Just. By formulas and clear simple rules. At the first stage

necessary given equation lead to 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 most important thing is right

determine all coefficients a, b and c.

Formula for finding the roots of a quadratic equation.

The expression under the root sign is called discriminant . As you can see, to find x, we

use only a, b and c. Those. odds from quadratic equation. Just carefully insert

values a, b and c into this formula and count. Substitute with their signs!

for example, in the equation:

a =1; b = 3; c = -4.

Substitute the values ​​and write:

Example almost solved:

This is the answer.

The most common mistakes are confusion with the signs of values a, b and with. Rather, with substitution

negative values into the formula for calculating the roots. Here the detailed formula saves

with specific numbers. If there are problems with calculations, do it!

Suppose we need to solve the following example:

Here a = -6; b = -5; c = -1

We paint everything in detail, carefully, without missing anything with all the signs and brackets:

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

Now take note practical techniques which drastically reduce the number of errors.

First reception. Don't be lazy before solving a quadratic equation bring it to 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:

Get rid of the minus. How? 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! By Vieta's theorem.

To solve the given quadratic equations, i.e. if the coefficient

x2+bx+c=0,

thenx 1 x 2 =c

x1 +x2 =−b

For a complete quadratic equation in which a≠1:

x 2 +bx+c=0,

divide the whole equation by a:

→ →

where x 1 and x 2 - roots of the equation.

Reception third. If your equation has fractional odds, - get rid of the fractions! Multiply

equation for common denominator.

Conclusion. 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 everything

equations for -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 clean, the coefficient at it equal to one, the solution can be easily verified by

Yakupova M.I. 1

Smirnova Yu.V. one

1 Municipal budgetary educational institution average comprehensive school № 11

The text of the work is placed without images and formulas.
Full version work is available in the "Files of work" tab in PDF format

History of quadratic equations

Babylon

The need to solve equations not only of the first degree, but also of the second in ancient times was caused by the need to solve problems related to finding areas land plots, with the development of astronomy and mathematics itself. Quadratic equations were able to solve about 2000 BC. e. Babylonians. The rules for solving these equations, set forth in the Babylonian texts, essentially coincide with modern ones, but these texts lack the concept negative number and common methods solutions of quadratic equations.

Ancient Greece

The solution of quadratic equations was also carried out in Ancient Greece scientists such as Diophantus, Euclid and Heron. Diophantus Diophantus of Alexandria was an ancient Greek mathematician who presumably lived in the 3rd century AD. The main work of Diophantus is "Arithmetic" in 13 books. Euclid. Euclid is an ancient Greek mathematician, the author of the first theoretical treatise on mathematics that has come down to us, Heron. Heron - Greek mathematician and engineer for the first time in Greece in the 1st century AD. gives a purely algebraic way of solving the quadratic equation

India

Problems for quadratic equations are already found in the astronomical treatise Aryabhattam, compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scholar, Brahmagupta (7th century), expounded general rule solutions of quadratic equations reduced to a single canonical form: ax2 + bх = с, a> 0. (1) In equation (1) the coefficients, can be negative. Brahmagupta's rule essentially coincides with ours. In 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 glory in popular assemblies, offering 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.

"A frisky flock of monkeys

And twelve along the vines

They began to jump, hanging

Them squared part eight

How many monkeys were

Having fun in the meadow

You tell me, in this flock?

Bhaskara's solution indicates that the author was aware of the two-valuedness of the roots of quadratic equations. Bhaskar writes the equation corresponding to the problem under the form x2 - 64x = - 768 and, in order to complete the left side of this equation to a square, he adds 322 to both parts, then getting: x2 - b4x + 322 = -768 + 1024, (x - 32) 2 \u003d 256, x - 32 \u003d ± 16, x1 \u003d 16, x2 \u003d 48.

Quadratic equations in Europe XVII century

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, is distinguished by 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. 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 others scientists way solving quadratic equations takes a modern form.

Definition of a quadratic equation

An equation of the form ax 2 + bx + c = 0, where a, b, c are numbers, is called a square equation.

Coefficients of a quadratic equation

The numbers a, b, c are the coefficients of the quadratic equation. a is the first coefficient (before x²), a ≠ 0; b is the second coefficient (before x); c is the free term (without x).

Which of these equations are not quadratic?

1. 4x² + 4x + 1 \u003d 0; 2. 5x - 7 \u003d 0; 3. - x² - 5x - 1 \u003d 0; 4. 2/x² + 3x + 4 = 0;5. ¼ x² - 6x + 1 \u003d 0; 6. 2x² = 0;

7. 4x² + 1 \u003d 0; 8. x² - 1 / x \u003d 0; 9. 2x² - x \u003d 0; 10. x² -16 = 0;11. 7x² + 5x = 0;12. -8х²= 0;13. 5x³ +6x -8= 0.

