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Types of quadratic equations and their solutions. Methods for solving quadratic equations

Let's work with quadratic equations. These are very popular equations! In the very general view the quadratic equation looks like this:

For example:

Here A =1; b = 3; c = -4

Here A =2; b = -0,5; c = 2,2

Here A =-3; b = 6; c = -18

Well, you understand...

How to solve quadratic equations? If you have a quadratic equation in front of you in this form, then everything is simple. Remember the magic word discriminant . Rarely a high school student has not heard this word! The phrase “we solve through a discriminant” inspires confidence and reassurance. Because there is no need to expect tricks from the discriminant! It is simple and trouble-free to use. So, the formula for finding the roots quadratic equation looks like that:

The expression under the sign of the root is the one discriminant. As you can see, to find X, we use only a, b and c. Those. coefficients from a quadratic equation. Just carefully substitute the values a, b and c This is the formula we calculate. Let's substitute with your own signs! For example, for the first equation A =1; b = 3; c= -4. Here we write it down:

The example is almost solved:

That's all.

What cases are possible when using this formula? There are only three cases.

1. The discriminant is positive. This means the root can be extracted from it. Whether the root is extracted well or poorly is another question. What is important is what is extracted in principle. Then your quadratic equation has two roots. Two different solutions.

2. The discriminant is zero. Then you have one solution. Strictly speaking, this is not one root, but two identical. But this plays a role in inequalities, where we will study the issue in more detail.

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

Everything is very simple. And what, you think it’s impossible to make a mistake? Well, yes, how...
The most common mistakes are confusion with sign values a, b and c. Or rather, not with their signs (where to get confused?), but with substitution negative values into the formula for calculating the roots. What helps here is a detailed recording of the formula with specific numbers. If there are problems with calculations, do that!

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 about 30 seconds to write an extra line. And the number of errors will decrease sharply. So we write in detail, with all the brackets and signs:

It seems incredibly difficult to write out so carefully. But it only seems so. Give it a try. Well, or choose. What's better, fast or right? Besides, I will make you happy. After a while, there will be no need to write everything down so carefully. It will work out right on its own. Especially if you use practical techniques that are described below. This evil example with a bunch of minuses can be solved easily and without errors!

So, how to solve quadratic equations through the discriminant we remembered. Or they learned, which is also good. You know how to correctly determine 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?

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

This incomplete quadratic equations . They can also be solved through a discriminant. You just need to understand correctly what they are equal to here. a, b and c.

Have you figured it out? In the first example a = 1; b = -4; A c? It's not there at all! Well yes, that's right. In mathematics this means that c = 0 ! That's all. Substitute zero into the formula instead c, and we will succeed. Same with the second example. Only we don’t have zero here With, A b !

But incomplete quadratic equations can be solved much more simply. Without any discrimination. Let's consider the first incomplete equation. What can you do on the left side? You can take X out of brackets! Let's take it out.

And what from this? And the fact that the product equals zero if and only if any of the factors equals zero! Don't believe me? Okay, then come up with two non-zero numbers that, when multiplied, will give zero!
Does not work? That's it...
Therefore, we can confidently write: x = 0, or x = 4

The second equation can also be solved simply. Move 9 to right side. We get:

All that remains is to extract the root from 9, and that’s it. It will turn out:

Also two roots . x = +3 and x = -3.

This is how all incomplete quadratic equations are solved. Either by placing X out of brackets, or by simply moving the number to the right and then extracting the root.
It is extremely difficult to confuse these techniques. Simply because in the first case you will have to extract the root of X, which is somehow incomprehensible, and in the second case there is nothing to take out of brackets...

Now take note of practical techniques that dramatically reduce the number of errors. The same ones that are due to inattention... For which it later becomes painful and offensive...

First appointment. Don’t be lazy before solving a quadratic equation and bring it to standard form. What does this mean?
Let's say that after all the transformations you get the following equation:

Don't rush to write the root formula! You'll almost certainly get the odds mixed up a, b and c. Construct the example correctly. First, X squared, then without square, then the free term. Like this:

And again, don’t rush! A minus in front of an X squared can really upset you. It's easy to forget... Get rid of the minus. How? Yes, as taught in the previous topic! We need to multiply the entire equation by -1. We get:

But now you can safely write down the formula for the roots, calculate the discriminant and finish solving the example. Decide for yourself. You should now have roots 2 and -1.

