First level

Quadratic equations. Exhaustive Guide (2019)

In terms of "Square Equation", the key is the word "square". This means that the variable must be present in the equation (the same ix) in the square, and there should be no ICs in the third (and greater) degree.

The solution of many equations is reduced to solving precisely square equations.

Let's learn how to determine that we have a square equation, and not any other.

Example 1.

Each member of the equation on the denominator and dominated

We transfer everything to the left and place members in descending order of degrees of ICA

Now you can say with confidence that this equation is square!

Example 2.

Domestic left and right side on:

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

Example 3.

Doming all on:

Scary? The fourth and second degree ... However, if we replace, then we will see that we have a simple square equation:

Example 4.

It seems to be, but let's look attentively. We transfer everything to the left:

See, decreased - and now it is a simple linear equation!

Now try to determine which of the following equations are square, and which no:

Examples:

Answers:

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

Mathematics conventionally divide all square equations on the type:

• Full square equations - equations in which coefficients and, as well as a free member are not equal to zero (as in the example). In addition, among full square equations allocate presented - These are equations in which the coefficient (equation from the example one is not only complete, but also given!)

• Incomplete square equations - equations in which the coefficient and free member is zero:

  Incompletely, because they lack some kind of item. But the equation should always be present in the square !!! Otherwise, it will not be square, but some other equation.

Why did you come up with such a division? It would seem that there is X in the square, and okay. Such division is due to the methods of solutions. Consider each of them in more detail.

Decision of incomplete square equations

To begin with, we will stop at solving incomplete square equations - they are much simpler!

Incomplete square equations are types:

1. In this equation, the coefficient is equal.
2. In this equation, a free member is equal.
3. In this equation, the coefficient and free member are equal.

1. and. As we know how to extract a square root, let's express from this equation

The expression can be both negative and positive. The number erected into the square cannot be negative, because with multiplying two negative or two positive numbers - the result will always be a positive number, so that if, the equation does not have solutions.

And if you get two roots. These formulas do not need to memorize. The main thing you should know and remember always that it may not be less.

Let's try to solve a few examples.

Example 5:

Decide equation

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

Answer:

Never forget about roots with a negative sign !!!

Example 6:

Decide equation

Answer:

Example 7:

Decide equation

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

no roots!

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

Answer:

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

Decide equation

I will summarize the brackets:

In this way,

This equation has two roots.

Answer:

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

Here we will do without examples.

Solving full square equations

We remind you that the full square equation is the equation of the equation where

The solution of complete square equations is a bit more complicated (very slightly) than the above.

Remember, any square equation can be solved with the help of discriminant! Even incomplete.

The rest of the ways will help make it faster, but if you have problems with square equations, to begin with, the solution is called with the help of discriminant.

1. The solution of square equations with the help of discriminant.

The solution of square equations this way is very simple, the main thing is to remember the sequence of actions and a couple of formulas.

If, the equation has a root of particular attention to pay a step. Discriminant () indicates us on the number of roots of the equation.

• If, then the formula is reduced to. Thus, the equation will have a whole root.
• If, we will not be able to extract the root from the discriminant in step. This indicates that the equation does not have roots.

Let's return to our equations and consider several examples.

Example 9:

Decide equation

Step 1 We skip.

Step 2.

We find discriminant:

So the equation has two roots.

Step 3.

Answer:

Example 10:

Decide equation

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

Step 2.

We find discriminant:

So the equation has one root.

Answer:

Example 11:

Decide equation

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

Step 2.

We find discriminant:

It will not be able to extract the root from the discriminant. The roots of the equation does not exist.

Now we know how to write such answers to correctly.

Answer:No roots

2. Solution of square equations using the Vieta Theorem.

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

Such equations are very easy to solve using the Vieta theorem:

The sum of the roots specified The square equation is equal, and the product of the roots is equal.

Example 12:

Decide equation

This equation is suitable for solving using the Vieta Theorem, because .

The amount of the roots of the equation is equal, i.e. We get the first equation:

And the work is:

We will also decide the system:

• and. The amount is equal;
• and. The amount is equal;
• and. The amount is equal.

and are the solution of the system:

Answer: ; .

