For example, for the trinomial \(3x^2+2x-7\), the discriminant will be equal to \(2^2-4\cdot3\cdot(-7)=4+84=88\). And for the trinomial \(x^2-5x+11\), it will be equal to \((-5)^2-4\cdot1\cdot11=25-44=-19\).

The discriminant is denoted by the letter \(D\) and is often used in solving. Also, by the value of the discriminant, you can understand what the graph approximately looks like (see below).

Discriminant and roots of the equation

The discriminant value shows the number of quadratic equations:
- if \(D\) is positive, the equation will have two roots;
- if \(D\) is equal to zero – there is only one root;
- if \(D\) is negative, there are no roots.

This does not need to be taught, it is not difficult to come to such a conclusion, simply knowing that from the discriminant (that is, \(\sqrt(D)\) is included in the formula for calculating the roots of the equation: \(x_(1)=\)\(\ frac(-b+\sqrt(D))(2a)\) and \(x_(2)=\)\(\frac(-b-\sqrt(D))(2a)\) Let's look at each case in more detail. .

If the discriminant is positive

In this case, the root of it is some positive number, which means \(x_(1)\) and \(x_(2)\) will have different meanings, because in the first formula \(\sqrt(D)\) is added, and in the second it is subtracted. And we have two different roots.

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

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

If the discriminant is zero

And how many roots will there be if the discriminant equal to zero? Let's reason.

The root formulas look like this: \(x_(1)=\)\(\frac(-b+\sqrt(D))(2a)\) and \(x_(2)=\)\(\frac(-b- \sqrt(D))(2a)\) . And if the discriminant is zero, then its root is also zero. Then it turns out:

\(x_(1)=\)\(\frac(-b+\sqrt(D))(2a)\) \(=\)\(\frac(-b+\sqrt(0))(2a)\) \(=\)\(\frac(-b+0)(2a)\) \(=\)\(\frac(-b)(2a)\)

\(x_(2)=\)\(\frac(-b-\sqrt(D))(2a)\) \(=\)\(\frac(-b-\sqrt(0))(2a) \) \(=\)\(\frac(-b-0)(2a)\) \(=\)\(\frac(-b)(2a)\)

That is, the values ​​of the roots of the equation will be the same, because adding or subtracting zero does not change anything.

Example : Find the roots of the equation \(x^2-4x+4=0\)
Solution :

Answer : \(x=2\)

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.

Speaking mathematical language, a quadratic equation is an equation of the form:

Here a, b and c- some numbers. b and c- absolutely any, but A– anything 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 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 given equation lead to standard view, 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 substitution negative values into the formula for calculating the roots. Saves here detailed entry formulas 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!

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.

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 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.

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 1 = 0, x 2 = 4.

All. These will be the roots of our equation. Both are suitable. When substituting any of them into 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 - it is 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 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.

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...

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 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.

Honestly speaking, when simple solution quadratic equations, the concept of a discriminant is not particularly required. 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 difficult 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. Did you understand that keyword Here - 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:

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 afraid, 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 you’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 odds, - get rid of fractions! Multiply the equation by common denominator, as described in the lesson "How to solve equations? Identical 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 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 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 the decision different 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.

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

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

D = b 2 – 4ac.

Depending on the value of the discriminant, 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 = (-b)/2a. When the discriminant is a 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 = 4 2 – 4 4 = 0

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

Answer: 2.

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

D = 1 2 – 4 2 3 = – 23

Answer: no roots.

Solve Equation 2 x 2 + 5x – 7 = 0.

D = 5 2 – 4 2 (–7) = 81

x 1 = (-5 - √81)/(2 2)= (-5 - 9)/4= – 3.5

x 2 = (-5 + √81)/(2 2) = (-5 + 9)/4=1

Answer: – 3.5; 1.

So let’s imagine the solution of complete quadratic equations using the diagram in Figure 1.

