Discriminant is an ambiguous term. This article will focus on the discriminant of a polynomial, which allows you to determine whether a given polynomial has real solutions. The formula for a square polynomial is found in the school course in algebra and analysis. How to find the discriminant? What is needed to solve the equation?
A quadratic polynomial or an equation of the second degree is called i * w ^ 2 + j * w + k equal to 0, where "i" and "j" are the first and second coefficients, respectively, "k" is a constant, sometimes called the "intercept", and "w" is a variable. Its roots will be all values of the variable at which it turns into an identity. Such an equality can be rewritten as the product of i, (w - w1) and (w - w2) equal to 0. In this case, it is obvious that if the coefficient "i" does not vanish, then the function on the left side will become zero only if if x takes the value w1 or w2. These values are the result of setting the polynomial to zero.
To find the value of a variable at which the square polynomial vanishes, an auxiliary construction is used, built on its coefficients and called the discriminant. This construction is calculated according to the formula D equals j * j - 4 * i * k. Why is it being used?
- She says if there are valid results.
- She helps to calculate them.
How this value shows the presence of real roots:
- If it is positive, then you can find two roots in the region of real numbers.
- If the discriminant is zero, then both solutions are the same. We can say that there is only one solution, and it is from the realm of real numbers.
- If the discriminant is less than zero, then the polynomial has no real roots.
Calculation options for fixing the material
For sum (7 * w^2; 3 * w; 1) equal to 0 we calculate D by the formula 3 * 3 - 4 * 7 * 1 = 9 - 28 we get -19. A discriminant value below zero indicates that there are no results on the real line.
If we consider 2 * w ^ 2 - 3 * w + 1 equivalent to 0, then D is calculated as (-3) squared minus the product of numbers (4; 2; 1) and equals 9 - 8, that is, 1. A positive value indicates two results on the real line.
If we take the sum (w^2; 2 * w; 1) and equate to 0, D is calculated as two squared minus the product of numbers (4; 1; 1). This expression will simplify to 4 - 4 and turn to zero. It turns out that the results are the same. If you look closely at this formula, it will become clear that this is " full square". This means that the equality can be rewritten in the form (w + 1) ^ 2 = 0. It became obvious that the result in this problem is “-1”. In a situation where D is equal to 0, the left side of the equality can always be collapsed according to the formula “square of the sum”.
Using the Discriminant to Calculate Roots
This auxiliary construction not only shows the number of real solutions, but also helps to find them. General formula calculation for the equation of the second degree is as follows:
w = (-j +/- d) / (2 * i), where d is the discriminant to the power of 1/2.
Suppose the discriminant is below zero, then d is imaginary and the results are imaginary.
D is zero, then d equal to D to the power of 1/2 is also zero. Solution: -j / (2 * i). Considering 1 * w ^ 2 + 2 * w + 1 = 0 again, we find results equivalent to -2 / (2 * 1) = -1.
Suppose D > 0, so d is a real number, and the answer here splits into two parts: w1 = (-j + d) / (2 * i) and w2 = (-j - d) / (2 * i) . Both results will be valid. Let's look at 2 * w ^ 2 - 3 * w + 1 = 0. Here the discriminant and d are ones. So w1 is (3 + 1) divided by (2 * 2) or 1, and w2 is (3 - 1) divided by 2 * 2 or 1/2.
The result of equating a square expression to zero is calculated according to the algorithm:
- Determining the number of valid solutions.
- Calculation d = D^(1/2).
- Finding the result according to the formula (-j +/- d) / (2 * i).
- Substitution of the received result in initial equality for check.
Some special cases
Depending on the coefficients, the solution can be somewhat simplified. Obviously, if the coefficient in front of the variable to the second power is zero, then a linear equality is obtained. When the coefficient in front of the variable is zero to the first power, then two options are possible:
- the polynomial expands into the difference of squares with a negative free term;
- for a positive constant, real solutions cannot be found.