Reception second. Check the roots! According to Vieta's theorem. Don't be scared, I'll explain everything! Checking last thing the equation. Those. the one we used to write down the root formula. If (as in this example) the coefficient a = 1, checking the roots is easy. It is enough to multiply them. The result should be a free member, i.e. in our case -2. Please note, not 2, but -2! Free member with your sign . If it doesn’t work out, it means they’ve already screwed up somewhere. Look for the error. If it works, you need to add the roots. Last and final check. The coefficient should be b With opposite familiar. In our case -1+2 = +1. A coefficient b, which is before the X, is equal to -1. So, everything is correct!
It’s a pity that this 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 and fewer errors.

Reception third. If your equation has fractional coefficients, get rid of the fractions! Multiply the equation by a common denominator as described in the previous section. When working with fractions, errors keep creeping in for some reason...

By the way, I promised to simplify the evil example with a bunch of minuses. Please! Here he is.

In order not to get confused by the minuses, we multiply the equation by -1. We get:

That's all! Solving is a pleasure!

So, let's summarize the topic.

Practical advice:

1. Before solving, we bring the quadratic equation to standard form and build it Right.

2. If there is a negative coefficient in front of the X squared, 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, its coefficient is equal to one, the solution can be easily verified using Vieta’s theorem. Do it!

Fractional equations. ODZ.

We continue to master the equations. We already know how to work with linear and quadratic equations. The last view left - fractional equations. Or they are also called much more respectably - fractional rational equations . It is the same.

Fractional equations.

As the name implies, these equations necessarily contain fractions. But not just fractions, but fractions that have unknown in denominator. At least in one. For example:

Let me remind you that if the denominators are only numbers, these are linear equations.

How to decide fractional equations? First of all, get rid of fractions! After this, the equation most often turns into linear or quadratic. And then we know what to do... In some cases it can turn into an identity, such as 5=5 or an incorrect expression, such as 7=2. But this rarely happens. I will mention this below.

But how to get rid of fractions!? Very simple. Applying the same identical transformations.

We need to multiply the entire equation by the same expression. So that all denominators are reduced! Everything will immediately become easier. Let me explain with an example. Let us need to solve the equation:

How were you taught in elementary school? We move everything to one side, bring it to a common denominator, etc. Forget how horrible dream! This is what you need to do when you add or subtract fractions. Or you work with inequalities. And in equations, we immediately multiply both sides by an expression that will give us the opportunity to reduce all the denominators (i.e., in essence, by a common denominator). And what is this expression?

On the left side, reducing the denominator requires multiplying by x+2. And on the right, multiplication by 2 is required. This means that the equation must be multiplied by 2(x+2). Multiply:

This is a common multiplication of fractions, but I’ll describe it in detail:

Please note that I am not opening the bracket yet (x + 2)! So, in its entirety, I write it:

On the left side it contracts entirely (x+2), and on the right 2. Which is what was required! After reduction we get linear the equation:

And everyone can solve this equation! x = 2.

Let's solve another example, a little more complicated:

If we remember that 3 = 3/1, and 2x = 2x/ 1, we can write:

And again we get rid of what we don’t really like - fractions.

We see that to reduce the denominator with X, we need to multiply the fraction by (x – 2). And a few are not a hindrance to us. Well, let's multiply. All left side and all right side:

Parentheses again (x – 2) I'm not revealing. I work with the bracket as a whole as if it were one number! This must always be done, otherwise nothing will be reduced.

With a feeling of deep satisfaction we reduce (x – 2) and we get an equation without any fractions, with a ruler!

Now let’s open the brackets:

We bring similar ones, move everything to the left side and get:

Classic quadratic equation. But the minus ahead is not good. You can always get rid of it by multiplying or dividing by -1. But if you look closely at the example, you will notice that it is best to divide this equation by -2! In one fell swoop, the minus will disappear, and the odds will become more attractive! Divide by -2. On the left side - term by term, and on the right - simply divide zero by -2, zero and we get:

We solve through the discriminant and check using Vieta’s theorem. We get x = 1 and x = 3. Two roots.

As you can see, in the first case the equation after the transformation became linear, but here it becomes quadratic. It happens that after getting rid of fractions, all the X's are reduced. Something remains, like 5=5. It means that x can be anything. Whatever it is, it will still be reduced. And it will work out pure truth, 5=5. But, after getting rid of fractions, it may turn out to be completely untrue, like 2=7. And this means that no solutions! Any X turns out to be untrue.

Realized main way solutions fractional equations? It is simple and logical. We change the original expression so that everything we don’t like disappears. Or it interferes. IN in this case these are fractions. We will do the same with all kinds of complex examples with logarithms, sines and other horrors. We Always Let's get rid of all this.

However, we need to change the original expression in the direction we need according to the rules, yes... The mastery of which is preparation for the Unified State Exam in mathematics. So we are mastering it.

Now we will learn how to bypass one of main ambushes on the Unified State Exam! But first, let's see whether you fall into it or not?

Let's look at a simple example:

The matter is already familiar, we multiply both sides by (x – 2), we get:

I remind you, with brackets (x – 2) We work as if with one, integral expression!

Here I no longer wrote one in the denominators, it’s undignified... And I didn’t draw brackets in the denominators, except for x – 2 there is nothing, you don’t have to draw. Let's shorten:

Open the parentheses, move everything to the left, and give similar ones:

We solve, check, we get two roots. x = 2 And x = 3. Great.

Suppose the assignment says to write down the root, or their sum if there is more than one root. What are we going to write?

If you decide the answer is 5, you were ambushed. And the task will not be credited to you. They worked in vain... Correct answer is 3.

What's the matter?! And you try to do a check. Substitute the values of the unknown into original example. And if at x = 3 everything will grow together wonderfully, we get 9 = 9, then when x = 2 It will be division by zero! What you absolutely cannot do. Means x = 2 is not a solution, and is not taken into account in the answer. This is the so-called extraneous or extra root. We simply discard it. The final root is one. x = 3.

How so?! – I hear indignant exclamations. We were taught that an equation can be multiplied by an expression! This is an identical transformation!

Yes, identical. Under a small condition - the expression by which we multiply (divide) - different from zero. A x – 2 at x = 2 equals zero! So everything is fair.

And now what i can do?! Don't multiply by expression? Should I check every time? Again it’s unclear!

Calmly! Don't panic!

In this difficult situation, three magic letters will save us. I know what you're thinking. Right! This ODZ . Area of Acceptable Values.

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. P. 17-20..03.2019).

Our project is about ways to solve quadratic equations. Goal of the project: learn to solve quadratic equations in ways not included in the school curriculum. Task: find everything possible ways solving quadratic equations and learning how to use them yourself and introducing these methods to your classmates.

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.

Who was the first to “invent” quadratic equations?

Some algebraic techniques for solving linear and quadratic equations were known 4000 years ago in Ancient Babylon. The discovery of ancient Babylonian clay tablets, dating from somewhere between 1800 and 1600 BC, provides 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, even in ancient times, was caused by the need to solve problems related to finding the areas of land plots and with excavation work of a military nature, as well as with the development of astronomy and mathematics itself.

The rule for solving these equations, set out in the Babylonian texts, essentially coincides with the modern one, but it is not known how the Babylonians arrived at this rule. Almost all cuneiform texts found so far provide only problems with solutions laid out in the form of recipes, with no indication as to how they were found. Despite high level 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 BC. used the square's complement method to solve equations with positive roots. Around 300 BC Euclid came up with a more general geometric solution method. The first mathematician who found solutions to equations with negative roots in the form of an algebraic formula was an Indian scientist Brahmagupta(India, 7th century AD).

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

ax2 + bx = c, a>0

The coefficients in this equation can also be negative. Brahmagupta's rule is essentially the same as ours.

Public competitions in solving difficult problems were common in India. One of the old Indian books says the following about such competitions: “As the sun eclipses the stars with its brilliance, so learned man will eclipse the glory in people's assemblies, proposing and solving algebraic problems.” Problems were often presented in poetic form.

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

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

2) “Squares are equal to numbers,” 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 the 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 and not subtractables. In this case, equations that do not have positive solutions are obviously not taken into account. The author sets out methods for solving these equations using the techniques of al-jabr and al-mukabal. His decision, of course, does not completely coincide with ours. Not to mention 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 until the 17th century, does not take into account the zero solution, probably because in specific practical it doesn't matter in tasks. 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 following the model of Al-Khwarizmi in Europe were first set forth in the “Book of the Abacus,” written in 1202. Italian mathematician Leonard Fibonacci. The author independently developed some new algebraic examples solving problems and was the first in Europe to introduce 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 problems from this book were used in almost all European textbooks of the 14th-17th centuries. General rule the solution of quadratic equations reduced to a single canonical form x2 + bх = с for all possible combinations of signs and coefficients b, c was formulated in Europe in 1544. M. Stiefel.

The derivation of the formula for solving a quadratic equation in general form is available from Viète, but Viète recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli among the first in the 16th century. In addition to positive ones, negative roots are also taken into account. Only in the 17th century. thanks to the efforts Girard, Descartes, Newton and other scientists, the method of solving quadratic equations takes a modern form.

Let's look at several ways to solve quadratic equations.

Standard methods for solving quadratic equations from school curriculum:

Factoring the left side of the equation.
Method for selecting a complete square.
Solving quadratic equations using the formula.
Graphical solution of a quadratic equation.
Solving equations using Vieta's theorem.

Let us dwell in more detail on the solution of reduced and unreduced quadratic equations using Vieta’s theorem.

Let us recall that to solve the above quadratic equations it is enough to find two numbers such that the product of them is equal to the free term, and the sum is equal to the second coefficient with opposite sign.

Example.x 2 -5x+6=0

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

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

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

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

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 equal to - 15, and whose sum is equal to - 2. These numbers are 5 and 3. To find the roots of the original equation, divide the resulting roots by the first coefficient.

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

6. Solving equations using the "throw" method.

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

Multiplying both sides by a, we obtain 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, equivalent to the given one. We find its roots for 1 and 2 using Vieta’s theorem.

We finally 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 “thrown” to it, which is why it is called the “throw” method. This method is used when the roots of the equation can be easily found using Vieta's theorem and, most importantly, when the discriminant is an exact square.

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

Let’s “throw” the coefficient 2 to the free term and make a substitution and 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 coefficients of a quadratic equation.

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

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

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

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

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

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

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

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

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

There are other properties of the coefficients of a quadratic equation. but their use is more complex.

8. Solving quadratic equations using a nomogram.

Fig 1. Nomogram

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

Table XXII. Nomogram for solving the equation z 2 + pz + q = 0. This nomogram allows, without solving a quadratic equation, to determine the roots of the equation from its coefficients.

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

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

which, after substitutions and simplifications, yields the equation z 2 + pz + q = 0, and the letter z means the mark of any point on a curved scale.

Rice. 2 Solving quadratic equations 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) Using a nomogram, we solve the equation

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 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 constructed on its sides so that the other side of each of them is 2.5, therefore the area of each is 2.5x. The resulting figure is then supplemented to a new square ABCD, building four equal squares in the corners, the side of each of them is 2.5, and the area is 6.25

Rice. 3 Graphic method solutions to the equation x 2 + 10x = 39

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

10. Solving equations using Bezout's theorem.

Bezout's theorem. The remainder of 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 a 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 to solve more complex equations, For example, fractional rational equations, equations higher degrees, biquadratic equations, and in high school trigonometric, exponential and logarithmic equations. Having studied all the found methods for solving quadratic equations, we can advise our classmates, in addition to the standard methods, to solve by the transfer method (6) and solve equations using the property of coefficients (7), since they are more accessible to understanding.

Literature:

Bradis V.M. Four-digit math tables. - M., Education, 1990.
Algebra 8th grade: textbook for 8th grade. general education institutions Makarychev Yu. N., Mindyuk N. G., Neshkov K. I., Suvorova S. B. ed. S. A. Telyakovsky 15th ed., revised. - M.: Education, 2015
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
Glazer G.I. History of mathematics at school. Manual for teachers. / Ed. V.N. Younger. - M.: Education, 1964.

Quadratic equations. Discriminant. Solution, examples.

Attention!
There are additional
materials in Special Section 555.
For those who are very "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 the keyword is "square". This means that in the equation Necessarily there must be an x squared. In addition to it, the equation may (or may not!) contain just X (to the first power) and just a number (free member). And there should be no X's to a power 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 other than 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 understand...

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

Such quadratic equations are called full.

And if b= 0, what do we get? We have X will be lost to the first power. This happens when multiplied by zero.) It turns out, for example:

5x 2 -25 = 0,

2x 2 -6x=0,

-x 2 +4x=0

And so on. And if both coefficients b And c are equal to zero, then it’s even simpler:

2x 2 =0,

-0.3x 2 =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 equal to zero? And you substitute instead A zero.) Our X squared will disappear! The equation will become linear. And the solution is completely different...

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

Solving quadratic equations.

Solving 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 a standard form, i.e. to the form:

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 a quadratic equation. Just carefully substitute the values a, b and c We calculate into this formula. Let's substitute with your own signs! For example, in the equation:

A =1; b = 3; c= -4. Here we write it down:

The example is almost solved:

This is the answer.

Everything is very simple. And what, you think it’s impossible to make a mistake? Well, yes, how...

The most common mistakes are confusion with sign values a, b and c. Or rather, not with their signs (where to get confused?), but with the substitution of negative values into the formula for calculating the roots. What helps here is a detailed recording of the formula with specific numbers. If there are problems with calculations, do that!

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 about 30 seconds to write an extra line. And the number of errors will decrease sharply. So we write in detail, with all the brackets and signs:

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

Did you recognize it?) Yes! This incomplete quadratic equations.

Solving incomplete quadratic equations.

They can also be solved using a general formula. You just need to understand correctly what they are equal to here. a, b and c.

But incomplete quadratic equations can be solved much more simply. Without any formulas. Let's consider the first incomplete equation. What can you do on the left side? You can take X out of brackets! Let's take it out.

All. These will be the roots of our equation. Both are suitable. 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 using the general formula. Let me note, by the way, which X will be the first and which will be the second - absolutely indifferent. It is convenient to write in order, x 1- what is smaller and x 2- that which is greater.

The second equation can also be solved simply. Move 9 to the right side. We get:

All that remains is to extract the root from 9, and that’s it. It will turn out:

Also two roots . x 1 = -3, x 2 = 3.

Discriminant. Discriminant formula.

Magic word discriminant ! Rarely a high school student has not heard this word! The phrase “we solve through a discriminant” inspires confidence and reassurance. Because there is no need to expect tricks from the discriminant! It is simple and trouble-free to use.) I remind you of the most general formula for solutions any quadratic equations:

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

D = b 2 - 4ac

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

Here's the thing. When solving a quadratic equation using this formula, it is possible only three cases.

1. The discriminant is positive. This means the root can be extracted from it. Whether the root is extracted well or poorly is another question. What is important is what is extracted in principle. Then your quadratic equation has two roots. Two different solutions.

2. The discriminant is zero. Then you will have one solution. Since adding or subtracting zero in the numerator does not change anything. Strictly speaking, this is not one root, but two identical. But, in a simplified version, it is customary to talk about one solution.

3. The discriminant is negative. The square root of a negative number cannot be taken. Well, okay. This means there are no solutions.

To be honest, when simply solving quadratic equations, the concept of a discriminant is not really needed. We substitute the values of the coefficients into the formula and count. Everything happens there by itself, two roots, one, and none. However, when solving more complex tasks, without knowledge meaning and formula of the discriminant not enough. Especially in equations with parameters. Such equations are aerobatics for the State Examination and the Unified State Examination!)

So, how to solve quadratic equations through the discriminant you remembered. Or you learned, which is also not bad.) You know how to correctly determine a, b and c. Do you know how? attentively substitute them into the root formula and attentively count the result. You understand that the key word here is attentively?

Now take note of practical techniques that dramatically reduce the number of errors. The same ones that are due to inattention... For which it later becomes painful and offensive...

First appointment . Don’t be lazy before solving a quadratic equation and bring it to standard form. What does this mean?
Let's say that after all the transformations you get the following equation:

But now you can safely write down the formula for the roots, calculate the discriminant and finish solving the example. Decide for yourself. You should now have roots 2 and -1.

Reception second. Check the roots! According to Vieta's theorem. Don't be scared, I'll explain everything! Checking last thing the equation. Those. the one we used to write down the root formula. If (as in this example) the coefficient a = 1, checking the roots is easy. It is enough to multiply them. The result should be a free member, i.e. in our case -2. Please note, not 2, but -2! Free member with your sign . If it doesn’t work out, it means they’ve already screwed up somewhere. Look for the error.

If it works, you need to add the roots. Last and final check. The coefficient should be b With opposite familiar. In our case -1+2 = +1. A coefficient b, which is before the X, is equal to -1. So, everything is correct!
It’s a pity that this 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 and fewer errors.

Reception third . If your equation has fractional coefficients, get rid of the fractions! Multiply the equation by a common denominator as described in the lesson "How to solve equations? Identity transformations." When working with fractions, errors keep creeping in for some reason...

By the way, I promised to simplify the evil example with a bunch of minuses. Please! Here he is.