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

A x 2 + bx + c, otherwise you may 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 = 3 2 – 4 1 2 = 1 and then the equation has two roots. And this is not true. (See solution to example 2 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 highest indicator degrees, that is A x 2 , then with less – bx and then a free member With.

When solving the reduced quadratic equation and a quadratic equation with an even coefficient in the second term, you can use other formulas. Let's get acquainted with these formulas. If in a complete quadratic equation the second term has an even coefficient (b = 2k), then you can solve the equation using the formulas shown in the diagram in Figure 2.

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

Figure 3 shows a diagram for solving the reduced square
equations. Let's look at an 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 the diagram in Figure 1.

D = 6 2 – 4 3 (– 6) = 36 + 72 = 108

√D = √108 = √(36 3) = 6√3

x 1 = (-6 - 6√3)/(2 3) = (6 (-1- √(3)))/6 = –1 – √3

x 2 = (-6 + 6√3)/(2 3) = (6 (-1+ √(3)))/6 = –1 + √3

Answer: –1 – √3; –1 + √3

You can notice that the coefficient of x in this equation even number, that is, b = 6 or b = 2k, whence k = 3. Then let’s try to solve the equation using the formulas given in the diagram of the figure D 1 = 3 2 – 3 · (– 6) = 9 + 18 = 27

√(D 1) = √27 = √(9 3) = 3√3

x 1 = (-3 - 3√3)/3 = (3 (-1 - √(3)))/3 = – 1 – √3

x 2 = (-3 + 3√3)/3 = (3 (-1 + √(3)))/3 = – 1 + √3

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

D 2 = 2 2 – 4 (– 2) = 4 + 8 = 12

√(D 2) = √12 = √(4 3) = 2√3

x 1 = (-2 - 2√3)/2 = (2 (-1 - √(3)))/2 = – 1 – √3

x 2 = (-2 + 2√3)/2 = (2 (-1+ √(3)))/2 = – 1 + √3

Answer: –1 – √3; –1 + √3.

As we see, when solving this equation by various formulas we received the same answer. Therefore, having thoroughly mastered the formulas shown in the diagram in Figure 1, you will always be able to solve any complete quadratic equation.

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Quadratic equation - easy to solve! *Hereinafter referred to as “KU”. Friends, it would seem that there could be nothing simpler in mathematics than solving such an equation. But something told me that many people have problems with him. I decided to see how many on-demand impressions Yandex gives out per month. Here's what happened, look:

What does it mean? This means that about 70,000 people per month are searching for this information, what does this summer have to do with it, and what will happen among school year— there will be twice as many requests. This is not surprising, because those guys and girls who graduated from school a long time ago and are preparing for the Unified State Exam are looking for this information, and schoolchildren also strive to refresh their memory.

Despite the fact that there are a lot of sites that tell you how to solve this equation, I decided to also contribute and publish the material. Firstly, I want visitors to come to my site based on this request; secondly, in other articles, when the topic of “KU” comes up, I will provide a link to this article; thirdly, I’ll tell you a little more about his solution than is usually stated on other sites. Let's get started! The content of the article:

A quadratic equation is an equation of the form:

where coefficients a,band with arbitrary numbers, where a≠0.

IN school course the material is given in the following form - the equations are conditionally divided into three classes:

1. They have two roots.

2. *Have only one root.

3. They have no roots. It is worth especially noting here that they do not have real roots

How are roots calculated? Just!

We calculate the discriminant. Underneath this “terrible” word lies a very simple formula:

The root formulas are as follows:

*You need to know these formulas by heart.

You can immediately write down and solve:

Example:

1. If D > 0, then the equation has two roots.

2. If D = 0, then the equation has one root.

3. If D< 0, то уравнение не имеет действительных корней.

Let's look at the equation:

In this regard, when the discriminant is equal to zero, the school course says that one root is obtained, here it is equal to nine. Everything is correct, it is so, but...

This idea is somewhat incorrect. In fact, there are two roots. Yes, yes, don’t be surprised, you get two equal roots, and to be mathematically precise, then the answer should write two roots:

x 1 = 3 x 2 = 3

But this is so - a small digression. At school you can write it down and say that there is one root.

Now the next example:

As we know, the root of a negative number cannot be taken, so the solutions in in this case No.

That's the whole decision process.

Quadratic function.

This shows what the solution looks like geometrically. This is extremely important to understand (in the future, in one of the articles we will analyze in detail the solution to the quadratic inequality).

This is a function of the form:

where x and y are variables

a, b, c – given numbers, where a ≠ 0

The graph is a parabola:

That is, it turns out that solving a quadratic equation at “y” equal to zero we find the intersection points of the parabola with the x axis. There can be two of these points (the discriminant is positive), one (the discriminant is zero) and none (the discriminant is negative). Details about quadratic function You can view article by Inna Feldman.

Let's look at examples:

Example 1: Solve 2x 2 +8 x–192=0

a=2 b=8 c= –192

D=b 2 –4ac = 8 2 –4∙2∙(–192) = 64+1536 = 1600

Answer: x 1 = 8 x 2 = –12

*It was possible to immediately divide the left and right sides of the equation by 2, that is, simplify it. The calculations will be easier.

Example 2: Decide x 2–22 x+121 = 0

a=1 b=–22 c=121

D = b 2 –4ac =(–22) 2 –4∙1∙121 = 484–484 = 0

We found that x 1 = 11 and x 2 = 11

It is permissible to write x = 11 in the answer.

Answer: x = 11

Example 3: Decide x 2 –8x+72 = 0

a=1 b= –8 c=72

D = b 2 –4ac =(–8) 2 –4∙1∙72 = 64–288 = –224

The discriminant is negative, there is no solution in real numbers.

Answer: no solution

The discriminant is negative. There is a solution!

Here we will talk about solving the equation in the case when it turns out negative discriminant. Do you know anything about complex numbers? I will not talk in detail here about why and where they arose and what they are specific role and the need for mathematics, this is a topic for a large separate article.

The concept of a complex number.

A little theory.

A complex number z is a number of the form

z = a + bi

where a and b are real numbers, i is the so-called imaginary unit.

a+bi – this is a SINGLE NUMBER, not an addition.

The imaginary unit is equal to the root of minus one:

Now consider the equation:

We get two conjugate roots.

Incomplete quadratic equation.

Let's consider special cases, this is when the coefficient “b” or “c” is equal to zero (or both are equal to zero). They can be solved easily without any discriminants.

Case 1. Coefficient b = 0.

The equation becomes:

Let's transform:

Example:

4x 2 –16 = 0 => 4x 2 =16 => x 2 = 4 => x 1 = 2 x 2 = –2

Case 2. Coefficient c = 0.

The equation becomes:

Let's transform and factorize:

*The product is equal to zero when at least one of the factors is equal to zero.

Example:

9x 2 –45x = 0 => 9x (x–5) =0 => x = 0 or x–5 =0

x 1 = 0 x 2 = 5

Case 3. Coefficients b = 0 and c = 0.

Here it is clear that the solution to the equation will always be x = 0.

Useful properties and patterns of coefficients.

There are properties that allow you to solve equations with large coefficients.

Ax 2 + bx+ c=0 equality holds

a + b+ c = 0, That

- if for the coefficients of the equation Ax 2 + bx+ c=0 equality holds

a+ s =b, That

These properties help to decide a certain type equations

Example 1: 5001 x 2 –4995 x – 6=0

The sum of the odds is 5001+( – 4995)+(– 6) = 0, which means

Example 2: 2501 x 2 +2507 x+6=0

Equality holds a+ s =b, Means

Regularities of coefficients.