If the free term is zero, then the roots will be (0; -j)
But there are other special cases that simplify finding a solution.
Reduced Second Degree Equation
The given is called such square trinomial, where the coefficient in front of the leading term is one. For this situation, the Vieta theorem is applicable, which says that the sum of the roots is equal to the coefficient of the variable to the first power, multiplied by -1, and the product corresponds to the constant "k".
Therefore, w1 + w2 is equal to -j and w1 * w2 is equal to k if the first coefficient is one. To verify the correctness of such a representation, we can express w2 = -j - w1 from the first formula and substitute it into the second equality w1 * (-j - w1) = k. The result is the original equality w1 ^ 2 + j * w1 + k = 0.
It is important to note that i * w ^ 2 + j * w + k = 0 can be reduced by dividing by "i". The result will be: w^2 + j1 * w + k1 = 0 where j1 is equal to j/i and k1 is equal to k/i.
Let's look at the already solved 2 * w ^ 2 - 3 * w + 1 = 0 with the results w1 = 1 and w2 = 1/2. It is necessary to divide it in half, as a result, w ^ 2 - 3/2 * w + 1/2 = 0. Let's check that the conditions of the theorem are true for the results found: 1 + 1/2 = 3/2 and 1 * 1/2 = 1 /2.
Even second factor
If the factor of the variable to the first power (j) is divisible by 2, then it will be possible to simplify the formula and look for a solution through a quarter of the discriminant D / 4 \u003d (j / 2) ^ 2 - i * k. it turns out w = (-j +/- d/2) / i, where d/2 = D/4 to the power of 1/2.
If i = 1 and coefficient j is even, then the solution is the product of -1 and half of the coefficient in the variable w, plus/minus the root of the square of this half, minus the constant "k". Formula: w = -j / 2 +/- (j ^ 2 / 4 - k) ^ 1/2.
Higher order discriminant
The second-degree discriminant considered above is the most commonly used special case. In the general case, the discriminant of a polynomial is the multiplied squares of the differences of the roots of this polynomial. Therefore, the discriminant zero indicates the presence of at least two multiple solutions.
Consider i * w ^ 3 + j * w ^ 2 + k * w + m = 0.
D \u003d j ^ 2 * k ^ 2 - 4 * i * k ^ 3 - 4 * i ^ 3 * k - 27 * i ^ 2 * m ^ 2 + 18 * i * j * k * m.
Let's say the discriminant is greater than zero. This means that there are three roots in the region of real numbers. At zero, there are multiple solutions. If D< 0, то два корня комплексно-сопряженные, которые дают отрицательное значение при возведении в квадрат, а также один корень — вещественный.
Video
Our video will tell you in detail about the calculation of the discriminant.
Didn't get an answer to your question? Suggest a topic to the authors.
Let's work with quadratic equations. These are very popular equations! In the very general view 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 simple. We remember Magic word discriminant . A rare high school student has not heard this word! The phrase “decide through the discriminant” is reassuring and reassuring. Because there is no need to wait for 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 consider. Substitute with your signs! For example, for the first equation a =1; b = 3; c= -4. Here we write:
Example almost solved:
That's all.
What cases are possible when using this formula? There are only three cases.
1. The discriminant is positive. This means that you can extract the root from it. Whether the root is extracted well or badly is 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 a single root, but two identical. But this plays a role in inequalities, where we will study the issue in more detail.
3. The discriminant is negative. A negative number does not take the square root. Well, okay. This means there are no solutions.
Everything is very simple. And what do you think, you can't go wrong? Well, yes, how...
The most common mistakes are confusion with the signs of values a, b and c. Or rather, not with their signs (where is there to be confused?), But with the substitution of negative values into the formula for calculating the roots. Here, a detailed record of the formula with specific numbers saves. If there are problems with calculations, so do it!
