Equally 0 in a square equation. Quadratic equations. Solving full square equations. How to solve square equations

We continue to study the topic " solving equations" We have already met with linear equations and go to acquaintance with square equations.

First we will analyze what a square equation is written in general, and give the associated definitions. After that, in detail we analyze in detail how incomplete square equations are solved. Further, we turn to solve the full equations, we obtain the root formula, we will get acquainted with the discriminant of a square equation and consider solutions of characteristic examples. Finally, trace the connection between the roots and coefficients.

Navigating page.

What is a square equation? Their species

First you need to clearly understand what a square equation is. Therefore, the conversation about the square equations is logically started from the definition of a square equation, as well as the associated definitions. After that, it is possible to consider the main types of square equations: those applied and unpaid, as well as complete and incomplete equations.

Definition and examples of square equations

Definition.

Quadratic equation - this is the equation of type a · x 2 + b · x + c \u003d 0 where X is a variable, a, b and c - some numbers, and a variety of zero.

Immediately, let's say that the square equations are often called the equations of the second degree. This is due to the fact that the square equation is algebraic equation second degree.

The voiced definition allows you to give examples of square equations. So 2 · x 2 + 6 · x + 1 \u003d 0, 0.2 · x 2 + 2.5 · x + 0.03 \u003d 0, etc. - These are square equations.

Definition.

Numbers a, b and c called coefficients of a square equation a · x 2 + b · x + c \u003d 0, and the coefficient A is called the first, or the older, or the coefficient at x 2, b is the second coefficient, or a coefficient with x, and with a free member.

For example, we take a square equation of the form 5 · x 2 -2 · x-3 \u003d 0, here the senior coefficient is 5, the second coefficient is -2, and the free member is -3. Note when the coefficients B and / or C are negative as in the example above, it uses a brief form of recording of the square equation of the form 5 · x 2 -2 · x-3 \u003d 0, and not 5 · x 2 + (- 2 ) · X + (- 3) \u003d 0.

It is worth noting that when the coefficients A and / or B are equal to 1 or -1, then they are usually not present in the recording of a square equation explicitly, which is associated with the features of the record of such. For example, in the square equation Y 2 -Y + 3 \u003d 0, the senior coefficient is a unit, and the coefficient at y is -1.

Specified and unmarried square equations

Depending on the value of the older coefficient, the above and unpaid square equations are distinguished. Let's give the appropriate definitions.

Definition.

Square equation in which the older coefficient is 1, called given square equation. Otherwise the square equation is nude.

According to this definition, the square equations x 2 -3 · x + 1 \u003d 0, x 2 -x-2/3 \u003d 0, etc. - These, in each of them the first coefficient is equal to one. A 5 · x 2 -X-1 \u003d 0, and the like. - Non-valid square equations, their older coefficients are different from 1.

From any unpaid square equation by dividing it both parts on the senior coefficient, you can go to the given one. This action is equivalent to the transformation, that is, the reduced square equation obtained by this method has the same roots as the original unparally square equation, or, as well as it does not have roots.

We will analyze on the example, as the transition from an integral square equation to the given one is performed.

Example.

From equation 3 · x 2 + 12 · X-7 \u003d 0 Go to the corresponding presented square equation.

Decision.

It is enough for us to divide both parts of the initial equation on the senior coefficient 3, it is different from zero, so we can perform this action. We have (3 · x 2 + 12 · x-7): 3 \u003d 0: 3, which is the same, (3 · x 2): 3+ (12 · x): 3-7: 3 \u003d 0, and further (3: 3) · x 2 + (12: 3) · X-7: 3 \u003d 0, from where. So we obtained a given square equation, equivalent to the source.

Answer:

Full and incomplete square equations

In the definition of the square equation there is a condition A ≠ 0. This condition is necessary in order for the equation a · x 2 + b · x + c \u003d 0 to be precisely square, since at a \u003d 0 it actually becomes the linear equation of the form B · x + c \u003d 0.

As for the coefficients B and C, they can be zero, and both separately and together. In these cases, the square equation is called incomplete.

Definition.

Square equation A · x 2 + b · x + c \u003d 0 called incompleteIf at least one of the coefficients B, C is zero.

In turn

Definition.

Full square equation - This is an equation that all coefficients are different from zero.

Such titles are not accidental. From the following reasoning, it will become clear.

If the coefficient B is zero, the square equation takes the form A · x 2 + 0 · X + C \u003d 0, and it is equivalent to equation A · x 2 + c \u003d 0. If C \u003d 0, that is, the square equation has the form a · x 2 + b · x + 0 \u003d 0, then it can be rewritten as a · x 2 + b · x \u003d 0. And for b \u003d 0 and c \u003d 0, we obtain a square equation A · x 2 \u003d 0. The obtained equations differ from the total square equation by the fact that their left parts do not contain either the component from the variable x, or the free member or both. Hence their name - incomplete square equations.

So equations x 2 + x + 1 \u003d 0 and -2 · x 2 -5 · x + 0.2 \u003d 0 are examples of complete square equations, and x 2 \u003d 0, -2 · x 2 \u003d 0, 5 · x 2 + 3 \u003d 0, -x 2 -5 · x \u003d 0 are incomplete square equations.

Decision of incomplete square equations

From the information of the previous point it follows that there is three types of incomplete square equations:

• a · x 2 \u003d 0, the coefficients B \u003d 0 and C \u003d 0 correspond to it;
• a · x 2 + c \u003d 0, when b \u003d 0;
• and a · x 2 + b · x \u003d 0, when C \u003d 0.

We will analyze in order how incomplete square equations of each of these species are solved.

a · x 2 \u003d 0

Let's start with the solution of incomplete square equations in which the coefficients B and C are zero, that is, from the equations of the form A · x 2 \u003d 0. The equation A · x 2 \u003d 0 is equivalent to equation x 2 \u003d 0, which is obtained from the initial division of its both parts to the number a different from zero. Obviously, the equation x 2 \u003d 0 is zero, as 0 2 \u003d 0. This equation does not have any other roots, as explained, indeed, for any different number of the number P, the inequality P 2\u003e 0, from where it follows that at p ≠ 0, the equality P 2 \u003d 0 is never achieved.

So, the incomplete square equation A · x 2 \u003d 0 has the only root x \u003d 0.

As an example, we give the solution of an incomplete square equation -4 · x 2 \u003d 0. It is equivalent to the equation x 2 \u003d 0, its only root is x \u003d 0, therefore, the initial equation has the only root of zero.

A brief solution in this case can be issued as follows:
-4 · x 2 \u003d 0,
x 2 \u003d 0,
x \u003d 0.

a · x 2 + c \u003d 0

Now consider how incomplete square equations are solved in which the coefficient B is zero, and C ≠ 0, that is, the equations of the form A · x 2 + c \u003d 0. We know that the transfer of the term from one part of the equation to another with the opposite sign, as well as the division of both parts of the equation for different from zero, the number is given equivalent equation. Therefore, it is possible to carry out the following equivalent transformations of an incomplete square equation A · X 2 + C \u003d 0:

• transfer C into the right-hand part, which gives equation A · x 2 \u003d -c,
• and split both parts of it on a, we get.

The resulting equation allows you to draw conclusions about its roots. Depending on the values \u200b\u200bA and C, the expression value may be negative (for example, if A \u003d 1 and C \u003d 2, then) or positive, (for example, if a \u003d -2 and c \u003d 6, then), it is not zero Since under the condition C ≠ 0. Separately we analyze cases and.

If, the equation does not have roots. This statement follows from the fact that the square of any number is the number non-negative. It follows from this that when, for any number p, equality can not be correct.

If, then the root of the equation is different. In this case, if you remember, the root of the equation immediately becomes immediately becoming the number, since. It is easy to guess that the number is also the root of the equation, really,. This equation does not have other roots, which can be shown, for example, by the method from the opposite. Let's do it.