Example 13:

Decide equation

Answer:

Example 14:

Decide equation

The equation is given, and therefore:

Answer:

QUADRATIC EQUATIONS. AVERAGE LEVEL

What is a square equation?

In other words, the square equation is the equation of the species where the unknown is some numbers, and.

The number is called elder or first coefficient square equation - the second coefficient, but - free member.

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

At the same time, and can be zero. In this stool, the equation is called incomplete. If all the components are in place, that is, the equation is complete.

Solutions of various types of square equations

Methods for solving incomplete square equations:

To begin with, we will analyze the methods of solutions of incomplete square equations - they are easier.

You can select the type of such equations:

I., In this equation, the coefficient and free member are equal.

II. In this equation, the coefficient is equal.

III. In this equation, a free member is equal.

Now consider the solution of each of these subtypes.

Obviously, this equation always has only one root:

The number erected into the square cannot be negative, because with multiplying two negative or two positive numbers, the result will always be a positive number. Therefore:

if, the equation does not have solutions;

if we have learned two roots

These formulas do not need to memorize. The main thing to remember that it may not be less.

Examples:

Solutions:

Answer:

Never forget about roots with a negative sign!

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

no roots.

To briefly record that the task has no solutions, use an empty set icon.

Answer:

So, this equation has two roots: and.

Answer:

I will summarize the factory for brackets:

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

So, this square equation has two roots: and.

Example:

Decide equation.

Decision:

Spread the left part of the factory equation and find the roots:

Answer:

Methods of solving full square equations:

1. Discriminant

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

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

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

  Such roots are called double.

• If, the root of the discriminant is not removed. This indicates that the equation does not have roots.

Why is it possible to different number of roots? Let us turn to the geometrical meaning of the square equation. The function graph is parabola:

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

In addition, a coefficient is responsible for the direction of the branches of the parabola. If, the parabola branches are directed upwards, and if it is down.

Examples:

Solutions:

Answer:

Answer:.

Answer:

So, there are no solutions.

Answer:.

2. Vieta Theorem

Vieta's theorem is very easy to use: you just need to pick up such a couple of numbers, the product of which is equal to a free member of the equation, and the amount is the second coefficient taken with the opposite sign.

It is important to remember that the theorem of the Vieta can only be used in reduced square equations ().

Consider a few examples:

Example number 1:

Decide equation.

Decision:

This equation is suitable for solving using the Vieta Theorem, because . The remaining coefficients :; .

The amount of the roots of the equation is:

And the work is:

We will select such pairs of numbers, the product of which is equal, and check whether their sum is equal:

• and. The amount is equal;
• and. The amount is equal;
• and. The amount is equal.

and are the solution of the system:

Thus, the roots of our equation.

Answer:; .

Example number 2:

Decision:

We will select such pairs of numbers that are given in the work, and then check whether their sum is equal:

and: in the amount they give.

and: in the amount they give. To get enough just to change the signs of the alleged roots: and, because the work.

Answer:

Example number 3:

Decision:

The free member of the equation is negative, which means the product of the roots - a negative number. This is possible only if one of the roots is negative, and the other is positive. Therefore the amount of the roots is equal the differences of their modules.

We will select such pairs of numbers that are given in the work, and the difference of which is equal to:

and: their difference is equal - not suitable;

and: - not suitable;

and: - not suitable;

and: - Suitable. It remains only to remember that one of the roots is negative. Since their amount should be equal, then a negative should be a smaller root module :. Check:

Answer:

Example number 4:

Decide equation.

Decision:

The equation is given, and therefore:

The free member is negative, and therefore 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 will select such pairs of numbers, the product of which is equal, and then we define which roots should have a negative sign:

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

Answer:

Example number 5:

Decide equation.

Decision:

The equation is given, and therefore:

The amount of the roots is negative, which means that at least one of the roots is negative. But since their work is positive, it means both roots with a minus sign.

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

Obviously, the roots are numbers and.

Answer:

Agree, it is very convenient - to invent roots orally, instead of considering this nasty discriminant. Try to use the theorem of the Vieta as much as possible.