Suppose we need to solve the following 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 drop sharply. So we write in detail, with all the brackets and signs:
It seems incredibly difficult to paint so carefully. But it only seems. 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 just turn out right. Especially if you apply practical techniques, which are described below. This evil example with a bunch of minuses will be solved easily and without errors!
So, how to solve quadratic equations through the discriminant we remembered. Or learned, which is also good. Can you correctly identify a, b and c. Do you know how carefully substitute them into the root formula and carefully count the result. Did you understand that keyword here - carefully?
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 correctly figure out what is equal here a, b and c.
Realized? In the first example a = 1; b = -4; a c? It doesn't exist at all! Well, yes, that's right. In mathematics, this means that c = 0 ! That's all. Substitute zero into the formula instead of c, and everything will work out for us. Similarly with the second example. Only zero we don't have here With, a b !
But incomplete quadratic equations can be solved much easier. Without any discrimination. Consider the first incomplete equation. What can be done on the left side? You can take the X out of brackets! 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? Well, then come up with two non-zero numbers that, when multiplied, will give zero!
Does not work? Something...
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 easily solved. We move 9 to the right side. We get:
It remains to extract the root from 9, and that's it. Get:
also two roots . x = +3 and x = -3.
This is how all incomplete quadratic equations are solved. Either by taking X out of brackets, or by simply transferring the number to the right, followed by extracting the root.
It is extremely difficult to confuse these methods. Simply because in the first case you will have to extract the root from X, which is somehow incomprehensible, and in the second case there is nothing to take out of brackets ...
Now take note of the practical techniques that dramatically reduce the number of errors. The very ones that are due to inattention ... For which it is then painful and insulting ...
First reception. Do not be lazy before solving a quadratic equation to bring it to a standard form. What does this mean?
Suppose, after any transformations, you get the following equation:
Do not rush to write the formula of the roots! You will almost certainly mix up the odds a, b and c. Build the example correctly. First, x squared, then without a square, then a free member. Like this:
And again, do not rush! The minus before the x squared can upset you a lot. Forgetting it is easy... Get rid of the minus. How? Yes, as taught in the previous topic! We need to multiply the whole equation by -1. We get:
And now you can safely write down the formula for the roots, calculate the discriminant and complete the example. Decide on your own. You should end up with roots 2 and -1.
Second reception. Check your roots! According to Vieta's theorem. Don't worry, I'll explain everything! Checking last thing the equation. Those. the one by which we wrote down the formula of the roots. If (as in this example) the coefficient a = 1, check the roots easily. It is enough to multiply them. You should get a free term, i.e. in our case -2. Pay attention, not 2, but -2! free member with your sign
. If it didn’t work out, it means they already messed up somewhere. Look for an error. If it worked out, you need to fold the roots. Last and final check. Should be a ratio b With opposite
sign. In our case -1+2 = +1. A coefficient b, which is before the x, is equal to -1. So, everything is correct!
It is a pity that it is so simple only for examples where x squared is pure, with a coefficient a = 1. But at least check in such equations! There will be fewer mistakes.
Reception third. If your equation has fractional coefficients, get rid of the fractions! Multiply the equation by the common denominator as described in the previous section. When working with fractions, errors, for some reason, climb ...
By the way, I promised an evil example with a bunch of minuses to simplify. Please! Here he is.
In order not to get confused in the minuses, we multiply the equation by -1. We get:
That's all! Deciding is fun!
So let's recap the topic.
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 corresponding factor.
4. If x squared is pure, the coefficient for it is equal to one, the solution can be easily checked 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 view remains fractional equations. Or they are also called much more solid - fractional rational equations. This is the same.
Fractional equations.
As the name implies, these equations necessarily contain fractions. But not just fractions, but fractions that have unknown in the denominator. At least in one. For example:
Let me remind you, if in the denominators only numbers, these are linear equations.
