How to write a quadratic equation as a product. Solution of quadratic equations, formula of roots, examples

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 analyze what a quadratic equation is, how it is written in general view, 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 the complete equations, get the formula for the roots, get acquainted with the discriminant quadratic equation and consider the solutions of 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 of 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, the short form of the quadratic equation of the form 5 x 2 −2 x−3=0 is used, 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.

Solution.

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 the 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.

Solution.

We take x out of brackets, this gives the equation. It is equivalent to two equations x=0 and . We solve the received linear equation: , and after dividing the mixed number by an ordinary 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 a negative number, which takes us beyond the scope of the 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 a school algebra course, it is usually we are talking not about complex, but about 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 .

Solution.

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 .

Solution.

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 .

Solution.

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 use 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 .

Solution.

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 leading 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.

Quadratic equations are studied in grade 8, so there is nothing complicated here. The ability to solve them is essential.

A quadratic equation is an equation of the form ax 2 + bx + c = 0, where the coefficients a , b and c are arbitrary numbers, and a ≠ 0.

Before studying specific solution methods, we note that all quadratic equations can be divided into three classes:

1. Have no roots;
2. They have exactly one root;
3. They have two different roots.

This is an important difference between quadratic and linear equations, where the root always exists and is unique. How to determine how many roots an equation has? There is a wonderful thing for this - discriminant.

Discriminant

Let the quadratic equation ax 2 + bx + c = 0 be given. Then the discriminant is simply the number D = b 2 − 4ac .

This formula must be known by heart. Where it comes from is not important now. Another thing is important: by the sign of the discriminant, you can determine how many roots a quadratic equation has. Namely:

1. If D< 0, корней нет;
2. If D = 0, there is exactly one root;
3. If D > 0, there will be two roots.

Please note: the discriminant indicates the number of roots, and not at all their signs, as for some reason many people think. Take a look at the examples and you will understand everything yourself:

A task. How many roots do quadratic equations have:

1. x 2 - 8x + 12 = 0;
2. 5x2 + 3x + 7 = 0;
3. x 2 − 6x + 9 = 0.

We write the coefficients for the first equation and find the discriminant:
a = 1, b = −8, c = 12;
D = (−8) 2 − 4 1 12 = 64 − 48 = 16

So, the discriminant is positive, so the equation has two different roots. We analyze the second equation in the same way:
a = 5; b = 3; c = 7;
D \u003d 3 2 - 4 5 7 \u003d 9 - 140 \u003d -131.

The discriminant is negative, there are no roots. The last equation remains:
a = 1; b = -6; c = 9;
D = (−6) 2 − 4 1 9 = 36 − 36 = 0.

The discriminant is equal to zero - the root will be one.

Note that coefficients have been written out for each equation. Yes, it's long, yes, it's tedious - but you won't mix up the odds and don't make stupid mistakes. Choose for yourself: speed or quality.

By the way, if you “fill your hand”, after a while you will no longer need to write out all the coefficients. You will perform such operations in your head. Most people start doing this somewhere after 50-70 solved equations - in general, not so much.

The roots of a quadratic equation

Now let's move on to the solution. If the discriminant D > 0, the roots can be found using the formulas:

The basic formula for the roots of a quadratic equation

When D = 0, you can use any of these formulas - you get the same number, which will be the answer. Finally, if D< 0, корней нет — ничего считать не надо.

1. x 2 - 2x - 3 = 0;
2. 15 - 2x - x2 = 0;
3. x2 + 12x + 36 = 0.

First equation:
x 2 - 2x - 3 = 0 ⇒ a = 1; b = −2; c = -3;
D = (−2) 2 − 4 1 (−3) = 16.

D > 0 ⇒ the equation has two roots. Let's find them:

Second equation:
15 − 2x − x 2 = 0 ⇒ a = −1; b = −2; c = 15;
D = (−2) 2 − 4 (−1) 15 = 64.

D > 0 ⇒ the equation again has two roots. Let's find them

\[\begin(align) & ((x)_(1))=\frac(2+\sqrt(64))(2\cdot \left(-1 \right))=-5; \\ & ((x)_(2))=\frac(2-\sqrt(64))(2\cdot \left(-1 \right))=3. \\ \end(align)\]

Finally, the third equation:
x 2 + 12x + 36 = 0 ⇒ a = 1; b = 12; c = 36;
D = 12 2 − 4 1 36 = 0.

D = 0 ⇒ the equation has one root. Any formula can be used. For example, the first one:

As you can see from the examples, everything is very simple. If you know the formulas and be able to count, there will be no problems. Most often, errors occur when negative coefficients are substituted into the formula. Here, again, the technique described above will help: look at the formula literally, paint each step - and get rid of mistakes very soon.

Incomplete quadratic equations

It happens that the quadratic equation is somewhat different from what is given in the definition. For example:

1. x2 + 9x = 0;
2. x2 − 16 = 0.

It is easy to see that one of the terms is missing in these equations. Such quadratic equations are even easier to solve than standard ones: they do not even need to calculate the discriminant. So let's introduce a new concept:

The equation ax 2 + bx + c = 0 is called an incomplete quadratic equation if b = 0 or c = 0, i.e. the coefficient of the variable x or the free element is equal to zero.

Of course, it is entirely possible Hard case, when both of these coefficients are equal to zero: b \u003d c \u003d 0. In this case, the equation takes the form ax 2 \u003d 0. Obviously, such an equation has a single root: x \u003d 0.

Let's consider other cases. Let b \u003d 0, then we get an incomplete quadratic equation of the form ax 2 + c \u003d 0. Let's slightly transform it:

Since the arithmetic square root exists only from a non-negative number, the last equality only makes sense when (−c / a ) ≥ 0. Conclusion:

1. If an incomplete quadratic equation of the form ax 2 + c = 0 satisfies the inequality (−c / a ) ≥ 0, there will be two roots. The formula is given above;
2. If (−c / a )< 0, корней нет.

As you can see, the discriminant was not required - there are no complex calculations at all in incomplete quadratic equations. In fact, it is not even necessary to remember the inequality (−c / a ) ≥ 0. It is enough to express the value of x 2 and see what is on the other side of the equal sign. If there is a positive number, there will be two roots. If negative, there will be no roots at all.

Now let's deal with equations of the form ax 2 + bx = 0, in which the free element is equal to zero. Everything is simple here: there will always be two roots. It is enough to factorize the polynomial:

Taking the common factor out of the bracket

The product is equal to zero when at least one of the factors is equal to zero. This is where the roots come from. In conclusion, we will analyze several of these equations:

A task. Solve quadratic equations:

1. x2 − 7x = 0;
2. 5x2 + 30 = 0;
3. 4x2 − 9 = 0.

x 2 − 7x = 0 ⇒ x (x − 7) = 0 ⇒ x 1 = 0; x2 = −(−7)/1 = 7.

5x2 + 30 = 0 ⇒ 5x2 = -30 ⇒ x2 = -6. There are no roots, because the square cannot be equal to a negative number.

4x 2 − 9 = 0 ⇒ 4x 2 = 9 ⇒ x 2 = 9/4 ⇒ x 1 = 3/2 = 1.5; x 2 \u003d -1.5.

Kopyevskaya rural secondary school

10 Ways to Solve Quadratic Equations

Head: Patrikeeva Galina Anatolyevna,

mathematic teacher

s.Kopyevo, 2007

1. History of the development of quadratic equations

1.1 Quadratic equations in ancient Babylon

1.2 How Diophantus compiled and solved quadratic equations

1.3 Quadratic equations in India

1.4 Quadratic equations in al-Khwarizmi

1.5 Quadratic equations in Europe XIII - XVII centuries

1.6 About Vieta's theorem

2. Methods for solving quadratic equations

Conclusion

Literature

1. History of the development of quadratic equations

1.1 Quadratic equations in ancient Babylon

The need to solve equations not only of the first, but also of the second degree in ancient times was caused by the need to solve problems related to finding the areas of land and earthworks of a military nature, as well as the development of astronomy and mathematics itself. Quadratic equations were able to solve about 2000 BC. e. Babylonians.

Applying modern algebraic notation, we can say that in their cuneiform texts there are, in addition to incomplete ones, such, for example, complete quadratic equations:

X 2 + X = ¾; X 2 - X = 14,5

The rule for solving these equations, stated in the Babylonian texts, coincides essentially with the modern one, but it is not known how the Babylonians came to this rule. Almost all the cuneiform texts found so far give only problems with solutions stated in the form of recipes, with no indication of how they were found.

Despite the high level of development of algebra in Babylon, in cuneiform texts there is no concept of a negative number and common methods solutions of quadratic equations.

1.2 How Diophantus compiled and solved quadratic equations.

Diophantus' Arithmetic does not contain a systematic exposition of algebra, but it contains a systematic series of problems, accompanied by explanations and solved by drawing up equations of various degrees.

When compiling equations, Diophantus skillfully chooses unknowns to simplify the solution.

Here, for example, is one of his tasks.

Task 11."Find two numbers knowing that their sum is 20 and their product is 96"

Diophantus argues as follows: it follows from the condition of the problem that the desired numbers are not equal, since if they were equal, then their product would not be 96, but 100. Thus, one of them will be more than half of their sum, i.e. . 10+x, the other is smaller, i.e. 10's. The difference between them 2x .

Hence the equation:

(10 + x)(10 - x) = 96

100 - x 2 = 96

x 2 - 4 = 0 (1)

From here x = 2. One of the desired numbers is 12 , other 8 . Solution x = -2 for Diophantus does not exist, since Greek mathematics knew only positive numbers.

If we solve this problem by choosing one of the desired numbers as the unknown, then we will come to the solution of the equation

y(20 - y) = 96,

y 2 - 20y + 96 = 0. (2)

It is clear that Diophantus simplifies the solution by choosing the half-difference of the desired numbers as the unknown; he manages to reduce the problem to solving an incomplete quadratic equation (1).

1.3 Quadratic equations in India

Problems for quadratic equations are already found in the astronomical tract "Aryabhattam", compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scholar, Brahmagupta (7th century), expounded general rule solutions of quadratic equations reduced to a single canonical form:

ah 2+ b x = c, a > 0. (1)

In equation (1), the coefficients, except for a, can also be negative. Brahmagupta's rule essentially coincides with ours.

AT ancient india public competitions in solving difficult problems were common. In one of the old Indian books, the following is said about such competitions: “As the sun outshines the stars with its brilliance, so a learned person will outshine the glory of another in public meetings, proposing and solving algebraic problems.” Tasks were often dressed in poetic form.

Here is one of the problems of the famous Indian mathematician of the XII century. Bhaskara.

Task 13.

“A frisky flock of monkeys And twelve in vines ...

Having eaten power, had fun. They began to jump, hanging ...

Part eight of them in a square How many monkeys were there,

Having fun in the meadow. You tell me, in this flock?

Bhaskara's solution indicates that he knew about the two-valuedness of the roots of quadratic equations (Fig. 3).

The equation corresponding to problem 13 is:

( x /8) 2 + 12 = x

Bhaskara writes under the guise of:

x 2 - 64x = -768

and, to complete the left side of this equation to a square, he adds to both sides 32 2 , getting then:

x 2 - 64x + 32 2 = -768 + 1024,

(x - 32) 2 = 256,

x - 32 = ± 16,

x 1 = 16, x 2 = 48.

1.4 Quadratic equations in al-Khorezmi

Al-Khorezmi's algebraic treatise gives a classification of linear and quadratic equations. The author lists 6 types of equations, expressing them as follows:

1) "Squares are equal to roots", i.e. ax 2 + c = b X.

2) "Squares are equal to number", i.e. ax 2 = s.

3) "The roots are equal to the number", i.e. ah = s.

4) "Squares and numbers are equal to roots", i.e. ax 2 + c = b X.

5) "Squares and roots are equal to the number", i.e. ah 2+ bx = s.

6) "Roots and numbers are equal to squares", i.e. bx + c \u003d ax 2.

For al-Khwarizmi, who avoided the use of negative numbers, the terms of each of these equations are addends, not subtractions. In this case, equations that do not have positive solutions are obviously not taken into account. The author outlines the methods for solving these equations, using the methods of al-jabr and al-muqabala. His decisions, of course, do not completely coincide with ours. Not to mention the fact that it is purely rhetorical, it should be noted, for example, that when solving an incomplete quadratic equation of the first type

al-Khorezmi, like all mathematicians before the 17th century, does not take into account the zero solution, probably because it does not matter in specific practical problems. When solving complete quadratic equations, al-Khorezmi sets out the rules for solving, and then geometric proofs, using particular numerical examples.

