Quadratic equation 1 root. Solving incomplete quadratic equations

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We continue to study the topic solution of equations". We have already got acquainted with linear equations and now we are going to get acquainted with quadratic equations.

First, we will discuss what a quadratic equation is, how it is written in general form, and give related definitions. After that, using examples, we will analyze in detail how incomplete quadratic equations are solved. Next, let's move on to solving complete equations, get the formula for the roots, get acquainted with the discriminant of a quadratic equation, and consider solutions to typical examples. Finally, we trace the connections between roots and coefficients.

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What is a quadratic equation? Their types

First you need to clearly understand what a quadratic equation is. Therefore, it is logical to start talking about quadratic equations with the definition of a quadratic equation, as well as definitions related to it. After that, you can consider the main types quadratic equations: reduced and non-reduced, as well as complete and incomplete equations.

Definition and examples of quadratic equations

Definition.

Quadratic equation is an equation of the form a x 2 +b x+c=0, where x is a variable, a , b and c are some numbers, and a is different from zero.

Let's say right away that quadratic equations are often called equations of the second degree. This is because the quadratic equation is algebraic equation second degree.

The sounded definition allows us to give examples of quadratic equations. So 2 x 2 +6 x+1=0, 0.2 x 2 +2.5 x+0.03=0, etc. are quadratic equations.

Definition.

Numbers a , b and c are called coefficients of the quadratic equation a x 2 + b x + c \u003d 0, and the coefficient a is called the first, or senior, or coefficient at x 2, b is the second coefficient, or coefficient at x, and c is a free member.

For example, let's take a quadratic equation of the form 5 x 2 −2 x−3=0, here the leading coefficient is 5, the second coefficient is −2, and the free term is −3. Note that when the coefficients b and/or c are negative, as in the example just given, then short form writing a quadratic equation of the form 5 x 2 −2 x−3=0 , and not 5 x 2 +(−2) x+(−3)=0 .

It is worth noting that when the coefficients a and / or b are equal to 1 or −1, then they are usually not explicitly present in the notation of the quadratic equation, which is due to the peculiarities of the notation of such . For example, in the quadratic equation y 2 −y+3=0, the leading coefficient is one, and the coefficient at y is −1.

Reduced and non-reduced quadratic equations

Depending on the value of the leading coefficient, reduced and non-reduced quadratic equations are distinguished. Let us give the corresponding definitions.

Definition.

A quadratic equation in which the leading coefficient is 1 is called reduced quadratic equation. Otherwise, the quadratic equation is unreduced.

According to this definition, quadratic equations x 2 −3 x+1=0 , x 2 −x−2/3=0, etc. - reduced, in each of them the first coefficient is equal to one. And 5 x 2 −x−1=0 , etc. - unreduced quadratic equations, their leading coefficients are different from 1 .

From any non-reduced quadratic equation, by dividing both of its parts by the leading coefficient, you can go to the reduced one. This action is an equivalent transformation, that is, the reduced quadratic equation obtained in this way has the same roots as the original non-reduced quadratic equation, or, like it, has no roots.

Let's take an example of how the transition from an unreduced quadratic equation to a reduced one is performed.

Example.

From the equation 3 x 2 +12 x−7=0, go to the corresponding reduced quadratic equation.

Decision.

It is enough for us to perform the division of both parts of the original equation by the leading coefficient 3, it is non-zero, so we can perform this action. We have (3 x 2 +12 x−7):3=0:3 , which is the same as (3 x 2):3+(12 x):3−7:3=0 , and so on (3:3) x 2 +(12:3) x−7:3=0 , whence . So we got the reduced quadratic equation, which is equivalent to the original one.

Answer:

Complete and incomplete quadratic equations

There is a condition a≠0 in the definition of a quadratic equation. This condition is necessary in order for the equation a x 2 +b x+c=0 to be exactly square, since with a=0 it actually becomes a linear equation of the form b x+c=0 .

As for the coefficients b and c, they can be equal to zero, both separately and together. In these cases, the quadratic equation is called incomplete.

Definition.

The quadratic equation a x 2 +b x+c=0 is called incomplete, if at least one of the coefficients b , c is equal to zero.

In its turn

Definition.

Complete quadratic equation is an equation in which all coefficients are different from zero.

These names are not given by chance. This will become clear from the following discussion.

If the coefficient b is equal to zero, then the quadratic equation takes the form a x 2 +0 x+c=0 , and it is equivalent to the equation a x 2 +c=0 . If c=0 , that is, the quadratic equation has the form a x 2 +b x+0=0 , then it can be rewritten as a x 2 +b x=0 . And with b=0 and c=0 we get the quadratic equation a·x 2 =0. The resulting equations differ from the full quadratic equation in that their left-hand sides do not contain either a term with the variable x, or a free term, or both. Hence their name - incomplete quadratic equations.

So the equations x 2 +x+1=0 and −2 x 2 −5 x+0,2=0 are examples of complete quadratic equations, and x 2 =0, −2 x 2 =0, 5 x 2 +3=0 , −x 2 −5 x=0 are incomplete quadratic equations.

Solving incomplete quadratic equations

It follows from the information of the previous paragraph that there is three kinds of incomplete quadratic equations:

  • a x 2 =0 , the coefficients b=0 and c=0 correspond to it;
  • a x 2 +c=0 when b=0 ;
  • and a x 2 +b x=0 when c=0 .