How to decide fractional equations? First of all, get rid of the fractions! After that, the equation, most often, turns into a linear or quadratic one. And then we know what to do... In some cases, it can turn into an identity, like 5=5 or an incorrect expression, like 7=2. But this rarely happens. Below I will mention it.
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 decrease! Everything will immediately become easier. I explain with an example. Let's say we need to solve the equation:
How were they taught in elementary school? We transfer everything in one direction, reduce it to a common denominator, etc. Forget how bad dream! This is what you need to do when you add or subtract fractional expressions. Or work with inequalities. And in equations, we immediately multiply both parts by an expression that will give us the opportunity to reduce all denominators (ie, in essence, by a common denominator). And what is this expression?
On the left side, to reduce the denominator, you need to multiply by x+2. And on the right, multiplication by 2 is required. So, the equation must be multiplied by 2(x+2). We multiply:
This is the usual multiplication of fractions, but I will write in detail:
Please note that I am not opening 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. As required! After reduction we get linear the equation:
Anyone can solve this equation! x = 2.
Let's solve another example, a little more complicated:
If we remember that 3 = 3/1, and 2x = 2x/ 1 can be written:
And again we get rid of what we don’t really like - from fractions.
We see that to reduce the denominator with x, it is necessary to multiply the fraction by (x - 2). And units are not a hindrance to us. Well, let's multiply. All left side and all right side:
Brackets again (x - 2) I don't reveal. I work with the bracket as a whole, as if it were one number! This must 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 coefficients will become prettier! We divide by -2. On the left side - term by term, and on the right - just divide zero by -2, zero and get:
We solve through the discriminant and check according to the Vieta 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, and here it is quadratic. It happens that after getting rid of fractions, all x's are reduced. There is something left, like 5=5. It means that x can be anything. Whatever it is, it will still be reduced. And get the pure truth, 5=5. But, after getting rid of fractions, it may turn out to be completely untrue, such as 2=7. And this means that no solutions! With any x, it turns out to be false.
Realized main way solutions fractional equations? It is simple and logical. We change the original expression so that everything that we don't like disappears. Or interfere. In this case, it's fractions. We will do the same with all 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 ... The development of which is the preparation for the exam in mathematics. Here we are learning.
Now we will learn how to bypass one of the the main ambushes on the exam! But first, let's see if you fall into it or not?
Let's take a simple example:
The matter is already familiar, we multiply both parts by (x - 2), we get:
Remember, with brackets (x - 2) working with one whole expression!
Here I no longer wrote the one in the denominators, undignified ... And I didn’t draw brackets in the denominators, except for x - 2 there is nothing, you can not draw. We shorten:
We open the brackets, move everything to the left, we give similar ones:
We solve, check, we get two roots. x = 2 and x = 3. Excellent.
Suppose the task says to write down the root, or their sum, if there are more than one root. What will we write?
If you decide the answer is 5, you were ambushed. And the task will not be counted for you. They worked in vain ... The correct answer is 3.
What's the matter?! And you try to check. Substitute the values of the unknown into original example. And if at x = 3 everything grows together wonderfully, we get 9 = 9, then with x = 2 divide by zero! What absolutely cannot be done. 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 discard it. There is only one final root. x = 3.
How so?! I hear outraged exclamations. We were taught that an equation can be multiplied by an expression! it identity transformation!
Yes, identical. Under a small condition - the expression by which we multiply (divide) - different from zero. BUT x - 2 at x = 2 equals zero! So it's all fair.
And now what i can do?! Do not multiply by expression? Do you check every time? Again unclear!
Calmly! No panic!
In this difficult situation, three magic letters will save us. I know what you were thinking. Correctly! it ODZ . Area of Valid Values.
Important! At roots of even multiplicity, the function does not change sign.
Note! Any non-linear inequality of a school algebra course must be solved using the method of intervals.
I offer you a detailed algorithm for solving inequalities by the interval method, following which you can avoid errors when solving nonlinear inequalities.