Denote by the only voiced roots of the equation as x 1 and -x 1. Suppose that the equation has another root x 2, different from the indicated roots x 1 and -x 1. It is known that the substitution to the equation instead of X its roots draws the equation to the right numerical equality. For x 1 and -x 1 we have, and for x 2 we have. The properties of numerical equalities can be allowed to perform the soil subtraction of faithful numeric equalities, so the subtraction of the corresponding parts of the equalities and gives x 1 2 -x 2 2 \u003d 0. Properties of actions with numbers allow you to rewrite the obtained equality as (x 1 -x 2) · (x 1 + x 2) \u003d 0. We know that the work of two numbers is zero if and only if at least one of them is zero. Consequently, from the resulting equality it follows that x 1 -x 2 \u003d 0 and / or x 1 + x 2 \u003d 0, which is the same, x 2 \u003d x 1 and / or x 2 \u003d -x 1. So we came to contradiction, since at first we said that the root of the equation x 2 is different from x 1 and -x 1. This is proved that the equation does not have other roots, except.

Summarizing information of this item. Incomplete square equation A · x 2 + c \u003d 0 is equivalent to equation that

• does not have roots if
• it has two roots and, if.

Consider the examples of the solution of incomplete square equations of the form A · x 2 + c \u003d 0.

Let's start with a square equation 9 · x 2 + 7 \u003d 0. After transferring a free member to the right-hand part of the equation, it will take the form 9 · x 2 \u003d -7. Dividing both parts of the obtained equation by 9, come to. Since the negative number turned out in the right part, this equation does not have roots, therefore, the initial incomplete square equation 9 · x 2 + 7 \u003d 0 does not have roots.

I decide another incomplete square equation -x 2 + 9 \u003d 0. We carry the nine to the right side: -x 2 \u003d -9. Now we divide both parts on -1, we obtain x 2 \u003d 9. The right part contains a positive number, where we conclude that or. After writing the final answer: an incomplete square equation -x 2 + 9 \u003d 0 has two roots x \u003d 3 or x \u003d -3.

a · x 2 + b · x \u003d 0

It remains to deal with the solution of the last species of incomplete square equations at C \u003d 0. Incomplete square equations of the form A · x 2 + b · x \u003d 0 allows you to solve method of decomposition of multipliers. Obviously, we can, located in the left part of the equation, for which it is enough to bear the general multiplier X for brackets. This allows you to move from the initial incomplete square equation to an equivalent equation of the form x · (a · x + b) \u003d 0. And this equation is equivalent to the totality of two equations x \u003d 0 and a · x + b \u003d 0, the last of which is linear and has a root x \u003d -b / a.

So, an incomplete square equation A · x 2 + b · x \u003d 0 has two roots x \u003d 0 and x \u003d -b / a.

To secure the material, we will analyze the solution of a specific example.

Example.

Decide equation.

Decision.

We carry out x for brackets, it gives the equation. It is equivalent to two equations x \u003d 0 and. We solve the resulting linear equation:, and by performing a division of a mixed number on an ordinary fraction, we find. Consequently, the roots of the initial equation are x \u003d 0 and.

After receiving the necessary practice, solutions such equations can be recorded briefly:

Answer:

x \u003d 0 ,.

Discriminant, the roots formula of the square equation

To solve square equations, there are formula roots. We write formula of the roots of the square equation:, where D \u003d b 2 -4 · a · c - so-called discriminant square equation. The record in essence means that.

It is useful to know how the roots formula was obtained, and how it is used when the roots of square equations is found. Tell me.

The output of the roots of the square equation

Let us need to solve the square equation a · x 2 + b · x + c \u003d 0. Perform some equivalent transformations:

• Both parts of this equation we can divide the number A different from zero, as a result we obtain the reduced square equation.
• Now we highlight a full square In his left part :. After that, the equation will take the form.
• At this stage, you can transfer the last two components to the right side with the opposite sign, we have.
• And we are still transforming the expression that turned out to be in the right part :.

As a result, we arrive at an equation that is equivalent to the original square equation A · x 2 + b · x + c \u003d 0.

We have already solved similar in the form of the equation, when they disassembled. This allows the following conclusions relating to the roots of the equation:

• if, the equation has no valid solutions;
• if, the equation has the form, therefore, where his only root is visible;
• if, then or that the same or, that is, the equation has two roots.

Thus, the presence or absence of the roots of the equation, which means the initial square equation depends on the sign of the expression standing in the right-hand side. In turn, the sign of this expression is determined by the number of the numerator, since the denominator 4 · a 2 is always positive, that is, the sign of expression B 2 -4 · a · c. This expression B 2 -4 · A · C, called discriminant square equation and identified the letter D.. From here, the essence of the discriminant is clear - according to its value and the sign is concluded, whether the square equation has a valid root, and if it has, what is their number - one or two.

We return to the equation, rewrite it using the discriminant designation :. And we draw conclusions:

• if D<0 , то это уравнение не имеет действительных корней;
• if D \u003d 0, then this equation has the only root;
• finally, if D\u003e 0, the equation has two roots or, which can be rewritten in the form of or, and after disclosure and bring fractions to a common denominator.

So we derived the formula of the roots of the square equation, they have the form where the discriminant D is calculated by the formula d \u003d b 2 -4 · a · c.

With their help, with a positive discriminant, both valid roots of the square equation can be calculated. With an equal zero discriminant, both formulas give the same root value corresponding to the only solution of the square equation. And with a negative discriminator, when attempting to use the formula of the roots of the square equation, we face the extraction of a square root from a negative number, which displays us beyond the framework and school curriculum. With a negative discriminant, the square equation does not have valid roots, but has a couple comprehensively conjugate The roots that can be found on the same root formulas we received.

Algorithm for solving square equations on root formulas

In practice, when solving square equations, you can immediately use the root formula, with which it is possible to calculate their values. But it is more refer to the finding of complex roots.

However, in the school year, algebra usually we are not talking about complex, but on the valid roots of the square equation. In this case, it is advisable before using the formulas of the roots of the square equation to pre-find the discriminant, make sure that it is nonnegative (otherwise it can be concluded that the equation does not have valid roots), and after that, to calculate the roots values.

The above reasoning allow you to record algorithm Solutions of the Square Equation. To solve the square equation A · x 2 + b · x + c \u003d 0, it is necessary:

• according to the formula of the discriminant d \u003d b 2 -4 · a · C calculate its value;
• conclude that the square equation does not have valid roots if the discriminant is negative;
• calculate the only root of the equation by the formula if D \u003d 0;
• find two valid roots of the square equation on the roots formula if the discriminant is positive.

Here you only note that with an equal zero discriminant you can use the formula, it will give the same meaning as.

You can proceed to the examples of the algorithm for solving square equations.

Examples of solutions of square equations

Consider solutions of three square equations with a positive, negative and equal zero discriminant. Having understood with their solution, it will be possible to solve any other square equation by analogy. Let's start.

Example.

Find the roots of the equation x 2 + 2 · x-6 \u003d 0.

Decision.

In this case, we have the following coefficients of the square equation: a \u003d 1, b \u003d 2 and c \u003d -6. According to the algorithm, you must first calculate the discriminant, for this we substitute these A, B and C in the discriminant formula, we have D \u003d b 2 -4 · a · c \u003d 2 2 -4 · 1 · (-6) \u003d 4 + 24 \u003d 28. Since 28\u003e 0, that is, the discriminant is greater than zero, the square equation has two valid roots. We find them by the formula of the roots, we get, here you can simplify the expressions obtained by performing multiplier for the root sign With the subsequent cutting of the fraction:

Answer:

Go to the next characteristic example.

Example.

Decide the square equation -4 · x 2 + 28 · x-49 \u003d 0.

Decision.

We start with finding discriminant: D \u003d 28 2 -4 · (-4) · (-49) \u003d 784-784 \u003d 0. Consequently, this square equation has the only root that we find as, that is,

Answer:

x \u003d 3.5.

It remains to consider the solution of square equations with a negative discriminant.

Example.

Decide equation 5 · y 2 + 6 · y + 2 \u003d 0.

Decision.

Here such coefficients of the square equation: a \u003d 5, b \u003d 6 and c \u003d 2. We substitute these values \u200b\u200bin the discriminant formula, we have D \u003d b 2 -4 · a · c \u003d 6 2 -4 · 5 · 2 \u003d 36-40 \u003d -4. The discriminant is negative, therefore, this square equation does not have valid roots.

If you need to specify complex roots, we use the well-known formula of the roots of the square equation, and perform actions with complex numbers:

Answer:

there are no valid roots, complex roots are as follows :.

