Vieta's theorem (more precisely, the theorem converse theorem Vieta) allows you to reduce the time for solving quadratic equations. You just need to know how to use it. How to learn to solve quadratic equations using Vieta's theorem? It's easy if you think a little.

Now we will only talk about the solution of the reduced quadratic equation by the Vieta theorem. quadratic equation is an equation in which a, that is, the coefficient in front of x², is equal to one. Not given quadratic equations can also be solved using the Vieta theorem, but there already at least one of the roots is not an integer. They are harder to guess.

The theorem converse to Vieta's theorem says: if the numbers x1 and x2 are such that

then x1 and x2 are the roots of the quadratic equation

When solving a quadratic equation using the Vieta theorem, only 4 options are possible. If you remember the course of reasoning, you can learn to find whole roots very quickly.

I. If q is a positive number,

this means that the roots x1 and x2 are numbers of the same sign (because only when multiplying numbers with the same signs, a positive number is obtained).

I.a. If -p is a positive number, (respectively, p<0), то оба корня x1 и x2 — положительные числа (поскольку складывали числа одного знака и получили положительное число).

I.b. If -p - a negative number, (respectively, p>0), then both roots are negative numbers (they added numbers of the same sign, got a negative number).

II. If q is a negative number,

this means that the roots x1 and x2 have different signs (when multiplying numbers, a negative number is obtained only when the signs of the factors are different). In this case, x1 + x2 is no longer a sum, but a difference (after all, when adding numbers with different signs we subtract the smaller from the larger modulo). Therefore, x1 + x2 shows how much the roots x1 and x2 differ, that is, how much one root is more than the other (modulo).

II.a. If -p is a positive number, (i.e. p<0), то больший (по модулю) корень — положительное число.

II.b. If -p is a negative number, (p>0), then the larger (modulo) root is a negative number.

Consider the solution of quadratic equations according to Vieta's theorem using examples.

Solve the given quadratic equation using Vieta's theorem:

Here q=12>0, so the roots x1 and x2 are numbers of the same sign. Their sum is -p=7>0, so both roots are positive numbers. We select integers whose product is 12. These are 1 and 12, 2 and 6, 3 and 4. The sum is 7 for the pair 3 and 4. Hence, 3 and 4 are the roots of the equation.

In this example, q=16>0, which means that the roots x1 and x2 are numbers of the same sign. Their sum -p=-10<0, поэтому оба корня — отрицательные числа. Подбираем числа, произведение которых равно 16. Это 1 и 16, 2 и 8, 4 и 4. Сумма 2 и 8 равна 10, а раз нужны отрицательные числа, то искомые корни — это -2 и -8.

Here q=-15<0, что означает, что корни x1 и x2 — числа разных знаков. Поэтому 2 — это уже не их сумма, а разность, то есть числа отличаются на 2. Подбираем числа, произведение которых равно 15, отличающиеся на 2. Произведение равно 15 у 1 и 15, 3 и 5. Отличаются на 2 числа в паре 3 и 5. Поскольку -p=2>0, then the larger number is positive. So the roots are 5 and -3.

q=-36<0, значит, корни x1 и x2 имеют разные знаки. Тогда 5 — это то, насколько отличаются x1 и x2 (по модулю, то есть пока что без учета знака). Среди чисел, произведение которых равно 36: 1 и 36, 2 и 18, 3 и 12, 4 и 9 — выбираем пару, в которой числа отличаются на 5. Это 4 и 9. Осталось определить их знаки. Поскольку -p=-5<0, бОльшее число имеет знак минус. Поэтому корни данного уравнения равны -9 и 4.

In mathematics, there are special tricks with which many quadratic equations are solved very quickly and without any discriminants. Moreover, with proper training, many begin to solve quadratic equations verbally, literally "at a glance."

Unfortunately, in the modern course of school mathematics, such technologies are almost not studied. And you need to know! And today we will consider one of these techniques - Vieta's theorem. First, let's introduce a new definition.

A quadratic equation of the form x 2 + bx + c = 0 is called reduced. Please note that the coefficient at x 2 is equal to 1. There are no other restrictions on the coefficients.

1. x 2 + 7x + 12 = 0 is the reduced quadratic equation;
2. x 2 − 5x + 6 = 0 - also reduced;
3. 2x 2 − 6x + 8 = 0 - but this is nothing reduced, since the coefficient at x 2 is 2.