Let us analyze in order how the incomplete quadratic equations of each of these types are solved.

a x 2 \u003d 0

Let's start by solving incomplete quadratic equations in which the coefficients b and c are equal to zero, that is, with equations of the form a x 2 =0. The equation a·x 2 =0 is equivalent to the equation x 2 =0, which is obtained from the original by dividing its both parts by a non-zero number a. Obviously, the root of the equation x 2 \u003d 0 is zero, since 0 2 \u003d 0. This equation has no other roots, which is explained, indeed, for any non-zero number p, the inequality p 2 >0 takes place, which implies that for p≠0, the equality p 2 =0 is never achieved.

So, the incomplete quadratic equation a x 2 \u003d 0 has a single root x \u003d 0.

As an example, we give the solution of an incomplete quadratic equation −4·x 2 =0. It is equivalent to the equation x 2 \u003d 0, its only root is x \u003d 0, therefore, the original equation has a single root zero.

A short solution in this case can be issued as follows:
−4 x 2 \u003d 0,
x 2 \u003d 0,
x=0 .

a x 2 +c=0

Now consider how incomplete quadratic equations are solved, in which the coefficient b is equal to zero, and c≠0, that is, equations of the form a x 2 +c=0. We know that the transfer of a term from one side of the equation to the other with the opposite sign, as well as the division of both sides of the equation by a non-zero number, give an equivalent equation. Therefore, the following equivalent transformations of the incomplete quadratic equation a x 2 +c=0 can be carried out:

  • move c to right side, which gives the equation a x 2 =−c ,
  • and divide both its parts by a , we get .

The resulting equation allows us to draw conclusions about its roots. Depending on the values ​​of a and c, the value of the expression can be negative (for example, if a=1 and c=2 , then ) or positive, (for example, if a=−2 and c=6 , then ), it is not equal to zero , because by condition c≠0 . We will separately analyze the cases and .

If , then the equation has no roots. This statement follows from the fact that the square of any number is a non-negative number. It follows from this that when , then for any number p the equality cannot be true.

If , then the situation with the roots of the equation is different. In this case, if we recall about, then the root of the equation immediately becomes obvious, it is the number, since. It is easy to guess that the number is also the root of the equation , indeed, . This equation has no other roots, which can be shown, for example, by contradiction. Let's do it.

Let's denote the just 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 substitution into the equation instead of x of its roots turns the equation into a true numerical equality. For x 1 and −x 1 we have , and for x 2 we have . The properties of numerical equalities allow us to perform term-by-term subtraction of true numerical equalities, so subtracting the corresponding parts of the equalities gives x 1 2 − x 2 2 =0. The properties of operations with numbers allow us to rewrite the resulting equality as (x 1 − x 2)·(x 1 + x 2)=0 . We know that the product of two numbers is equal to zero if and only if at least one of them is equal to zero. Therefore, it follows from the obtained equality that x 1 −x 2 =0 and/or x 1 +x 2 =0 , which is the same, x 2 =x 1 and/or x 2 = −x 1 . So we have come to a contradiction, since at the beginning we said that the root of the equation x 2 is different from x 1 and −x 1 . This proves that the equation has no other roots than and .

Let's summarize the information in this paragraph. The incomplete quadratic equation a x 2 +c=0 is equivalent to the equation , which

  • has no roots if ,
  • has two roots and if .

Consider examples of solving incomplete quadratic equations of the form a·x 2 +c=0 .

Let's start with the quadratic equation 9 x 2 +7=0 . After transferring the free term to the right side of the equation, it will take the form 9·x 2 =−7. Dividing both sides of the resulting equation by 9 , we arrive at . Since a negative number is obtained on the right side, this equation has no roots, therefore, the original incomplete quadratic equation 9 x 2 +7=0 has no roots.

Let's solve one more incomplete quadratic equation −x 2 +9=0. We transfer the nine to the right side: -x 2 \u003d -9. Now we divide both parts by −1, we get x 2 =9. The right side contains a positive number, from which we conclude that or . After we write down the final answer: the incomplete quadratic equation −x 2 +9=0 has two roots x=3 or x=−3.

a x 2 +b x=0

It remains to deal with the solution of the last type of incomplete quadratic equations for c=0 . Incomplete quadratic equations of the form a x 2 +b x=0 allows you to solve factorization method. Obviously, we can, located on the left side of the equation, for which it is enough to take the common factor x out of brackets. This allows us to move from the original incomplete quadratic equation to an equivalent equation of the form x·(a·x+b)=0 . And this equation is equivalent to the set of two equations x=0 and a x+b=0 , the last of which is linear and has a root x=−b/a .

So, the incomplete quadratic equation a x 2 +b x=0 has two roots x=0 and x=−b/a.

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

Example.

Solve the equation.

Decision.

We take x out of brackets, this gives the equation. It is equivalent to two equations x=0 and . We solve the resulting linear equation: , and after dividing mixed number on the common fraction, we find . Therefore, the roots of the original equation are x=0 and .

After getting the necessary practice, the solutions of such equations can be written briefly:

Answer:

x=0 , .

Discriminant, formula of the roots of a quadratic equation

To solve quadratic equations, there is a root formula. Let's write down the formula of the roots of the quadratic equation: , where D=b 2 −4 a c- so-called discriminant of a quadratic equation. The notation essentially means that .

It is useful to know how the root formula was obtained, and how it is applied in finding the roots of quadratic equations. Let's deal with this.

