Formulas for the roots of a quadratic equation. The cases of real, multiple and complex roots are considered. Factorization of a square trinomial. Geometric interpretation. Examples of determining roots and factorization.

Basic formulas

Consider the quadratic equation:
(1) .
The roots of a quadratic equation(1) are determined by the formulas:
; .
These formulas can be combined like this:
.
When the roots of the quadratic equation are known, then the polynomial of the second degree can be represented as a product of factors (factored):
.

Further, we assume that are real numbers.
Consider discriminant of a quadratic equation:
.
If the discriminant is positive, then the quadratic equation (1) has two different real roots:
; .
Then the factorization of the square trinomial has the form:
.
If the discriminant is zero, then the quadratic equation (1) has two multiple (equal) real roots:
.
Factorization:
.
If the discriminant is negative, then the quadratic equation (1) has two complex conjugate roots:
;
.
Here is the imaginary unit, ;
and are the real and imaginary parts of the roots:
; .
Then

.

Graphic interpretation

If we graph the function
,
which is a parabola, then the points of intersection of the graph with the axis will be the roots of the equation
.
When , the graph intersects the abscissa axis (axis) at two points.
When , the graph touches the x-axis at one point.
When , the graph does not cross the x-axis.

Below are examples of such graphs.

Useful Formulas Related to Quadratic Equation

(f.1) ;
(f.2) ;
(f.3) .

Derivation of the formula for the roots of a quadratic equation

We perform transformations and apply formulas (f.1) and (f.3):




,
where
; .

So, we got the formula for the polynomial of the second degree in the form:
.
From this it can be seen that the equation

performed at
and .
That is, and are the roots of the quadratic equation
.

Examples of determining the roots of a quadratic equation

Example 1

(1.1) .

Decision

.
Comparing with our equation (1.1), we find the values ​​of the coefficients:
.
Finding the discriminant:
.
Since the discriminant is positive, the equation has two real roots:
;
;
.

From here we obtain the decomposition of the square trinomial into factors:

.

Graph of the function y = 2 x 2 + 7 x + 3 crosses the x-axis at two points.

Let's plot the function
.
The graph of this function is a parabola. It crosses the x-axis (axis) at two points:
and .
These points are the roots of the original equation (1.1).

Answer

;
;
.

Example 2

Find the roots of a quadratic equation:
(2.1) .

Decision

We write the quadratic equation in general form:
.
Comparing with the original equation (2.1), we find the values ​​of the coefficients:
.
Finding the discriminant:
.
Since the discriminant is zero, the equation has two multiple (equal) roots:
;
.

Then the factorization of the trinomial has the form:
.

Graph of the function y = x 2 - 4 x + 4 touches the x-axis at one point.

Let's plot the function
.
The graph of this function is a parabola. It touches the x-axis (axis) at one point:
.
This point is the root of the original equation (2.1). Since this root is factored twice:
,
then such a root is called a multiple. That is, they consider that there are two equal roots:
.

Answer

;
.

Example 3

Find the roots of a quadratic equation:
(3.1) .

Decision

We write the quadratic equation in general form:
(1) .
Let us rewrite the original equation (3.1):
.
Comparing with (1), we find the values ​​of the coefficients:
.
Finding the discriminant:
.
The discriminant is negative, . Therefore, there are no real roots.

You can find complex roots:
;
;
.

Then


.

The graph of the function does not cross the x-axis. There are no real roots.

Let's plot the function
.
The graph of this function is a parabola. It does not cross the abscissa (axis). Therefore, there are no real roots.

Answer

There are no real roots. Complex roots:
;
;
.

Bibliographic description: Gasanov A. R., Kuramshin A. A., Elkov A. A., Shilnenkov N. V., Ulanov D. D., Shmeleva O. V. Methods for solving quadratic equations // Young scientist. - 2016. - No. 6.1. - S. 17-20..02.2019).