Of course, any quadratic equation of the form ax 2 + bx + c = 0 can be made reduced - it is enough to divide all the coefficients by the number a. We can always do this, since it follows from the definition of a quadratic equation that a ≠ 0.

True, these transformations will not always be useful for finding roots. A little lower, we will make sure that this should be done only when in the final squared equation all the coefficients are integer. For now, let's look at some simple examples:

Task. Convert quadratic equation to reduced:

1. 3x2 − 12x + 18 = 0;
2. −4x2 + 32x + 16 = 0;
3. 1.5x2 + 7.5x + 3 = 0;
4. 2x2 + 7x − 11 = 0.

Let's divide each equation by the coefficient of the variable x 2 . We get:

1. 3x 2 - 12x + 18 = 0 ⇒ x 2 - 4x + 6 = 0 - divided everything by 3;
2. −4x 2 + 32x + 16 = 0 ⇒ x 2 − 8x − 4 = 0 - divided by −4;
3. 1.5x 2 + 7.5x + 3 \u003d 0 ⇒ x 2 + 5x + 2 \u003d 0 - divided by 1.5, all coefficients became integers;
4. 2x 2 + 7x - 11 \u003d 0 ⇒ x 2 + 3.5x - 5.5 \u003d 0 - divided by 2. In this case, fractional coefficients arose.

As you can see, the given quadratic equations can have integer coefficients even if the original equation contained fractions.

Now we formulate the main theorem, for which, in fact, the concept of a reduced quadratic equation was introduced:

Vieta's theorem. Consider the reduced quadratic equation of the form x 2 + bx + c \u003d 0. Suppose that this equation has real roots x 1 and x 2. In this case, the following statements are true:

1. x1 + x2 = −b. In other words, the sum of the roots of the given quadratic equation is equal to the coefficient of the variable x, taken with the opposite sign;
2. x 1 x 2 = c. The product of the roots of a quadratic equation is equal to the free coefficient.

Examples. For simplicity, we will consider only the given quadratic equations that do not require additional transformations:

1. x 2 − 9x + 20 = 0 ⇒ x 1 + x 2 = − (−9) = 9; x 1 x 2 = 20; roots: x 1 = 4; x 2 \u003d 5;
2. x 2 + 2x − 15 = 0 ⇒ x 1 + x 2 = −2; x 1 x 2 \u003d -15; roots: x 1 = 3; x 2 \u003d -5;
3. x 2 + 5x + 4 = 0 ⇒ x 1 + x 2 = −5; x 1 x 2 = 4; roots: x 1 \u003d -1; x 2 \u003d -4.

Vieta's theorem gives us additional information about the roots of a quadratic equation. At first glance, this may seem complicated, but even with minimal training, you will learn to "see" the roots and literally guess them in a matter of seconds.

Task. Solve the quadratic equation:

1. x2 − 9x + 14 = 0;
2. x 2 - 12x + 27 = 0;
3. 3x2 + 33x + 30 = 0;
4. −7x2 + 77x − 210 = 0.

Let's try to write down the coefficients according to the Vieta theorem and "guess" the roots:

1. x 2 − 9x + 14 = 0 is the reduced quadratic equation.
   By the Vieta theorem, we have: x 1 + x 2 = −(−9) = 9; x 1 x 2 = 14. It is easy to see that the roots are the numbers 2 and 7;
2. x 2 − 12x + 27 = 0 - also reduced.
   By the Vieta theorem: x 1 + x 2 = −(−12) = 12; x 1 x 2 = 27. Hence the roots: 3 and 9;
3. 3x 2 + 33x + 30 = 0 - this equation is not reduced. But we will fix this now by dividing both sides of the equation by the coefficient a \u003d 3. We get: x 2 + 11x + 10 \u003d 0.
   We solve according to the Vieta theorem: x 1 + x 2 = −11; x 1 x 2 = 10 ⇒ roots: −10 and −1;
4. −7x 2 + 77x − 210 = 0 - again the coefficient at x 2 is not equal to 1, i.e. equation not given. We divide everything by the number a = −7. We get: x 2 - 11x + 30 = 0.
   By the Vieta theorem: x 1 + x 2 = −(−11) = 11; x 1 x 2 = 30; from these equations it is easy to guess the roots: 5 and 6.

From the above reasoning, it can be seen how Vieta's theorem simplifies the solution of quadratic equations. No complicated calculations, no arithmetic roots and fractions. And even the discriminant (see the lesson " Solving quadratic equations") We did not need.