Derivation of the formula of the roots of a quadratic equation

Let us need to solve the quadratic equation a·x 2 +b·x+c=0 . Let's perform some equivalent transformations:

  • We can divide both parts of this equation by a non-zero number a, as a result we get the reduced quadratic equation.
  • Now select a full square on its left side: . After that, the equation will take the form .
  • At this stage, it is possible to carry out the transfer of the last two terms to the right side with the opposite sign, we have .
  • And let's also transform the expression on the right side: .

As a result, we arrive at the equation , which is equivalent to the original quadratic equation a·x 2 +b·x+c=0 .

We have already solved equations similar in form in the previous paragraphs when we analyzed . This allows us to draw the following conclusions regarding the roots of the equation:

  • if , then the equation has no real solutions;
  • if , then the equation has the form , therefore, , from which its only root is visible;
  • if , then or , which is the same as or , that is, the equation has two roots.

Thus, the presence or absence of the roots of the equation, and hence the original quadratic equation, depends on the sign of the expression on the right side. In turn, the sign of this expression is determined by the sign of the numerator, since the denominator 4 a 2 is always positive, that is, the sign of the expression b 2 −4 a c . This expression b 2 −4 a c is called discriminant of a quadratic equation and marked with the letter D. From here, the essence of the discriminant is clear - by its value and sign, it is concluded whether the quadratic equation has real roots, and if so, what is their number - one or two.

We return to the equation , rewrite it using the notation of the discriminant: . And we conclude:

  • if D<0 , то это уравнение не имеет действительных корней;
  • if D=0, then this equation has a single root;
  • finally, if D>0, then the equation has two roots or , which can be rewritten in the form or , and after expanding and reducing the fractions to a common denominator, we get .

So we derived the formulas for the roots of the quadratic equation, they look like , where the discriminant D is calculated by the formula D=b 2 −4 a c .

With their help, with a positive discriminant, you can calculate both real roots of a quadratic equation. When the discriminant is equal to zero, both formulas give the same root value corresponding to the only solution of the quadratic equation. And with a negative discriminant, when trying to use the formula for the roots of a quadratic equation, we are faced with extracting the square root from negative number which takes us out of bounds and school curriculum. With a negative discriminant, the quadratic equation has no real roots, but has a pair complex conjugate roots, which can be found using the same root formulas we obtained.

Algorithm for solving quadratic equations using root formulas

In practice, when solving a quadratic equation, you can immediately use the root formula, with which to calculate their values. But this is more about finding complex roots.

However, in school course algebra is usually not about complex, but about the real roots of a quadratic equation. In this case, it is advisable to first find the discriminant before using the formulas for the roots of the quadratic equation, make sure that it is non-negative (otherwise, we can conclude that the equation has no real roots), and after that calculate the values ​​of the roots.

The above reasoning allows us to write algorithm for solving a quadratic equation. To solve the quadratic equation a x 2 + b x + c \u003d 0, you need:

  • using the discriminant formula D=b 2 −4 a c calculate its value;
  • conclude that the quadratic equation has no real roots if the discriminant is negative;
  • calculate the only root of the equation using the formula if D=0 ;
  • find two real roots of a quadratic equation using the root formula if the discriminant is positive.

Here we only note that if the discriminant is equal to zero, the formula can also be used, it will give the same value as .

You can move on to examples of applying the algorithm for solving quadratic equations.

Examples of solving quadratic equations

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

Example.

Find the roots of the equation x 2 +2 x−6=0 .

Decision.

In this case, we have the following coefficients of the quadratic equation: a=1 , b=2 and c=−6 . According to the algorithm, you first need to calculate the discriminant, for this we substitute the indicated a, b and c into the discriminant formula, we have D=b 2 −4 a c=2 2 −4 1 (−6)=4+24=28. Since 28>0, that is, the discriminant is greater than zero, the quadratic equation has two real roots. Let's find them by the formula of roots , we get , here we can simplify the expressions obtained by doing factoring out the sign of the root followed by fraction reduction:

Answer:

Let's move on to the next typical example.

Example.

Solve the quadratic equation −4 x 2 +28 x−49=0 .

Decision.

We start by finding the discriminant: D=28 2 −4 (−4) (−49)=784−784=0. Therefore, this quadratic equation has a single root, which we find as , that is,

Answer:

x=3.5 .

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

Example.

Solve the equation 5 y 2 +6 y+2=0 .

Decision.

Here are the coefficients of the quadratic equation: a=5 , b=6 and c=2 . Substituting these values ​​into the discriminant formula, we have D=b 2 −4 a c=6 2 −4 5 2=36−40=−4. The discriminant is negative, therefore, this quadratic equation has no real roots.

If you need to specify complex roots, then we apply the well-known formula for the roots of the quadratic equation , and perform operations with complex numbers:

Answer:

there are no real roots, the complex roots are: .

Once again, we note that if the discriminant of the quadratic equation is negative, then the school usually immediately writes down the answer, in which they indicate that there are no real roots, and they do not find complex roots.

Root formula for even second coefficients

The formula for the roots of a quadratic equation , where D=b 2 −4 a c allows you to get a more compact formula that allows you to solve quadratic equations with an even coefficient at x (or simply with a coefficient that looks like 2 n, for example, or 14 ln5=2 7 ln5 ). Let's take her out.