Task 14.“The square and the number 21 are equal to 10 roots. Find the root" (assuming the root of the equation x 2 + 21 = 10x).

The author's solution goes something like this: divide the number of roots in half, you get 5, multiply 5 by itself, subtract 21 from the product, 4 remains. Take the root of 4, you get 2. Subtract 2 from 5, you get 3, this will be the desired root. Or add 2 to 5, which will give 7, this is also a root.

Treatise al - Khorezmi is the first book that has come down to us, in which the classification of quadratic equations is systematically stated and formulas for their solution are given.

1.5 Quadratic equations in Europe XIII - XVII centuries

Formulas for solving quadratic equations on the model of al - Khorezmi in Europe were first set forth in the "Book of the Abacus", written in 1202 by the Italian mathematician Leonardo Fibonacci. This voluminous work, which reflects the influence of mathematics, both the countries of Islam and Ancient Greece, differs in both completeness and clarity of presentation. The author independently developed some new algebraic examples of problem solving and was the first in Europe to approach the introduction of negative numbers. His book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many tasks from the "Book of the Abacus" passed into almost all European textbooks of the 16th - 17th centuries. and partly XVIII.

The general rule for solving quadratic equations reduced to a single canonical form:

x 2+ bx = with,

for all possible combinations of signs of the coefficients b , With was formulated in Europe only in 1544 by M. Stiefel.

Vieta has a general derivation of the formula for solving a quadratic equation, but Vieta recognized only positive roots. The Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 16th century. Take into account, in addition to positive, and negative roots. Only in the XVII century. Thanks to the work of Girard, Descartes, Newton and other scientists, the way to solve quadratic equations takes on a modern look.

1.6 About Vieta's theorem

The theorem expressing the relationship between the coefficients of a quadratic equation and its roots, bearing the name of Vieta, was formulated by him for the first time in 1591 as follows: “If B + D multiplied by A - A 2 , equals BD, then A equals AT and equal D ».

To understand Vieta, one must remember that BUT, like any vowel, meant for him the unknown (our X), the vowels AT, D- coefficients for the unknown. In the language of modern algebra, Vieta's formulation above means: if

(a + b )x - x 2 = ab ,

x 2 - (a + b )x + a b = 0,

x 1 = a, x 2 = b .

Expressing the relationship between the roots and coefficients of equations by general formulas written using symbols, Viet established uniformity in the methods of solving equations. However, the symbolism of Vieta is still far from modern look. He did not recognize negative numbers, and therefore, when solving equations, he considered only cases where all roots are positive.

2. Methods for solving quadratic equations

Quadratic equations are the foundation on which the majestic edifice of algebra rests. Quadratic equations find wide application when solving trigonometric, exponential, logarithmic, irrational and transcendental equations and inequalities. We all know how to solve quadratic equations with school bench(grade 8), before graduation.

The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Equations have been used by man since ancient times and since then their use has only increased. The discriminant allows you to solve any quadratic equations using the general formula, which has the following form:

The discriminant formula depends on the degree of the polynomial. The above formula is suitable for solving quadratic equations of the following form:

The discriminant has the following properties that you need to know:

* "D" is 0 when the polynomial has multiple roots (equal roots);

* "D" is a symmetric polynomial with respect to the roots of the polynomial and therefore is a polynomial in its coefficients; moreover, the coefficients of this polynomial are integers, regardless of the extension in which the roots are taken.

Suppose we are given a quadratic equation of the following form:

1 equation

According to the formula we have:

Since \, then the equation has 2 roots. Let's define them:

Where can I solve the equation through the discriminant online solver?

You can solve the equation on our website https: // site. Free online solver will allow you to solve an online equation of any complexity in seconds. All you have to do is just enter your data into the solver. You can also watch the video instruction and learn how to solve the equation on our website. And if you have any questions, you can ask them in our Vkontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.

Quadratic equation - easy to solve! *Further in the text "KU". Friends, it would seem that in mathematics it can be easier than solving such an equation. But something told me that many people have problems with him. I decided to see how many impressions Yandex gives per request per month. Here's what happened, take a look:

What does it mean? This means that about 70,000 people a month are looking for this information, and this is summer, 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 solve:

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:

By 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 this is so - a 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 to a certain kind equations.

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.