Our project is dedicated to the ways of solving quadratic equations. The purpose of the project: to learn how to solve quadratic equations in ways that are not included in the school curriculum. Task: find all possible ways to solve quadratic equations and learn how to use them yourself and introduce classmates to these methods.

What are "quadratic equations"?

Quadratic equation- equation of the form ax2 + bx + c = 0, where a, b, c- some numbers ( a ≠ 0), x- unknown.

The numbers a, b, c are called the coefficients of the quadratic equation.

• a is called the first coefficient;
• b is called the second coefficient;
• c - free member.

And who was the first to "invent" quadratic equations?

Some algebraic techniques for solving linear and quadratic equations were known as early as 4000 years ago in Ancient Babylon. The found ancient Babylonian clay tablets, dated somewhere between 1800 and 1600 BC, are the earliest evidence of the study of quadratic equations. The same tablets contain methods for solving certain types of quadratic equations.

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.

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.

Babylonian mathematicians from about the 4th century B.C. used the square complement method to solve equations with positive roots. Around 300 B.C. Euclid came up with a more general geometric solution method. The first mathematician who found solutions to an equation with negative roots in the form of an algebraic formula was an Indian scientist. Brahmagupta(India, 7th century AD).

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

ax2 + bx = c, a>0

In this equation, 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 a learned person will outshine the glory in public meetings, proposing and solving algebraic problems.” Tasks were often dressed in poetic form.

In an algebraic treatise Al-Khwarizmi a classification of linear and quadratic equations is given. The author lists 6 types of equations, expressing them as follows:

1) “Squares are equal to roots”, i.e. ax2 = bx.

2) “Squares are equal to number”, i.e. ax2 = c.

3) "The roots are equal to the number", i.e. ax2 = c.

4) “Squares and numbers are equal to roots”, i.e. ax2 + c = bx.

5) “Squares and roots are equal to number”, i.e. ax2 + bx = c.

6) “Roots and numbers are equal to squares”, i.e. bx + c == ax2.

For Al-Khwarizmi, who avoided the use of negative numbers, the terms of each of these equations are addends, not subtractions. In this case, equations that do not have positive solutions are obviously not taken into account. The author outlines the methods for solving these equations, using the methods of al-jabr and al-muqabala. His decision, of course, does 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-Khwarizmi, like all mathematicians before the 17th century, does not take into account the zero solution, probably because in specific practical tasks, it does not matter. When solving complete quadratic equations, Al-Khwarizmi sets out the rules for solving them using particular numerical examples, and then their geometric proofs.

Forms for solving quadratic equations on the model of Al-Khwarizmi in Europe were first described in the "Book of the Abacus", written in 1202. Italian mathematician Leonard Fibonacci. The author independently developed some new algebraic examples of problem solving and was the first in Europe to approach the introduction of negative numbers.

This book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many tasks from this book were transferred to almost all European textbooks of the 14th-17th centuries. The general rule for solving quadratic equations reduced to a single canonical form x2 + bx = c with all possible combinations of signs and coefficients b, c, was formulated in Europe in 1544. M. Stiefel.

Vieta has a general derivation of the formula for solving a quadratic equation, but Vieta recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli 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 Girard, Descartes, Newton and other scientists, the way of solving quadratic equations takes on a modern form.

Consider several ways to solve quadratic equations.

Standard ways to solve quadratic equations from the school curriculum:

1. Factorization of the left side of the equation.
2. Full square selection method.
3. Solution of quadratic equations by formula.
4. Graphical solution of a quadratic equation.
5. Solution of equations using Vieta's theorem.

Let us dwell in more detail on the solution of reduced and non-reduced quadratic equations using the Vieta theorem.

Recall that to solve the given quadratic equations, it is enough to find two numbers such that the product of which is equal to the free term, and the sum is equal to the second coefficient with the opposite sign.

Example.x 2 -5x+6=0

You need to find numbers whose product is 6 and the sum is 5. These numbers will be 3 and 2.

Answer: x 1 =2,x 2 =3.