Let's say we need to solve a quadratic equation of the form a x 2 +2 n x + c=0 . Let's find its roots using the formula known to us. To do this, we calculate the discriminant D=(2 n) 2 −4 a c=4 n 2 −4 a c=4 (n 2 −a c), and then we use the root formula:

Denote the expression n 2 −a c as D 1 (sometimes it is denoted D "). Then the formula for the roots of the considered quadratic equation with the second coefficient 2 n takes the form , where D 1 =n 2 −a c .

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

So, to solve a quadratic equation with the second coefficient 2 n, you need

  • Calculate D 1 =n 2 −a·c ;
  • If D 1<0 , то сделать вывод, что действительных корней нет;
  • If D 1 =0, then calculate the only root of the equation using the formula;
  • If D 1 >0, then find two real roots using the formula.

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

Example.

Solve the quadratic equation 5 x 2 −6 x−32=0 .

Decision.

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

Note that it was possible to use the usual formula for the roots of a quadratic equation, but in this case, more computational work would have to be done.

Answer:

Simplification of the form of quadratic equations

Sometimes, before embarking on the calculation of the roots of a quadratic equation using formulas, it does not hurt to ask the question: “Is it possible to simplify the form of this equation”? Agree that in terms of calculations it will be easier to solve the quadratic equation 11 x 2 −4 x −6=0 than 1100 x 2 −400 x−600=0 .

Usually, a simplification of the form of a quadratic equation is achieved by multiplying or dividing both sides of it by some number. For example, in the previous paragraph, we managed to achieve a simplification of the equation 1100 x 2 −400 x −600=0 by dividing both sides by 100 .

A similar transformation is carried out with quadratic equations, the coefficients of which are not . In this case, both parts of the equation are usually divided by the absolute values ​​of its coefficients. For example, let's take the quadratic equation 12 x 2 −42 x+48=0. absolute values ​​of its coefficients: gcd(12, 42, 48)= gcd(gcd(12, 42), 48)= gcd(6, 48)=6 . Dividing both parts of the original quadratic equation by 6 , we arrive at the equivalent quadratic equation 2 x 2 −7 x+8=0 .

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

In conclusion of this paragraph, we note that almost always get rid of the minus at the highest coefficient of the quadratic equation by changing the signs of all terms, which corresponds to multiplying (or dividing) both parts by −1. For example, usually from the quadratic equation −2·x 2 −3·x+7=0 go to the solution 2·x 2 +3·x−7=0 .

Relationship between roots and coefficients of a quadratic equation

The formula for the roots of a quadratic equation expresses the roots of an equation in terms of its coefficients. Based on the formula of the roots, you can get other relationships between the roots and coefficients.

The most well-known and applicable formulas from the Vieta theorem of the form and . In particular, for the given quadratic equation, the sum of the roots is equal to the second coefficient with the opposite sign, and the product of the roots is the free term. For example, by the form of the quadratic equation 3 x 2 −7 x+22=0, we can immediately say that the sum of its roots is 7/3, and the product of the roots is 22/3.

Using the already written formulas, you can get a number of other relationships between the roots and coefficients of the quadratic equation. For example, you can express the sum of the squares of the roots of a quadratic equation in terms of its coefficients: .

Bibliography.

  • Algebra: textbook for 8 cells. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
  • Mordkovich A. G. Algebra. 8th grade. At 2 pm Part 1. A textbook for students of educational institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemozina, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.

It is known that it is a particular version of the equality ax 2 + in + c \u003d o, where a, b and c are real coefficients for unknown x, and where a ≠ o, and b and c will be zeros - simultaneously or separately. For example, c = o, v ≠ o or vice versa. We almost remembered the definition of a quadratic equation.

A trinomial of the second degree is equal to zero. Its first coefficient a ≠ o, b and c can take on any values. The value of the variable x will then be when the substitution turns it into the correct numerical equality. Let us dwell on real roots, although the solutions of the equation can also be complete. It is customary to call an equation in which none of the coefficients is equal to o, a ≠ o, b ≠ o, c ≠ o.
Let's solve an example. 2x2 -9x-5 = oh, we find
D \u003d 81 + 40 \u003d 121,
D is positive, so there are roots, x 1 = (9+√121): 4 = 5, and the second x 2 = (9-√121): 4 = -o.5. Checking will help make sure they are correct.

Here step by step solution quadratic equation

Through the discriminant, you can solve any equation, on the left side of which there is a known square trinomial with a ≠ o. In our example. 2x 2 -9x-5 \u003d 0 (ax 2 + in + c \u003d o)

Consider what are incomplete equations of the second degree

  1. ax 2 + in = o. The free term, the coefficient c at x 0, is zero here, in ≠ o.
    How to solve an incomplete quadratic equation of this kind? Let's take x out of brackets. Remember when the product of two factors is zero.
    x(ax+b) = o, this can be when x = o or when ax+b = o.
    Solving the 2nd, we have x = -v/a.
    As a result, we have roots x 1 \u003d 0, according to calculations x 2 \u003d -b / a.
  2. Now the coefficient of x is o, but c is not equal to (≠) o.
    x 2 + c \u003d o. We transfer c to the right side of the equality, we get x 2 \u003d -c. This equation only has real roots when -c is a positive number (c ‹ o),
    x 1 is then equal to √(-c), respectively, x 2 is -√(-c). Otherwise, the equation has no roots at all.
  3. The last option: b \u003d c \u003d o, that is, ax 2 \u003d o. Naturally, such a simple equation has one root, x = o.

Special cases

We have considered how to solve an incomplete quadratic equation, and now we will take any kind.