1. If in the equation ax 2 + bx + c = 0 the coefficient “b” is equal to (a 2 +1), and the coefficient “c” is numerically equal to the coefficient “a”, then its roots are equal

ax 2 + (a 2 +1)∙x+ a= 0 = > x 1 = –a x 2 = –1/a.

Example. Consider the equation 6x 2 + 37x + 6 = 0.

x 1 = –6 x 2 = –1/6.

2. If in the equation ax 2 – bx + c = 0 the coefficient “b” is equal to (a 2 +1), and the coefficient “c” is numerically equal to the coefficient “a”, then its roots are equal

ax 2 – (a 2 +1)∙x+ a= 0 = > x 1 = a x 2 = 1/a.

Example. Consider the equation 15x 2 –226x +15 = 0.

x 1 = 15 x 2 = 1/15.

3. If in Eq. ax 2 + bx – c = 0 coefficient “b” is equal to (a 2 – 1), and coefficient “c” is numerically equal to the coefficient “a”, then its roots are equal

ax 2 + (a 2 –1)∙x – a= 0 = > x 1 = – a x 2 = 1/a.

Example. Consider the equation 17x 2 +288x – 17 = 0.

x 1 = – 17 x 2 = 1/17.

4. If in the equation ax 2 – bx – c = 0 the coefficient “b” is equal to (a 2 – 1), and the coefficient c is numerically equal to the coefficient “a”, then its roots are equal

ax 2 – (a 2 –1)∙x – a= 0 = > x 1 = a x 2 = – 1/a.

Example. Consider the equation 10x 2 – 99x –10 = 0.

x 1 = 10 x 2 = – 1/10

Vieta's theorem.

Vieta's theorem is named after the famous French mathematician Francois Vieta. Using Vieta's theorem, we can express the sum and product of the roots of an arbitrary KU in terms of its coefficients.

45 = 1∙45 45 = 3∙15 45 = 5∙9.

In total, the number 14 gives only 5 and 9. These are the roots. With a certain skill, using the presented theorem, you can solve many quadratic equations orally immediately.

Vieta's theorem, in addition. is convenient in that after solving a quadratic equation in the usual way (through a discriminant), the resulting roots can be checked. I recommend doing this always.

TRANSPORTATION METHOD

With this method, the coefficient “a” is multiplied by the free term, as if “thrown” to it, which is why it is called "transfer" method. This method is used when you can easily find the roots of the equation using Vieta's theorem and, most importantly, when the discriminant is an exact square.

If A± b+c≠ 0, then the transfer technique is used, for example:

2X 2 – 11x+ 5 = 0 (1) => X 2 – 11x+ 10 = 0 (2)

Using Vieta's theorem in equation (2), it is easy to determine that x 1 = 10 x 2 = 1

The resulting roots of the equation must be divided by 2 (since the two were “thrown” from x 2), we get

x 1 = 5 x 2 = 0.5.

What is the rationale? Look what's happening.

The discriminants of equations (1) and (2) are equal:

If you look at the roots of the equations, you only get different denominators, and the result depends precisely on the coefficient of x 2:

The second (modified) one has roots that are 2 times larger.

Therefore, we divide the result by 2.

*If we reroll the three, we will divide the result by 3, etc.

Answer: x 1 = 5 x 2 = 0.5

Sq. ur-ie and Unified State Examination.

I’ll tell you briefly about its importance - YOU MUST BE ABLE TO DECIDE quickly and without thinking, you need to know the formulas of roots and discriminants by heart. Many problems included in the Unified State Examination tasks come down to solving a quadratic equation (geometric ones included).

Something worth noting!

1. The form of writing an equation can be “implicit”. For example, the following entry is possible:

15+ 9x 2 - 45x = 0 or 15x+42+9x 2 - 45x=0 or 15 -5x+10x 2 = 0.

You need to bring it to a standard form (so as not to get confused when solving).

2. Remember that x is an unknown quantity and it can be denoted by any other letter - t, q, p, h and others.