Solution quadratic equations with negative discriminants
As we know,
i 2 = - 1.
However,
(- 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 - 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 the coefficients of the imaginary parts are equal. That's why
| { | and 2 - b 2 = - 1 ab = 0 (1) |
According to the second equation of system (1), at least one of the numbers a and b should equal zero. If a b = 0, then the first equation yields a 2 = - 1. Number a real, and therefore a 2 > 0. Non-negative number a 2 cannot equal negative number- 1. Therefore, equality b = 0 is impossible in this case. It remains to be recognized that a = 0, but then from the first equation of the system we get: - b 2 = - 1, b = ± 1.
Therefore, the only complex numbers whose squares are -1 are the numbers i and - i , This is conditionally written as:
√-1 = ± i .
By similar reasoning, students can verify that there are exactly two numbers whose squares are equal to a negative number - a . These numbers are √ ai and -√ ai . Conventionally, it is written like this:
√- a = ± √ ai .
Under √ a here the arithmetic, that is, positive, root is meant. For example, √4 = 2, √9 =.3; that's why
√-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 by formulas known to us. Let, for example, given the equation x 2 + 2X + 5 = 0; then
X 1.2 = - 1 ± √1 -5 = - 1 ± √-4 = - 1 ± 2 i .
So this equation has two roots: X 1 = - 1 +2i , X 2 = - 1 - 2i . These roots are mutually conjugate. It is interesting to note that their sum is equal to - 2, and the product is 5, so Vieta's theorem is fulfilled.
The concept of a complex number
A complex number is an expression of the form a + ib, where a and b are any real numbers, i is a special number, which is called the imaginary unit. For such expressions, the concepts of equality and the operations of addition and multiplication are introduced as follows:
- Two complex numbers a + ib and c + id are said to be equal if and only if
a = b and c = d . - The sum of two complex numbers a + ib and c + id is a complex number
a + c + i (b + d). - The product of two complex numbers a + ib and c + id is a complex number
ac - bd + i (ad + bc).
Complex numbers are often denoted by a single letter, such as z = a + ib. The real number a is called the real part of the complex number z, the real part is denoted a = Re z . The real number b is called the imaginary part of the complex number z, the imaginary part is denoted b = Im z . Such names are chosen in connection with the following special properties of complex numbers.
Note that arithmetic operations on complex numbers of the form z = a + i · 0 are carried out in exactly the same way as on real numbers. Really,
Therefore, complex numbers of the form a + i · 0 are naturally identified with real numbers. Because of this, complex numbers of this kind are called simply real. So, the set of real numbers is contained in the set of complex numbers. The set of complex numbers is denoted by . We have established that, namely
Unlike real numbers, numbers of the form 0 + ib are called purely imaginary. Often just write bi , for example, 0 + i 3 = 3 i . A purely imaginary number i1 = 1 i = i has a surprising property:
In this way,
№ 4 .1. In mathematics, a number function is a function whose domains and values are subsets of number sets—generally the set of real numbers or the set of complex numbers.
Function Graph
Function Graph Fragment 
Ways to set a function
[edit] Analytical method
Typically, a function is defined using a formula that includes variables, operations, and elementary functions. Perhaps a piecewise assignment, that is, different for different values of the argument.
[edit] Tabular way
A function can be defined by listing all of its possible arguments and their values. After that, if necessary, the function can be extended for arguments that are not in the table, by interpolation or extrapolation. Examples are a program guide, a train schedule, or a table of values for a Boolean function:
[edit] Graphical way
The oscillogram sets the value of some function graphically.
A function can be specified graphically by displaying a set of points of its graph on a plane. This can be a rough sketch of what the function should look like, or readings taken from an instrument such as an oscilloscope. This specification may suffer from a lack of precision, but in some cases other specification methods cannot be applied at all. In addition, this way of setting is one of the most representative, easy to understand and high-quality heuristic analysis of the function.