  • In the full quadratic equation, the second coefficient of x is an even number.
    Let k = o,5b. We have formulas for calculating the discriminant and roots.
    D / 4 \u003d k 2 - ac, the roots are calculated as follows x 1,2 \u003d (-k ± √ (D / 4)) / a for D › o.
    x = -k/a at D = o.
    There are no roots for D ‹ o.
  • There are reduced quadratic equations, when the coefficient of x squared is 1, it is customary to write them x 2 + px + q \u003d o. All of the above formulas apply to them, but the calculations are somewhat simpler.
    Example, x 2 -4x-9 \u003d 0. We calculate D: 2 2 +9, D \u003d 13.
    x 1 = 2+√13, x 2 = 2-√13.
  • In addition, it is easy to apply to the given ones. It says that the sum of the roots of the equation is equal to -p, the second coefficient with a minus (meaning the opposite sign), and the product of these same roots will be equal to q, the free term. Check out how easy it would be to verbally determine the roots of this equation. For non-reduced (for all coefficients that are not equal to zero), this theorem is applicable as follows: the sum x 1 + x 2 is equal to -v / a, the product x 1 x 2 is equal to c / a.

The sum of the free term c and the first coefficient a is equal to the coefficient b. In this situation, the equation has at least one root (it is easy to prove), the first one is necessarily equal to -1, and the second - c / a, if it exists. How to solve an incomplete quadratic equation, you can check it yourself. As easy as pie. Coefficients can be in some ratios among themselves

  • x 2 + x \u003d o, 7x 2 -7 \u003d o.
  • The sum of all coefficients is o.
    The roots of such an equation are 1 and c / a. Example, 2x 2 -15x + 13 = o.
    x 1 \u003d 1, x 2 \u003d 13/2.

There are a number of other ways to solve different equations of the second degree. Here, for example, is a method for extracting a full square from a given polynomial. There are several graphic ways. When you often deal with such examples, you will learn to “click” them like seeds, because all methods come to mind automatically.

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

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

A quadratic equation is an equation of the form:

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

In the school course, the material is given in the following form - the division of equations into three classes is conditionally done:

1. Have two roots.

2. * Have only one root.

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

How are roots calculated? Just!

We calculate the discriminant. Under this "terrible" word lies a very simple formula:

The root formulas are as follows:

*These formulas must be known by heart.

You can immediately write down and decide:

Example:


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

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

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

Let's look at the equation:


On this occasion, when the discriminant is zero, the school course says that one root is obtained, here it is equal to nine. That's right, it is, but...

This representation is somewhat incorrect. In fact, there are two roots. Yes, yes, do not be surprised, it turns out two equal roots, and to be mathematically accurate, then two roots should be written in the answer:

x 1 = 3 x 2 = 3

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

Now the following example:


As we know, the root of a negative number is not extracted, so the solutions in this case no.

That's the whole decision process.

Quadratic function.

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

This is a function of the form:

where x and y are variables

a, b, c are given numbers, where a ≠ 0

The graph is a parabola:

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

Consider examples:

Example 1: Decide 2x 2 +8 x–192=0

a=2 b=8 c= -192

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

Answer: x 1 = 8 x 2 = -12

* You could immediately divide the left and right sides of the equation by 2, that is, simplify it. The calculations will be easier.

Example 2: Decide x2–22 x+121 = 0

a=1 b=-22 c=121

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

We got that x 1 \u003d 11 and x 2 \u003d 11

In the answer, it is permissible to write x = 11.

Answer: x = 11

Example 3: Decide x 2 –8x+72 = 0

a=1 b= -8 c=72

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

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

Answer: no solution

The discriminant is negative. There is a solution!

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

The concept of a complex number.

A bit of theory.

A complex number z is a number of the form

z = a + bi

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

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

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

Now consider the equation:


Get two conjugate roots.

Incomplete quadratic equation.

Consider special cases, this is when the coefficient "b" or "c" is equal to zero (or both are equal to zero). They are solved easily without any discriminants.

Case 1. Coefficient b = 0.

The equation takes the form:

Let's transform:

Example:

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

Case 2. Coefficient c = 0.

The equation takes the form:

Transform, factorize:

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

Example:

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

x 1 = 0 x 2 = 5

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

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

Useful properties and patterns of coefficients.

There are properties that allow solving equations with large coefficients.

ax 2 + bx+ c=0 equality

a + b+ c = 0, then

— if for the coefficients of the equation ax 2 + bx+ c=0 equality

a+ with =b, then

These properties help solve a certain kind of equation.

Example 1: 5001 x 2 –4995 x – 6=0

The sum of the coefficients is 5001+( 4995)+( 6) = 0, so

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

Equality a+ with =b, means

Regularities of coefficients.

1. If in the equation ax 2 + bx + c \u003d 0 the coefficient "b" is (a 2 +1), and the coefficient "c" is numerically equal to the coefficient "a", then its roots are

ax 2 + (a 2 +1) ∙ x + a \u003d 0 \u003d\u003e x 1 \u003d -a x 2 \u003d -1 / a.

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

x 1 \u003d -6 x 2 \u003d -1/6.

2. If in the equation ax 2 - bx + c \u003d 0, the coefficient "b" is (a 2 +1), and the coefficient "c" is numerically equal to the coefficient "a", then its roots are

ax 2 - (a 2 + 1) ∙ x + a \u003d 0 \u003d\u003e x 1 \u003d a x 2 \u003d 1 / a.

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

x 1 = 15 x 2 = 1/15.