But you can use this method for equations with the first coefficient not equal to one.

Example.3x 2 +2x-5=0

We take the first coefficient and multiply it by the free term: x 2 +2x-15=0

The roots of this equation will be numbers whose product is - 15, and the sum is - 2. These numbers are 5 and 3. To find the roots of the original equation, we divide the obtained roots by the first coefficient.

Answer: x 1 =-5/3, x 2 =1

6. Solution of equations by the method of "transfer".

Consider the quadratic equation ax 2 + bx + c = 0, where a≠0.

Multiplying both its parts by a, we get the equation a 2 x 2 + abx + ac = 0.

Let ax = y, whence x = y/a; then we arrive at the equation y 2 + by + ac = 0, which is equivalent to the given one. We find its roots at 1 and at 2 using the Vieta theorem.

Finally we get x 1 = y 1 /a and x 2 = y 2 /a.

With this method, the coefficient a is multiplied by the free term, as if "transferred" to it, therefore it is called the "transfer" method. This method is used when it is easy to find the roots of an equation using Vieta's theorem and, most importantly, when the discriminant is an exact square.

Example.2x 2 - 11x + 15 = 0.

Let's "transfer" the coefficient 2 to the free term and making the replacement we get the equation y 2 - 11y + 30 = 0.

According to Vieta's inverse theorem

y 1 = 5, x 1 = 5/2, x 1 = 2.5; y 2 ​​= 6, x 2 = 6/2, x 2 = 3.

Answer: x 1 =2.5; X 2 = 3.

7. Properties of the coefficients of a quadratic equation.

Let the quadratic equation ax 2 + bx + c \u003d 0, a ≠ 0 be given.

1. If a + b + c \u003d 0 (i.e., the sum of the coefficients of the equation is zero), then x 1 \u003d 1.

2. If a - b + c \u003d 0, or b \u003d a + c, then x 1 \u003d - 1.

Example.345x 2 - 137x - 208 = 0.

Since a + b + c \u003d 0 (345 - 137 - 208 \u003d 0), then x 1 \u003d 1, x 2 \u003d -208/345.

Answer: x 1 =1; X 2 = -208/345 .

Example.132x 2 + 247x + 115 = 0

Because a-b + c \u003d 0 (132 - 247 + 115 \u003d 0), then x 1 \u003d - 1, x 2 \u003d - 115/132

Answer: x 1 = - 1; X 2 =- 115/132

There are other properties of the coefficients of a quadratic equation. but their usage is more complicated.

8. Solving quadratic equations using a nomogram.

Fig 1. Nomogram

This is an old and currently forgotten method for solving quadratic equations, placed on p. 83 of the collection: Bradis V.M. Four-digit mathematical tables. - M., Education, 1990.

Table XXII. Nomogram for Equation Solving z2 + pz + q = 0. This nomogram allows, without solving the quadratic equation, to determine the roots of the equation by its coefficients.

The curvilinear scale of the nomogram is built according to the formulas (Fig. 1):

Assuming OS = p, ED = q, OE = a(all in cm), from Fig. 1 similarity of triangles SAN and CDF we get the proportion

whence, after substitutions and simplifications, the equation follows z 2 + pz + q = 0, and the letter z means the label of any point on the curved scale.

Rice. 2 Solving a quadratic equation using a nomogram

Examples.

1) For the equation z 2 - 9z + 8 = 0 the nomogram gives the roots z 1 = 8.0 and z 2 = 1.0

Answer: 8.0; 1.0.

2) Solve the equation using the nomogram

2z 2 - 9z + 2 = 0.

Divide the coefficients of this equation by 2, we get the equation z 2 - 4.5z + 1 = 0.

The nomogram gives the roots z 1 = 4 and z 2 = 0.5.

Answer: 4; 0.5.

9. Geometric method for solving quadratic equations.

Example.X 2 + 10x = 39.

In the original, this problem is formulated as follows: "The square and ten roots are equal to 39."

