answer is 5, - you hit the ambush. And the task is not counted. In vain worked ... the correct answer is 3.
What's the matter?! And you try checking. Substitute the values \u200b\u200bof the unknown in source example. And if for x \u003d 3. We all wonderfully grow together, we get 9 \u003d 9, then when x \u003d 2. It will be divided into zero! What can not be done categorically. So x \u003d 2. The decision is not, and in response is not taken into account. This is the so-called stranger or excess root. We just throw it away. The final root is one. x \u003d 3..
How so?! - I hear indignant exclamations. We were taught that the equation can be multiplied by the expression! This is a identical conversion!
Yes, identical. With a small condition - an expression on which we multiply (divide) - full from zero. BUT x - 2. for x \u003d 2. Equally zero! So everything is honest.
And now what i can do?! Do not multiply an expression? Each time checking to do? Again it is not clear!
Calm! Without panic!
In this difficult situation, we will save three magical letters. I know what you thought about. Right! it Odd . Area of \u200b\u200bpermissible values.
In modern society, the ability to perform actions with the equations containing the variable raised to the square can be useful in many areas of activity and is widely used in practice in scientific and technical developments. Evidence of this can serve the design of marine and river vessels, aircraft and missiles. With the help of such calculations, the trajectories of the movement of various bodies, including space objects. Examples with a solution of square 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 in tourist campaigns, in sports, in shopping stores and in other very common situations.
We break the expression on the components of multipliers
The degree of equation is determined by the maximum value of the degree in the variable, which contains this expression. In the event that it is 2, then such an equation is just called square.
If the language of the formulas is expressing, then the indicated expressions, no matter how they look, can always be caused by the form when the left part of the expression consists of three terms. Among them: AX 2 (that is, the variable erected into a square with its coefficient), BX (unknown without a square with its coefficient) and C (free component, that is, the usual number). All this in the right side is equal to 0. In the case when there is no one of its components of the terms, with the exception of AX 2, it is called an incomplete square equation. Examples with solving such tasks, the value of the variables in which it is easy to find, should be considered first.
If the expression appears in the form looks in such a way that two, more precisely, AX 2 and BX, the expression on the expression on the expression on the right side, is easiest to find a variable for brackets. Now our equation will look like this: x (ax + b). Next, it becomes obvious that or x \u003d 0, or the task is reduced to finding a variable from the following expression: AX + B \u003d 0. The specified dictated one of the multiplication properties. The rule says that the product of two factors gives as a result of 0 only if one of them is zero.
Example
x \u003d 0 or 8x - 3 \u003d 0
As a result, we obtain two roots of the equation: 0 and 0.375.
The equations of this kind can describe the movement of bodies under the influence of gravity, which began movement from a certain point adopted at the beginning of the coordinates. Here, the mathematical record takes the following form: Y \u003d V 0 T + GT 2/2. Substituting the necessary values, equating the right side 0 and finding possible unknowns, you can find out the time passing from the moment of the body's rise until its fall, as well as many other values. But we will talk about it later.
Decomposition of the expression on multipliers
The rule described above makes it possible to solve the specified tasks and in more complex cases. Consider examples with solving square equations of this type.
X 2 - 33x + 200 \u003d 0
This square triple is complete. To begin with, we transform the expression and decompose it for multipliers. They are obtained two: (x-8) and (x-25) \u003d 0. As a result, we have two roots 8 and 25.
Examples with solving square equations in grade 9 allow this method to find a variable in expressions not only the second, but even the third and fourth orders.
For example: 2x 3 + 2x 2 - 18x - 18 \u003d 0. With the decomposition of the right part of the multipliers with a variable, they are obtained three, that is, (x + 1), (x-3) and (x + 3).
As a result, it becomes obvious that this equation has three roots: -3; -one; 3.
Extract square root
Another case of the incomplete equation of the second order is the expression, in the language of the letters presented in such a way that the right side is built from the components of AX 2 and C. Here, for the value of the variable, the free member is transferred to the right side, and then a square root is extracted from both parts of equality. It should be noted that in this case the roots of the equation usually two. An exception can only be equal to equality, generally not containing the term C, where the variable is zero, as well as the options for expressions, when the right side turns out to be negative. In the latter case, the solutions do not exist at all, since the above action cannot be produced with roots. Examples of solutions of square equations of this type must be considered.
In this case, the roots of the equation will be -4 and 4.
Calculation of a land plot
The need for such calculations appeared in deep antiquity, because the development of mathematics in many respects in those distant times was due to the need to determine the most accuracy of the area and the perimeter of land plots.
Examples with solving square equations drawn up on the basis of tasks of this kind should be considered to us.
So, let's say there is a rectangular plot of land, the length of which is 16 meters more than the width. It should be found a length, width and perimeter of the site, if it is known that its area is equal to 612 m 2.
Starting a matter, first make the necessary equation. Denote by x the width of the site, then its length will be (x + 16). From the written it follows 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) \u003d 612.
The solution of complete square equations, and this expression is precisely such, cannot be carried out by the same way. Why? Although the left side of it still contains two factors, the product is not at all equal to 0, so other methods are used here.