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

ax 2 + (a 2 -1) ∙ x - a \u003d 0 \u003d\u003e x 1 \u003d - a x 2 \u003d 1 / a.

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

x 1 \u003d - 17 x 2 \u003d 1/17.

4. If in the equation ax 2 - bx - c \u003d 0, the coefficient "b" is equal to (a 2 - 1), and the coefficient c is numerically equal to the coefficient "a", then its roots are

ax 2 - (a 2 -1) ∙ x - a \u003d 0 \u003d\u003e x 1 \u003d a x 2 \u003d - 1 / a.

Example. Consider the equation 10x2 - 99x -10 = 0.

x 1 \u003d 10 x 2 \u003d - 1/10

Vieta's theorem.

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

45 = 1∙45 45 = 3∙15 45 = 5∙9.

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

Vieta's theorem, moreover. convenient because after solving the quadratic equation in the usual way (through the discriminant), the resulting roots can be checked. I recommend doing this all the time.

TRANSFER METHOD

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

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

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

According to the Vieta theorem in equation (2), it is easy to determine that x 1 \u003d 10 x 2 \u003d 1

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

x 1 \u003d 5 x 2 \u003d 0.5.

What is the rationale? See what's happening.

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

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


The second (modified) roots are 2 times larger.

Therefore, we divide the result by 2.

*If we roll three of a kind, then we divide the result by 3, and so on.

Answer: x 1 = 5 x 2 = 0.5

sq. ur-ie and the exam.

I will say briefly about its importance - YOU SHOULD BE ABLE TO DECIDE quickly and without thinking, you need to know the formulas of the roots and the discriminant by heart. A lot of the tasks that are part of the USE tasks come down to solving a quadratic equation (including geometric ones).

What is worth noting!

1. The form of the equation can be "implicit". For example, the following entry is possible:

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

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

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

Let's work with quadratic equations. These are very popular equations! In its most general form, the quadratic equation looks like this:

For example:

Here a =1; b = 3; c = -4

Here a =2; b = -0,5; c = 2,2

Here a =-3; b = 6; c = -18

Well, you get the idea...

How to solve quadratic equations? If you have a quadratic equation in this form, then everything is simple. Remember the magic word discriminant . A rare high school student has not heard this word! The phrase “decide through the discriminant” is reassuring and reassuring. Because there is no need to wait for tricks from the discriminant! It is simple and trouble-free to use. So, the formula for finding the roots of a quadratic equation looks like this:

The expression under the root sign is the same discriminant. As you can see, to find x, we use only a, b and c. Those. coefficients from the quadratic equation. Just carefully substitute the values a, b and c into this formula and consider. Substitute with your signs! For example, for the first equation a =1; b = 3; c= -4. Here we write:

Example almost solved:

That's all.

What cases are possible when using this formula? There are only three cases.

1. The discriminant is positive. This means that you can extract the root from it. Whether the root is extracted well or badly is another question. It is important what is extracted in principle. Then your quadratic equation has two roots. Two different solutions.

2. The discriminant is zero. Then you have one solution. Strictly speaking, this is not a single root, but two identical. But this plays a role in inequalities, where we will study the issue in more detail.

3. The discriminant is negative. From a negative number Square root is not extracted. Well, okay. This means there are no solutions.

Everything is very simple. And what do you think, you can't go wrong? Well, yes, how...
The most common mistakes are confusion with the signs of values a, b and c. Or rather, not with their signs (where is there to be confused?), But with the substitution of negative values ​​​​into the formula for calculating the roots. Here, a detailed record of the formula with specific numbers saves. If there are problems with calculations, so do it!



Suppose we need to solve the following example:

Here a = -6; b = -5; c=-1

Let's say you know that you rarely get answers the first time.

Well, don't be lazy. It will take 30 seconds to write an extra line. And the number of errors will drop sharply. So we write in detail, with all the brackets and signs:

It seems incredibly difficult to paint so carefully. But it only seems. Try it. Well, or choose. Which is better, fast, or right? Besides, I will make you happy. After a while, there will be no need to paint everything so carefully. It will just turn out right. Especially if you apply practical techniques, which are described below. This evil example with a bunch of minuses will be solved easily and without errors!

So, how to solve quadratic equations through the discriminant we remembered. Or learned, which is also good. Can you correctly identify a, b and c. Do you know how attentively substitute them into the root formula and attentively count the result. Did you understand that the key word here is - attentively?

However, quadratic equations often look slightly different. For example, like this:

This is incomplete quadratic equations . They can also be solved through the discriminant. You just need to correctly figure out what is equal here a, b and c.

Realized? In the first example a = 1; b = -4; a c? It doesn't exist at all! Well, yes, that's right. In mathematics, this means that c = 0 ! That's all. Substitute zero into the formula instead of c, and everything will work out for us. Similarly with the second example. Only zero we don't have here with, a b !

But incomplete quadratic equations can be solved much easier. Without any discrimination. Consider the first incomplete equation. What can be done on the left side? You can take the X out of brackets! Let's take it out.

And what from this? And the fact that the product is equal to zero if, and only if any of the factors is equal to zero! Don't believe? Well, then come up with two non-zero numbers that, when multiplied, will give zero!
Does not work? Something...
Therefore, we can confidently write: x = 0, or x = 4

Everything. These will be the roots of our equation. Both fit. When substituting any of them into the original equation, we get the correct identity 0 = 0. As you can see, the solution is much simpler than through the discriminant.