But the Vieta Theorem is needed in order to facilitate and speed up the finding of the roots. To help you use it, you must bring action to automatism. And for this, slander more heels of examples. But not a scaling: the discriminant cannot be used! Only Vieta theorem:

Task solutions for independent work:

Task 1. ((x) ^ (2)) - 8x + 12 \u003d 0

On the Vieta Theorem:

As usual, we begin the selection of the work:

Does not fit because the amount;

: Amount - what you need.

Answer:; .

Task 2.

And again, our favorite Vieta Theorem: in the amount should turn out, and the work is equal.

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

Answer:; .

Task 3.

Hmm ... and where is what?

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

The amount of the roots is equal, the work.

So, stop! The equation is not given. But the Vieta theorem is applicable only in the above equations. So first you need to bring the equation. If you do not work, throw this idea and decide in a different way (for example, through discriminant). Let me remind you that bring the square equation - it means to make a senior coefficient to:

Excellent. Then the amount of the roots is equal, and the work.

Here it is easier to pick up simple: after all, a simple number (sorry for tautology).

Answer:; .

Task 4.

Free member is negative. What's special in this? And the fact that the roots will be different signs. And now during the selection, we do not check the amount of the roots, but the difference between their modules: this difference is equal, and the work.

So, the roots are equal and, but one of them with a minus. The Vieta Theorem tells us that the amount of the roots is equal to the second coefficient with the opposite sign, that is. So minus will be at a smaller root: and, since.

Answer:; .

Task 5.

What needs to be done first? Right, bring the equation:

Again: We select the multipliers of the number, and their difference should be equal:

The roots are equal and, but one of them with a minus. What? Their amount should be equal, it means that the minus will be larger root.

Answer:; .

I will summarize:

1. Vieta theorem is used only in the given square equations.
2. Using the Vieta theorem you can find the roots by the selection, orally.
3. If the equation is not given or there is no suitable pair of multipliers of a free member, which means there are no whole roots, and it is necessary to solve another method (for example, through discriminant).

3. Method of allocation of a full square

If all the terms comprising an unknown, to present in the form of the components of the abbreviated multiplication of the sum of the sum or difference, then after replacing the variables, an equation in the form of an incomplete square equation of type can be represented.

For example:

Example 1:

Decide equation :.

Decision:

Answer:

Example 2:

Decide equation :.

Decision:

Answer:

In general, the transformation will look like this:

This implies: .

Nothing reminds? This is the discriminant! That's it, the formula of the discriminant and got.

QUADRATIC EQUATIONS. Briefly about the main thing

Quadratic equation- This is the equation of the species, where - the unknown, - the coefficients of the square equation, is a free member.

Full square equation - equation in which the coefficients are not equal to zero.

The reduced square equation - equation in which the coefficient, that is :.

Incomplete square equation - equation in which the coefficient and free member is zero:

• if the coefficient, the equation is:,
• if a free member, the equation has the form :,
• if, the equation has the form :.

1. Algorithm solving incomplete square equations

1.1. An incomplete square equation of the species where,:

1) Express the Unknown:

2) Checking the sign of expression:

• if, the equation does not have solutions,
• if, the equation has two roots.

1.2. An incomplete square equation of the species where,:

1) I will summarize the factory for brackets:

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

1.3. An incomplete square equation of the species, where:

This equation always has only one root :.

2. Algorithm for solving full square equations of the species where

2.1. Solution with the help of discriminant

1) We give the equation to the standard form :,

2) Calculate the discriminant according to the formula: which indicates the number of roots of the equation:

3) Find the roots of the equation:

• if, the equation has a root that are in the formula:
• if, the equation has the root, which is by the formula:
• if, the equation does not have roots.

2.2. Solution using the Vieta Theorem

The sum of the roots of the reduced square equation (equation of the form, where) is equal, and the product of the roots is equal, i.e. , but.

2.3. Solving a full square allocation method

Vieta's theorem (more precisely, theorem, reverse theorem of Vieta) reduces the time to solve square equations. Just need to be able to use it. How to learn to solve square equations on the Vieta theorem? It is easy, if we ripen a little.

Now we will only talk about the solution on the theorem of the Vieta of the present square equation. The delivered square equation is an equation in which A, that is, the coefficient in front of x² is equal to one. You can also not solve the square equations on the Vieta theorem, but there are already at least one of the roots is not an integer. It is harder to guess.