Consider a square with side x, rectangles are built on its sides so that the other side of each of them is 2.5, therefore, the area of ​​\u200b\u200beach is 2.5x. The resulting figure is then supplemented to a new square ABCD, completing four equal squares in the corners, the side of each of them is 2.5, and the area is 6.25

Rice. 3 Graphical way to solve the equation x 2 + 10x = 39

The area S of square ABCD can be represented as the sum of the areas: the original square x 2, four rectangles (4∙2.5x = 10x) and four attached squares (6.25∙4 = 25), i.e. S \u003d x 2 + 10x \u003d 25. Replacing x 2 + 10x with the number 39, we get that S \u003d 39 + 25 \u003d 64, which implies that the side of the square ABCD, i.e. segment AB \u003d 8. For the desired side x of the original square, we get

10. Solution of equations using Bezout's theorem.

Bezout's theorem. The remainder after dividing the polynomial P(x) by the binomial x - α is equal to P(α) (that is, the value of P(x) at x = α).

If the number α is the root of the polynomial P(x), then this polynomial is divisible by x -α without remainder.

Example.x²-4x+3=0

Р(x)= x²-4x+3, α: ±1,±3, α=1, 1-4+3=0. Divide P(x) by (x-1): (x²-4x+3)/(x-1)=x-3

x²-4x+3=(x-1)(x-3), (x-1)(x-3)=0

x-1=0; x=1, or x-3=0, x=3; Answer: x1 =2, x2 =3.

Conclusion: The ability to quickly and rationally solve quadratic equations is simply necessary for solving more complex equations, for example, fractional rational equations, equations of higher powers, biquadratic equations, and in high school trigonometric, exponential and logarithmic equations. Having studied all the methods found for solving quadratic equations, we can advise classmates, in addition to standard methods, to solve by the transfer method (6) and solve equations by the property of coefficients (7), since they are more accessible for understanding.

Literature:

1. Bradis V.M. Four-digit mathematical tables. - M., Education, 1990.
2. Algebra grade 8: textbook for grade 8. general education institutions Makarychev Yu. N., Mindyuk N. G., Neshkov K. I., Suvorova S. B. ed. S. A. Telyakovsky 15th ed., revised. - M.: Enlightenment, 2015
3. https://ru.wikipedia.org/wiki/%D0%9A%D0%B2%D0%B0%D0%B4%D1%80%D0%B0%D1%82%D0%BD%D0%BE%D0 %B5_%D1%83%D1%80%D0%B0%D0%B2%D0%BD%D0%B5%D0%BD%D0%B8%D0%B5
4. Glazer G.I. History of mathematics at school. A guide for teachers. / Ed. V.N. Younger. - M.: Enlightenment, 1964.

The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Equations have been used by man since ancient times and since then their use has only increased. The discriminant allows you to solve any quadratic equations using the general formula, which has the following form:

The discriminant formula depends on the degree of the polynomial. The above formula is suitable for solving quadratic equations of the following form:

The discriminant has the following properties that you need to know:

* "D" is 0 when the polynomial has multiple roots (equal roots);

* "D" is a symmetric polynomial with respect to the roots of the polynomial and therefore is a polynomial in its coefficients; moreover, the coefficients of this polynomial are integers, regardless of the extension in which the roots are taken.

Suppose we are given a quadratic equation of the following form:

1 equation

According to the formula we have:

Since \, then the equation has 2 roots. Let's define them:

Where can I solve the equation through the discriminant online solver?

You can solve the equation on our website https: // site. Free online solver will allow you to solve an online equation of any complexity in seconds. All you have to do is just enter your data into the solver. You can also watch the video instruction and learn how to solve the equation on our website. And if you have any questions, you can ask them in our Vkontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.

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 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 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 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. 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 attentively substitute them into the root formula and attentively count the result. Did you understand that the key word here is - 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 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 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 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.

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 the 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 the coefficient \(c\) is equal to zero. We write the equation as complete.