Once again, we note that if the discriminant is negative, then the school usually immediately records the answer, which indicates that there are no valid roots, and there are no complex roots.

Formula roots for even second coefficients

The root formula of the square equation, where d \u003d b 2 -4 · A · C allows to obtain a formula of a more compact form that allows you to solve square equations with an even coefficient at x (or simply with a factor having a form 2 · n, for example, or 14 · ln5 \u003d 2 · 7 · ln5). Give it.

Suppose we need to solve the square equation of the form A · x 2 + 2 · n · x + c \u003d 0. Find his roots using the formula known to us. To do this, calculate the discriminant D \u003d (2 · n) 2 -4 · a · c \u003d 4 · n 2 -4 · a · c \u003d 4 · (n 2 -a · c), and then use the root formula:

Denote the expression N 2 -a · C as D 1 (sometimes D "). Then the core formula of the square equation under consideration with the second coefficient 2 · n will take the form , where d 1 \u003d n 2 -a · c.

It is easy to see that d \u003d 4 · d 1, or d 1 \u003d d / 4. In other words, D 1 is the fourth part of the discriminant. It is clear that the sign D 1 is the same as the D sign. That is, the sign D 1 is also an indicator of the presence or absence of the roots of the square equation.

So, to solve the square equation with the second coefficient 2 · n, it is necessary

• Calculate D 1 \u003d N 2 -a · C;
• If D 1.<0 , то сделать вывод, что действительных корней нет;
• If D 1 \u003d 0, then calculate the only root of the equation by the formula;
• If D 1\u003e 0, then find two valid roots by the formula.

Consider the solution of the example using the root formula obtained in this paragraph.

Example.

Decide the square equation 5 · x 2 -6 · x-32 \u003d 0.

Decision.

The second coefficient of this equation can be represented as 2 · (-3). That is, you can rewrite the original square equation in the form 5 · x 2 + 2 · (-3) · x-32 \u003d 0, here a \u003d 5, n \u003d -3 and c \u003d -32, and calculate the fourth part of the discriminant: D 1 \u003d N 2 -a · C \u003d (- 3) 2 -5 · (-32) \u003d 9 + 160 \u003d 169. Since its value is positive, the equation has two valid roots. Find them using the corresponding root formula:

Note that it was possible to use the usual formula of the roots of the square equation, but in this case there would have to perform a larger volume of computational operation.

Answer:

Simplification of the species of square equations

Sometimes, before going to the calculation of the roots of the square equation according to the formulas, it will not prevent the question: "Is it possible to simplify the appearance of this equation"? Agree that in terms of calculations it is easier to solve the square equation 11 · x 2 -4 · x-6 \u003d 0 than 1100 · x 2 -400 · x-600 \u003d 0.

Usually simplification of the species of the square equation is achieved by multiplying or dividing it both parts by a number. For example, in the previous paragraph, it was possible to simplify the equation 1100 · x 2 -400 · x-600 \u003d 0, separating both parts by 100.

Such transformation is carried out with square equations whose coefficients are not. At the same time, both part of the equation on the absolute values \u200b\u200bof its coefficients are usually divided. For example, take a square equation 12 · x 2 -42 · x + 48 \u003d 0. absolute values \u200b\u200bof its coefficients: node (12, 42, 48) \u003d node (node \u200b\u200b(12, 42), 48) \u003d node (6, 48) \u003d 6. By dividing both parts of the original square equation by 6, we will come to an equivalent square equation 2 · x 2 -7 · x + 8 \u003d 0.

And the multiplication of both parts of the square equation is usually made to get rid of fractional coefficients. In this case, multiplication is carried out on the denominators of its coefficients. For example, if both parts of the square equation are multiplied by NOC (6, 3, 1) \u003d 6, then it will take a simpler form x 2 + 4 · x-18 \u003d 0.

In conclusion of this paragraph, we note that almost always get rid of the minus with the senior coefficient of the square equation, changing the signs of all members, which corresponds to multiplication (or division) of both parts by -1. For example, usually from a square equation -2 · x 2 -3 · x + 7 \u003d 0, they go to the solution 2 · x 2 + 3 · x-7 \u003d 0.

Communication between roots and coefficients of the square equation

The formula of the roots of the square equation expresses the roots of the equation through its coefficients. Stripping from the root formula, you can get other dependencies between the roots and coefficients.

The most famous and applicable formulas from the Vieta view theorem and are most well. In particular, for the reduced square equation, the amount of the roots is equal to the second coefficient with the opposite sign, and the product of the roots is a free member. For example, according to the species of the square equation 3 · x 2 -7 · x + 22 \u003d 0, it is possible to immediately say that the sum of its roots is 7/3, and the product of the roots is 22/3.

Using already recorded formulas, a number of other connections between the roots and the coefficients of the square equation can be obtained. For example, you can express the sum of the squares of the roots of the square equation through its coefficients :.

Bibliography.

• Algebra: studies. For 8 cl. general education. institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorov]; Ed. S. A. Telikovsky. - 16th ed. - M.: Enlightenment, 2008. - 271 p. : IL. - ISBN 978-5-09-019243-9.
• Mordkovich A. G. Algebra. 8th grade. In 2 tsp. 1. Tutorial for students of general educational institutions / A. Mordkovich. - 11th ed., Ched. - M.: Mnemozina, 2009. - 215 p.: Il. ISBN 978-5-346-01155-2.

In continuation of the topic "Decision of equations", the material of this article will introduce you to square equations.

Consider everything in detail: the essence and record of the square equation, set the accompanying terms, we will analyze the scheme for the solution of incomplete and complete equations, get acquainted with the formula of roots and discriminant, establish links between roots and coefficients, and of course we give a visual solution of practical examples.

Square equation, its types

Definition 1.

Quadratic equation - This is the equation recorded as a · x 2 + b · x + c \u003d 0where X. - variable, a, b and C. - Some numbers, while a.no zero.

Often, square equations are also called the name of the second degree equations, since in essence the square equation is the algebraic equation of the second degree.

We give an example to illustrate a given definition: 9 · x 2 + 16 · x + 2 \u003d 0; 7, 5 · x 2 + 3, 1 · x + 0, 11 \u003d 0, etc. - These are square equations.

Definition 2.

Numbers a, b and C. - these are the coefficients of the square equation a · x 2 + b · x + c \u003d 0, with the coefficient A. It is called the first, or older, or the coefficient at x 2, b - the second coefficient, or the coefficient when X., but C. Call free member.

For example, in a square equation 6 · x 2 - 2 · x - 11 \u003d 0 Senior coefficient is 6, the second coefficient is − 2 and free member is equal − 11 . Pay attention to the fact that when the coefficients B.and / or C are negative, then a brief form of a view recording is used. 6 · x 2 - 2 · x - 11 \u003d 0, but not 6 · x 2 + (- 2) · x + (- 11) \u003d 0.

We also clarify this aspect: if the coefficients A. and / or B. equal 1 or − 1 , then explicit participation in the recording of the square equation, they may not be taken, which is explained by the features of the record of these numeric coefficients. For example, in a square equation Y 2 - Y + 7 \u003d 0 Senior coefficient is 1, and the second coefficient is − 1 .

Specified and unmarried square equations

By the value of the first coefficient, the square equations are divided into the above and unpaid.

Definition 3.

The reduced square equation - This is a square equation where the older coefficient is equal to 1. For other values \u200b\u200bof the older coefficient, the square equation is non-invalid.

We give examples: square equations x 2 - 4 · x + 3 \u003d 0, x 2 - x - 4 5 \u003d 0 are presented in each of which the older coefficient is 1.

9 · x 2 - x - 2 \u003d 0 - an integral square equation, where the first coefficient is different from 1 .

Any unpassed square equation is possible to convert into a given equation if it is divided from both parts to the first coefficient (equivalent transformation). The transformed equation will have the same roots as the specified intelligent equation or not to have roots at all.

Consideration of a specific example will allow us to clearly demonstrate the transition from an integral square equation to the given one.

Example 1.

The equation is set 6 · x 2 + 18 · x - 7 \u003d 0 . It is necessary to convert the initial equation in the above form.