Theorem, the reverse theorem of the Vieta, says: if the numbers x1 and x2 are such that

then x1 and x2 - the roots of the square equation

When solving a square equation on the theorem, the Vieta is possible only 4 options. If you remember the course of reasoning, finding whole roots can be learned very quickly.

I. If q is a positive number,

this means that the roots of the X1 and X2 are the numbers of the same sign (since only the multiplication of numbers with the same signs is a positive number).

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 is a negative number, (respectively, P\u003e 0), then both roots are negative numbers (there were numbers of one sign, a negative number was obtained).

II. If q is a negative number,

this means that the roots of X1 and X2 have different signs (with multiplication of numbers a negative number is obtained only in the case when there are different signs from multipliers). In this case, the X1 + X2 is no longer the amount, but by the difference (because when adding numbers with different signs, we will deduct less from the smaller module). Therefore, X1 + X2 shows how much the roots of X1 and X2 are different, that is, how much one root is greater than the other (by module).

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\u003e 0), the larger (module) root is a negative number.

Consider the solution of square equations on the Vieta theorem on the examples.

Solve the reduced square equation on the Vieta Theorem:

Here Q \u003d 12\u003e 0, so the roots of X1 and X2 are the numbers of one sign. Their amount is -p \u003d 7\u003e 0, so both roots are positive numbers. We select integers whose product is 12. This is 1 and 12, 2 and 6, 3 and 4. The amount is 7 in the pair 3 and 4. So, 3 and 4 are the roots of the equation.

In this example, Q \u003d 16\u003e 0, it means that the roots are X1 and X2 - the number of one sign. Their amount is -p \u003d -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 \u003d -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, the longer number is positive. So, the roots are 5 and -3.

q \u003d -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 techniques with which many square equations are solved very quickly and without any discriminants. Moreover, with proper training, many begin to solve square equations orally, literally "at first glance."

Unfortunately, in the modern course of school mathematics, such technologies are almost not studied. And you need to know! And today we will look at one of these techniques - Vieta theorem. To begin with, we introduce a new definition.

The square equation of the form x 2 + bx + c \u003d 0 is called the above. Note: The coefficient at x 2 is 1. No other restrictions on the coefficients are not superimposed.

1. x 2 + 7x + 12 \u003d 0 is a given square equation;
2. x 2 - 5x + 6 \u003d 0 - also given;
3. 2x 2 - 6x + 8 \u003d 0 - But this is not the nifiga, since the coefficient at x 2 is 2.

Of course, any square equation of the type AX 2 + BX + C \u003d 0 can be made given - it is enough to divide all the coefficients to the number a. We can always do that, since it follows from the definition of the square equation, that A ≠ 0.

True, not always these transformations will be useful for finding roots. Just below, we make sure that it is necessary only when in the final square of the equation all the coefficients will be integer. In the meantime, consider the simplest examples:

A task. Convert a square equation to the above:

1. 3x 2 - 12x + 18 \u003d 0;
2. -4x 2 + 32x + 16 \u003d 0;
3. 1.5x 2 + 7.5x + 3 \u003d 0;
4. 2x 2 + 7x - 11 \u003d 0.

We divide each equation on the coefficient with a variable x 2. We get:

1. 3x 2 - 12x + 18 \u003d 0 ⇒ x 2 - 4x + 6 \u003d 0 - divided by all by 3;
2. -4x 2 + 32x + 16 \u003d 0 ⇒ x 2 - 8x - 4 \u003d 0 - divided into -4;
3. 1.5x 2 + 7.5X + 3 \u003d 0 ⇒ x 2 + 5x + 2 \u003d 0 - divided by 1.5, all coefficients became integer;
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 presented square equations may have entire coefficients even if the initial equation contained the fraction.

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

Vieta theorem. Consider the given square equation of the form x 2 + BX + C \u003d 0. Suppose that this equation has valid roots x 1 and x 2. In this case, the following statements are true:

1. x 1 + x 2 \u003d -b. In other words, the sum of the roots of the present square equation is equal to the coefficient with a variable X, taken with the opposite sign;
2. x 1 · x 2 \u003d c. The product of the roots of the square equation is equal to the free coefficient.