Of course, in all our reflections, we proceeded from two important assumptions, which, generally speaking, are not always fulfilled in real problems:

1. The quadratic equation is reduced, i.e. the coefficient at x 2 is 1;
2. The equation has two different roots. From the point of view of algebra, in this case the discriminant D > 0 - in fact, we initially assume that this inequality is true.

However, in typical mathematical problems these conditions are met. If, as a result of the calculations, a “bad” quadratic equation is obtained (the coefficient at x 2 is different from 1), this is easy to fix - take a look at the examples at the very beginning of the lesson. I am generally silent about the roots: what kind of task is this in which there is no answer? Of course there will be roots.

Thus, the general scheme for solving quadratic equations according to the Vieta theorem is as follows:

1. Reduce the quadratic equation to the given one, if this has not already been done in the condition of the problem;
2. If the coefficients in the above quadratic equation turned out to be fractional, we solve through the discriminant. You can even go back to the original equation to work with more "convenient" numbers;
3. In the case of integer coefficients, we solve the equation using the Vieta theorem;
4. If within a few seconds it was not possible to guess the roots, we score on the Vieta theorem and solve through the discriminant.

Task. Solve the equation: 5x 2 − 35x + 50 = 0.

So, we have an equation that is not reduced, because coefficient a \u003d 5. Divide everything by 5, we get: x 2 - 7x + 10 \u003d 0.

All coefficients of the quadratic equation are integer - let's try to solve using the Vieta theorem. We have: x 1 + x 2 = −(−7) = 7; x 1 x 2 \u003d 10. In this case, the roots are easy to guess - these are 2 and 5. You do not need to count through the discriminant.

Task. Solve the equation: -5x 2 + 8x - 2.4 = 0.

We look: -5x 2 + 8x - 2.4 = 0 - this equation is not reduced, we divide both sides by the coefficient a = -5. We get: x 2 - 1.6x + 0.48 = 0 - an equation with fractional coefficients.

It is better to return to the original equation and count through the discriminant: −5x 2 + 8x − 2.4 = 0 ⇒ D = 8 2 − 4 (−5) (−2.4) = 16 ⇒ ... ⇒ x 1 = 1.2; x 2 \u003d 0.4.

Task. Solve the equation: 2x 2 + 10x − 600 = 0.

To begin with, we divide everything by the coefficient a \u003d 2. We get the equation x 2 + 5x - 300 \u003d 0.

This is the reduced equation, according to the Vieta theorem we have: x 1 + x 2 = −5; x 1 x 2 \u003d -300. It is difficult to guess the roots of the quadratic equation in this case - personally, I seriously "froze" when I solved this problem.

We will have to look for roots through the discriminant: D = 5 2 − 4 1 (−300) = 1225 = 35 2 . If you don't remember the root of the discriminant, I'll just note that 1225: 25 = 49. Therefore, 1225 = 25 49 = 5 2 7 2 = 35 2 .

Now that the root of the discriminant is known, solving the equation is not difficult. We get: x 1 \u003d 15; x 2 \u003d -20.

Vieta's theorem is often used to test already found roots. If you have found the roots, you can use the formulas \(\begin(cases)x_1+x_2=-p \\x_1 \cdot x_2=q\end(cases)\) to calculate the values ​​\(p\) and \(q\ ). And if they turn out to be the same as in the original equation, then the roots are found correctly.

For example, let's use , solve the equation \(x^2+x-56=0\) and get the roots: \(x_1=7\), \(x_2=-8\). Let's check if we made a mistake in the process of solving. In our case, \(p=1\), and \(q=-56\). By Vieta's theorem we have:

\(\begin(cases)x_1+x_2=-p \\x_1 \cdot x_2=q\end(cases)\) \(\Leftrightarrow\) \(\begin(cases)7+(-8)=-1 \\7\cdot(-8)=-56\end(cases)\) \(\Leftrightarrow\) \(\begin(cases)-1=-1\\-56=-56\end(cases)\ )

Both statements converged, which means that we solved the equation correctly.

This test can be done orally. It will take 5 seconds and save you from stupid mistakes.

Inverse Vieta theorem

If \(\begin(cases)x_1+x_2=-p \\x_1 \cdot x_2=q\end(cases)\), then \(x_1\) and \(x_2\) are the roots of the quadratic equation \(x^ 2+px+q=0\).

Or in a simple way: if you have an equation of the form \(x^2+px+q=0\), then by solving the system \(\begin(cases)x_1+x_2=-p \\x_1 \cdot x_2=q\ end(cases)\) you will find its roots.

