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Quadratic equations. The Comprehensive Guide (2019)

Important! In roots of even multiplicity the function does not change sign.

Note! Any nonlinear inequality in a school algebra course must be solved using the interval method.

I offer you a detailed algorithm for solving inequalities using the interval method, following which you can avoid mistakes when solving nonlinear inequalities.

Solving quadratic equations with negative discriminants

As we know,

i 2 = - 1.

At the same time

(- i ) 2 = (- 1 i ) 2 = (- 1) 2 i 2 = -1.

Thus, there are at least two values of the square root of - 1, namely i And - i . But maybe there are some more complex numbers, whose squares are equal to - 1?

To clarify this question, suppose that the square of a complex number a + bi is equal to - 1. Then

(a + bi ) 2 = - 1,

A 2 + 2аbi - b 2 = - 1

Two complex numbers are equal if and only if their real parts and coefficients of their imaginary parts are equal. That's why

{	and 2 - b 2 = - 1 ab = 0 (1)

According to the second equation of system (1), at least one of the numbers A And b must be zero. If b = 0, then from the first equation we get A 2 = - 1. Number A real, and therefore A 2 > 0. Non-negative number A 2 cannot equal a negative number - 1. Therefore, the equality b = 0 V in this case impossible. It remains to admit that A = 0, but then from the first equation of the system we obtain: - b 2 = - 1, b = ± 1.

Therefore, the only complex numbers whose squares are -1 are i And - i , Conventionally, this is written in the form:

√-1 = ± i .

Using similar reasoning, students can be convinced that there are exactly two numbers whose squares are equal to a negative number - A . Such numbers are √ ai and -√ ai . Conventionally, it is written like this:

√- A = ± √ ai .

Under √ a here we mean an arithmetic, that is, positive, root. For example, √4 = 2, √9 =.3; That's why

√-4 = + 2i , √-9= ± 3 i

If earlier, when considering quadratic equations with negative discriminants, we said that such equations have no roots, now we can no longer say that. Quadratic equations with negative discriminants have complex roots. These roots are obtained according to the formulas known to us. Let, for example, be given the equation x 2 + 2X + 5 = 0; Then

X 1.2 = - 1 ± √1 -5 = - 1 ± √-4 = - 1 ± 2 i .

So, given equation has two roots: X 1 = - 1 +2i , X 2 = - 1 - 2i . These roots are mutually conjugate. It is interesting to note that their sum is - 2, and their product is 5, so Vieta's theorem holds.

Concept of a complex number

A complex number is an expression of the form a + ib, where a and b are any real numbers, i is a special number called the imaginary unit. For such expressions, the concepts of equality and the operations of addition and multiplication are introduced as follows:

Two complex numbers a + ib and c + id are said to be equal if and only if
a = b and c = d.
The sum of two complex numbers a + ib and c + id is a complex number
a + c + i (b + d).
The product of two complex numbers a + ib and c + id is a complex number
ac – bd + i (ad + bc).

Complex numbers are often denoted by a single letter, for example z = a + ib. A real number a is called the real part of a complex number z, the real part is denoted a = Re z. The real number b is called the imaginary part of the complex number z, the imaginary part is denoted b = Im z. These names were chosen due to the following special properties of complex numbers.

Note that arithmetic operations on complex numbers of the form z = a + i · 0 are carried out in exactly the same way as on real numbers. Really,

Consequently, complex numbers of the form a + i · 0 are naturally identified with real numbers. Because of this, complex numbers of this type are simply called real. So, a lot real numbers is contained in the set of complex numbers. The set of complex numbers is denoted by . We have established that, namely

Unlike real numbers, numbers of the form 0 + ib are called purely imaginary. Often they simply write bi, for example, 0 + i 3 = 3 i. The purely imaginary number i1 = 1 i = i has an amazing property:
Thus,

№ 4 .1. In mathematics, a number function is a function whose domains and values are subsets of number sets—typically the set of real numbers or the set of complex numbers.

Graph of a function

Fragment of a function graph

Methods for specifying a function

[edit] Analytical method

Typically, a function is specified using a formula that includes variables, operations, and elementary functions. Perhaps a piecewise task, that is, different for different meanings argument.

[edit] Tabular method

A function can be specified by listing all its possible arguments and their values. After this, if necessary, the function can be further defined for arguments that are not in the table, by interpolation or extrapolation. Examples include a program guide, a train schedule, or a table of Boolean function values:

[edit] Graphic method

An oscillogram sets the value of a certain function graphically.