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

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

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

A task. Solve the square equation:

1. x 2 - 9x + 14 \u003d 0;
2. x 2 - 12x + 27 \u003d 0;
3. 3x 2 + 33x + 30 \u003d 0;
4. -7x 2 + 77x - 210 \u003d 0.

Let's try to write the coefficients on the theorem of Vieta and "Guess" the roots:

1. x 2 - 9x + 14 \u003d 0 is a given square equation.
   By the theorem, we have: x 1 + x 2 \u003d - (- 9) \u003d 9; x 1 · x 2 \u003d 14. It is easy to notice that the roots are numbers 2 and 7;
2. x 2 - 12x + 27 \u003d 0 - also given.
   By the Vieta Theorem: x 1 + x 2 \u003d - (- 12) \u003d 12; x 1 · x 2 \u003d 27. Hence the roots: 3 and 9;
3. 3x 2 + 33x + 30 \u003d 0 - this equation is not given. But we will correct it now, dividing both sides of the equation to the coefficient a \u003d 3. We obtain: x 2 + 11x + 10 \u003d 0.
   We decide on the VEETORE Theorem: x 1 + x 2 \u003d -11; x 1 · x 2 \u003d 10 ⇒ roots: -10 and -1;
4. -7x 2 + 77x - 210 \u003d 0 - again the coefficient at x 2 is not equal to 1, i.e. The equation is not given. We divide everything for the number a \u003d -7. We obtain: x 2 - 11x + 30 \u003d 0.
   By the Vieta Theorem: x 1 + x 2 \u003d - (- 11) \u003d 11; x 1 · x 2 \u003d 30; Of these equations, it is easy to guess the roots: 5 and 6.

From the above reasoning, it is seen how the Vieta theorem simplifies the solution of square equations. No compound calculations, no arithmetic roots and fractions. And even discriminant (see the lesson "Solution of Square Equations") we did not need.

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

1. The square equation is given, 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 discriminant D\u003e 0 - in fact, we initially assume that this inequality is true.

However, in typical mathematical tasks, these conditions are performed. If, as a result of the calculations, it turned out a "bad" square equation (the coefficient at x 2 is different from 1), it is easy to fix - take a look at the examples at the very beginning of the lesson. About the roots at all silent: what is this task in which there is no answer? Of course, the roots will be.

Thus, the general scheme of solving square equations on the Vieta Theorem looks like this:

1. Reduce the square equation to the given one if it has not yet been done in the condition of the problem;
2. If the coefficients in the specified square equation turned out fractional, solve through the discriminant. You can even return to the initial equation to work with more "convenient" numbers;
3. In the case of integer coefficients, we solve the equation on the Vieta Theorem;
4. If within a few seconds it did not manage to guess the roots, scored on the Vieta theorem and solve through the discriminant.

A task. Decide equation: 5x 2 - 35x + 50 \u003d 0.

So, in front of us, the equation that is not given, because The coefficient a \u003d 5. We divide all by 5, we obtain: x 2 - 7x + 10 \u003d 0.

All the coefficients of the square equation are integer - we will try to decide on the Vieta theorem. We have: x 1 + x 2 \u003d - (- 7) \u003d 7; x 1 · x 2 \u003d 10. In this case, the roots are guessed easily - this is 2 and 5. It is not necessary to count through the discriminant.

A task. Decide the equation: -5x 2 + 8x - 2.4 \u003d 0.

We look: -5x 2 + 8x - 2.4 \u003d 0 - this equation is not given, we divide both sides to the coefficient a \u003d -5. We obtain: x 2 - 1.6x + 0.48 \u003d 0 - Equation with fractional coefficients.

It is better to return to the initial equation and count through the discriminant: -5x 2 + 8x - 2.4 \u003d 0 ⇒ d \u003d 8 2 - 4 · (-5) · (-2.4) \u003d 16 ⇒ ... ⇒ x 1 \u003d 1.2; x 2 \u003d 0.4.

A task. Solve the equation: 2x 2 + 10x - 600 \u003d 0.

To begin with, we divide everything to the coefficient A \u003d 2. It turns out equation x 2 + 5x - 300 \u003d 0.