Thanks to this theorem, you can quickly find the roots of a quadratic equation, especially if these roots are . This skill is important as it saves a lot of time.

Example . Solve the equation \(x^2-5x+6=0\).

Solution : Using the inverse Vieta theorem, we get that the roots satisfy the conditions: \(\begin(cases)x_1+x_2=5 \\x_1 \cdot x_2=6\end(cases)\).
Look at the second equation of the \(x_1 \cdot x_2=6\) system. Into what two can the number \(6\) be decomposed? On \(2\) and \(3\), \(6\) and \(1\) or \(-2\) and \(-3\), and \(-6\) and \(- 1\). And which pair to choose, the first equation of the system will tell: \(x_1+x_2=5\). \(2\) and \(3\) are similar, because \(2+3=5\).
Answer : \(x_1=2\), \(x_2=3\).

Examples . Using the inverse of Vieta's theorem, find the roots of the quadratic equation:
a) \(x^2-15x+14=0\); b) \(x^2+3x-4=0\); c) \(x^2+9x+20=0\); d) \(x^2-88x+780=0\).

Solution :
a) \(x^2-15x+14=0\) - what factors does \(14\) decompose into? \(2\) and \(7\), \(-2\) and \(-7\), \(-1\) and \(-14\), \(1\) and \(14\ ). What pairs of numbers add up to \(15\)? Answer: \(1\) and \(14\).

b) \(x^2+3x-4=0\) - into what factors does \(-4\) decompose? \(-2\) and \(2\), \(4\) and \(-1\), \(1\) and \(-4\). What pairs of numbers add up to \(-3\)? Answer: \(1\) and \(-4\).

c) \(x^2+9x+20=0\) – into what factors does \(20\) decompose? \(4\) and \(5\), \(-4\) and \(-5\), \(2\) and \(10\), \(-2\) and \(-10\ ), \(-20\) and \(-1\), \(20\) and \(1\). What pairs of numbers add up to \(-9\)? Answer: \(-4\) and \(-5\).

d) \(x^2-88x+780=0\) - into what factors does \(780\) decompose? \(390\) and \(2\). Do they add up to \(88\)? No. What other multipliers does \(780\) have? \(78\) and \(10\). Do they add up to \(88\)? Yes. Answer: \(78\) and \(10\).

It is not necessary to decompose the last term into all possible factors (as in the last example). You can immediately check whether their sum gives \(-p\).

Important! Vieta's theorem and the converse theorem only work with , that is, one whose coefficient in front of \(x^2\) is equal to one. If we initially have a non-reduced equation, then we can make it reduced by simply dividing by the coefficient in front of \ (x ^ 2 \).

For example, let the equation \(2x^2-4x-6=0\) be given and we want to use one of Vieta's theorems. But we can't, because the coefficient before \(x^2\) is equal to \(2\). Let's get rid of it by dividing the whole equation by \(2\).

\(2x^2-4x-6=0\) \(|:2\)
\(x^2-2x-3=0\)

Ready. Now we can use both theorems.

Answers to frequently asked questions

Question: By Vieta's theorem, you can solve any ?
Answer: Unfortunately no. If there are not integers in the equation or the equation has no roots at all, then Vieta's theorem will not help. In this case, you need to use discriminant . Fortunately, 80% of the equations in the school math course have integer solutions.

The discriminant, as well as quadratic equations, begin to be studied in the algebra course in grade 8. You can solve a quadratic equation through the discriminant and using the Vieta theorem. The methodology for studying quadratic equations, as well as the discriminant formula, is rather unsuccessfully instilled in schoolchildren, like much in real education. Therefore, school years pass, education in grades 9-11 replaces "higher education" and everyone is again looking for - "How to solve a quadratic equation?", "How to find the roots of an equation?", "How to find the discriminant?" And...

Discriminant Formula

The discriminant D of the quadratic equation a*x^2+bx+c=0 is D=b^2–4*a*c.
The roots (solutions) of the quadratic equation depend on the sign of the discriminant (D):
D>0 - the equation has 2 different real roots;
D=0 - the equation has 1 root (2 coinciding roots):
D<0 – не имеет действительных корней (в школьной теории). В ВУЗах изучают комплексные числа и уже на множестве комплексных чисел уравнение с отрицательным дискриминантом имеет два комплексных корня.
The formula for calculating the discriminant is quite simple, so many sites offer an online discriminant calculator. We have not figured out this kind of scripts yet, so who knows how to implement this, please write to the mail This email address is being protected from spambots. You must have JavaScript enabled to view. .