The second equation can also be easily solved. We move 9 to the right side. We get:

It remains to extract the root from 9, and that's it. Get:

also two roots . x = +3 and x = -3.

This is how all incomplete quadratic equations are solved. Either by taking x out of brackets, or simple transfer numbers to the right, followed by root extraction.
It is extremely difficult to confuse these methods. Simply because in the first case you will have to extract the root from X, which is somehow incomprehensible, and in the second case there is nothing to take out of brackets ...

Now take note of the practical techniques that dramatically reduce the number of errors. The very ones that are due to inattention ... For which it is then painful and insulting ...

First reception. Do not be lazy before solving a quadratic equation to bring it to a standard form. What does this mean?
Suppose, after any transformations, you get the following equation:

Do not rush to write the formula of the roots! You will almost certainly mix up the odds a, b and c. Build the example correctly. First, x squared, then without a square, then a free member. Like this:

And again, do not rush! The minus before the x squared can upset you a lot. Forgetting it is easy... Get rid of the minus. How? Yes, as taught in the previous topic! We need to multiply the whole equation by -1. We get:

And now you can safely write down the formula for the roots, calculate the discriminant and complete the example. Decide on your own. You should end up with roots 2 and -1.

Second reception. Check your roots! According to Vieta's theorem. Don't worry, I'll explain everything! Checking last thing the equation. Those. the one by which we wrote down the formula of the roots. If (as in this example) the coefficient a = 1, check the roots easily. It is enough to multiply them. You should get a free term, i.e. in our case -2. Pay attention, not 2, but -2! free member with your sign . If it didn’t work out, it means they already messed up somewhere. Look for an error. If it worked out, you need to fold the roots. Last and final check. Should be a ratio b with opposite sign. In our case -1+2 = +1. A coefficient b, which is before the x, is equal to -1. So, everything is right!
It is a pity that it is so simple only for examples where x squared is pure, with a coefficient a = 1. But at least check in such equations! There will be fewer mistakes.

Reception third. If your equation has fractional coefficients, get rid of the fractions! Multiply the equation by the common denominator as described in the previous section. When working with fractions, errors, for some reason, climb ...

By the way, I promised an evil example with a bunch of minuses to simplify. You are welcome! There he is.

In order not to get confused in the minuses, we multiply the equation by -1. We get:

That's all! Deciding is fun!

So let's recap the topic.

Practical Tips:

1. Before solving, we bring the quadratic equation to the standard form, build it right.

2. If there is a negative coefficient in front of the x in the square, we eliminate it by multiplying the entire equation by -1.

3. If the coefficients are fractional, we eliminate the fractions by multiplying the entire equation by the corresponding factor.

4. If x squared is pure, the coefficient for it is equal to one, the solution can be easily checked by Vieta's theorem. Do it!

Fractional equations. ODZ.

We continue to master the equations. We already know how to work with linear and quadratic equations. Remained last viewfractional equations. Or they are also called much more solid - fractional rational equations. This is the same.

Fractional equations.

As the name implies, these equations necessarily contain fractions. But not just fractions, but fractions that have unknown in the denominator. At least in one. For example:

Let me remind you, if in the denominators only numbers, these are linear equations.

How to decide fractional equations? First of all, get rid of the fractions! After that, the equation, most often, turns into a linear or quadratic one. And then we know what to do... In some cases, it can turn into an identity, like 5=5 or an incorrect expression, like 7=2. But this rarely happens. Below I will mention it.

But how to get rid of fractions!? Very simple. Applying all the same identical transformations.

We need to multiply the whole equation by the same expression. So that all denominators decrease! Everything will immediately become easier. I explain with an example. Let's say we need to solve the equation:

How were they taught in elementary school? We transfer everything in one direction, reduce it to a common denominator, etc. Forget how bad dream! This is what you need to do when you add or subtract fractional expressions. Or work with inequalities. And in equations, we immediately multiply both parts by an expression that will give us the opportunity to reduce all denominators (ie, in essence, by a common denominator). And what is this expression?

On the left side, to reduce the denominator, you need to multiply by x+2. And on the right, multiplication by 2 is required. So, the equation must be multiplied by 2(x+2). We multiply:

This is the usual multiplication of fractions, but I will write in detail:

Please note that I am not opening the parenthesis yet. (x + 2)! So, in its entirety, I write it:

On the left side, it is reduced entirely (x+2), and in the right 2. As required! After reduction we get linear the equation:

Anyone can solve this equation! x = 2.

Let's solve another example, a little more complicated:

If we remember that 3 = 3/1, and 2x = 2x/ 1 can be written:

And again we get rid of what we don’t really like - from fractions.

We see that to reduce the denominator with x, it is necessary to multiply the fraction by (x - 2). And units are not a hindrance to us. Well, let's multiply. All left side and all right side:

Brackets again (x - 2) I don't reveal. I work with the bracket as a whole, as if it were one number! This must always be done, otherwise nothing will be reduced.

With a feeling of deep satisfaction, we cut (x - 2) and we get the equation without any fractions, in a ruler!

And now we open the brackets:

We give similar ones, transfer everything to the left side and get:

Classical quadratic equation. But the minus ahead is not good. You can always get rid of it by multiplying or dividing by -1. But if you look closely at the example, you will notice that it is best to divide this equation by -2! In one fell swoop, the minus will disappear, and the coefficients will become prettier! We divide by -2. On the left side - term by term, and on the right - just divide zero by -2, zero and get:

We solve through the discriminant and check according to the Vieta theorem. We get x=1 and x=3. Two roots.