Decision

The scheme of the specified above is separated by both parts of the initial equation on the senior coefficient 6. Then we get: (6 · x 2 + 18 · x - 7): 3 \u003d 0: 3And this is the same as: (6 · x 2): 3 + (18 · x): 3 - 7: 3 \u003d 0 And further: (6: 6) · x 2 + (18: 6) · x - 7: 6 \u003d 0. From here: x 2 + 3 · X - 1 1 6 \u003d 0. Thus, an equation is considered to be specified.

Answer: x 2 + 3 · x - 1 1 6 \u003d 0.

Full and incomplete square equations

Turn to the definition of the square equation. In it we clarified that A ≠ 0. Such a condition is necessary to equation a · x 2 + b · x + c \u003d 0 it was exactly square because a \u003d 0. It is essentially converted into linear equation b · x + c \u003d 0.

In the case when the coefficients B. and C.equal to zero (which is possible, both individually and together), the square equation is called incomplete.

Definition 4.

Incomplete square equation - such a square equation a · x 2 + b · x + c \u003d 0,where at least one of the coefficients B.and C.(or both) is zero.

Full square equation - a square equation in which all numeric coefficients are not zero.

We indulge in why the types of square equations are given exactly the names.

For b \u003d 0 square equation takes the form A · x 2 + 0 · X + C \u003d 0that the same thing is that a · x 2 + c \u003d 0. For C \u003d 0. Square equation is recorded as a · x 2 + b · x + 0 \u003d 0That is equivalent a · x 2 + b · x \u003d 0. For B \u003d 0. and C \u003d 0. The equation will take the view a · x 2 \u003d 0. The equations that we have received are different from the full square equation in that their left parts are not contained either a component from the X variable or a free member or both at once. Actually, this fact was asked the name of such a type of equations - incomplete.

For example, x 2 + 3 · x + 4 \u003d 0 and - 7 · x 2 - 2 · x + 1, 3 \u003d 0 are complete square equations; x 2 \u003d 0, - 5 · x 2 \u003d 0; 11 · x 2 + 2 \u003d 0, - x 2 - 6 · x \u003d 0 - incomplete square equations.

Decision of incomplete square equations

The above definition makes it possible to distinguish the following types of incomplete square equations:

• a · x 2 \u003d 0, this equation corresponds to the coefficients B \u003d 0. and c \u003d 0;
• a · x 2 + c \u003d 0 for b \u003d 0;
• a · x 2 + b · x \u003d 0 at c \u003d 0.

Consider the decision of each type of incomplete square equation.

Solution of the equation A · x 2 \u003d 0

As mentioned above, the equation corresponds to the coefficients B. and C.equal to zero. The equation a · x 2 \u003d 0 It is possible to convert equation to equivalent to it x 2 \u003d 0which we get, sharing both parts of the source equation for the number A.not equal to zero. Obvious fact that the root of the equation x 2 \u003d 0 this is zero because 0 2 = 0 . Other roots, this equation has no, which is explained by the properties of the degree: for any number p,not equal to zero, faithful inequality P 2\u003e 0what follows that when P ≠ 0 equality P 2 \u003d 0will never be achieved.

Definition 5.

Thus, for an incomplete square equation A · x 2 \u003d 0 there is the only root x \u003d 0..

Example 2.

For example, we solve an incomplete square equation - 3 · x 2 \u003d 0. It is equivalent to the equation x 2 \u003d 0, his only root is x \u003d 0., Then the initial equation has the only root - zero.

Briefly the decision is made up so:

- 3 · x 2 \u003d 0, x 2 \u003d 0, x \u003d 0.

Solution of the equation A · x 2 + c \u003d 0

On the queue - the solution of incomplete square equations, where B \u003d 0, C ≠ 0, that is, the equations of the form a · x 2 + c \u003d 0. We transform this equation carried out the term from one part of the equation to another, changing the sign to the opposite and dividing both parts of the equation to the number, not equal to zero:

• transfer C. in the right part, which gives the equation A · x 2 \u003d - C;
• we divide both parts of the equation on A., I get in the end x \u003d - c a.

Our transformations are equivalent, respectively, the resulting equation is also equivalent to the source, and this fact makes it possible to conclude the roots of the equation. From what is the meanings A. and C.the value of the expression depends - C A: it may have a minus sign (let's say if a \u003d 1. and C \u003d 2., then - c a \u003d - 2 1 \u003d - 2) or a plus sign (for example, if A \u003d - 2 and C \u003d 6., then - C a \u003d - 6 - 2 \u003d 3); it is not zero because C ≠ 0. Let us dwell in more detail in situations when - C a< 0 и - c a > 0 .

In the case when - C a< 0 , уравнение x 2 = - c a не будет иметь корней. Утверждая это, мы опираемся на то, что квадратом любого числа является число неотрицательное. Из сказанного следует, что при - c a < 0 ни для какого числа P. Equality P 2 \u003d - C A cannot be true.

All otherwise, when - C a\u003e 0: Recall the square root, and it will be obvious that the equation x 2 \u003d - C A will be the number - C A, since - C a 2 \u003d - C a. It is not difficult to understand that the number is - C A is also the root of the equation x 2 \u003d - C a: indeed, - - C a 2 \u003d - C a.

Other roots equation will not have. We can demonstrate it using the nasty method. To begin with, set the designations found above the roots as x 1 and - X 1.. I will suggest that equation x 2 \u003d - C A is also root x 2which differs from the roots x 1 and - X 1.. We know that, substituting into the equation instead X. Its roots, we transform the equation into a fair numerical equality.

For x 1 and - X 1. We write: x 1 2 \u003d - C A, and for x 2 - x 2 2 \u003d - C a. Relying on the properties of numerical equalities, replete one right equality from another, which will give us: x 1 2 - x 2 2 \u003d 0. Use the properties of actions with numbers to rewrite the latest equality as (x 1 - x 2) · (x 1 + x 2) \u003d 0. It is known that the work of two numbers is zero then and only if at least one of the numbers is zero. From said it follows that x 1 - x 2 \u003d 0 and / or x 1 + x 2 \u003d 0that the same thing x 2 \u003d x 1 and / or x 2 \u003d - x 1. There was an obvious contradiction, because at first it was agreed that the root of the equation x 2 differs from x 1 and - X 1.. So, we proved that the equation does not have other roots, except x \u003d - c a and x \u003d - - c a.

We summarize all reasoning above.

Definition 6.

Incomplete square equation a · x 2 + c \u003d 0 equivalent to equation x 2 \u003d - c a, which:

• will not have roots when - c a< 0 ;
• there will be two roots X \u003d - C A and X \u003d - - C A with - C a\u003e 0.

We give examples of solving equations a · x 2 + c \u003d 0.

Example 3.

The square equation is specified 9 · x 2 + 7 \u003d 0.It is necessary to find his decision.

Decision

We transfer a free member to the right-hand part of the equation, then the equation will take the form 9 · x 2 \u003d - 7.
We divide both parts of the obtained equation on 9 , come to x 2 \u003d - 7 9. In the right part, we see a number with a minus sign, which means: the specified equation has no roots. Then the original incomplete square equation 9 · x 2 + 7 \u003d 0 Will not have roots.

Answer: the equation 9 · x 2 + 7 \u003d 0it does not have roots.

Example 4.

It is necessary to solve the equation - x 2 + 36 \u003d 0.

Decision

We move 36 to the right side: - x 2 \u003d - 36.
We split both parts on − 1 , get x 2 \u003d 36. In the right part - a positive number, from here we can conclude that x \u003d 36 or X \u003d - 36.
Remove the root and write down the final result: an incomplete square equation - x 2 + 36 \u003d 0 It has two roots x \u003d 6. or x \u003d - 6.

Answer: x \u003d 6. or x \u003d - 6.

Solution of the equation A · x 2 + b · x \u003d 0

We will examine the third type of incomplete square equations when C \u003d 0.. To find a decision of an incomplete square equation a · x 2 + b · x \u003d 0, We use the method of decomposition on multipliers. Spread on multipliers of the polynomial, which is in the left part of the equation, by making a general multiplier for brackets X.. This step will provide an opportunity to convert the original incomplete square equation to the equivalent x · (a · x + b) \u003d 0. And this equation, in turn, is equivalent to the totality of equations x \u003d 0. and A · X + B \u003d 0. The equation A · X + B \u003d 0 Linear, and its root: x \u003d - b a.

