Vieta's theorem (more precisely, the theorem inverse to Vieta's theorem) allows us to reduce the time to solve 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 using the Vieta theorem. The reduced 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.

Today it deserves to be sung in verse
On the properties of roots, Vieta's theorem.
Which is better, say, the constancy of this:
You multiplied the roots - and the fraction is ready
In the numerator With, in the denominator A.
And the sum of the roots is also a fraction
Even with a minus this fraction
What's the trouble
In numerators V, in the denominator A.
(From school folklore)

In the epigraph, the remarkable theorem of François Vieta is given not quite exactly. Indeed, we can write down a quadratic equation that has no roots and write down their sum and product. For example, the equation x 2 + 2x + 12 = 0 has no real roots. But, approaching formally, we can write down their product (x 1 x 2 \u003d 12) and the sum (x 1 + x 2 \u003d -2). Our the verses will correspond to the theorem with the caveat: "if the equation has roots", i.e. D ≥ 0.

The first practical application of this theorem is the compilation of a quadratic equation that has given roots. Second: it allows you to verbally solve many quadratic equations. At the development of these skills, first of all, attention is drawn to school textbooks.

Here we will consider more complex problems solved using the Vieta theorem.

Example 1

One of the roots of the equation 5x 2 - 12x + c \u003d 0 is three times larger than the second. Find with.

Solution.

Let the second root be x2.

Then the first root x1 = 3x2.

According to the Vieta theorem, the sum of the roots is 12/5 = 2.4.

Let's make the equation 3x2 + x2 = 2.4.

Hence x 2 \u003d 0.6. Therefore x 1 \u003d 1.8.

Answer: c \u003d (x 1 x 2) a \u003d 0.6 1.8 5 \u003d 5.4.

Example 2

It is known that x 1 and x 2 are the roots of the equation x 2 - 8x + p = 0, and 3x 1 + 4x 2 = 29. Find p.

Solution.

According to the Vieta theorem x 1 + x 2 = 8, and by condition 3x 1 + 4x 2 = 29.

Having solved the system of these two equations, we find the value x 1 \u003d 3, x 2 \u003d 5.

And therefore p = 15.

Answer: p = 15.

Example 3

Without calculating the roots of the equation 3x 2 + 8 x - 1 \u003d 0, find x 1 4 + x 2 4

Solution.

Note that according to the Vieta theorem x 1 + x 2 = -8/3 and x 1 x 2 = -1/3 and transform the expression

a) x 1 4 + x 2 4 = (x 1 2 + x 2 2) 2 - 2x 1 2 x 2 2 = ((x 1 + x 2) 2 - 2x 1 x 2) 2 - 2 (x 1 x 2) 2 \u003d ((-8/3) 2 - 2 (-1/3)) 2 - 2 (-1/3) 2 \u003d 4898/9

Answer: 4898/9.

Example 4

At what values ​​of the parameter a is the difference between the largest and smallest roots of the equation
2x 2 - (a + 1) x + (a - 1) \u003d 0 is equal to their product.

Solution.

This is a quadratic equation. It will have 2 different roots if D > 0. In other words, (a + 1) 2 - 8 (a - 1) > 0 or (a - 3) 2 > 0. Therefore, we have 2 roots for all a, for except a = 3.

For definiteness, we assume that x 1 > x 2 and get x 1 + x 2 \u003d (a + 1) / 2 and x 1 x 2 \u003d (a - 1) / 2. Based on the condition of the problem x 1 - x 2 \u003d (a - 1) / 2. All three conditions must be met simultaneously. Consider the first and last equations as a system. It is easily solved by the algebraic addition method.

We get x 1 \u003d a / 2, x 2 \u003d 1/2. Let's check for what A the second equality will be fulfilled: x 1 x 2 \u003d (a - 1) / 2. Let's substitute the received values ​​and we will have: а/4 = (а – 1)/2. Then, a = 2. It is obvious that if a = 2, then all conditions are satisfied.

Answer: when a = 2.

Example 5

What is the smallest value of a for which the sum of the roots of the equation
x 2 - 2a (x - 1) - 1 \u003d 0 is equal to the sum of the squares of its roots.

Solution.

First of all, let's bring the equation to the canonical form: x 2 - 2ax + 2a - 1 \u003d 0. It will have roots if D / 4 ≥ 0. Therefore: a 2 - (2a - 1) ≥ 0. Or (a - 1 ) 2 ≥ 0. And this condition is valid for any a.