Discriminant
First of all, we will produce the necessary conversion, then the appearance of this expression will look like this: x 2 + 16x - 612 \u003d 0. This means we got an expression in the form corresponding to the previously specified standard, where a \u003d 1, b \u003d 16, c \u003d -612.
This can be an example of solving square equations through discriminant. Here, the required calculations are made according to the scheme: d \u003d b 2 - 4ac. This auxiliary value does not just make it possible to find the desired values \u200b\u200bin the second order equation, it determines the number of possible options. In case D\u003e 0, there are two; When d \u003d 0, there is one root. In case 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) \u003d 2704. This suggests that the answer from our task exists. If you know, K, the solution of square equations must be continued using the formula below. It allows you to calculate the roots.
This means that in the case presented: x 1 \u003d 18, x 2 \u003d -34. The second version in this dilemma cannot be a solution, because the dimensions of the land can not be measured in negative values, it means x (i.e. the width of the site) is 18 m. From here, we calculate the length: 18 + 16 \u003d 34, and perimeter 2 (34+ 18) \u003d 104 (m 2).
Examples and objectives
We continue to study square equations. Examples and a detailed solution of several of them will be given later.
1) 15x 2 + 20x + 5 \u003d 12x 2 + 27x + 1
We transfer everything to the left part of equality, we will make a transformation, that is, we obtain the form of the equation that is called standard, and equalize it with zero.
15x 2 + 20x + 5 - 12x 2 - 27x - 1 \u003d 0
After folding like, we define the discriminant: d \u003d 49 - 48 \u003d 1. So, our equation will have two roots. We calculate them according to the above formula, which means that the first one of them is 4/3, and the second one.
2) Now reveal the riddles of another kind.
Find out, is there any roots here x 2 - 4x + 5 \u003d 1? To obtain a comprehensive response, we give a polynomial to the appropriate familiarity and calculate the discriminant. In the specified example, the solution of the square equation is not necessary, because the essence of the task is not at all this. In this case, D \u003d 16 - 20 \u003d4, which means there are really no roots.
Vieta theorem
Square equations are conveniently solved through the above formulas and discriminant when the square root is extracted from the last value. But it happens not always. However, there are many ways to obtain variables in this case. Example: solutions of square equations on the Vieta Theorem. She is named after which lived in the XVI century in France and made a brilliant career due to his mathematical talent and courtyards. Portrait of it can be seen in the article.
The pattern that the famous French noted was as follows. He proved that the roots of the equation in the amount are numerically equal to -p \u003d b / a, and their product corresponds to q \u003d c / a.
Now consider specific tasks.
3x 2 + 21x - 54 \u003d 0
For simplicity, we transform the expression:
x 2 + 7x - 18 \u003d 0
We use the Vieta theorem, it will give us the following: the amount of the roots is -7, and their work -18. From here, we obtain that the roots of the equation are numbers -9 and 2. Having made a check, make sure that these values \u200b\u200bof variables are really suitable in the expression.
Graph and Parabola equation
Concepts The quadratic function and square equations are closely connected. Examples of this have already been shown earlier. Now consider some mathematical riddles a little more. Any equation of the described type can be imagined. A similar dependence drawn in the form of a graph is called a parabola. Her various types are shown in the figure below.
Any parabola has a vertex, that is, the point from which its branches come out. In case a\u003e 0, they leave high in 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 images of functions help solve any equations, including square. This method is called graphic. And the value of the variable x is the coordinate of the abscissa at points where the graph of the graph is crossing from 0x. The coordinates of the vertices can be found according to the only given formula X 0 \u003d -B / 2A. And, substituting the resulting value to the initial equation of the function, you can learn Y 0, that is, the second coordinate of the pearabol vertex belonging to the ordinate axis.
Crossing the branches of parabola with the abscissa axis
Examples with solutions of square equations are very much, but there are general patterns. Consider them. It is clear that the intersection of the graph with the axis 0x at a\u003e 0 is only possible if 0 receives 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.
According to the chart, the parabolas can be identified and roots. The opposite is also true. That is, if you get a visual image of a quadratic function is not easy, 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 build a schedule.
From the history
With the help of equations containing the variable raised to the square, in the old days not only made mathematical calculations and determined the area of \u200b\u200bgeometric figures. Similar calculations of the ancient were needed for grand discoveries in the field of physics and astronomy, as well as to compile astrological forecasts.
As the modern science figures suggest, among the first solutions of square equations, residents of Babylon took up. It happened in four centuries before the onset of our era. Of course, their calculations in the root differed from now adopted and turned out to be much primitive. For example, Mesopotamian mathematicians had no idea about the existence of negative numbers. The strangers also had other subtleties from those who know any student of our time.
Perhaps even earlier scientists of Babylon, the solution of square equations, a sage of India Budhoyama was engaged. It happened in about eight centuries before the era of Christ. True, the equation of the second order, the methods of solving which he led was the most simultaneous. In addition to him, such questions were interested in old and Chinese mathematicians. In Europe, the square equations began to solve only in the early XIII century, but later they were used in their work such great scientists as Newton, Descartes and many others.