Definition 7.

Thus, an incomplete square equation a · x 2 + b · x \u003d 0 will have two roots x \u003d 0. and x \u003d - b a.

Fasten the material by an example.

Example 5.

It is necessary to find the solution of the equation 2 3 · x 2 - 2 2 7 · x \u003d 0.

Decision

Let's lead X. For brackets and obtain equation x · 2 3 · x - 2 2 7 \u003d 0. This equation is equivalent to equations x \u003d 0. and 2 3 · x - 2 2 7 \u003d 0. Now it is necessary to solve the resulting linear equation: 2 3 · x \u003d 2 2 7, x \u003d 2 2 7 2 3.

Briefly solving the equation to write this way:

2 3 · x 2 - 2 2 7 · x \u003d 0 x · 2 3 · x - 2 2 7 \u003d 0

x \u003d 0 or 2 3 · x - 2 2 7 \u003d 0

x \u003d 0 or x \u003d 3 3 7

Answer: x \u003d 0, x \u003d 3 3 7.

Discriminant, the roots formula of the square equation

To find a solution of square equations, there is a formula for roots:

Definition 8.

x \u003d - b ± d 2 · a where D \u003d b 2 - 4 · a · c - the so-called discriminant of a square equation.

Recording X \u003d - B ± D 2 · A in essence means that x 1 \u003d - b + d 2 · a, x 2 \u003d - b - d 2 · a.

It will be useful to understand how the specified formula was derived and how to apply it.

The output of the roots of the square equation

Let us be challenged to solve the square equation a · x 2 + b · x + c \u003d 0. Perform a number of equivalent transformations:

• we split both parts of the equation for the number a.other than zero, we obtain the reduced square equation: x 2 + b a · x + c a \u003d 0;
• we highlight the full square in the left side of the received equation:
  x 2 + ba · x + ca \u003d x 2 + 2 · b 2 · a · x + b 2 · a 2 - b 2 · a 2 + ca \u003d x + b 2 · a 2 - b 2 · a 2 + CA.
  After that, the equation will take the form: X + B 2 · A 2 - B 2 · A 2 + C a \u003d 0;
• now it is possible to make the transfer of the last two terms into the right-hand side, changing the sign to the opposite, after which we obtain: x + b 2 · a 2 \u003d b 2 · a 2 - C a;
• finally, we transform the expression recorded on the right side of the last equality:
  B 2 · A 2 - C a \u003d b 2 4 · a 2 - C a \u003d b 2 4 · a 2 - 4 · A · C 4 · A 2 \u003d B 2 - 4 · A · C 4 · A 2.

Thus, we came to the equation X + B 2 · a 2 \u003d B 2 - 4 · A · C 4 · A 2, equivalent source equation a · x 2 + b · x + c \u003d 0.

We understood the solution of such equations in previous paragraphs (decision of incomplete square equations). The experience gained makes it possible to conclude regarding the roots of the equation X + B 2 · A 2 \u003d B 2 - 4 · A · C 4 · A 2:

• at B 2 - 4 · A · C 4 · A 2< 0 уравнение не имеет действительных решений;
• for B 2 - 4 · A · C 4 · A 2 \u003d 0, the equation has the form x + b 2 · a 2 \u003d 0, then x + b 2 · a \u003d 0.

Hence the only root X \u003d - B 2 · A is obvious;

• for b 2 - 4 · a · C 4 · A 2\u003e 0, it will be correct: X + B 2 · A \u003d B 2 - 4 · A · C 4 · A 2 or X \u003d B 2 · A - B 2 - 4 · a · C 4 · a 2, which is the same as X + - B 2 · A \u003d B 2 - 4 · A · C 4 · A 2 or X \u003d - B 2 · A - B 2 - 4 · A · C 4 · A 2, i.e. The equation has two roots.

It is possible to conclude that the presence or absence of the roots of the equation X + B 2 · A 2 \u003d B 2 - 4 · A · C 4 · A 2 (and hence the initial equation) depends on the sign of expression B 2 - 4 · A · C 4 · A 2, recorded on the right side. And the sign of this expression is set by the number of the numerator, (denominator 4 · a 2 will always be positive), that is, a sign of expression B 2 - 4 · A · C. This expression B 2 - 4 · A · C The name is the discriminant of a square evacuation and is defined as its designation of the letter D. Here you can record the essence of the discriminant - by its value and the sign are concluded whether the square equation will have valid roots, and if it is, what is the number of roots - one or two.

Returning to the equation X + B 2 · A 2 \u003d B 2 - 4 · A · C 4 · A 2. I rewrite it using the discriminant designation: X + B 2 · a 2 \u003d D 4 · A 2.

We will formulate conclusions again:

Definition 9.

• for D.< 0 The equation has no valid roots;
• for D \u003d 0. The equation has the only root X \u003d - B 2 · A;
• for D\u003e 0 The equation has two roots: x \u003d - b 2 · a + d 4 · a 2 or x \u003d - b 2 · a - d 4 · a 2. These roots based on the properties of the radicals can be written in the form: x \u003d - b 2 · a + d 2 · a or - b 2 · a - d 2 · a. And when we reveal the modules and give the fractions to the common denominator, we obtain: x \u003d - b + d 2 · a, x \u003d - b - d 2 · a.

Thus, the result of our reasoning was the removal of the formula of the roots of the square equation:

x \u003d - b + d 2 · a, x \u003d - b - d 2 · a, discriminant D. Calculated by formula D \u003d b 2 - 4 · a · c.

These formulas make it possible when discriminated is larger to determine both valid roots. When the discriminant is zero, the use of both formulas will give the same root as the sole solution of the square equation. In the case when the discriminant is negative, trying to use the root formula of the square equation, we will face the need to remove the square root from the negative number, which will lead us beyond the actual numbers. With a negative discriminant, the square equation will not be valid roots, but a pair of comprehensively conjugate roots, determined by the same root formulas obtained by us are possible.

Algorithm for solving square equations on root formulas

It is possible to solve the square equation, immediately cycling the formula of the roots, but basically they do, if necessary, find complex roots.

In the main mass of cases, it is usually implied for the search for non-complex, but valid roots of the square equation. Then optimally before using the formulas of the roots of the square equation, first determine the discriminant and make sure it is not negative (otherwise we conclude that the equation has no valid roots), and then proceed to calculate the roots value.

The arguments above provide the ability to formulate an algorithm for solving a square equation.

Definition 10.

To solve a square equation a · x 2 + b · x + c \u003d 0, it is necessary:

• according to the formula D \u003d b 2 - 4 · a · c find the value of the discriminant;
• with D.< 0 сделать вывод об отсутствии у квадратного уравнения действительных корней;
• at d \u003d 0 find the only root of the equation according to the formula X \u003d - B 2 · A;
• for d\u003e 0, determine the two valid roots of the square equation according to the formula X \u003d - B ± d 2 · a.

Note that when the discriminant is zero, you can use the formula x \u003d - b ± d 2 · a, it will give the same result as the formula X \u003d - B 2 · a.

Consider examples.

Examples of solutions of square equations

We present the solution of examples at different values \u200b\u200bof the discriminant.

Example 6.

It is necessary to find the roots of the equation x 2 + 2 · x - 6 \u003d 0.

Decision

We write the number coefficients of the square equation: a \u003d 1, b \u003d 2 and C \u003d - 6. Next, we act on the algorithm, i.e. We will proceed to calculate the discriminant, for which we will substitute the coefficients A, B and C. In the formula of the discriminant: D \u003d B 2 - 4 · A · C \u003d 2 2 - 4 · 1 · (- 6) \u003d 4 + 24 \u003d 28.

So, we obtained D\u003e 0, and this means that the initial equation will have two valid roots.
To find them, we use the root formula x \u003d - b ± d 2 · a and, substituting the corresponding values, we obtain: x \u003d - 2 ± 28 2 · 1. We simplify the resulting expression, making a multiplier for the root sign, followed by the cutting of the fraction:

x \u003d - 2 ± 2 · 7 2

x \u003d - 2 + 2 · 7 2 or x \u003d - 2 - 2 · 7 2

x \u003d - 1 + 7 or x \u003d - 1 - 7

Answer: x \u003d - 1 + 7, x \u003d - 1 - 7.