We apply the Vieta theorem: x 1 + x 2 \u003d 2a, x 1 x 2 \u003d 2a - 1. We calculate

x 1 2 + x 2 2 \u003d (x 1 + x 2) 2 - 2x 1 x 2. Or after substitution x 1 2 + x 2 2 \u003d (2a) 2 - 2 (2a - 1) \u003d 4a 2 - 4a + 2. It remains to make an equality that corresponds to the condition of the problem: x 1 + x 2 \u003d x 1 2 + x 2 2 . We get: 2a \u003d 4a 2 - 4a + 2. This quadratic equation has 2 roots: a 1 \u003d 1 and a 2 \u003d 1/2. The smallest of them is -1/2.

Answer: 1/2.

Example 6

Find the relationship between the coefficients of the equation ax 2 + inx + c \u003d 0 if the sum of the cubes of its roots is equal to the product of the squares of these roots.

Solution.

We will proceed from the fact that this equation has roots and, therefore, Vieta's theorem can be applied to it.

Then the condition of the problem will be written as follows: x 1 3 + x 2 3 = x 1 2 x 2 2. Or: (x 1 + x 2) (x 1 2 - x 1 x 2 + x 2 2) \u003d (x 1 x 2) 2.

You need to convert the second factor. x 1 2 - x 1 x 2 + x 2 2 \u003d ((x 1 + x 2) 2 - 2x 1 x 2) - x 1 x 2.

We get (x 1 + x 2) ((x 1 + x 2) 2 - 3x 1 x 2) \u003d (x 1 x 2) 2. It remains to replace the sums and products of the roots through the coefficients.

(-b/a)((b/a) 2 – 3 c/a) = (c/a) 2 . This expression can easily be converted to the form b (3ac - b 2) / a \u003d c 2. The ratio is found.

Comment. It should be taken into account that the resulting relation makes sense to consider only after the other is fulfilled: D ≥ 0.

Example 7

Find the value of the variable a for which the sum of the squares of the roots of the equation x 2 + 2ax + 3a 2 - 6a - 2 \u003d 0 is the largest value.

Solution.

If this equation has roots x 1 and x 2, then their sum x 1 + x 2 \u003d -2a, and the product x 1 x 2 \u003d 3a 2 - 6a - 2.

We calculate x 1 2 + x 2 2 \u003d (x 1 + x 2) 2 - 2x 1 x 2 \u003d (-2a) 2 - 2 (3a 2 - 6a - 2) \u003d -2a 2 + 12a + 4 \u003d -2 (a – 3) 2 + 22.

It is now obvious that this expression takes highest value for a = 3.

It remains to check whether the original quadratic equation really has roots at a \u003d 3. We check by substitution and we get: x 2 + 6x + 7 \u003d 0 and for it D \u003d 36 - 28\u003e 0.

Therefore, the answer is: for a = 3.

Example 8

The equation 2x 2 - 7x - 3 \u003d 0 has roots x 1 and x 2. Find the triple sum of the coefficients of the given quadratic equation, the roots of which are the numbers X 1 \u003d 1 / x 1 and X 2 \u003d 1 / x 2. (*)

Solution.

Obviously, x 1 + x 2 \u003d 7/2 and x 1 x 2 \u003d -3/2. We compose the second equation by its roots in the form x 2 + px + q \u003d 0. To do this, we use the assertion inverse to the Vieta theorem. We get: p \u003d - (X 1 + X 2) and q \u003d X 1 X 2.

After substituting into these formulas, based on (*), then: p \u003d - (x 1 + x 2) / (x 1 x 2) \u003d 7/3 and q \u003d 1 / (x 1 x 2) \u003d - 2/3.

The desired equation will take the form: x 2 + 7/3 x - 2/3 = 0. Now we can easily calculate the tripled sum of its coefficients:

3(1 + 7/3 - 2/3) = 8. Answer received.

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Any complete quadratic equation ax2 + bx + c = 0 can be brought to mind x 2 + (b/a)x + (c/a) = 0, if we first divide each term by the coefficient a before x2. And if we introduce new notation (b/a) = p And (c/a) = q, then we will have the equation x 2 + px + q = 0, which in mathematics is called reduced quadratic equation.

The roots of the reduced quadratic equation and the coefficients p And q interconnected. It's confirmed Vieta's theorem, named after the French mathematician Francois Vieta, who lived in late XVI century.

Theorem. The sum of the roots of the reduced quadratic equation x 2 + px + q = 0 equal to the second coefficient p, taken with the opposite sign, and the product of the roots - to the free term q.

We write these ratios in the following form:

Let x 1 And x2 various roots of the reduced equation x 2 + px + q = 0. According to Vieta's theorem x1 + x2 = -p And x 1 x 2 = q.

To prove this, let's substitute each of the roots x 1 and x 2 into the equation. We get two true equalities:

x 1 2 + px 1 + q = 0

x 2 2 + px 2 + q = 0

Subtract the second from the first equality. We get:

x 1 2 – x 2 2 + p(x 1 – x 2) = 0

We expand the first two terms according to the difference of squares formula:

(x 1 - x 2) (x 1 - x 2) + p (x 1 - x 2) = 0

By condition, the roots x 1 and x 2 are different. Therefore, we can reduce the equality by (x 1 - x 2) ≠ 0 and express p.

(x 1 + x 2) + p = 0;

(x 1 + x 2) = -p.

The first equality is proved.

To prove the second equality, we substitute into the first equation

x 1 2 + px 1 + q \u003d 0 instead of the coefficient p, its equal number is (x 1 + x 2):

x 1 2 - (x 1 + x 2) x 1 + q \u003d 0

Having transformed left side equations, we get:

x 1 2 - x 2 2 - x 1 x 2 + q \u003d 0;

x 1 x 2 = q, which was to be proved.

Vieta's theorem is good because, even without knowing the roots of the quadratic equation, we can calculate their sum and product .

Vieta's theorem helps to determine the integer roots of the given quadratic equation. But for many students, this causes difficulties due to the fact that they do not know a clear algorithm of action, especially if the roots of the equation have different signs.

So, the given quadratic equation has the form x 2 + px + q \u003d 0, where x 1 and x 2 are its roots. According to the Vieta theorem x 1 + x 2 = -p and x 1 x 2 = q.

We can draw the following conclusion.

If in the equation there is a minus sign before the last term, then the roots x 1 and x 2 have various signs. In addition, the sign of the smaller root is the same as the sign of the second coefficient in the equation.

Based on the fact that when adding numbers with different signs, their modules are subtracted, and the sign of the larger number is put in front of the result, you should proceed as follows:

1. determine such factors of the number q so that their difference is equal to the number p;
2. put the sign of the second coefficient of the equation in front of the smaller of the obtained numbers; the second root will have the opposite sign.

Let's look at some examples.

Example 1.

Solve the equation x 2 - 2x - 15 = 0.

Solution.

Let's try to solve this equation using the rules proposed above. Then we can say for sure that this equation will have two different roots, because D \u003d b 2 - 4ac \u003d 4 - 4 (-15) \u003d 64\u003e 0.

Now, from all the factors of the number 15 (1 and 15, 3 and 5), we select those whose difference is equal to 2. These will be the numbers 3 and 5. We put a minus sign in front of the smaller number, i.e. the sign of the second coefficient of the equation. Thus, we get the roots of the equation x 1 \u003d -3 and x 2 \u003d 5.

Answer. x 1 = -3 and x 2 = 5.

Example 2.

Solve the equation x 2 + 5x - 6 = 0.

Solution.

Let's check if this equation has roots. To do this, we find the discriminant:

D \u003d b 2 - 4ac \u003d 25 + 24 \u003d 49\u003e 0. The equation has two different roots.

The possible factors of the number 6 are 2 and 3, 6 and 1. The difference is 5 for a pair of 6 and 1. In this example, the coefficient of the second term has a plus sign, so the smaller number will have the same sign. But before the second number there will be a minus sign.

Answer: x 1 = -6 and x 2 = 1.

Vieta's theorem can also be written for a complete quadratic equation. So if the quadratic equation ax2 + bx + c = 0 has roots x 1 and x 2 , then they satisfy the equalities

x 1 + x 2 = -(b/a) And x 1 x 2 = (c/a). However, the application of this theorem in the full quadratic equation is rather problematic, since if there are roots, at least one of them is a fractional number. And working with the selection of fractions is quite difficult. But still there is a way out.