[edit] Recursive way
A function can be defined recursively, that is, through itself. In this case, some values of the function are determined through its other values.
- factorial;
- Fibonacci numbers;
- Ackerman function.
[edit] verbal way
A function can be described in natural language words in some unambiguous way, for example, by describing its input and output values, or the algorithm by which the function assigns correspondences between these values. Along with graphically, sometimes this is the only way to describe a function, although natural languages are not as deterministic as formal ones.
- a function that returns a digit in the notation of pi by its number;
- a function that returns the number of atoms in the universe at a given point in time;
- a function that takes a person as an argument and returns the number of people who will be born into the world after his birth
AT modern society the ability to operate with equations containing a squared variable can be useful in many areas of activity and is widely used in practice in scientific and technical developments. This can be evidenced by the design of sea and river vessels, aircraft and missiles. With the help of such calculations, the trajectories of movement of various bodies are determined, including space objects. 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 component 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 a quadratic equation.
If we speak 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 (an unknown without a square with its coefficient) and c (free component, that is, an ordinary number). All this on the right side is equal to 0. In the case when such a polynomial does not have 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, in which the value of the variables is not difficult 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 bracketing the variable. 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 says that the product of two factors results in 0 only if one of them is 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. By substituting the necessary values, equating the right side to 0 and finding possible unknowns, you can find out the time elapsed from the moment the body rises to the moment it falls, as well as many other quantities. But we will talk about this later.
Factoring an Expression
The rule described above makes it possible to solve these problems in more complex cases. Consider examples with the solution of quadratic equations of this type.
X2 - 33x + 200 = 0
This square trinomial is complete. First, we transform the expression and decompose it into factors. 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; -one; 3.
Extracting the square root
Another case incomplete equation second order is an expression expressed in the language of letters in such a way that right part is built 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 after that, 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 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 land
The need for this kind of calculations appeared in ancient times, because the development of mathematics in those distant times was largely due to the need to determine the areas and perimeters of land plots with the greatest accuracy.
We should also consider examples with the solution of quadratic equations compiled on the basis of problems of this kind.
So, let's say there is a rectangular piece of land, the length of which is 16 meters more than the width. You should find the length, width and perimeter of the site, if it is known that its area is 612 m 2.
Getting down to business, at first we will make the necessary equation. Let's denote the width of the section as x, 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) \u003d 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 of them is not equal to 0 at all, 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 that we have received an expression in the form corresponding to the previously specified standard, where a=1, b=16, c= -612.
This can be an example of solving quadratic equations through the discriminant. Here necessary calculations produced 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 options. In case D>0, there are two of them; for D=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) = 2704. This indicates that our problem has an answer. If you know, to, 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 size of the land plot cannot be measured in negative values, which means that 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 the study of quadratic equations. Examples and a detailed solution of several of them will be given below.
1) 15x2 + 20x + 5 = 12x2 + 27x + 1
Let's 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 the standard one, and equate it to zero.
15x 2 + 20x + 5 - 12x 2 - 27x - 1 = 0
Having added similar ones, we determine 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 of them will be equal to 4/3, and the second 1.
2) Now we will reveal riddles of a different kind.
Let's find out if there are roots x 2 - 4x + 5 = 1 here at all? To obtain an exhaustive answer, we bring the polynomial to the corresponding familiar form and calculate the discriminant. In this example, it is not necessary to solve the quadratic equation, because the essence of the problem is not at all in this. In this case, D \u003d 16 - 20 \u003d -4, which means that there really are no roots.
Vieta's theorem
It is convenient to solve quadratic equations through the above formulas and the discriminant, when the square root is extracted from the value of the latter. But this does not always happen. However, there are many ways to get the values of variables in this case. Example: solving quadratic equations using Vieta's theorem. It is named after a man who lived in 16th-century France and had a brilliant career thanks to his mathematical talent and connections at court. His portrait can be seen in the article.