Example 7.

It is necessary to solve the square equation - 4 · x 2 + 28 · x - 49 \u003d 0.

Decision

Determine the discriminant: D \u003d 28 2 - 4 · (- 4) · (- 49) \u003d 784 - 784 \u003d 0. With this discriminant value, the initial equation will have only one root defined by the formula X \u003d - B 2 · a.

x \u003d - 28 2 · (- 4) x \u003d 3, 5

Answer: x \u003d 3, 5.

Example 8.

It is necessary to solve the equation 5 · Y 2 + 6 · Y + 2 \u003d 0

Decision

The numeric coefficients of this equation will be: a \u003d 5, b \u003d 6 and c \u003d 2. We use these values \u200b\u200bto find a discriminant: d \u003d b 2 - 4 · a · C \u003d 6 2 - 4 · 5 · 2 \u003d 36 - 40 \u003d - 4. The calculated discriminant is negative, thus, the initial square equation does not have valid roots.

In the case when the task is to specify complex roots, apply the root formula, performing actions with complex numbers:

x \u003d - 6 ± - 4 2 · 5,

x \u003d - 6 + 2 · i 10 or x \u003d - 6 - 2 · i 10,

x \u003d - 3 5 + 1 5 · i or x \u003d - 3 5 - 1 5 · i.

Answer: There are no valid roots; Complex roots are as follows: - 3 5 + 1 5 · I, - 3 5 - 1 5 · I.

In the school program, there is still no requirement to look for complex roots, so if during the solution discriminant is defined as a negative, the answer is immediately recorded that there are no valid roots.

Formula roots for even second coefficients

The formula of the roots x \u003d - b ± d 2 · a (d \u003d b 2 - 4 · a · c) makes it possible to obtain another formula, more compact, allowing to find solutions of square equations with an even coefficient at x (or with a type 2 coefficient · n, for example, 2 · 3 or 14 · ln 5 \u003d 2 · 7 · ln 5). We show how this formula is displayed.

Let us be the task of finding the solution of the square equation A · x 2 + 2 · n · x + c \u003d 0. We act on the algorithm: Determine the discriminant d \u003d (2 · n) 2 - 4 · a · c \u003d 4 · n 2 - 4 · a · c \u003d 4 · (n 2 - a · c), and then use the root formula:

x \u003d - 2 · n ± d 2 · a, x \u003d - 2 · n ± 4 · n 2 - a · C 2 · a, x \u003d - 2 · n ± 2 n 2 - a · C 2 · a, x \u003d - n ± N 2 - A · Ca.

Let the expression N 2 - A · C be indicated as D 1 (sometimes D "). Then the formula of the roots of the square equation under consideration with the second coefficient 2 · n will take the form:

x \u003d - n ± d 1 A, where d 1 \u003d n 2 - a · c.

It is easy to see that that d \u003d 4 · d 1, or d 1 \u003d d 4. In other words, D 1 is a quarter of the discriminant. It is obvious that the sign D 1 is the same as the sign D, which means the sign D 1 can also serve as an indicator of the presence or absence of the roots of the square equation.

Definition 11.

Thus, to find the solution of the square equation with the second coefficient 2 · n, it is necessary:

• find D 1 \u003d N 2 - A · C;
• with D 1.< 0 сделать вывод, что действительных корней нет;
• for d 1 \u003d 0, determine the only root of the equation according to the formula X \u003d - N A;
• for d 1\u003e 0, determine the two valid roots according to the formula x \u003d - n ± d 1 a.

Example 9.

It is necessary to solve the square equation 5 · x 2 - 6 · x - 32 \u003d 0.

Decision

The second coefficient of the specified equation can be represented as 2 · (- 3). Then rewrite the specified square equation as 5 · x 2 + 2 · (- 3) · x - 32 \u003d 0, where a \u003d 5, n \u003d - 3 and c \u003d - 32.

We calculate the fourth part of the discriminant: D 1 \u003d N 2 - A · C \u003d (- 3) 2 - 5 · (- 32) \u003d 9 + 160 \u003d 169. The value obtained positively, this means that the equation has two valid roots. We define them according to the corresponding root formula:

x \u003d - n ± d 1 a, x \u003d - - 3 ± 169 5, x \u003d 3 ± 13 5,

x \u003d 3 + 13 5 or x \u003d 3 - 13 5

x \u003d 3 1 5 or x \u003d - 2

It would be possible to make calculations and by the usual formula of the roots of the square equation, but in this case the solution would be more cumbersome.

Answer: x \u003d 3 1 5 or x \u003d - 2.

Simplification of the species of square equations

Sometimes it is possible to optimize the type of source equation, which will simplify the process of calculating the roots.

For example, a square equation 12 · x 2 - 4 · X - 7 \u003d 0 is clearly more convenient for solving than 1200 · x 2 - 400 · x - 700 \u003d 0.

More often simplification of the species of the square equation is performed by the multiplication or division of its both parts into a kind of number. For example, we showed a simplified record of the equation 1200 · x 2 - 400 · x - 700 \u003d 0, obtained by dividing both parts by 100.

Such a conversion is possible when the coefficients of the square equation are not mutually simple numbers. Then there are usually dividing both parts of the equation to the largest common divisor of absolute values \u200b\u200bof its coefficients.

As an example, use a square equation 12 · x 2 - 42 · x + 48 \u003d 0. We define the node of absolute values \u200b\u200bof its coefficients: nodes (12, 42, 48) \u003d node (node \u200b\u200b(12, 42), 48) \u003d node (6, 48) \u003d 6. We will divide the two parts of the original square equation to 6 and we obtain the equivalent square equation 2 · x 2 - 7 · x + 8 \u003d 0.

The multiplication of both parts of the square equation is usually get rid of fractional coefficients. At the same time multiplied by the smallest general multiple denominator of its coefficients. For example, if each part of the square equation is 1 6 · x 2 + 2 3 · x - 3 \u003d 0 multiply from the NOC (6, 3, 1) \u003d 6, then it will become recorded in a simpler form x 2 + 4 · X - 18 \u003d 0.

Finally, we note that almost always get rid of the minus at the first coefficient of the square equation, changing the signs of each member of the equation, which is achieved by multiplying (or divisions) of both parts of 1. For example, from a square equation - 2 · x 2 - 3 · x + 7 \u003d 0, you can go to its simplified version 2 · x 2 + 3 · x - 7 \u003d 0.

Communication between roots and coefficients

The formula of the roots of square equations x \u003d - b ± d 2 · A already known to us expresses the roots of the equation through its numerical coefficients. Relying on this formula, we have the opportunity to set other dependencies between the roots and coefficients.

The most famous and applicable are the formulas of the Vieta theorem:

x 1 + x 2 \u003d - b a and x 2 \u003d c a.

In particular, for the reduced square equation, the amount of the roots is the second coefficient with the opposite sign, and the product of the roots is free of charge. For example, according to the species of the square equation 3 · x 2 - 7 · x + 22 \u003d 0, it is possible to immediately determine that the sum of its roots is 7 3, and the product of the roots is 22 3.

You can also find a number of other links between the roots and coefficients of the square equation. For example, the sum of squares of the roots of the square equation can be expressed through the coefficients:

x 1 2 + x 2 2 \u003d (x 1 + x 2) 2 - 2 · x 1 · x 2 \u003d - Ba 2 - 2 · Ca \u003d B 2 A 2 - 2 · Ca \u003d B 2 - 2 · A · Ca 2.

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In modern society, the ability to perform actions with the equations containing the variable raised to the square can be useful in many areas of activity and is widely used in practice in scientific and technical developments. Evidence of this can serve the design of marine and river vessels, aircraft and missiles. With the help of such calculations, the trajectories of the movement of various bodies, including space objects. Examples with a solution of square equations are used not only in economic forecasting, in the design and construction of buildings, but also in the most ordinary everyday circumstances. They may be needed in tourist campaigns, in sports, in shopping stores and in other very common situations.