Consider the complete quadratic equation ax 2 + bx + c = 0. Multiply its left and right sides by the coefficient a. The equation will take the form (ax) 2 + b(ax) + ac = 0. Now let's introduce a new variable, for example t = ax.

In this case, the resulting equation turns into a reduced quadratic equation of the form t 2 + bt + ac = 0, the roots of which t 1 and t 2 (if any) can be determined by the Vieta theorem.

In this case, the roots of the original quadratic equation will be

x 1 = (t 1 / a) and x 2 = (t 2 / a).

Example 3.

Solve the equation 15x 2 - 11x + 2 = 0.

Solution.

We make an auxiliary equation. Let's multiply each term of the equation by 15:

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

We make the change t = 15x. We have:

t 2 - 11t + 30 = 0.

According to the Vieta theorem, the roots of this equation will be t 1 = 5 and t 2 = 6.

We return to the replacement t = 15x:

5 = 15x or 6 = 15x. Thus x 1 = 5/15 and x 2 = 6/15. We reduce and get the final answer: x 1 = 1/3 and x 2 = 2/5.

Answer. x 1 = 1/3 and x 2 = 2/5.

To master the solution of quadratic equations using the Vieta theorem, students need to practice as much as possible. This is precisely the secret of success.

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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 extract Square root, then 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 little 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 draw a 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 the Vieta 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 the Vieta 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 product.

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 view 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) We bring the equation to standard view: ,

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

I. Vieta's theorem for the reduced quadratic equation.

The sum of the roots of the reduced quadratic equation x 2 +px+q=0 is equal to the second coefficient, taken with the opposite sign, and the product of the roots is equal to the free term:

x 1 + x 2 \u003d-p; x 1 ∙ x 2 \u003d q.

Find the roots of the given quadratic equation using Vieta's theorem.

Example 1) x 2 -x-30=0. This is the reduced quadratic equation ( x 2 +px+q=0), the second coefficient p=-1, and the free term q=-30. First, make sure that the given equation has roots, and that the roots (if any) will be expressed as integers. For this, it is sufficient that the discriminant be the full square of an integer.

Finding the discriminant D=b 2 - 4ac=(-1) 2 -4∙1∙(-30)=1+120=121= 11 2 .

Now, according to the Vieta theorem, the sum of the roots must be equal to the second coefficient, taken with the opposite sign, i.e. ( -p), and the product is equal to the free term, i.e. ( q). Then:

x 1 + x 2 =1; x 1 ∙ x 2 \u003d -30. We need to choose such two numbers so that their product is equal to -30 , and the sum is unit. These are the numbers -5 And 6 . Answer: -5; 6.

Example 2) x 2 +6x+8=0. We have the reduced quadratic equation with the second coefficient p=6 and free member q=8. Make sure that there are integer roots. Let's find the discriminant D1 D1=3 2 -1∙8=9-8=1=1 2 . The discriminant D 1 is the perfect square of the number 1 , so the roots of this equation are integers. We choose the roots according to the Vieta theorem: the sum of the roots is equal to –p=-6, and the product of the roots is q=8. These are the numbers -4 And -2 .

Actually: -4-2=-6=-p; -4∙(-2)=8=q. Answer: -4; -2.

Example 3) x 2 +2x-4=0. In this reduced quadratic equation, the second coefficient p=2, and the free term q=-4. Let's find the discriminant D1, since the second coefficient is an even number. D1=1 2 -1∙(-4)=1+4=5. The discriminant is not a perfect square of a number, so we do conclusion: the roots of this equation are not integers and cannot be found using Vieta's theorem. So, we solve this equation, as usual, according to the formulas (in this case formulas). We get:

Example 4). Write a quadratic equation using its roots if x 1 \u003d -7, x 2 \u003d 4.

Solution. The desired equation will be written in the form: x 2 +px+q=0, moreover, based on the Vieta theorem –p=x1 +x2=-7+4=-3 →p=3; q=x 1 ∙x 2=-7∙4=-28 . Then the equation will take the form: x2 +3x-28=0.

Example 5). Write a quadratic equation using its roots if:

II. Vieta's theorem for the complete quadratic equation ax2+bx+c=0.

The sum of the roots is minus b divided by A, the product of the roots is With divided by A:

x 1 + x 2 \u003d -b / a; x 1 ∙ x 2 \u003d c / a.