General formula for finding the roots of a quadratic equation:

The roots of the equation are found by the formula
If the coefficient of the variable in the square is paired, then it is advisable to calculate not the discriminant, but its fourth part
In such cases, the roots of the equation are found by the formula

The second way to find roots is Vieta's Theorem.

The theorem is formulated not only for quadratic equations, but also for polynomials. You can read this on Wikipedia or other electronic resources. However, to simplify, consider that part of it that concerns the reduced quadratic equations, that is, equations of the form (a=1)
The essence of the Vieta formulas is that the sum of the roots of the equation is equal to the coefficient of the variable, taken with the opposite sign. The product of the roots of the equation is equal to the free term. The formulas of Vieta's theorem have a notation.
The derivation of the Vieta formula is quite simple. Let's write the quadratic equation in terms of prime factors
As you can see, everything ingenious is simple at the same time. It is effective to use the Vieta formula when the difference in the modulus of the roots or the difference in the modulus of the roots is 1, 2. For example, the following equations, according to the Vieta theorem, have roots




Up to 4 equation analysis should look like this. The product of the roots of the equation is 6, so the roots can be the values ​​(1, 6) and (2, 3) or pairs with the opposite sign. The sum of the roots is 7 (the coefficient of the variable with the opposite sign). From here we conclude that the solutions of the quadratic equation are equal to x=2; x=3.
It is easier to select the roots of the equation among the divisors of the free term, correcting their sign in order to fulfill the Vieta formulas. At the beginning, this seems difficult to do, but with practice on a number of quadratic equations, this technique will be more efficient than calculating the discriminant and finding the roots of the quadratic equation in the classical way.
As you can see, the school theory of studying the discriminant and ways to find solutions to the equation is devoid of practical meaning - "Why do schoolchildren need a quadratic equation?", "What is the physical meaning of the discriminant?".

Let's try to figure it out what does the discriminant describe?

In the course of algebra, they study functions, schemes for studying functions and plotting functions. Of all the functions, an important place is occupied by a parabola, the equation of which can be written in the form
So the physical meaning of the quadratic equation is the zeros of the parabola, that is, the points of intersection of the graph of the function with the abscissa axis Ox
I ask you to remember the properties of parabolas that are described below. The time will come to take exams, tests, or entrance exams and you will be grateful for the reference material. The sign of the variable in the square corresponds to whether the branches of the parabola on the graph will go up (a>0),

or a parabola with branches down (a<0) .

The vertex of the parabola lies midway between the roots

The physical meaning of the discriminant:

If the discriminant is greater than zero (D>0), the parabola has two points of intersection with the Ox axis.
If the discriminant is equal to zero (D=0), then the parabola at the top touches the x-axis.
And the last case, when the discriminant is less than zero (D<0) – график параболы принадлежит плоскости над осью абсцисс (ветки параболы вверх), или график полностью под осью абсцисс (ветки параболы опущены вниз).

Incomplete quadratic equations

First level

Quadratic equations. Comprehensive Guide (2019)

In the term "quadratic equation" the key word is "quadratic". This means that the equation must necessarily contain a variable (the same X) in the square, and at the same time there should not be Xs in the third (or greater) degree.

The solution of many equations is reduced to the solution of quadratic equations.

Let's learn to determine that we have a quadratic equation, and not some other.

Example 1

Get rid of the denominator and multiply each term of the equation by

Let's move everything to the left side and arrange the terms in descending order of powers of x

Now we can say with confidence that this equation is quadratic!

Example 2

Multiply the left and right sides by:

This equation, although it was originally in it, is not a square!

Example 3

Let's multiply everything by:

Scary? The fourth and second degrees ... However, if we make a replacement, we will see that we have a simple quadratic equation:

Example 4

It seems to be, but let's take a closer look. Let's move everything to the left side:

You see, it has shrunk - and now it's a simple linear equation!

Now try to determine for yourself which of the following equations are quadratic and which are not:

Examples:

Answers:

1. square;
2. square;
3. not square;
4. not square;
5. not square;
6. square;
7. not square;
8. square.

Mathematicians conditionally divide all quadratic equations into the following types:

• Complete quadratic equations- equations in which the coefficients and, as well as the free term c, are not equal to zero (as in the example). In addition, among the complete quadratic equations, there are given are equations in which the coefficient (the equation from example one is not only complete, but also reduced!)

• Incomplete quadratic equations- equations in which the coefficient and or free term c are equal to zero:

  They are incomplete because some element is missing from them. But the equation must always contain x squared !!! Otherwise, it will no longer be a quadratic, but some other equation.

Why did they come up with such a division? It would seem that there is an X squared, and okay. Such a division is due to the methods of solution. Let's consider each of them in more detail.

Solving incomplete quadratic equations

First, let's focus on solving incomplete quadratic equations - they are much simpler!

Incomplete quadratic equations are of types:

1. , in this equation the coefficient is equal.
2. , in this equation the free term is equal to.
3. , in this equation the coefficient and the free term are equal.

1. i. Since we know how to take the square root, let's express from this equation

The expression can be either negative or positive. A squared number cannot be negative, because when multiplying two negative or two positive numbers, the result will always be a positive number, so: if, then the equation has no solutions.

And if, then we get two roots. These formulas do not need to be memorized. The main thing is that you should always know and remember that it cannot be less.

Let's try to solve some examples.

Example 5:

Solve the Equation

Now it remains to extract the root from the left and right parts. After all, do you remember how to extract the roots?

Answer:

Never forget about roots with a negative sign!!!

Example 6:

Solve the Equation

Answer:

Example 7:

Solve the Equation

Oh! The square of a number cannot be negative, which means that the equation

no roots!

For such equations in which there are no roots, mathematicians came up with a special icon - (empty set). And the answer can be written like this:

Answer:

Thus, this quadratic equation has two roots. There are no restrictions here, since we did not extract the root.
Example 8:

Solve the Equation

Let's take the common factor out of brackets:

Thus,

This equation has two roots.

Answer:

The simplest type of incomplete quadratic equations (although they are all simple, right?). Obviously, this equation always has only one root:

Here we will do without examples.

Solving complete quadratic equations

We remind you that the complete quadratic equation is an equation of the form equation where

Solving full quadratic equations is a bit more complicated (just a little bit) than those given.

Remember, any quadratic equation can be solved using the discriminant! Even incomplete.

The rest of the methods will help you do it faster, but if you have problems with quadratic equations, first master the solution using the discriminant.

1. Solving quadratic equations using the discriminant.

Solving quadratic equations in this way is very simple, the main thing is to remember the sequence of actions and a couple of formulas.

If, then the equation has a root. Special attention should be paid to the step. The discriminant () tells us the number of roots of the equation.

• If, then the formula at the step will be reduced to. Thus, the equation will have only a root.
• If, then we will not be able to extract the root of the discriminant at the step. This indicates that the equation has no roots.

Let's go back to our equations and look at a few examples.

Example 9:

Solve the Equation

Step 1 skip.

Step 2

Finding the discriminant:

So the equation has two roots.

Step 3

Answer:

Example 10:

Solve the Equation

The equation is in standard form, so Step 1 skip.

Step 2

Finding the discriminant:

So the equation has one root.

Answer:

Example 11:

Solve the Equation

The equation is in standard form, so Step 1 skip.

Step 2

Finding the discriminant:

This means that we will not be able to extract the root from the discriminant. There are no roots of the equation.

Now we know how to write down such answers correctly.

Answer: no roots

2. Solution of quadratic equations using the Vieta theorem.

If you remember, then there is such a type of equations that are called reduced (when the coefficient a is equal to):

Such equations are very easy to solve using Vieta's theorem:

The sum of the roots given quadratic equation is equal, and the product of the roots is equal.

Example 12:

Solve the Equation

This equation is suitable for solution using Vieta's theorem, because .

The sum of the roots of the equation is, i.e. we get the first equation:

And the product is:

Let's create and solve the system:

• And. The sum is;
• And. The sum is;
• And. The amount is equal.

and are the solution of the system:

Answer: ; .

Example 13:

Solve the Equation

Answer:

Example 14:

Solve the Equation

The equation is reduced, which means:

Answer:

QUADRATIC EQUATIONS. AVERAGE LEVEL

What is a quadratic equation?

In other words, a quadratic equation is an equation of the form, where - unknown, - some numbers, moreover.

The number is called the highest or first coefficient quadratic equation, - second coefficient, A - free member.

Why? Because if, the equation will immediately become linear, because will disappear.

In this case, and can be equal to zero. In this stool equation is called incomplete. If all the terms are in place, that is, the equation is complete.

Solutions to various types of quadratic equations

Methods for solving incomplete quadratic equations:

To begin with, we will analyze the methods for solving incomplete quadratic equations - they are simpler.

The following types of equations can be distinguished:

I. , in this equation the coefficient and the free term are equal.

II. , in this equation the coefficient is equal.

III. , in this equation the free term is equal to.

Now consider the solution of each of these subtypes.

Obviously, this equation always has only one root:

A number squared cannot be negative, because when multiplying two negative or two positive numbers, the result will always be a positive number. That's why:

if, then the equation has no solutions;

if we have two roots

These formulas do not need to be memorized. The main thing to remember is that it cannot be less.

Examples:

Solutions:

Answer:

Never forget about roots with a negative sign!

The square of a number cannot be negative, which means that the equation

no roots.

To briefly write that the problem has no solutions, we use the empty set icon.

Answer:

So, this equation has two roots: and.

Answer:

Let's take the common factor out of brackets:

The product is equal to zero if at least one of the factors is equal to zero. This means that the equation has a solution when:

So, this quadratic equation has two roots: and.

Example:

Solve the equation.

Solution:

We factorize the left side of the equation and find the roots:

Answer:

Methods for solving complete quadratic equations:

1. Discriminant

Solving quadratic equations in this way is easy, the main thing is to remember the sequence of actions and a couple of formulas. Remember, any quadratic equation can be solved using the discriminant! Even incomplete.

Did you notice the root of the discriminant in the root formula? But the discriminant can be negative. What to do? We need to pay special attention to step 2. The discriminant tells us the number of roots of the equation.

• If, then the equation has a root:
• If, then the equation has the same root, but in fact, one root:

  Such roots are called double roots.

• If, then the root of the discriminant is not extracted. This indicates that the equation has no roots.

Why are there different numbers of roots? Let us turn to the geometric meaning of the quadratic equation. The graph of the function is a parabola:

In a particular case, which is a quadratic equation, . And this means that the roots of the quadratic equation are the points of intersection with the x-axis (axis). The parabola may not cross the axis at all, or it may intersect it at one (when the top of the parabola lies on the axis) or two points.

In addition, the coefficient is responsible for the direction of the branches of the parabola. If, then the branches of the parabola are directed upwards, and if - then downwards.

Examples:

Solutions:

Answer:

Answer: .

Answer:

This means there are no solutions.

Answer: .

2. Vieta's theorem

Using the Vieta theorem is very easy: you just need to choose a pair of numbers whose product is equal to the free term of the equation, and the sum is equal to the second coefficient, taken with the opposite sign.

It is important to remember that Vieta's theorem can only be applied to given quadratic equations ().

Let's look at a few examples:

Example #1:

Solve the equation.

Solution:

This equation is suitable for solution using Vieta's theorem, because . Other coefficients: ; .

The sum of the roots of the equation is:

And the product is:

Let's select such pairs of numbers, the product of which is equal, and check if their sum is equal:

• And. The sum is;
• And. The sum is;
• And. The amount is equal.

and are the solution of the system:

Thus, and are the roots of our equation.

Answer: ; .

Example #2:

Solution:

We select such pairs of numbers that give in the product, and then check whether their sum is equal:

and: give in total.

and: give in total. To get it, you just need to change the signs of the alleged roots: and, after all, the work.

Answer:

Example #3:

Solution:

The free term of the equation is negative, and hence the product of the roots is a negative number. This is possible only if one of the roots is negative and the other is positive. So the sum of the roots is differences of their modules.

We select such pairs of numbers that give in the product, and the difference of which is equal to:

and: their difference is - not suitable;

and: - not suitable;

and: - not suitable;

and: - suitable. It remains only to remember that one of the roots is negative. Since their sum must be equal, then the root, which is smaller in absolute value, must be negative: . We check:

Answer:

Example #4:

Solve the equation.

Solution:

The equation is reduced, which means:

The free term is negative, and hence the product of the roots is negative. And this is possible only when one root of the equation is negative and the other is positive.

We select such pairs of numbers whose product is equal, and then determine which roots should have a negative sign:

Obviously, only roots and are suitable for the first condition:

Answer:

Example #5:

Solve the equation.

Solution:

The equation is reduced, which means:

The sum of the roots is negative, which means that at least one of the roots is negative. But since their product is positive, it means both roots are minus.

We select such pairs of numbers, the product of which is equal to:

Obviously, the roots are the numbers and.