A function can be specified graphically by displaying a set of points on its graph on a plane. This could be a rough sketch of what the function should look like, or readings taken from a device such as an oscilloscope. This method of specifying may suffer from a lack of precision, but in some cases other methods of specifying cannot be applied at all. In addition, this method of specifying is one of the most representative, easy-to-understand and high-quality heuristic analysis of the function.

[edit] Recursive way

A function can be specified recursively, that is, through itself. In this case, some function values are determined through its other values.

factorial;
Fibonacci numbers;
Ackermann function.

[edit] Verbal method

A function can be described in natural language words in some unambiguous way, for example by describing its input and output values, or the algorithm by which the function defines correspondences between these values. Along with graphically, sometimes this is the only way to describe a function, although natural languages are not as deterministic as formal languages.

a function that returns a digit in pi by its number;
a function that returns the number of atoms in the universe at a certain point in time;
a function that takes a person as an argument and returns the number of people that will be born after that person is born

COMPLEX NUMBERS XI

§ 253. Extracting square roots from negative numbers.
Solving quadratic equations with negative discriminants

As we know,

i 2 = - 1.

At the same time

(- i ) 2 = (- 1 i ) 2 = (- 1) 2 i 2 = -1.

Thus, there are at least two values of the square root of - 1, namely i And - i . But maybe there are some other complex numbers whose squares are equal to - 1?

To clarify this question, suppose that the square of a complex number a + bi is equal to - 1. Then

(a + bi ) 2 = - 1,

A 2 + 2аbi - b 2 = - 1

Two complex numbers are equal if and only if their real parts and coefficients of their imaginary parts are equal. That's why

{	A 2 - b 2 = - 1 ab = 0 (1)

According to the second equation of system (1), at least one of the numbers A And b must be zero. If b = 0, then from the first equation we get A 2 = - 1. Number A real, and therefore A 2 > 0. Non-negative number A 2 cannot equal a negative number - 1. Therefore, the equality b = 0 is impossible in this case. It remains to admit that A = 0, but then from the first equation of the system we obtain: - b 2 = - 1, b = ± 1.

Therefore, the only complex numbers whose squares are -1 are i And - i , Conventionally, this is written in the form:

√-1 = ± i .

Using similar reasoning, students can be convinced that there are exactly two numbers whose squares are equal to a negative number - A . Such numbers are √ a i and -√ a i . Conventionally, it is written like this:

√- A = ± √ a i .

Under √ a here we mean an arithmetic, that is, positive, root. For example, √4 = 2, √9 =.3; That's why

√-4 = + 2i , √-9 = ± 3 i

X 1.2 = - 1 ± √1 -5 = - 1 ± √-4 = - 1 ± 2 i .

So, this equation has two roots: X 1 = - 1 +2i , X 2 = - 1 - 2i . These roots are mutually conjugate. It is interesting to note that their sum is - 2, and their product is 5, so Vieta's theorem holds.

Exercises

2022. (Set no.) Solve the equations:

A) x 2 = - 16; b) x 2 = - 2; at 3 x 2 = - 5.

2023. Find all complex numbers whose squares are equal:

A) i ; b) 1 / 2 - √ 3 / 2 i ;

2024. Solve quadratic equations:

A) x 2 - 2x + 2 = 0; b) 4 x 2 + 4x + 5 = 0; V) x 2 - 14x + 74 = 0.

Solve systems of equations (No. 2025, 2026):

{	*x+y* = 6 xy = 45

{	2x- 3y = 1 xy = 1

2027. Prove that the roots of a quadratic equation with real coefficients and a negative discriminant are mutually conjugate.

2028. Prove that Vieta's theorem is true for any quadratic equations, and not just for equations with a non-negative discriminant.

2029. Compose a quadratic equation with real coefficients, the roots of which are:

a) X 1 = 5 - i , X 2 = 5 + i ; b) X 1 = 3i , X 2 = - 3i .

2030. Compose a quadratic equation with real coefficients, one of the roots of which is equal to (3 - i ) (2i - 4).

2031. Compose a quadratic equation with real coefficients, one of the roots of which is equal to 32 - i
1- 3i .

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. Let's remember 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 of a quadratic equation looks like this:

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 Square root not extracted. 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 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:

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

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 a discriminant.

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 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 identity 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.

Among the whole course school curriculum In algebra, one of the most extensive topics is the topic of quadratic equations. In this case, a quadratic equation is understood as an equation of the form ax 2 + bx + c = 0, where a ≠ 0 (read: a multiplied by x squared plus be x plus ce is equal to zero, where a is not equal to zero). In this case, the main place is occupied by formulas for finding the discriminant of a quadratic equation specified type, which is understood as an expression that allows you to determine the presence or absence of roots in a quadratic equation, as well as their number (if any).