Types of quadratic equations

Name	General view of the equation	Feature (what coefficients)	Equation Examples
	ax2 + bx + c = 0	a, b, c - numbers other than 0	1/3x 2 + 5x - 1 = 0
Incomplete
			x 2 - 1/5x = 0
Given	x 2 + bx + c = 0		x 2 - 3x + 5 = 0

A reduced quadratic equation is called, in which the leading coefficient is equal to one. Such an equation can be obtained by dividing the entire expression by the leading coefficient a:

x 2 + px + q =0, p = b/a, q = c/a

A quadratic equation is said to be complete if all of its coefficients are nonzero.

Such a quadratic equation is called incomplete if at least one of the coefficients, except for the highest one (either the second coefficient or the free term), is equal to zero.

Ways to solve quadratic equations

I way. General formula for calculating roots

To find the roots of a quadratic equation ax 2 + b + c = 0 in general case the following algorithm should be used:

Calculate the value of the discriminant of the quadratic equation: this is the expression for it D= b 2 - 4ac

Derivation of the formula:

Note: it is obvious that the formula for a root of multiplicity 2 is a special case of the general formula, it is obtained by substituting the equality D=0 into it, and the conclusion about the absence of real roots at D0, and (displaystyle (sqrt (-1))=i) = i.

The described method is universal, but it is far from the only one. The solution of one equation can be approached in different ways, the preferences usually depend on the solver itself. In addition, often for this some of the methods turns out to be much more elegant, simpler, less time-consuming than the standard one.

II way. The roots of a quadratic equation with an even coefficient b III way. Solving incomplete quadratic equations

IV way. Using partial ratios of coefficients

There are special cases of quadratic equations in which the coefficients are in proportion to each other, which makes it much easier to solve them.

The roots of a quadratic equation in which the sum of the leading coefficient and the free term is equal to the second coefficient

If in a quadratic equation ax 2 + bx + c = 0 the sum of the first coefficient and the free term is equal to the second coefficient: a+b=c, then its roots are -1 and the number the opposite of free term to the leading coefficient ( -c/a).

Hence, before solving any quadratic equation, one should check the possibility of applying this theorem to it: compare the sum of the leading coefficient and the free term with the second coefficient.

The roots of a quadratic equation whose sum of all coefficients is zero

If in a quadratic equation the sum of all its coefficients is equal to zero, then the roots of such an equation are 1 and the ratio of the free term to the leading coefficient ( c/a).

Hence, before solving the equation by standard methods, one should check the applicability of this theorem to it: add all the coefficients given equation and see if this sum is zero.

V way. Decomposition of a square trinomial into linear factors

If a trinomial of the form (displaystyle ax^(2)+bx+c(anot =0))ax 2 + bx + c(a ≠ 0) can somehow be represented as a product of linear factors (displaystyle (kx+m)(lx+n)=0)(kx + m)(lx + n), then we can find the roots of the equation ax 2 + bx + c = 0- they will be -m / k and n / l, indeed, because (displaystyle (kx+m)(lx+n)=0Longleftrightarrow kx+m=0cup lx+n=0)(kx + m)(lx + n) = 0 kx + mUlx + n, and by solving the indicated linear equations, we get the above. Note that square trinomial is not always decomposed into linear factors with real coefficients: this is possible if the equation corresponding to it has real roots.

Consider some special cases

Using the formula for the square of the sum (difference)

If a square trinomial has the form (displaystyle (ax)^(2)+2abx+b^(2))ax 2 + 2abx + b 2 , then applying the above formula to it, we can factor it into linear factors and, therefore, find roots:

(ax) 2 + 2abx + b 2 = (ax + b) 2

Selection full square sums (differences)

Also, the named formula is used using the method called "selection of the full square of the sum (difference)". In relation to the given quadratic equation with the notation introduced earlier, this means the following:

Note: if you notice given formula coincides with the one proposed in the “Roots of the reduced quadratic equation” section, which, in turn, can be obtained from the general formula (1) by substituting the equality a=1. This fact is not just a coincidence: by the described method, after some additional reasoning, it is possible to deduce and general formula, as well as to prove the properties of the discriminant.

VI way. Using direct and inverse Vieta theorem

Vieta's direct theorem (see below in the section of the same name) and its inverse theorem allow us to solve the reduced quadratic equations orally without resorting to rather cumbersome calculations using formula (1).

According to converse theorem, any pair of numbers (number) (displaystyle x_(1),x_(2)) x 1 , x 2 being the solution of the system of equations below, are the roots of the equation

In the general case, that is, for a non-reduced quadratic equation ax 2 + bx + c = 0

x 1 + x 2 \u003d -b / a, x 1 * x 2 \u003d c / a

A direct theorem will help you verbally select numbers that satisfy these equations. With its help, you can determine the signs of the roots without knowing the roots themselves. To do this, follow the rule:

1) if the free term is negative, then the roots have different sign, and the greatest modulus of the roots is the sign opposite to the sign of the second coefficient of the equation;

2) if the free term is positive, then both roots have the same sign, and this is the opposite sign of the second coefficient.

7th way. Transfer method

The so-called "transfer" method makes it possible to reduce the solution of non-reduced and non-transformable equations to the form of equations reduced with integer coefficients by dividing them by the leading coefficient of equations to the solution of equations reduced with integer coefficients. It is as follows:

Next, the equation is solved orally in the manner described above, then they return to the original variable and find the roots of the equations (displaystyle y_(1)=ax_(1)) y 1 = ax 1 and y 2 = ax 2 .(displaystyle y_(2)=ax_(2))

geometric sense

The graph of a quadratic function is a parabola. The solutions (roots) of a quadratic equation are the abscissas of the points of intersection of the parabola with the abscissa axis. If the parabola described quadratic function, does not intersect with the x-axis, the equation has no real roots. If the parabola intersects the x-axis at one point (at the vertex of the parabola), the equation has one real root (the equation is also said to have two coinciding roots). If the parabola intersects the x-axis at two points, the equation has two real roots (see image on the right.)

If coefficient (displaystyle a) a positive, the branches of the parabola are directed upwards and vice versa. If the coefficient (display style b) bpositive (when positive (displaystyle a) a, if negative, vice versa), then the vertex of the parabola lies in the left half-plane and vice versa.

Application of quadratic equations in life

The quadratic equation is widespread. It is used in many calculations, structures, sports, and also around us.

Consider and give some examples of the application of the quadratic equation.

Sport. High jumps: during the run-up of the jumper for the most accurate hit on the repulsion bar and high flight use the calculations associated with the parabola.

Also, similar calculations are needed in throwing. The flight range of an object depends on a quadratic equation.

Astronomy. The trajectory of the planets can be found using a quadratic equation.

Airplane flight. The takeoff of an aircraft is the main component of the flight. Here the calculation is taken for a small resistance and takeoff acceleration.

Also, quadratic equations are used in various economic disciplines, in programs for processing sound, video, vector and raster graphics.

Conclusion

As a result of the work done, it turned out that quadratic equations attracted scientists in ancient times, they already encountered them when solving some problems and tried to solve them. Considering various ways solving quadratic equations, I came to the conclusion that not all of them are simple. In my opinion the most the best way solving quadratic equations is a solution by formulas. Formulas are easy to remember, this method is universal. The hypothesis that equations are widely used in life and mathematics was confirmed. Having studied the topic, I learned a lot interesting facts about quadratic equations, their use, application, types, solutions. And I will continue to study them with pleasure. I hope this will help me do well in my exams.

List of used literature

Site materials:

Wikipedia

Open lesson.rf

Handbook of elementary mathematics Vygodsky M. Ya.

The transformation of a complete quadratic equation into an incomplete one looks like this (for the case \(b=0\)):

For cases when \(c=0\) or when both coefficients are equal to zero, everything is similar.

Please note that \(a\) is not equal to zero, it cannot be equal to zero, since in this case it turns into:

Solution of incomplete quadratic equations.

First of all, you need to understand that the incomplete quadratic equation is still, therefore, it can be solved in the same way as the usual quadratic (through). To do this, we simply add the missing component of the equation with a zero coefficient.

Example : Find the roots of the equation \(3x^2-27=0\)
Decision :

		We have an incomplete quadratic equation with the coefficient \(b=0\). That is, we can write the equation in following form:
\(3x^2+0\cdot x-27=0\)		In fact, here is the same equation as at the beginning, but now it can be solved as an ordinary square. First we write down the coefficients.
\(a=3;\) \(b=0;\) \(c=-27;\)		Calculate the discriminant using the formula \(D=b^2-4ac\)
\(D=0^2-4\cdot3\cdot(-27)=\) \(=0+324=324\)		Let's find the roots of the equation using the formulas \(x_(1)=\)\(\frac(-b+\sqrt(D))(2a)\) and \(x_(2)=\)\(\frac(-b-\sqrt(D) )(2a)\)
\(x_(1)=\) \(\frac(-0+\sqrt(324))(2\cdot3)\)\(=\)\(\frac(18)(6)\) \(=3\) \(x_(2)=\) \(\frac(-0-\sqrt(324))(2\cdot3)\)\(=\)\(\frac(-18)(6)\) \(=-3\)		Write down the answer

Answer : \(x_(1)=3\); \(x_(2)=-3\)

Example : Find the roots of the equation \(-x^2+x=0\)
Decision :

		Again an incomplete quadratic equation, but now zero is equal to the coefficient\(c\). We write the equation as complete.

Quadratic equations. Discriminant. Solution, examples.

Attention!
There are additional
materials 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.

talking mathematical language, 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 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...

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 becomes 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 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!

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

Did you know?) Yes! This is 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 from this? 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. Transferring 9 to 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. From a negative number the square root is not taken. Well, okay. This means there are no solutions.

To be honest, at simple solution quadratic equations, the concept of discriminant is not particularly 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 difficult tasks, without knowledge meaning and discriminant formula not enough. Especially - in equations with parameters. Such equations are aerobatics at 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 attentively substitute them into the root formula and attentively count the result. Did you understand that keyword here - attentively?

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 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! There 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.

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 help you Section 555. There, all these examples are sorted by bones. Showing main errors in the solution. Of course, it also talks about the use identical transformations in decision various equations. 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.