We break the expression on the components of multipliers

The degree of equation is determined by the maximum value of the degree in the variable, which contains this expression. In the event that it is 2, then such an equation is just called square.

If the language of the formulas is expressing, then the indicated expressions, no matter how they look, can always be caused by the form when the left part of the expression consists of three terms. Among them: AX 2 (that is, the variable erected into a square with its coefficient), BX (unknown without a square with its coefficient) and C (free component, that is, the usual number). All this in the right side is equal to 0. In the case when there is no one of its components of the terms, with the exception of AX 2, it is called an incomplete square equation. Examples with solving such tasks, the value of the variables in which it is easy to find, should be considered first.

If the expression appears in the form looks in such a way that two, more precisely, AX 2 and BX, the expression on the expression on the expression on the right side, is easiest to find a variable for brackets. Now our equation will look like this: x (ax + b). Next, it becomes obvious that or x \u003d 0, or the task is reduced to finding a variable from the following expression: AX + B \u003d 0. The specified dictated one of the multiplication properties. The rule says that the product of two factors gives as a result of 0 only if one of them is zero.

Example

x \u003d 0 or 8x - 3 \u003d 0

As a result, we obtain two roots of the equation: 0 and 0.375.

The equations of this kind can describe the movement of bodies under the influence of gravity, which began movement from a certain point adopted at the beginning of the coordinates. Here, the mathematical record takes the following form: Y \u003d V 0 T + GT 2/2. Substituting the necessary values, equating the right side 0 and finding possible unknowns, you can find out the time passing from the moment of the body's rise until its fall, as well as many other values. But we will talk about it later.

Decomposition of the expression on multipliers

The rule described above makes it possible to solve the specified tasks and in more complex cases. Consider examples with solving square equations of this type.

X 2 - 33x + 200 \u003d 0

This square triple is complete. To begin with, we transform the expression and decompose it for multipliers. They are obtained two: (x-8) and (x-25) \u003d 0. As a result, we have two roots 8 and 25.

Examples with solving square equations in grade 9 allow this method to find a variable in expressions not only the second, but even the third and fourth orders.

For example: 2x 3 + 2x 2 - 18x - 18 \u003d 0. With the decomposition of the right part of the multipliers with a variable, they are obtained three, that is, (x + 1), (x-3) and (x + 3).

As a result, it becomes obvious that this equation has three roots: -3; -one; 3.

Extract square root

Another case of the incomplete equation of the second order is the expression, in the language of the letters presented in such a way that the right side is built from the components of AX 2 and C. Here, for the value of the variable, the free member is transferred to the right side, and then a square root is extracted from both parts of equality. It should be noted that in this case the roots of the equation usually two. An exception can only be equal to equality, generally not containing the term C, where the variable is zero, as well as the options for expressions, when the right side turns out to be negative. In the latter case, the solutions do not exist at all, since the above action cannot be produced with roots. Examples of solutions of square equations of this type must be considered.

In this case, the roots of the equation will be -4 and 4.

Calculation of a land plot

The need for such calculations appeared in deep antiquity, because the development of mathematics in many respects in those distant times was due to the need to determine the most accuracy of the area and the perimeter of land plots.

Examples with solving square equations drawn up on the basis of tasks of this kind should be considered to us.

So, let's say there is a rectangular plot of land, the length of which is 16 meters more than the width. It should be found a length, width and perimeter of the site, if it is known that its area is equal to 612 m 2.

Starting a matter, first make the necessary equation. Denote by x the width of the site, then its length will be (x + 16). From the written it follows that the area is determined by the expression x (x + 16), which, according to the condition of our problem, is 612. This means that x (x + 16) \u003d 612.

The solution of complete square equations, and this expression is precisely such, cannot be carried out by the same way. Why? Although the left side of it still contains two factors, the product is not at all equal to 0, so other methods are used here.

Discriminant

First of all, we will produce the necessary conversion, then the appearance of this expression will look like this: x 2 + 16x - 612 \u003d 0. This means we got an expression in the form corresponding to the previously specified standard, where a \u003d 1, b \u003d 16, c \u003d -612.

This can be an example of solving square equations through discriminant. Here, the required calculations are made according to the scheme: d \u003d b 2 - 4ac. This auxiliary value does not just make it possible to find the desired values \u200b\u200bin the second order equation, it determines the number of possible options. In case D\u003e 0, there are two; When d \u003d 0, there is one root. In case D<0, никаких шансов для решения у уравнения вообще не имеется.

About roots and their formula

In our case, the discriminant is: 256 - 4 (-612) \u003d 2704. This suggests that the answer from our task exists. If you know, K, the solution of square equations must be continued using the formula below. It allows you to calculate the roots.

This means that in the case presented: x 1 \u003d 18, x 2 \u003d -34. The second version in this dilemma cannot be a solution, because the dimensions of the land can not be measured in negative values, it means x (i.e. the width of the site) is 18 m. From here, we calculate the length: 18 + 16 \u003d 34, and perimeter 2 (34+ 18) \u003d 104 (m 2).

Examples and objectives

We continue to study square equations. Examples and a detailed solution of several of them will be given further.

1) 15x 2 + 20x + 5 \u003d 12x 2 + 27x + 1

We transfer everything to the left part of equality, we will make a transformation, that is, we obtain the form of the equation that is called standard, and equalize it with zero.

15x 2 + 20x + 5 - 12x 2 - 27X - 1 \u003d 0

After folding like, we define the discriminant: d \u003d 49 - 48 \u003d 1. So, our equation will have two roots. We calculate them according to the above formula, which means that the first one of them is 4/3, and the second one.

2) Now reveal the riddles of another kind.

Find out, is there any roots here x 2 - 4x + 5 \u003d 1? To obtain a comprehensive response, we give a polynomial to the appropriate familiarity and calculate the discriminant. In the specified example, the solution of the square equation is not necessary, because the essence of the task is not at all this. In this case, D \u003d 16 - 20 \u003d4, which means there are really no roots.

Vieta theorem

Square equations are conveniently solved through the above formulas and discriminant when the square root is extracted from the last value. But it happens not always. However, there are many ways to obtain variables in this case. Example: solutions of square equations on the Vieta Theorem. She is named after which lived in the XVI century in France and made a brilliant career due to his mathematical talent and courtyards. Portrait of it can be seen in the article.

The pattern that the famous French noted was as follows. He proved that the roots of the equation in the amount are numerically equal to -p \u003d b / a, and their product corresponds to q \u003d c / a.

Now consider specific tasks.

3x 2 + 21x - 54 \u003d 0

For simplicity, we transform the expression:

x 2 + 7x - 18 \u003d 0

We use the Vieta theorem, it will give us the following: the amount of the roots is -7, and their work -18. From here, we obtain that the roots of the equation are numbers -9 and 2. Having made a check, make sure that these values \u200b\u200bof variables are really suitable in the expression.

Graph and Parabola equation

Concepts The quadratic function and square equations are closely connected. Examples of this have already been shown earlier. Now consider some mathematical riddles a little more. Any equation of the described type can be imagined. A similar dependence drawn in the form of a graph is called a parabola. Her various types are shown in the figure below.

Any parabola has a vertex, that is, the point from which its branches come out. In case a\u003e 0, they leave high in infinity, and when a<0, они рисуются вниз. Простейшим примером подобной зависимости является функция y = x 2 . В данном случае в уравнении x 2 =0 неизвестное может принимать только одно значение, то есть х=0, а значит существует только один корень. Это неудивительно, ведь здесь D=0, потому что a=1, b=0, c=0. Выходит формула корней (точнее одного корня) квадратного уравнения запишется так: x = -b/2a.

Visual images of functions help solve any equations, including square. This method is called graphic. And the value of the variable x is the coordinate of the abscissa at points where the graph of the graph is crossing from 0x. The coordinates of the vertices can be found according to the only given formula X 0 \u003d -B / 2A. And, substituting the resulting value to the initial equation of the function, you can learn Y 0, that is, the second coordinate of the pearabol vertex belonging to the ordinate axis.

Crossing the branches of parabola with the abscissa axis

Examples with solutions of square equations are very much, but there are general patterns. Consider them. It is clear that the intersection of the graph with the axis 0x at a\u003e 0 is only possible if 0 receives negative values. And for A.<0 координата у 0 должна быть положительна. Для указанных вариантов D>0. Otherwise D<0. А когда D=0, вершина параболы расположена непосредственно на оси 0х.