In order not to get confused by the minuses, we multiply the equation by -1. We get:

That's all! Solving is a pleasure!

So, let's summarize the topic.

Practical tips:

1. Before solving, we bring the quadratic equation to standard form and build it Right.

2. If there is a negative coefficient in front of the X squared, 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, its coefficient is equal to one, the solution can be easily verified using Vieta’s theorem. Do it!

Now we 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 = -0.5

x - any number

x 1 = -3
x 2 = 3

no solutions

x 1 = 0.25
x 2 = 0.5

Does everything fit? Great! Quadratic equations are not your thing headache. The first three worked, but the rest didn’t? Then the problem is not with quadratic equations. The problem is in identical transformations of equations. Take a look at the link, it's helpful.

Doesn't quite work out? Or does it not work out at all? Then Section 555 will help you. All these examples are broken down there. Shown main errors in the solution. Of course, it also talks about the use identity transformations in solving 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. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

Formulas for the roots of a quadratic equation. The cases of real, multiple and complex roots. Factoring a quadratic trinomial. Geometric interpretation. Examples of determining roots and factoring.

Basic formulas

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

We further assume that - real numbers.
Let's 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 quadratic trinomial has the form:
.
If the discriminant is equal to 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 you build graph of a function
,
which is a parabola, then the points of intersection of the graph with the axis will be the roots of the equation
.
At , the graph intersects the x-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 carry out transformations and apply formulas (f.1) and (f.3):

,
Where
; .

So, we got the formula for a polynomial of the second degree in the form:
.
This shows 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) .

Solution

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

From here we obtain the factorization of the quadratic trinomial:

.

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

Let's plot the function
.
The graph of this function is a parabola. It crosses the abscissa 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) .

Solution

Let's write the quadratic equation in general form:
.
Comparing with the original equation (2.1), we find the values of the coefficients:
.
We find 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). Because this root is factored twice:
,
then such a root is usually called a multiple. That is, they believe that there are two equal roots:
.

Answer

;
.

Example 3

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

Solution

Let's write the quadratic equation in general form:
(1) .
Let's rewrite the original equation (3.1):
.
Comparing with (1), we find the values of the coefficients:
.
We find 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 intersect the x-axis (axis). Therefore there are no real roots.

Answer

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

The discriminant, like quadratic equations, begins to be studied in an algebra course in the 8th grade. You can solve a quadratic equation through a discriminant and using Vieta's theorem. The method of studying quadratic equations, as well as discriminant formulas, is rather unsuccessfully taught to schoolchildren, like many things in real education. Therefore they pass school years, education in grades 9-11 replaces " higher education"and everyone is looking again - “How to solve a quadratic equation?”, “How to find the roots of the equation?”, “How to find the discriminant?” And...

Discriminant formula

The discriminant D of the quadratic equation a*x^2+bx+c=0 is equal to D=b^2–4*a*c.
The roots (solutions) of a 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 matching 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 websites offer an online discriminant calculator. We haven’t figured out this kind of scripts yet, so if anyone knows how to implement this, please write to us by email This email address is being protected from spambots. You must have JavaScript enabled to view it. .

General formula for finding the roots of a quadratic equation:

We find the roots of the equation using the formula
If the coefficient of a squared variable 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 using 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, let’s consider the part that concerns the above quadratic equations, that is, equations of the form (a=1)
The essence of Vieta's 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. Vieta's theorem can be written in formulas.
The derivation of Vieta's formula is quite simple. Let's write the quadratic equation through simple factors
As you can see, everything ingenious is simple at the same time. It is effective to use Vieta’s formula when the difference in modulus of the roots or the difference in the moduli of the roots is 1, 2. For example, the following equations, according to Vieta’s theorem, have roots

Up to equation 4, the analysis should look like this. The product of the roots of the equation is 6, therefore the roots can be the values (1, 6) and (2, 3) or pairs with opposite signs. The sum of the roots is 7 (the coefficient of the variable with the opposite sign). From here we conclude that the solutions to the quadratic equation are x=2; x=3.
It is easier to select the roots of the equation among the divisors of the free term, adjusting their sign in order to fulfill the Vieta formulas. At first, this seems difficult to do, but with practice on a number of quadratic equations, this technique will turn out to be more effective 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 methods of finding 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 algebra course they study functions, schemes for studying functions and constructing a graph of 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 squared variable 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

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 zero (D=0) then the parabola at the vertex 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

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