I hope that after studying this article, you will learn how to find the roots of a complete quadratic equation.

With the help of the discriminant, only complete quadratic equations are solved; to solve incomplete quadratic equations use other methods that you will find in the article "Solving incomplete quadratic equations".

What quadratic equations are called complete? This 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 complete quadratic equation, you need to calculate the discriminant D.

D \u003d b 2 - 4ac.

Depending on what value the discriminant has, we will write down the answer.

If the discriminant is a negative number (D< 0),то корней нет.

If the discriminant zero, then x \u003d (-b) / 2a. When the discriminant positive number(D > 0),

then x 1 = (-b - √D)/2a, and x 2 = (-b + √D)/2a.

For instance. solve the equation x 2– 4x + 4= 0.

D \u003d 4 2 - 4 4 \u003d 0

x = (- (-4))/2 = 2

Answer: 2.

Solve Equation 2 x 2 + x + 3 = 0.

D \u003d 1 2 - 4 2 3 \u003d - 23

Answer: no roots.

Solve Equation 2 x 2 + 5x - 7 = 0.

D \u003d 5 2 - 4 2 (-7) \u003d 81

x 1 \u003d (-5 - √81) / (2 2) \u003d (-5 - 9) / 4 \u003d - 3.5

x 2 \u003d (-5 + √81) / (2 2) \u003d (-5 + 9) / 4 \u003d 1

Answer: - 3.5; one.

So let's imagine 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 the equation was written as a polynomial standard view

a x 2 + bx + c, otherwise you can make a mistake. For example, in writing the equation x + 3 + 2x 2 = 0, you can mistakenly decide that

a = 1, b = 3 and c = 2. Then

D \u003d 3 2 - 4 1 2 \u003d 1 and then the equation has two roots. And this is not true. (See example 2 solution 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 (the monomial with the largest exponent should be in the first place, that is a x 2 , then with less – bx, and then the free term With.

When solving the above quadratic equation and the quadratic equation with an even coefficient for the second term, other formulas can also be used. Let's get acquainted with these formulas. If in the full quadratic equation with the second term the coefficient is even (b = 2k), then the equation can be solved using the formulas shown in the diagram of Figure 2.

A complete quadratic equation is called reduced if the coefficient at x 2 equals unity and the equation takes the form x 2 + px + q = 0. Such an equation can be given to solve, or is obtained by dividing all the coefficients of the equation by the coefficient a standing at x 2 .

Figure 3 shows a diagram of the solution of the reduced square
equations. Consider the 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 Figure 1.

D \u003d 6 2 - 4 3 (- 6) \u003d 36 + 72 \u003d 108

√D = √108 = √(36 3) = 6√3

x 1 \u003d (-6 - 6 √ 3) / (2 3) \u003d (6 (-1- √ (3))) / 6 \u003d -1 - √ 3

x 2 \u003d (-6 + 6 √ 3) / (2 3) \u003d (6 (-1 + √ (3))) / 6 \u003d -1 + √ 3

Answer: -1 - √3; –1 + √3

You can see that the coefficient at x in this equation even number, that is, b \u003d 6 or b \u003d 2k, whence k \u003d 3. Then let's try to solve the equation using the formulas shown in the diagram of the figure D 1 \u003d 3 2 - 3 (- 6) \u003d 9 + 18 \u003d 27

√(D 1) = √27 = √(9 3) = 3√3

x 1 \u003d (-3 - 3√3) / 3 \u003d (3 (-1 - √ (3))) / 3 \u003d - 1 - √3

x 2 \u003d (-3 + 3√3) / 3 \u003d (3 (-1 + √ (3))) / 3 \u003d - 1 + √3

Answer: -1 - √3; –1 + √3. Noticing that all the coefficients in this quadratic equation are divisible by 3 and dividing, we get the reduced quadratic equation x 2 + 2x - 2 = 0 We solve this equation using the formulas for the reduced quadratic
equations figure 3.

D 2 \u003d 2 2 - 4 (- 2) \u003d 4 + 8 \u003d 12

√(D 2) = √12 = √(4 3) = 2√3

x 1 \u003d (-2 - 2√3) / 2 \u003d (2 (-1 - √ (3))) / 2 \u003d - 1 - √3

x 2 \u003d (-2 + 2 √ 3) / 2 \u003d (2 (-1 + √ (3))) / 2 \u003d - 1 + √ 3

Answer: -1 - √3; –1 + √3.