According to the chart, the parabolas can be defined and roots. The opposite is also true. That is, if you get a visual image of a quadratic function is not easy, you can equate the right side of the expression to 0 and solve the resulting equation. And knowing the intersection points with the 0x axis, it is easier to build a schedule.

From the history

With the help of equations containing the variable raised to the square, in the old days not only made mathematical calculations and determined the area of \u200b\u200bgeometric figures. Similar calculations of the ancient were needed for grand discoveries in the field of physics and astronomy, as well as to compile astrological forecasts.

As the modern science figures suggest, among the first solutions of square equations, residents of Babylon took up. It happened in four centuries before the onset of our era. Of course, their calculations in the root differed from now adopted and turned out to be much primitive. For example, Mesopotamian mathematicians had no idea about the existence of negative numbers. The strangers also had other subtleties from those who know any student of our time.

Perhaps even earlier scientists of Babylon, the solution of square equations, a sage of India Budhoyama was engaged. It happened in about eight centuries before the era of Christ. True, the equation of the second order, the methods of solving which he led was the most simultaneous. In addition to him, such questions were interested in old and Chinese mathematicians. In Europe, the square equations began to solve only in the early XIII century, but later they were used in their work such great scientists as Newton, Descartes and many others.

"That is, the equations of the first degree. In this lesson we will analyze what is called a square equation And how to solve it.

What is called a square equation

Important!

The degree of equation is determined by the greatest extent in which an unknown one is.

If the maximum degree in which the unknown is "2", it means that you are a square equation.

Examples of square equations

• 5x 2 - 14x + 17 \u003d 0
• -X 2 + x +

  1

  3

  = 0
• x 2 + 0.25x \u003d 0
• x 2 - 8 \u003d 0

Important! The general view of the square equation looks like this:

A x 2 + b x + c \u003d 0

"A", "B" and "C" - specified numbers.

• "A" is the first or senior coefficient;
• "B" - the second coefficient;
• "C" is a free member.

To find "a", "b" and "c" you need to compare your equation with a common view of the square equation "AX 2 + BX + C \u003d 0".

Let's take care of determining the coefficients "A", "B" and "C" in square equations.

5x 2 - 14x + 17 \u003d 0 -7x 2 - 13x + 8 \u003d 0 -X 2 + x +

The equation	Factors
a \u003d 5. b \u003d -14. c \u003d 17.
a \u003d -7. b \u003d -13. c \u003d 8.

1

3

= 0

• a \u003d -1.
• b \u003d 1.
• c \u003d.

  1

  3

x 2 + 0.25x \u003d 0

• a \u003d 1.
• b \u003d 0.25.
• c \u003d 0

x 2 - 8 \u003d 0

• a \u003d 1.
• b \u003d 0.
• c \u003d -8.

How to solve square equations

In contrast to linear equations for solving square equations, a special formula for finding roots.

Remember!

To solve the square equation you need:

• create a square equation to the general type "AX 2 + BX + C \u003d 0". That is, only "0" should remain in the right part;
• use the root formula:

Let's analyze on the example, how to apply the formula for finding the roots of the square equation. Let the square equation.

X 2 - 3x - 4 \u003d 0

The "X 2 - 3X - 4 \u003d 0" equation is already given to the total appearance of "AX 2 + BX + C \u003d 0" and does not require additional simplifications. To solve it, we have enough to apply the formula of finding the roots of the square equation.

We define the coefficients "A", "B" and "C" for this equation.

x 1; 2 \u003d
x 1; 2 \u003d
x 1; 2 \u003d
x 1; 2 \u003d

With it, any square equation is solved.

In the formula "x 1; 2 \u003d" often replace the guided expression
"B 2 - 4AC" on the letter "D" and is called discriminant. The concept of discriminant is considered in more detail in the lesson "What is discriminant".

Consider another example of a square equation.

x 2 + 9 + x \u003d 7x

In this form, determine the coefficients "A", "B" and "C" is quite difficult. Let's first give the equation to the general type "AX 2 + BX + C \u003d 0".

X 2 + 9 + x \u003d 7x
x 2 + 9 + x - 7x \u003d 0
x 2 + 9 - 6x \u003d 0
x 2 - 6x + 9 \u003d 0

Now you can use the root formula.

X 1; 2 \u003d
x 1; 2 \u003d
x 1; 2 \u003d
x 1; 2 \u003d
x \u003d.

6

2

x \u003d 3.
Answer: x \u003d 3

There are cases when there are no roots in square equations. This situation occurs when a negative number is under the root.

The formulas of the roots of the square equation. Cases of valid, multiple and complex roots are considered. Decomposition of square three-shred multipliers. Geometric interpretation. Examples of determining the roots and decomposition of multipliers.

Content

See also: Solution of square equations online

Basic formulas

Consider a square equation:
(1) .
Roots square equation (1) are determined by formulas:
; .
These formulas can be combined like this:
.
When the roots of the square equation are known, the second degree polynomial can be represented as a work of the factors (decompose on multipliers):
.

Next, we believe that - the actual numbers.
Consider discriminant square equation:
.
If the discriminant is positive, then the square equation (1) has two different valid root:
; .
Then the decomposition of the square three decreases on the factors has the form:
.
If the discriminant is zero, then the square equation (1) has two multiple (equal) valid root:
.
Factorization:
.
If the discriminant is negative, then the square equation (1) has two comprehensively conjugated root:
;
.
Here - the imaginary unit;
And - the actual and imaginary parts of the roots:
; .
Then

.

Graphic interpretation

If you build a chart function
,
which is parabola, then the point of intersection of the graph with the axis will be roots of the equation
.
When, the schedule crosses the abscissa axis (axis) at two points ().
When, the graph concerns the abscissa axis at one point ().
When, the schedule does not cross the abscissa axis ().

Useful formulas associated with a square equation

(F.1) ;
(F.2) ;
(F.3) .

The output of the formula for the roots of the square equation

We carry out transformations and apply formulas (F.1) and (F.3):




,
Where
; .

So, we got a formula for a polynomial of the second degree in the form:
.
From here it can be seen that the equation

performed at
and.
That is, the roots of the square equation are roots
.

Examples of determining the roots of the square equation

Example 1.

(1.1) .

.
Comparing with our equation (1.1), we find the values \u200b\u200bof the coefficients:
.
We find discriminant:
.
Since the discriminant is positive, the equation has two valid root:
;
;
.

From here we get a decomposition of a square three-stakes on multipliers:

.

Schedule function y \u003d 2 x 2 + 7 x + 3 Crosses the abscissa axis at two points.

We construct a function schedule
.
The schedule of this function is parabola. She places the abscissa axis (axis) at two points:
and.
These points are roots of the initial equation (1.1).

;
;
.

Example 2.

Find the roots of the square equation:
(2.1) .

We write the square equation in general form:
.
Comparing with the initial equation (2.1), we find the values \u200b\u200bof the coefficients:
.
We find discriminant:
.
Since the discriminant is zero, the equation has two multiple (equal) root:
;
.

Then the decomposition of three decisions on multipliers has the form:
.

Function graph y \u003d x 2 - 4 x + 4 Requests the abscissa axis at one point.

We construct a function schedule
.
The schedule of this function is parabola. It concerns the abscissa axis (axis) at one point:
.
This point is the root of the initial equation (2.1). Since this root enters the expansion of multipliers twice:
,
That such root is called multiple. That is, it is believed that there are two equal root:
.

;
.

Example 3.

Find the roots of the square equation:
(3.1) .

We write the square equation in general form:
(1) .
We rewrite the initial equation (3.1):
.
Compare C (1), we find the values \u200b\u200bof the coefficients:
.
We find discriminant:
.
Discriminant is negative. Therefore, there are no valid roots.

You can find complex roots:
;
;
.

Then


.

The function graph does not cross the abscissa axis. There are no valid roots.

We construct a function schedule
.
The schedule of this function is parabola. It does not intersect the abscissa axis (axis). Therefore, there are no valid roots.

There are no valid roots. Roings are integrated:
;
;
.

See also: