Next, we get a quadratic equation. Ways to solve quadratic equations

Quadratic equations. Discriminant. Solution, examples.

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

Types of quadratic equations

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

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

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

Here are all the main types 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, you need given equation lead to standard view, i.e. to the view:

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

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

The expression under the root sign is called discriminant. But more about him below. As you can see, to find x, we use only a, b and c. Those. coefficients from the quadratic equation. Just carefully substitute the values a, b and c into this formula and count. Substitute with your signs! For example, in the equation:

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

Example almost solved:

This is the answer.

Everything is very simple. And what do you think, you can't go wrong? Well, yes, how...

The most common mistakes are confusion with the signs of values a, b and c. Or rather, not with their signs (where is there to be confused?), But with the substitution of negative values into the formula for calculating the roots. Here, a detailed record of the formula with specific numbers saves. If there are problems with calculations, so do it!

Suppose we need to solve the following example:

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

Let's say you know that you rarely get answers the first time.

Well, don't be lazy. It will take 30 seconds to write an extra line. And the number of errors will drop sharply. So we write in detail, with all the brackets and signs:

It seems incredibly difficult to paint so carefully. But it only seems. Try it. Well, or choose. Which is better, fast, or right? Besides, I will make you happy. After a while, there will be no need to paint everything so carefully. It will just turn out right. Especially if you apply practical techniques, which are described below. This evil example with a bunch of minuses will be solved easily and without errors!

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

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

Solution of incomplete quadratic equations.

They can also be solved by the general formula. You just need to correctly figure out what is equal here a, b and c.

Realized? In the first example a = 1; b = -4; a c? It doesn't exist at all! Well, yes, that's right. In mathematics, this means that c = 0 ! That's all. Substitute zero into the formula instead of c, and everything will work out for us. Similarly with the second example. Only zero we don't have here with, a b !

But incomplete quadratic equations can be solved much easier. Without any formulas. Consider the first incomplete equation. What can be done on the left side? You can take the X out of brackets! Let's take it out.

And what from this? And the fact that the product is equal to zero if, and only if any of the factors is equal to zero! Don't believe? Well, then come up with two non-zero numbers that, when multiplied, will give zero!
Does not work? Something...
Therefore, we can confidently write: x 1 = 0, x 2 = 4.

Everything. These will be the roots of our equation. Both fit. When substituting any of them into the original equation, we get the correct identity 0 = 0. As you can see, the solution is much simpler than the general formula. I note, by the way, which X will be the first, and which the second - it is absolutely indifferent. Easy to write in order x 1- whichever is less x 2- that which is more.

The second equation can also be easily solved. Transferring 9 to right side. We get:

It remains to extract the root from 9, and that's it. Get:

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

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

Discriminant. Discriminant formula.

Magic word discriminant ! A rare high school student has not heard this word! The phrase “decide through the discriminant” is reassuring and reassuring. Because there is no need to wait for tricks from the discriminant! It is simple and trouble-free to use.) I remind you of the most general formula for solving any quadratic equations:

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

D = b 2 - 4ac

And what is so special about this expression? Why did it deserve 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. Discriminant 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 simplified version, it is customary to talk about one solution.

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

To be honest, at simple solution quadratic equations, the concept of discriminant is not particularly required. We substitute the values of the coefficients in the formula, and we consider. There everything turns out by itself, and two roots, and one, and not a single one. However, when solving more 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 keyword here - attentively?

Now take note of the practical techniques that dramatically reduce the number of errors. The very ones that are due to inattention ... For which it is then painful and insulting ...

First reception . Do not be lazy before solving a quadratic equation to bring it to a standard form. What does this mean?
Suppose, after any transformations, you get the following equation:

Do not rush to write the formula of the roots! You will almost certainly mix up the odds a, b and c. Build the example correctly. First, x squared, then without a square, then a free member. Like this:

And again, do not rush! The minus before the x squared can upset you a lot. Forgetting it is easy... Get rid of the minus. How? Yes, as taught in 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! Everything less mistakes will.

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, we also talk about the application of identical transformations in the solution various equations. Helps a lot!

If you like this site...

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

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

you can get acquainted with functions and derivatives.

Quadratic equations often appear during solution various tasks physics and mathematics. In this article, we will consider how to solve these equalities in a universal way "through the discriminant". Examples of using the acquired knowledge are also given in the article.

What equations are we talking about?

The figure below shows a formula in which x is an unknown variable, and the Latin characters a, b, c represent some known numbers.

Each of these symbols is called a coefficient. As you can see, the number "a" is in front of the squared variable x. This is the maximum power of the represented expression, which is why it is called a quadratic equation. Another name is often used: a second-order equation. The value of a itself is the square coefficient (squaring the variable), b is linear coefficient(it is next to the variable raised to the first power), finally, the number c is a free member.

Note that the form of the equation shown in the figure above is a general classical quadratic expression. In addition to it, there are other second-order equations in which the coefficients b, c can be zero.

When the task is set to solve the equality under consideration, this means that such values of the variable x must be found that would satisfy it. The first thing to remember here is the following: since the maximum power of x is 2, this type of expression cannot have more than 2 solutions. This means that if, when solving the equation, 2 x values \u200b\u200bthat satisfy it were found, then you can be sure that there is no 3rd number, substituting which instead of x, the equality would also be true. Solutions to an equation in mathematics are called its roots.

Methods for solving second-order equations

Solving equations of this type requires knowledge of some theory about them. In the school course of algebra, 4 are considered different methods solutions. Let's list them:

using factorization;
using the formula for the perfect square;
applying the graph of the corresponding quadratic function;
using the discriminant equation.

The advantage of the first method is its simplicity, however, it can not be applied to all equations. The second method is universal, but somewhat cumbersome. The third method is distinguished by its clarity, but it is not always convenient and applicable. And finally, using the discriminant equation is a universal and fairly simple way to find the roots of absolutely any second-order equation. Therefore, in the article we will consider only it.

Formula for obtaining the roots of the equation

Let us turn to the general form of the quadratic equation. Let's write it down: a*x²+ b*x + c =0. Before using the method of solving it "through the discriminant", equality should always be reduced to the written form. That is, it must consist of three terms (or less if b or c is 0).

For example, if there is an expression: x²-9*x+8 = -5*x+7*x², then first you should transfer all its members to one side of equality and add the terms containing the variable x in the same powers.

In this case, this operation will lead to the following expression: -6*x²-4*x+8=0, which is equivalent to the equation 6*x²+4*x-8=0 (here we have multiplied the left and right sides of the equation by -1) .

In the example above, a = 6, b=4, c=-8. Note that all terms of the considered equality are always summed among themselves, therefore, if the "-" sign appears, this means that the corresponding coefficient is negative, like the number c in this case.

Having analyzed this point, we now turn to the formula itself, which makes it possible to obtain the roots of a quadratic equation. It looks like the photo below.

As can be seen from this expression, it allows you to get two roots (you should pay attention to the "±" sign). To do this, it is enough to substitute the coefficients b, c, and a into it.

The concept of discriminant

In the previous paragraph, a formula was given that allows you to quickly solve any second-order equation. In it, the radical expression is called the discriminant, that is, D \u003d b²-4 * a * c.

Why is this part of the formula singled out, and does it even have its own name? The fact is that the discriminant connects all three coefficients of the equation into a single expression. Last fact means that it fully carries information about the roots, which can be expressed by the following list:

D>0: the equality has 2 different solutions, both of which are real numbers.
D=0: The equation has only one root, and it is a real number.

The task of determining the discriminant

Here is a simple example of how to find the discriminant. Let the following equality be given: 2*x² - 4+5*x-9*x² = 3*x-5*x²+7.

Let's bring it to the standard form, we get: (2*x²-9*x²+5*x²) + (5*x-3*x) + (- 4-7) = 0, from which we come to equality: -2*x² +2*x-11 = 0. Here a=-2, b=2, c=-11.

Now you can use the named formula for the discriminant: D \u003d 2² - 4 * (-2) * (-11) \u003d -84. The resulting number is the answer to the task. Since the discriminant in the example is less than zero, we can say that this quadratic equation has no real roots. Its solution will be only numbers of complex type.

An example of inequality through the discriminant

Let's solve problems of a slightly different type: the equality -3*x²-6*x+c = 0 is given. It is necessary to find such values of c for which D>0.

In this case, only 2 out of 3 coefficients are known, so it will not be possible to calculate the exact value of the discriminant, but it is known that it is positive. We use the last fact when compiling the inequality: D= (-6)²-4*(-3)*c>0 => 36+12*c>0. The solution of the obtained inequality leads to the result: c>-3.

Let's check the resulting number. To do this, we calculate D for 2 cases: c=-2 and c=-4. The number -2 satisfies the result (-2>-3), the corresponding discriminant will have the value: D = 12>0. In turn, the number -4 does not satisfy the inequality (-4Thus, any numbers c that are greater than -3 will satisfy the condition.

An example of solving an equation

Here is a problem that consists not only in finding the discriminant, but also in solving the equation. It is necessary to find the roots for the equality -2*x²+7-9*x = 0.

In this example, the discriminant is equal to the following value: D = 81-4*(-2)*7= 137. Then the roots of the equation are determined as follows: x = (9±√137)/(-4). This is exact values roots, if you calculate the approximate root, then you get the numbers: x \u003d -5.176 and x \u003d 0.676.

geometric problem

Let's solve a problem that will require not only the ability to calculate the discriminant, but also the use of abstract thinking skills and knowledge of how to write quadratic equations.

Bob had a 5 x 4 meter duvet. The boy wanted to sew a continuous strip of beautiful fabric around the entire perimeter. How thick will this strip be if it is known that Bob has 10 m² of fabric.

Let the strip have a thickness of x m, then the area of the fabric along the long side of the blanket will be (5 + 2 * x) * x, and since there are 2 long sides, we have: 2 * x * (5 + 2 * x). On the short side, the area of the sewn fabric will be 4*x, since there are 2 of these sides, we get the value 8*x. Note that 2*x has been added to the long side because the length of the quilt has increased by that number. The total area of fabric sewn to the blanket is 10 m². Therefore, we get the equality: 2*x*(5+2*x) + 8*x = 10 => 4*x²+18*x-10 = 0.

For this example, the discriminant is: D = 18²-4*4*(-10) = 484. Its root is 22. Using the formula, we find the desired roots: x = (-18±22)/(2*4) = (- 5; 0.5). Obviously, of the two roots, only the number 0.5 is suitable for the condition of the problem.

Thus, the strip of fabric that Bob sews to his blanket will be 50 cm wide.

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

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

What quadratic equations are called complete? This is equations of the form ax 2 + b x + c = 0, where the coefficients a, b and c are not equal to zero. So, to solve the complete quadratic equation, you need to calculate the discriminant D.

D \u003d b 2 - 4ac.

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

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

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

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

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

D \u003d 4 2 - 4 4 \u003d 0

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

Answer: 2.

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

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

Answer: no roots.

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

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

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

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

Answer: - 3.5; one.

So let's imagine the solution of complete quadratic equations by the scheme in Figure 1.

These formulas can be used to solve any complete quadratic equation. You just need to be careful to the equation was written as a polynomial of standard form

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

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

D \u003d 3 2 - 4 1 2 \u003d 1 and then the equation has two roots. And this is not true. (See example 2 solution above).

Therefore, if the equation is not written as a polynomial of the standard form, first the complete quadratic equation must be written as a polynomial of the standard form (the monomial with the largest exponent should be in the first place, that is a x 2 , then with less – bx, and then the free term with.

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

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

Figure 3 shows a diagram of the solution of the reduced square
equations. Consider the example of the application of the formulas discussed in this article.

Example. solve the equation

3x 2 + 6x - 6 = 0.

Let's solve this equation using the formulas shown in Figure 1.

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

√D = √108 = √(36 3) = 6√3

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

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

Answer: -1 - √3; –1 + √3

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

√(D 1) = √27 = √(9 3) = 3√3

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

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

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

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

√(D 2) = √12 = √(4 3) = 2√3

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

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

Answer: -1 - √3; –1 + √3.

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

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Discriminant - ambiguous term. This article will focus on the discriminant of a polynomial, which allows you to determine whether a given polynomial has real solutions. The formula for a square polynomial is found in the school course in algebra and analysis. How to find the discriminant? What is needed to solve the equation?

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

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

She says if there are valid results.
She helps to calculate them.

How this value shows the presence of real roots:

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

Calculation options for fixing the material

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

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

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

Using the Discriminant to Calculate Roots

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

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

Suppose the discriminant is below zero, then d is imaginary and the results are imaginary.

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

Suppose D > 0, so d is a real number, and the answer here splits into two parts: w1 = (-j + d) / (2 * i) and w2 = (-j - d) / (2 * i) . Both results will be valid. Let's look at 2 * w ^ 2 - 3 * w + 1 = 0. Here the discriminant and d are units. So w1 is (3 + 1) divided by (2 * 2) or 1, and w2 is (3 - 1) divided by 2 * 2 or 1/2.

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

Determining the number of valid solutions.
Calculation d = D^(1/2).
Finding the result according to the formula (-j +/- d) / (2 * i).
Substitution of the received result in initial equality for check.

Some special cases

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

the polynomial expands into the difference of squares with a negative free term;
for a positive constant, real solutions cannot be found.

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

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

Reduced Second Degree Equation

The given is called such a square trinomial, where the coefficient in front of the highest term is one. For this situation, the Vieta theorem is applicable, which says that the sum of the roots is equal to the coefficient of the variable to the first power, multiplied by -1, and the product corresponds to the constant "k".

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

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

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

Even second factor

If the factor of the variable to the first power (j) is divisible by 2, then it will be possible to simplify the formula and look for a solution through a quarter of the discriminant D / 4 \u003d (j / 2) ^ 2 - i * k. it turns out w = (-j +/- d/2) / i, where d/2 = D/4 to the power of 1/2.

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

Higher order discriminant

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

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

D \u003d j ^ 2 * k ^ 2 - 4 * i * k ^ 3 - 4 * i ^ 3 * k - 27 * i ^ 2 * m ^ 2 + 18 * i * j * k * m.

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

Video

Our video will tell you in detail about the calculation of the discriminant.

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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 get the idea...

How to solve quadratic equations? If you have a quadratic equation in this form, then everything is simple. We remember 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. So, the formula for finding the roots of a quadratic equation looks like this:

The expression under the root sign is the same discriminant. 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 consider. Substitute with your signs! For example, for the first equation a =1; b = 3; c= -4. Here we write:

Example 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 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. Strictly speaking, this is not a single 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 a negative number Square root is not extracted. Well, okay. This means there are no solutions.

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:

So, how to solve quadratic equations through the discriminant we remembered. Or learned, which is also good. Can you 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?

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

This is incomplete quadratic equations . They can also be solved through the discriminant. You just need to correctly figure out what is equal here a, b and c.

But incomplete quadratic equations can be solved much easier. Without any discrimination. 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.

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 = +3 and x = -3.

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:

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 previous section. 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!

Fractional equations. ODZ.

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

Fractional equations.

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

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

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

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

We need to multiply the whole equation by the same expression. So that all denominators decrease! Everything will immediately become easier. I explain with an example. Let's say we need to solve the equation:

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

On the left side, to reduce the denominator, you need to multiply by x+2. And on the right, multiplication by 2 is required. So, the equation must be multiplied by 2(x+2). We multiply:

This is the usual multiplication of fractions, but I will write in detail:

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

On the left side, it is reduced entirely (x+2), and in the right 2. As required! After reduction we get linear the equation:

Anyone 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 can be written:

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

We see that to reduce the denominator with x, it is necessary to multiply the fraction by (x - 2). And units are not a hindrance to us. Well, let's multiply. All left side and all right side:

Brackets again (x - 2) I don't reveal. 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 cut (x - 2) and we get the equation without any fractions, in a ruler!

And now we open the brackets:

We give similar ones, transfer everything to the left side and get:

Classical 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 coefficients will become prettier! We divide by -2. On the left side - term by term, and on the right - just divide zero by -2, zero and get:

We solve through the discriminant and check according to the Vieta 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, and here it is quadratic. It happens that after getting rid of fractions, all x's are reduced. There is something left, like 5=5. It means that x can be anything. Whatever it is, it will still be reduced. And get the pure truth, 5=5. But, after getting rid of fractions, it may turn out to be completely untrue, such as 2=7. And this means that no solutions! With any x, it turns out to be false.

Realized main way solutions fractional equations? It is simple and logical. We change the original expression so that everything that we don't like disappears. Or interfere. In this case, it's fractions. We will do the same with all sorts of complex examples with logarithms, sines and other horrors. We always we will get rid of all this.

However, we need to change the original expression in the direction we need according to the rules, yes ... The development of which is the preparation for the exam in mathematics. Here we are learning.

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

Let's take a simple example:

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

Remember, with brackets (x - 2) we work as with one, integral expression!

Here I no longer wrote the one in the denominators, undignified ... And I didn’t draw brackets in the denominators, except for x - 2 there is nothing, you can not draw. We shorten:

We open the brackets, move everything to the left, we give similar ones:

We solve, check, we get two roots. x = 2 and x = 3. Fine.

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

If you decide the answer is 5, you were ambushed. And the task will not be counted for you. They worked in vain ... The correct answer is 3.

What's the matter?! And you try to check. Substitute the values of the unknown into initial example. And if at x = 3 everything grows together wonderfully, we get 9 = 9, then with x = 2 divide by zero! What absolutely cannot be done. 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 just discard it. There is only one final root. x = 3.

How so?! I hear outraged exclamations. We were taught that an equation can be multiplied by an expression! This is the same transformation!

Yes, identical. At small condition- the expression by which we multiply (divide) - different from zero. BUT x - 2 at x = 2 equals zero! So it's all fair.

And now what i can do?! Do not multiply by expression? Do you check every time? Again unclear!

Calmly! No panic!

In this difficult situation, three magic letters will save us. I know what you were thinking. Correctly! This is ODZ . Area of Valid Values.