As you can see, when solving this equation using different formulas, we got the same answer. Therefore, having well mastered the formulas shown in the diagram of Figure 1, you can always solve any complete quadratic equation.

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Discriminant - 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?

1. She says if there are valid results.
2. 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, then it will become clear that this is a “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. The general formula for calculating 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:

1. Determining the number of valid solutions.
2. Calculation d = D^(1/2).
3. Finding the result according to the formula (-j +/- d) / (2 * i).
4. 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:

1. the polynomial expands into the difference of squares with a negative free term;
2. 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 a square trinomial, where the coefficient in front of the highest 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, a discriminant equal to 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.

The discriminant, as well as quadratic equations, begin to be studied in the algebra course in grade 8. You can solve a quadratic equation through the discriminant and using the Vieta theorem. The methodology for studying quadratic equations, as well as the discriminant formula, is rather unsuccessfully instilled in schoolchildren, like much in real education. Therefore pass school years, training in grades 9-11 replaces " higher education"and everyone is looking again - "How to solve a quadratic equation?", "How to find the roots of an equation?", "How to find the discriminant?" and...

Discriminant Formula

The discriminant D of the quadratic equation a*x^2+bx+c=0 is D=b^2–4*a*c.
The roots (solutions) of the quadratic equation depend on the sign of the discriminant (D):
D>0 - the equation has 2 different real roots;
D=0 - the equation has 1 root (2 coinciding roots):
D<0 – не имеет действительных корней (в школьной теории). В ВУЗах изучают комплексные числа и уже на множестве комплексных чисел уравнение с отрицательным дискриминантом имеет два комплексных корня.
The formula for calculating the discriminant is quite simple, so many sites offer an online discriminant calculator. We have not figured out this kind of scripts yet, so who knows how to implement this, please write to the mail This email address is being protected from spambots. You must have JavaScript enabled to view. .

General formula for finding the roots of a quadratic equation:

The roots of the equation are found by the formula
If the coefficient of the variable squared is paired, then it is advisable to calculate not the discriminant, but its fourth part
In such cases, the roots of the equation are found by the formula

The second way to find roots is Vieta's Theorem.

The theorem is formulated not only for quadratic equations, but also for polynomials. You can read this on Wikipedia or other electronic resources. However, to simplify, consider that part of it that concerns the reduced quadratic equations, that is, equations of the form (a=1)
The essence of the Vieta formulas is that the sum of the roots of the equation is equal to the coefficient of the variable, taken with the opposite sign. The product of the roots of the equation is equal to the free term. The formulas of Vieta's theorem have a notation.
The derivation of the Vieta formula is quite simple. Let's write the quadratic equation in terms of prime factors
As you can see, everything ingenious is simple at the same time. It is effective to use the Vieta formula when the difference in the modulus of the roots or the difference in the modulus of the roots is 1, 2. For example, the following equations, according to the Vieta theorem, have roots




Up to 4 equation analysis should look like this. The product of the equation's roots is 6, so the roots can be the values ​​(1, 6) and (2, 3) or pairs with the opposite sign. The sum of the roots is 7 (the coefficient of the variable with the opposite sign). From here we conclude that the solutions of the quadratic equation are equal to x=2; x=3.
It is easier to select the roots of the equation among the divisors of the free term, correcting their sign in order to fulfill the Vieta formulas. At the beginning, this seems difficult to do, but with practice on a number of quadratic equations, this technique will be more efficient than calculating the discriminant and finding the roots of the quadratic equation in the classical way.
As you can see, the school theory of studying the discriminant and ways to find solutions to the equation is devoid of practical meaning - "Why do schoolchildren need a quadratic equation?", "What is the physical meaning of the discriminant?".

Let's try to figure it out what does the discriminant describe?

In the course of algebra, they study functions, schemes for studying functions and plotting functions. Of all the functions, the parabola occupies an important place, the equation of which can be written in the form
So the physical meaning of the quadratic equation is the zeros of the parabola, that is, the points of intersection of the graph of the function with the abscissa axis Ox
I ask you to remember the properties of parabolas that are described below. The time will come to take exams, tests, or entrance exams and you will be grateful for the reference material. The sign of the variable in the square corresponds to whether the branches of the parabola on the graph will go up (a>0),

or a parabola with branches down (a<0) .