Answer:

Agree, it is very convenient - to invent roots orally, instead of counting this nasty discriminant. Try to use Vieta's theorem as often as possible.

But the Vieta theorem is needed in order to facilitate and speed up finding the roots. To make it profitable for you to use it, you must bring the actions to automatism. And for this, solve five more examples. But don't cheat: you can't use the discriminant! Only Vieta's theorem:

Solutions for tasks for independent work:

Task 1. ((x)^(2))-8x+12=0

According to Vieta's theorem:

As usual, we start the selection with the product:

Not suitable because the amount;

: the amount is what you need.

Answer: ; .

Task 2.

And again, our favorite Vieta theorem: the sum should work out, but the product is equal.

But since it should be not, but, we change the signs of the roots: and (in total).

Answer: ; .

Task 3.

Hmm... Where is it?

It is necessary to transfer all the terms into one part:

The sum of the roots is equal to the product.

Yes, stop! The equation is not given. But Vieta's theorem is applicable only in the given equations. So first you need to bring the equation. If you can’t bring it up, drop this idea and solve it in another way (for example, through the discriminant). Let me remind you that to bring a quadratic equation means to make the leading coefficient equal to:

Great. Then the sum of the roots is equal, and the product.

It's easier to pick up here: after all - a prime number (sorry for the tautology).

Answer: ; .

Task 4.

The free term is negative. What's so special about it? And the fact that the roots will be of different signs. And now, during the selection, we check not the sum of the roots, but the difference between their modules: this difference is equal, but the product.

So, the roots are equal and, but one of them is with a minus. Vieta's theorem tells us that the sum of the roots is equal to the second coefficient with the opposite sign, that is. This means that the smaller root will have a minus: and, since.

Answer: ; .

Task 5.

What needs to be done first? That's right, give the equation:

Again: we select the factors of the number, and their difference should be equal to:

The roots are equal and, but one of them is minus. Which? Their sum must be equal, which means that with a minus there will be a larger root.

Answer: ; .

Let me summarize:

1. Vieta's theorem is used only in the given quadratic equations.
2. Using the Vieta theorem, you can find the roots by selection, orally.
3. If the equation is not given or no suitable pair of factors of the free term was found, then there are no integer roots, and you need to solve it in another way (for example, through the discriminant).

3. Full square selection method

If all the terms containing the unknown are represented as terms from the formulas of abbreviated multiplication - the square of the sum or difference - then after the change of variables, the equation can be represented as an incomplete quadratic equation of the type.

For example:

Example 1:

Solve the equation: .

Solution:

Answer:

Example 2:

Solve the equation: .

Solution:

Answer:

In general, the transformation will look like this:

This implies: .

Doesn't it remind you of anything? It's the discriminant! That's exactly how the discriminant formula was obtained.

QUADRATIC EQUATIONS. BRIEFLY ABOUT THE MAIN

Quadratic equation is an equation of the form, where is the unknown, are the coefficients of the quadratic equation, is the free term.

Complete quadratic equation- an equation in which the coefficients are not equal to zero.

Reduced quadratic equation- an equation in which the coefficient, that is: .

Incomplete quadratic equation- an equation in which the coefficient and or free term c are equal to zero:

• if the coefficient, the equation has the form: ,
• if a free term, the equation has the form: ,
• if and, the equation has the form: .

1. Algorithm for solving incomplete quadratic equations

1.1. An incomplete quadratic equation of the form, where, :

1) Express the unknown: ,

2) Check the sign of the expression:

• if, then the equation has no solutions,
• if, then the equation has two roots.

1.2. An incomplete quadratic equation of the form, where, :

1) Let's take the common factor out of brackets: ,

2) The product is equal to zero if at least one of the factors is equal to zero. Therefore, the equation has two roots:

1.3. An incomplete quadratic equation of the form, where:

This equation always has only one root: .

2. Algorithm for solving complete quadratic equations of the form where

2.1. Solution using the discriminant

1) Let's bring the equation to the standard form: ,

2) Calculate the discriminant using the formula: , which indicates the number of roots of the equation:

3) Find the roots of the equation:

• if, then the equation has a root, which are found by the formula:
• if, then the equation has a root, which is found by the formula:
• if, then the equation has no roots.

2.2. Solution using Vieta's theorem

The sum of the roots of the reduced quadratic equation (an equation of the form, where) is equal, and the product of the roots is equal, i.e. , A.

2.3. Full square solution