Formula (equation) of the discriminant of a quadratic equation

The generally accepted formula for the discriminant of a quadratic equation is as follows: D = b 2 – 4ac. By calculating the discriminant using the specified formula, you can not only determine the presence and number of roots of a quadratic equation, but also choose a method for finding these roots, of which there are several depending on the type of quadratic equation.

What does it mean if the discriminant is zero \ Formula for the roots of a quadratic equation if the discriminant is zero

The discriminant, as follows from the formula, is denoted by the Latin letter D. In the case when the discriminant is equal to zero, it should be concluded that a quadratic equation of the form ax 2 + bx + c = 0, where a ≠ 0, has only one root, which is calculated by simplified formula. This formula applies only when the discriminant is zero and looks like this: x = –b/2a, where x is the root of the quadratic equation, b and a are the corresponding variables of the quadratic equation. To find the root of a quadratic equation you need negative meaning variable b divided by twice the value of variable a. The resulting expression will be the solution to a quadratic equation.

Solving a quadratic equation using a discriminant

If, when calculating the discriminant using the above formula, a positive value is obtained (D is greater than zero), then the quadratic equation has two roots, which are calculated using the following formulas: x 1 = (–b + vD)/2a, x 2 = (–b – vD) /2a. Most often, the discriminant is not calculated separately, but the radical expression in the form of a discriminant formula is simply substituted into the value D from which the root is extracted. If the variable b has an even value, then to calculate the roots of a quadratic equation of the form ax 2 + bx + c = 0, where a ≠ 0, you can also use following formulas: x 1 = (–k + v(k2 – ac))/a, x 2 = (–k + v(k2 – ac))/a, where k = b/2.

In some cases, to practically solve quadratic equations, you can use Vieta’s Theorem, which states that for the sum of the roots of a quadratic equation of the form x 2 + px + q = 0 the value x 1 + x 2 = –p will be true, and for the product of the roots of the specified equation – expression x 1 x x 2 = q.

Can the discriminant be less than zero?

When calculating the discriminant value, you may encounter a situation that does not fall under any of the described cases - when the discriminant has a negative value (that is, less than zero). In this case, it is generally accepted that a quadratic equation of the form ax 2 + bx + c = 0, where a ≠ 0, has no real roots, therefore, its solution will be limited to calculating the discriminant, and the above formulas for the roots of a quadratic equation will not apply in this case there will be. At the same time, in the answer to the quadratic equation it is written that “the equation has no real roots.”

Explanatory video:

Discriminant is a multi-valued term. In this article we will talk about the discriminant of a polynomial, which allows you to determine whether a given polynomial has valid solutions. The formula for a quadratic polynomial appears in school course algebra and analysis. How to find a discriminant? What is needed to solve the equation?

A quadratic polynomial or equation of the second degree is called i * w ^ 2 + j * w + k equals 0, where “i” and “j” are the first and second coefficients, respectively, “k” is a constant, sometimes called the “dismissive term,” and “w” is a variable. Its roots will be all the values of the variable at which it turns into an identity. Such an equality can be rewritten as the product of i, (w - w1) and (w - w2) equal to 0. In this case, it is obvious that if the coefficient “i” does not become zero, then the function on the left side will become zero only if if x takes the value w1 or w2. These values are the result of setting the polynomial equal to zero.

To find the value of a variable at which a quadratic polynomial vanishes, an auxiliary construction is used, built on its coefficients and called a discriminant. This design is calculated according to the formula D equals j * j - 4 * i * k. Why is it used?

It tells whether there are valid results.
She helps calculate them.

How does this value show the presence of real roots:

If it is positive, then two roots can be found in the region of real numbers.
If the discriminant is zero, then both solutions are the same. We can say that there is only one solution, and it is from the field of real numbers.
If the discriminant is less than zero, then the polynomial has no real roots.

Calculation options for securing material

For the sum (7 * w^2; 3 * w; 1) equal to 0 We calculate D using the formula 3 * 3 - 4 * 7 * 1 = 9 - 28, we get -19. A discriminant value below zero indicates that there are no results on the actual line.

If we consider 2 * w^2 - 3 * w + 1 equivalent to 0, then D is calculated as (-3) squared minus the product of numbers (4; 2; 1) and equals 9 - 8, that is, 1. Positive value says there are two results on the real line.

If we take the sum (w ^ 2; 2 * w; 1) and equate it to 0, D is calculated as two squared minus the product of the numbers (4; 1; 1). This expression will simplify to 4 - 4 and go to zero. It turns out that the results are the same. If you look closely at this formula, it will become clear that this is a “complete square”. This means that the equality can be rewritten in the form (w + 1) ^ 2 = 0. It became obvious that the result in this problem is “-1”. In a situation where D is equal to 0, the left side of the equality can always be collapsed using the “square of the sum” formula.

Using a discriminant in calculating roots

This auxiliary construction not only shows the number of real solutions, but also helps to find them. General formula The calculation for the second degree equation is:

w = (-j +/- d) / (2 * i), where d is the discriminant to the power of 1/2.

Let's say the discriminant is below zero, then d is imaginary and the results are imaginary.

D is zero, then d equal to D to the power of 1/2 is also zero. Solution: -j / (2 * i). Again considering 1 * w ^ 2 + 2 * w + 1 = 0, we find results equivalent to -2 / (2 * 1) = -1.

Suppose D > 0, then d is a real number, and the answer here breaks down into two parts: w1 = (-j + d) / (2 * i) and w2 = (-j - d) / (2 * i) . Both results will be valid. Let's look at 2 * w ^ 2 - 3 * w + 1 = 0. Here the discriminant and d are ones. It turns out that w1 is equal to (3 + 1) divided by (2 * 2) or 1, and w2 is equal to (3 - 1) divided by 2 * 2 or 1/2.

The result of equating a quadratic expression to zero is calculated according to the algorithm:

Determining the number of valid solutions.
Calculation d = D^(1/2).
Finding the result according to the formula (-j +/- d) / (2 * i).
Substituting the obtained result into the original equality for verification.

Some special cases

Depending on the coefficients, the solution may be somewhat simplified. Obviously, if the coefficient of a variable to the second power is zero, then a linear equality is obtained. When the coefficient of a variable to the first power is zero, then two options are possible:

the polynomial is expanded into a difference of squares when the free term is negative;
for a positive constant, no real solutions can be found.

If the free term is zero, then the roots will be (0; -j)

But there are other special cases that simplify finding a solution.

Reduced second degree equation

The given is called such a quadratic trinomial, where the coefficient of the leading term is one. For this situation, Vieta’s theorem is applicable, which states that the sum of the roots is equal to the coefficient of the variable to the first power, multiplied by -1, and the product corresponds to the constant “k”.

Therefore, w1 + w2 equals -j and w1 * w2 equals k if the first coefficient is one. To verify the correctness of this representation, you can express w2 = -j - w1 from the first formula and substitute it into the second equality w1 * (-j - w1) = k. The result is the original equality w1 ^ 2 + j * w1 + k = 0.

It is important to note, that i * w ^ 2 + j * w + k = 0 can be achieved by dividing by “i”. The result will be: w^2 + j1 * w + k1 = 0, where j1 is equal to j/i and k1 is equal to k/i.

Let's look at the already solved 2 * w^2 - 3 * w + 1 = 0 with the results w1 = 1 and w2 = 1/2. We need to divide it in half, as a result w ^ 2 - 3/2 * w + 1/2 = 0. Let's check that the conditions of the theorem are true for the results found: 1 + 1/2 = 3/2 and 1*1/2 = 1 /2.

Even second factor

If the factor of a variable to the first power (j) is divisible by 2, then it will be possible to simplify the formula and look for a solution through a quarter of the discriminant D/4 = (j / 2) ^ 2 - i * k. it turns out w = (-j +/- d/2) / i, where d/2 = D/4 to the power of 1/2.

If i = 1, and the coefficient j is even, then the solution will be the product of -1 and half the coefficient of the variable w, plus/minus the root of the square of this half minus the constant “k”. Formula: w = -j/2 +/- (j^2/4 - k)^1/2.

Higher discriminant order

The discriminant of the second degree trinomial discussed above is the most commonly used special case. In the general case, the discriminant of a polynomial is multiplied squares of the differences of the roots of this polynomial. Therefore, a discriminant equal to zero indicates the presence of at least two multiple solutions.

Consider i * w^3 + j * w^2 + k * w + m = 0.

D = j^2 * k^2 - 4 * i * k^3 - 4 * i^3 * k - 27 * i^2 * m^2 + 18 * i * j * k * m.

Suppose the discriminant exceeds zero. This means that there are three roots in the region of real numbers. At zero there are multiple solutions. If D< 0, то два корня комплексно-сопряженные, которые дают отрицательное значение при возведении в квадрат, а также один корень — вещественный.

Video

Our video will tell you in detail about calculating the discriminant.

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