As you can see, in the first case, the equation after the transformation became linear, and here it is quadratic. It happens that after getting rid of fractions, all x's are reduced. There is something left, like 5=5. It means that x can be anything. Whatever it is, it will still be reduced. And get the pure truth, 5=5. But, after getting rid of fractions, it may turn out to be completely untrue, such as 2=7. And this means that no solutions! With any x, it turns out to be false.

Realized main way solutions fractional equations? It is simple and logical. We change the original expression so that everything that we don't like disappears. Or interfere. In this case, it's fractions. We will do the same with all complex examples with logarithms, sines and other horrors. We always we will get rid of all this.

However, we need to change the original expression in the direction we need according to the rules, yes ... The development of which is the preparation for the exam in mathematics. Here we are learning.

Now we will learn how to bypass one of the the main ambushes on the exam! But first, let's see if you fall into it or not?

Let's take a simple example:

The matter is already familiar, we multiply both parts by (x - 2), we get:

Remember, with brackets (x - 2) we work as with one, integral expression!

Here I no longer wrote the one in the denominators, undignified ... And I didn’t draw brackets in the denominators, except for x - 2 there is nothing, you can not draw. We shorten:

We open the brackets, move everything to the left, we give similar ones:

We solve, check, we get two roots. x = 2 and x = 3. Fine.

Suppose the task says to write down the root, or their sum, if there are more than one root. What will we write?

If you decide the answer is 5, you were ambushed. And the task will not be counted for you. They worked in vain ... The correct answer is 3.

What's the matter?! And you try to check. Substitute the values ​​of the unknown into original example. And if at x = 3 everything grows together wonderfully, we get 9 = 9, then with x = 2 divide by zero! What absolutely cannot be done. Means x = 2 is not a solution, and is not taken into account in the answer. This is the so-called extraneous or extra root. We just discard it. There is only one final root. x = 3.

How so?! I hear outraged exclamations. We were taught that an equation can be multiplied by an expression! This is the same transformation!

Yes, identical. Under a small condition - the expression by which we multiply (divide) - different from zero. BUT x - 2 at x = 2 equals zero! So it's all fair.

And now what i can do?! Do not multiply by expression? Do you check every time? Again unclear!

Calmly! No panic!

In this difficult situation, three magic letters will save us. I know what you were thinking. Correctly! This is ODZ . Area of ​​Valid Values.

”, that is, equations of the first degree. In this lesson, we will explore what is a quadratic equation and how to solve it.

What is a quadratic equation

Important!

The degree of an equation is determined by the highest degree to which the unknown stands.

If the maximum degree to which the unknown stands is “2”, then you have a quadratic equation.

Examples of quadratic equations

  • 5x2 - 14x + 17 = 0
  • −x 2 + x +
    1
    3
    = 0
  • x2 + 0.25x = 0
  • x 2 − 8 = 0

Important! The general form of the quadratic equation looks like this:

A x 2 + b x + c = 0

"a", "b" and "c" - given numbers.
  • "a" - 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 the general form of the quadratic equation "ax 2 + bx + c \u003d 0".

Let's practice determining the coefficients "a", "b" and "c" in quadratic equations.

5x2 - 14x + 17 = 0 −7x 2 − 13x + 8 = 0 −x 2 + x +
The equation Odds
  • a=5
  • b = −14
  • c = 17
  • a = −7
  • b = −13
  • c = 8
1
3
= 0
  • a = −1
  • b = 1
  • c =
    1
    3
x2 + 0.25x = 0
  • a = 1
  • b = 0.25
  • c = 0
x 2 − 8 = 0
  • a = 1
  • b = 0
  • c = −8

How to solve quadratic equations

Unlike linear equations to solve quadratic equations, a special formula for finding roots.

Remember!

To solve a quadratic equation you need:

  • bring the quadratic equation to general view"ax 2 + bx + c = 0". That is, only "0" should remain on the right side;
  • use the formula for roots:

Let's use an example to figure out how to apply the formula to find the roots of a quadratic equation. Let's solve the quadratic equation.

X 2 - 3x - 4 = 0


The equation "x 2 - 3x - 4 = 0" has already been reduced to the general form "ax 2 + bx + c = 0" and does not require additional simplifications. To solve it, we need only apply formula for finding the roots of a quadratic equation.

Let's define the coefficients "a", "b" and "c" for this equation.


x 1;2 =
x 1;2 =
x 1;2 =
x 1;2 =

With its help, any quadratic equation is solved.

In the formula "x 1; 2 \u003d" the root expression is often replaced
"b 2 − 4ac" to the letter "D" and called discriminant. The concept of a discriminant is discussed in more detail in the lesson "What is a discriminant".

Consider another example of a quadratic equation.

x 2 + 9 + x = 7x

In this form, it is rather difficult to determine the coefficients "a", "b", and "c". Let's first bring the equation to the general form "ax 2 + bx + c \u003d 0".

X 2 + 9 + x = 7x
x 2 + 9 + x − 7x = 0
x2 + 9 - 6x = 0
x 2 − 6x + 9 = 0

Now you can use the formula for the roots.

X 1;2 =
x 1;2 =
x 1;2 =
x 1;2 =
x=

6
2

x=3
Answer: x = 3

There are times when there are no roots in quadratic equations. This situation occurs when a negative number appears in the formula under the root.

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