The pattern that the famous Frenchman noticed was as follows. He proved that the sum of the roots of the equation is equal to -p=b/a, and their product corresponds to q=c/a.
Now let's look at specific tasks.
3x2 + 21x - 54 = 0
For simplicity, let's 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 made a check, we will make sure that these values of the variables really fit into the expression.
Graph and Equation of a Parabola
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 little more detail. Any equation of the described type can be represented visually. Such a dependence, 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 come out. 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 graphic. And the value of the x variable is the abscissa coordinate at the points where the graph line intersects with 0x. The coordinates of the vertex can be found by the formula just given 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 parabola vertex belonging to the y-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.
From the graph of a parabola, you can also determine the roots. The reverse is also true. That is, if it is not easy to get a visual representation 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 plot.
From the history
With the help of equations containing a squared variable, in the old days, not only did mathematical calculations and determined the area of \u200b\u200bgeometric 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 suggest, the inhabitants of Babylon were among the first to solve 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 known to any student of our time.
Perhaps even earlier than the scientists of Babylon, the sage from India, Baudhayama, took up the solution of quadratic equations. This happened about eight centuries before the advent of the era of Christ. True, the second-order equations, the methods for 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 work by such great scientists as Newton, Descartes and many others.
Tasks for a quadratic equation are studied both in the school curriculum and in universities. They are understood as equations of the form a * x ^ 2 + b * x + c \u003d 0, where x- variable, a,b,c – constants; a<>0 . The problem is to find the roots of the equation.
The geometric meaning of the quadratic equation
The graph of a function that is represented by a quadratic equation is a parabola. The solutions (roots) of a quadratic equation are the points of intersection of the parabola with the x-axis. It follows that there are three possible cases:
1) the parabola has no points of intersection with the x-axis. This means that it is in the upper plane with branches up or the lower one with branches down. In such cases, the quadratic equation has no real roots (it has two complex roots).
2) the parabola has one point of intersection with the axis Ox. Such a point is called the vertex of the parabola, and the quadratic equation in it acquires its minimum or maximum value. In this case, the quadratic equation has one real root (or two identical roots).
3) The last case is more interesting in practice - there are two points of intersection of the parabola with the abscissa axis. This means that there are two real roots of the equation.
Based on the analysis of the coefficients at the powers of the variables, interesting conclusions can be drawn about the placement of the parabola.
1) If the coefficient a is greater than zero, then the parabola is directed upwards, if negative, the branches of the parabola are directed downwards.
2) If the coefficient b is greater than zero, then the vertex of the parabola lies in the left half-plane, if it takes a negative value, then in the right.
Derivation of a formula for solving a quadratic equation
Let's transfer the constant from the quadratic equation 
for the equal sign, we get the expression
Multiply both sides by 4a 
To get a full square on the left, add b ^ 2 in both parts and perform the transformation
From here we find
Formula of the discriminant and roots of the quadratic equation
The discriminant is the value of the radical expression. If it is positive, then the equation has two real roots, calculated by the formula 
When the discriminant is zero, the quadratic equation has one solution (two coinciding roots), which are easy to obtain from the above formula for D=0. When the discriminant is negative, there are no real roots. However, to study the solutions of the quadratic equation in the complex plane, and their value is calculated by the formula 
Vieta's theorem
Consider two roots of a quadratic equation and construct a quadratic equation on their basis. The Vieta theorem itself easily follows from the notation: if we have a quadratic equation of the form 
then the sum of its roots is equal to the coefficient p, taken with the opposite sign, and the product of the roots of the equation is equal to the free term q. The formula for the above will look like If the constant a in the classical equation is nonzero, then you need to divide the entire equation by it, and then apply the Vieta theorem.
Schedule of the quadratic equation on factors
Let the task be set: to decompose the quadratic equation into factors. To perform it, we first solve the equation (find the roots). Next, we substitute the found roots into the formula for expanding the quadratic equation. This problem will be solved.