The vertex of the parabola lies midway between the roots

The physical meaning of the discriminant:

If the discriminant is greater than zero (D>0), the parabola has two points of intersection with the Ox axis.
If the discriminant is equal to zero (D=0), then the parabola at the top touches the x-axis.
And the last case, when the discriminant is less than zero (D<0) – график параболы принадлежит плоскости над осью абсцисс (ветки параболы вверх), или график полностью под осью абсцисс (ветки параболы опущены вниз).

Incomplete quadratic equations

Quadratic equations. Discriminant. Solution, examples.

Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")

Types of quadratic equations

What is a quadratic equation? What does it look like? In term quadratic equation keyword is "square". It means that in the equation necessarily there must be an x ​​squared. In addition to it, in the equation there may be (or may not be!) Just x (to the first degree) and just a number (free member). And there should not be x's in a degree greater than two.

In mathematical terms, a quadratic equation is an equation of the form:

Here a, b and c- some numbers. b and c- absolutely any, but a- anything but zero. For instance:

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...

In these quadratic equations, on the left, there is full set members. x squared with coefficient a, x to the first power with coefficient b and free member of

Such quadratic equations are called complete.

And if b= 0, what will we get? We have X will disappear in the first degree. This happens from multiplying by zero.) It turns out, for example:

5x 2 -25 = 0,

2x 2 -6x=0,

-x 2 +4x=0

Etc. And if both coefficients b and c are equal to zero, then it is even simpler:

2x 2 \u003d 0,

-0.3x 2 \u003d 0

Such equations, where something is missing, are called incomplete quadratic equations. Which is quite logical.) Please note that x squared is present in all equations.

By the way why a can't be zero? And you substitute instead a zero.) The X in the square will disappear! The equation will become linear. And it's done differently...

That's all the main types of quadratic equations. Complete and incomplete.

Solution of quadratic equations.

Solution of complete quadratic equations.

Quadratic equations are easy to solve. According to formulas and clear simple rules. At the first stage, it is necessary to bring the given equation to the standard form, i.e. to the view:

If the equation is already given to you in this form, you do not need to do the first stage.) The main thing is to correctly determine all the coefficients, a, b and c.

The formula for finding the roots of a quadratic equation looks like this:

The expression under the root sign is called discriminant. But more about him below. 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, in the equation:

a =1; b = 3; c= -4. Here we write:

Example almost solved:

This is the answer.

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 get 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!

But, often, quadratic equations look slightly different. For example, like this:

Did you know?) Yes! This incomplete quadratic equations.

Solution of incomplete quadratic equations.

They can also be solved by the general formula. 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 formulas. 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 1 = 0, x 2 = 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 the general formula. I note, by the way, which X will be the first, and which the second - it is absolutely indifferent. Easy to write in order x 1- whichever is less x 2- that which is more.

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 1 = -3, x 2 = 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 ...

Discriminant. Discriminant formula.

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.) I remind you of the most general formula for solving any quadratic equations:

The expression under the root sign is called the discriminant. The discriminant is usually denoted by the letter D. Discriminant formula:

D = b 2 - 4ac

And what is so special about this expression? Why does it deserve a special name? What meaning of the discriminant? After all -b, or 2a in this formula they don’t specifically name ... Letters and letters.

The point is this. When solving a quadratic equation using this formula, it is possible 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. Since adding or subtracting zero in the numerator does not change anything. Strictly speaking, this is not a single root, but two identical. But, in a simplified version, it is customary to talk about one solution.

3. The discriminant is negative. A negative number does not take the square root. Well, okay. This means there are no solutions.

To be honest, with a simple solution of quadratic equations, the concept of a discriminant is not really required. We substitute the values ​​​​of the coefficients in the formula, and we consider. There everything turns out by itself, and two roots, and one, and not a single one. However, when solving more complex tasks, without knowledge meaning and discriminant formula not enough. Especially - in equations with parameters. Such equations are aerobatics for the GIA and the Unified State Examination!)

So, how to solve quadratic equations through the discriminant you remembered. Or learned, which is also not bad.) You know how to 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 the key word here is - carefully?

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 right!
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 lesson "How to solve equations? Identity transformations". When working with fractions, errors, for some reason, 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 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.

Practical Tips:

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!

Now you can decide.)

Solve Equations:

8x 2 - 6x + 1 = 0

x 2 + 3x + 8 = 0

x 2 - 4x + 4 = 0

(x+1) 2 + x + 1 = (x+1)(x+2)

Answers (in disarray):

x 1 = 0
x 2 = 5

x 1.2 =2

x 1 = 2
x 2 \u003d -0.5

x - any number

x 1 = -3
x 2 = 3

no solutions

x 1 = 0.25
x 2 \u003d 0.5

Does everything fit? Fine! Quadratic equations are not your headache. The first three turned out, but the rest did not? Then the problem is not in quadratic equations. The problem is in identical transformations of equations. Take a look at the link, it's helpful.

Doesn't quite work? Or does it not work at all? Then Section 555 will help you. There, all these examples are sorted by bones. Showing main errors in the solution. Of course, the application of identical transformations in solving various equations is also described. Helps a lot!

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

Kopyevskaya rural secondary school

10 Ways to Solve Quadratic Equations

Head: Patrikeeva Galina Anatolyevna,

mathematic teacher

s.Kopyevo, 2007

1. History of the development of quadratic equations

1.1 Quadratic equations in ancient Babylon

1.2 How Diophantus compiled and solved quadratic equations

1.3 Quadratic equations in India

1.4 Quadratic equations in al-Khwarizmi

1.5 Quadratic equations in Europe XIII - XVII centuries

1.6 About Vieta's theorem

2. Methods for solving quadratic equations

Conclusion

Literature

1. History of the development of quadratic equations

1.1 Quadratic equations in ancient Babylon

The need to solve equations not only of the first, but also of the second degree in ancient times was caused by the need to solve problems related to finding the areas of land and earthworks of a military nature, as well as the development of astronomy and mathematics itself. Quadratic equations were able to solve about 2000 BC. e. Babylonians.

Applying modern algebraic notation, we can say that in their cuneiform texts there are, in addition to incomplete ones, such, for example, complete quadratic equations:

X 2 + X = ¾; X 2 - X = 14,5

The rule for solving these equations, stated in the Babylonian texts, coincides essentially with the modern one, but it is not known how the Babylonians came to this rule. Almost all the cuneiform texts found so far give only problems with solutions stated in the form of recipes, with no indication of how they were found.

Despite the high level of development of algebra in Babylon, the cuneiform texts lack the concept of a negative number and general methods for solving quadratic equations.

1.2 How Diophantus compiled and solved quadratic equations.

Diophantus' Arithmetic does not contain a systematic exposition of algebra, but it contains a systematic series of problems, accompanied by explanations and solved by formulating equations of various degrees.

When compiling equations, Diophantus skillfully chooses unknowns to simplify the solution.

Here, for example, is one of his tasks.

Task 11."Find two numbers knowing that their sum is 20 and their product is 96"

Diophantus argues as follows: it follows from the condition of the problem that the desired numbers are not equal, since if they were equal, then their product would be equal not to 96, but to 100. Thus, one of them will be more than half of their sum, i.e. . 10+x, the other is smaller, i.e. 10's. The difference between them 2x .

Hence the equation:

(10 + x)(10 - x) = 96

100 - x 2 = 96

x 2 - 4 = 0 (1)

From here x = 2. One of the desired numbers is 12 , other 8 . Solution x = -2 for Diophantus does not exist, since Greek mathematics knew only positive numbers.

If we solve this problem by choosing one of the desired numbers as the unknown, then we will come to the solution of the equation

y(20 - y) = 96,

y 2 - 20y + 96 = 0. (2)

It is clear that Diophantus simplifies the solution by choosing the half-difference of the desired numbers as the unknown; he manages to reduce the problem to solving an incomplete quadratic equation (1).

1.3 Quadratic equations in India

Problems for quadratic equations are already found in the astronomical tract "Aryabhattam", compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scientist, Brahmagupta (7th century), outlined the general rule for solving quadratic equations reduced to a single canonical form:

ah 2+ b x = c, a > 0. (1)

In equation (1), the coefficients, except for a, can also be negative. Brahmagupta's rule essentially coincides with ours.

V ancient india public competitions in solving difficult problems were common. In one of the old Indian books, the following is said about such competitions: “As the sun outshines the stars with its brilliance, so scientist man eclipse the glory of another in public meetings, proposing and solving algebraic problems. Tasks were often dressed in poetic form.

Here is one of the problems of the famous Indian mathematician of the XII century. Bhaskara.

Task 13.

“A frisky flock of monkeys And twelve in vines ...

Having eaten power, had fun. They began to jump, hanging ...

Part eight of them in a square How many monkeys were there,

Having fun in the meadow. You tell me, in this flock?

Bhaskara's solution indicates that he knew about the two-valuedness of the roots of quadratic equations (Fig. 3).

The equation corresponding to problem 13 is:

( x /8) 2 + 12 = x

Bhaskara writes under the guise of:

x 2 - 64x = -768

and, to complete the left side of this equation to a square, he adds to both sides 32 2 , getting then:

x 2 - 64x + 32 2 = -768 + 1024,

(x - 32) 2 = 256,

x - 32 = ± 16,

x 1 = 16, x 2 = 48.

1.4 Quadratic equations in al-Khorezmi

Al-Khorezmi's algebraic treatise gives a classification of linear and quadratic equations. The author lists 6 types of equations, expressing them as follows:

1) "Squares are equal to roots", i.e. ax 2 + c = b X.

2) "Squares are equal to number", i.e. ax 2 = s.

3) "The roots are equal to the number", i.e. ah = s.

4) "Squares and numbers are equal to roots", i.e. ax 2 + c = b X.

5) "Squares and roots are equal to the number", i.e. ah 2+ bx = s.

6) "Roots and numbers are equal to squares", i.e. bx + c \u003d ax 2.

For al-Khorezmi, who avoided the use of negative numbers, the terms of each of these equations are summands, not subtrahends. In this case, equations that do not have positive solutions are obviously not taken into account. The author outlines ways to solve the indicated equations, using the techniques of al-jabr and al-muqabala. His decisions, of course, do not completely coincide with ours. Not to mention the fact that it is purely rhetorical, it should be noted, for example, that when solving an incomplete quadratic equation of the first type

al-Khorezmi, like all mathematicians before the 17th century, does not take into account the zero solution, probably because it does not matter in specific practical problems. When solving the complete quadratic equations of al - Khorezmi on partial numerical examples sets out the decision rules, and then the geometric proofs.

Task 14.“The square and the number 21 are equal to 10 roots. Find the root" (assuming the root of the equation x 2 + 21 = 10x).

The author's solution goes something like this: divide the number of roots in half, you get 5, multiply 5 by itself, subtract 21 from the product, 4 remains. Take the root of 4, you get 2. Subtract 2 from 5, you get 3, this will be the desired root. Or add 2 to 5, which will give 7, this is also a root.

Treatise al - Khorezmi is the first book that has come down to us, in which the classification of quadratic equations is systematically stated and formulas for their solution are given.

1.5 Quadratic equations in Europe XIII - XVII centuries

Formulas for solving quadratic equations on the model of al - Khorezmi in Europe were first set forth in the "Book of the Abacus", written in 1202 by the Italian mathematician Leonardo Fibonacci. This voluminous work, which reflects the influence of mathematics, both the countries of Islam and Ancient Greece, differs in both completeness and clarity of presentation. The author independently developed some new algebraic examples problem solving and was the first in Europe to approach the introduction of negative numbers. His book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many tasks from the "Book of the Abacus" passed into almost all European textbooks of the 16th - 17th centuries. and partly XVIII.

The general rule for solving quadratic equations reduced to a single canonical form:

x 2+ bx = with,

for all possible combinations of signs of the coefficients b , With was formulated in Europe only in 1544 by M. Stiefel.

Derivation of the formula for solving a quadratic equation in general view Viet has, but Viet recognized only positive roots. The Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 16th century. Take into account, in addition to positive, and negative roots. Only in the XVII century. Thanks to the work of Girard, Descartes, Newton and other scientists, the way to solve quadratic equations takes on a modern look.

1.6 About Vieta's theorem

The theorem expressing the relationship between the coefficients of a quadratic equation and its roots, bearing the name of Vieta, was formulated by him for the first time in 1591 as follows: “If B + D multiplied by A - A 2 , equals BD, then A equals V and equal D ».

To understand Vieta, one must remember that A, like any vowel, meant for him the unknown (our X), the vowels V, D- coefficients for the unknown. In the language of modern algebra, Vieta's formulation above means: if

(a + b )x - x 2 = ab ,

x 2 - (a + b )x + a b = 0,

x 1 = a, x 2 = b .

Expressing the relationship between the roots and coefficients of equations by general formulas written using symbols, Viet established uniformity in the methods of solving equations. However, the symbolism of Vieta is still far from modern look. He did not recognize negative numbers, and therefore, when solving equations, he considered only cases where all roots are positive.

2. Methods for solving quadratic equations

Quadratic equations are the foundation on which the majestic edifice of algebra rests. Quadratic equations are widely used in solving trigonometric, exponential, logarithmic, irrational and transcendental equations and inequalities. We all know how to solve quadratic equations from school (grade 8) until graduation.