This is the reduced equation, on the Vieta theorem, we have: x 1 + x 2 \u003d -5; x 1 · x 2 \u003d -300. Guess the roots of the square equation in this case is difficult - personally, I seriously "hung" when I solved this task.

We'll have to look for roots through discriminant: d \u003d 5 2 - 4 · 1 · (-300) \u003d 1225 \u003d 35 2. If you do not remember the root from the discriminant, I simply note that 1225: 25 \u003d 49. Therefore, 1225 \u003d 25 · 49 \u003d 5 2 · 7 2 \u003d 35 2.

Now that the root of the discriminant is known, the equation will not be solved. We obtain: x 1 \u003d 15; x 2 \u003d -20.

One of the methods of solutions of the square equation is the application vieta formulaswhich was called in honor of Francois Vieta.

He was a famous lawyer, and served in the 16th century in the French king. In his free time, engaged in astronomy and mathematics. It has established the relationship between the roots and coefficients of the square equation.

Advantages of formula:

1 . Applying the formula, you can quickly find a solution. Because you do not need to enter the second coefficient in the square, then subtract 4As from it, find a discriminate, to substitute its value in the formula for finding the roots.

2 . Without solutions, you can define roots signs, pick up the values \u200b\u200bof the roots.

3 . Deciding the system of two records, it is easy to find the roots themselves. In the present square equation, the amount of the roots is equal to the value of the second coefficient with a minus sign. The product of the roots in the specified square equation is equal to the value of the third coefficient.

4 . According to the roots, write a square equation, that is, to solve the opposite task. For example, this method is used in solving problems in theoretical mechanics.

5 . It is convenient to use the formula when the older coefficient is equal to one.

Disadvantages:

1 . Formula is not universal.

Vieta Theorem Grade 8

Formula
If x 1 and x 2 are the roots of the given square equation x 2 + px + q \u003d 0, then:

Examples
x 1 \u003d -1; x 2 \u003d 3 - the roots of the equation x 2 - 2x - 3 \u003d 0.

P \u003d -2, q \u003d -3.

X 1 + x 2 \u003d -1 + 3 \u003d 2 \u003d -p,

X 1 x 2 \u003d -1 3 \u003d -3 \u003d Q.

Reverse theorem

Formula
If the numbers x 1, x 2, p, q are related to the conditions:

That x 1 and x 2 are the roots of the equation x 2 + px + q \u003d 0.

Example
Let's make a square equation for its roots:

X 1 \u003d 2 -? 3 and x 2 \u003d 2 +? 3.

P \u003d x 1 + x 2 \u003d 4; p \u003d -4; Q \u003d x 1 x 2 \u003d (2 -? 3) (2 +? 3) \u003d 4 - 3 \u003d 1.

The desired equation is: x 2 - 4x + 1 \u003d 0.

When studying how to solve the second-order equations in the school year of algebra, consider the properties of the roots received. They are currently known under the name of the Vieta theorem. Examples of using it are given in this article.

Quadratic equation

The second order equation is equality that is shown in the photo below.

Here, the symbols A, B, C are some numbers wearing the name of the coefficients of the equation under consideration. To solve equality, it is necessary to find such x values \u200b\u200bthat make it true.

Note that since the maximum value of the degree into which the X is erected is to two, then the number of roots in the general case is also two.

To solve this type of equations there are several ways. In this article, consider one of them, which involves the use of the so-called Vieta theorem.

The wording of the Vieta Theorem

At the end of the XVI, the well-known mathematician Francois Vieta (French) noticed by analyzing the properties of the roots of various square equations that their specific combinations satisfy specific ratios. In particular, these combinations are their work and amount.

The Vieta Theorem sets the following: The roots of the square equation under their sum give the ratio of the linear coefficients to the quadratic taken with the opposite sign, and when they are produced, they lead to the ratio of a free member to the quadratic coefficient.

If the general view of the equation is recorded as presented in the photo in the previous section of the article, then mathematically this theorem can be recorded in the form of two equalities:

• r 2 + R 1 \u003d -B / A;
• r 1 x R 2 \u003d C / a.

Where R 1, R 2 is the value of the roots of the equation under consideration.

These equalities can be used to solve a number of various mathematical tasks. The use of the Vieta Theorem in the examples with the solution is given in the following sections of the article.