Tasks for a quadratic equation
Task 1. Find the roots of a quadratic equation
x^2-26x+120=0 .
Solution: Write down the coefficients and substitute in the discriminant formula
The root of this value is 14, it is easy to find it with a calculator, or remember it with frequent use, however, for convenience, at the end of the article I will give you a list of squares of numbers that can often be found in such tasks.
The found value is substituted into the root formula 
and we get 
Task 2. solve the equation
2x2+x-3=0.
Solution: We have a complete quadratic equation, write out the coefficients and find the discriminant 
Using well-known formulas, we find the roots of the quadratic equation 
Task 3. solve the equation
9x2 -12x+4=0.
Solution: We have a complete quadratic equation. Determine the discriminant
We got the case when the roots coincide. We find the values of the roots by the formula 
Task 4. solve the equation
x^2+x-6=0 .
Solution: In cases where there are small coefficients for x, it is advisable to apply the Vieta theorem. By its condition, we obtain two equations
From the second condition, we get that the product must be equal to -6. This means that one of the roots is negative. We have the following possible pair of solutions(-3;2), (3;-2) . Taking into account the first condition, we reject the second pair of solutions.
The roots of the equation are
Task 5. Find the lengths of the sides of a rectangle if its perimeter is 18 cm and area is 77 cm 2.
Solution: Half the perimeter of a rectangle is equal to the sum of the adjacent sides. Let's denote x - the larger side, then 18-x is its smaller side. The area of a rectangle is equal to the product of these lengths:
x(18x)=77;
or
x 2 -18x + 77 \u003d 0.
Find the discriminant of the equation
We calculate the roots of the equation 
If a x=11, then 18x=7 , vice versa is also true (if x=7, then 21-x=9).
Problem 6. Factorize the quadratic 10x 2 -11x+3=0 equation.
Solution: Calculate the roots of the equation, for this we find the discriminant 
We substitute the found value into the formula of the roots and calculate 
We apply the formula for expanding the quadratic equation in terms of roots 
Expanding the brackets, we get the identity.
Quadratic equation with parameter
Example 1. For what values of the parameter a , does the equation (a-3) x 2 + (3-a) x-1 / 4 \u003d 0 have one root?
Solution: By direct substitution of the value a=3, we see that it has no solution. Further, we will use the fact that with a zero discriminant, the equation has one root of multiplicity 2. Let's write out the discriminant 
simplify it and equate to zero
We have obtained a quadratic equation with respect to the parameter a, the solution of which is easy to obtain using the Vieta theorem. The sum of the roots is 7, and their product is 12. By simple enumeration, we establish that the numbers 3.4 will be the roots of the equation. Since we have already rejected the solution a=3 at the beginning of the calculations, the only correct one will be - a=4. Thus, for a = 4, the equation has one root.
Example 2. For what values of the parameter a , the equation a(a+3)x^2+(2a+6)x-3a-9=0 has more than one root?
Solution: Consider first the singular points, they will be the values a=0 and a=-3. When a=0, the equation will be simplified to the form 6x-9=0; x=3/2 and there will be one root. For a= -3 we get the identity 0=0 .
Calculate the discriminant 
and find the values of a for which it is positive 
From the first condition we get a>3. For the second, we find the discriminant and the roots of the equation 
Let's define the intervals where the function takes positive values. By substituting the point a=0 we get 3>0
.
So, outside the interval (-3; 1/3) the function is negative. Don't forget the dot a=0 which should be excluded, since the original equation has one root in it.
As a result, we obtain two intervals that satisfy the condition of the problem 
There will be many similar tasks in practice, try to deal with the tasks yourself and do not forget to take into account conditions that are mutually exclusive. Study well the formulas for solving quadratic equations, they are quite often needed in calculations in various problems and sciences.
