Quadratic equations d1. Solving quadratic equations: root formula, examples

Quadratic equations study in the 8th grade, 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:

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:

Because arithmetic Square root exists only from negative number, the last equality makes sense only for (−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:

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.

Yakupova M.I. 1

Smirnova Yu.V. one

1 Municipal budgetary educational institution secondary school No. 11

The text of the work is placed without images and formulas.
Full version work is available in the "Files of work" tab in PDF format

History of quadratic equations

Babylon

The need to solve equations not only of the first degree, but also of the second in ancient times was caused by the need to solve problems related to finding areas land plots, with the development of astronomy and mathematics itself. Quadratic equations were able to solve about 2000 BC. e. Babylonians. The rules for solving these equations set forth in the Babylonian texts essentially coincide with modern ones, but these texts lack the concept of a negative number and general methods for solving quadratic equations.

Ancient Greece

The solution of quadratic equations was also carried out in Ancient Greece scientists such as Diophantus, Euclid and Heron. Diophantus Diophantus of Alexandria was an ancient Greek mathematician who presumably lived in the 3rd century AD. The main work of Diophantus is "Arithmetic" in 13 books. Euclid. Euclid is an ancient Greek mathematician, the author of the first theoretical treatise on mathematics that has come down to us, Heron. Heron - Greek mathematician and engineer for the first time in Greece in the 1st century AD. gives a purely algebraic way of solving the quadratic equation

India

Problems for quadratic equations are already found in the astronomical treatise 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: ax2 + bx = c, a > 0. (1) In equation (1), the coefficients can also be negative. Brahmagupta's rule essentially coincides with ours. In 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 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.

"A frisky flock of monkeys

And twelve along the vines

They began to jump, hanging

Them squared part eight

How many monkeys were

Having fun in the meadow

You tell me, in this flock?

Bhaskara's solution indicates that the author was aware of the two-valuedness of the roots of quadratic equations. Bhaskar writes the equation corresponding to the problem under the form x2 - 64x = - 768 and, in order to complete the left side of this equation to a square, he adds 322 to both parts, then getting: x2 - b4x + 322 = -768 + 1024, (x - 32) 2 \u003d 256, x - 32 \u003d ± 16, x1 \u003d 16, x2 \u003d 48.

Quadratic equations in Europe XVII century

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, is distinguished by both completeness and clarity of presentation. The author independently developed some new algebraic examples 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. Derivation of the formula for solving a quadratic equation in general view Viet has, but Viet 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.

Definition of a quadratic equation

An equation of the form ax 2 + bx + c = 0, where a, b, c are numbers, is called a square equation.

Coefficients of a quadratic equation

The numbers a, b, c are the coefficients of the quadratic equation. a is the first coefficient (before x²), a ≠ 0; b is the second coefficient (before x); c is the free term (without x).

Which of these equations are not quadratic?

1. 4x² + 4x + 1 \u003d 0; 2. 5x - 7 \u003d 0; 3. - x² - 5x - 1 \u003d 0; 4. 2/x² + 3x + 4 = 0;5. ¼ x² - 6x + 1 \u003d 0; 6. 2x² = 0;

7. 4x² + 1 \u003d 0; 8. x² - 1 / x \u003d 0; 9. 2x² - x \u003d 0; 10. x² -16 = 0;11. 7x² + 5x = 0;12. -8х²= 0;13. 5x³ +6x -8= 0.

Types of quadratic equations

Name	General view of the equation	Feature (what coefficients)	Equation Examples
	ax2 + bx + c = 0	a, b, c - numbers other than 0	1/3x 2 + 5x - 1 = 0
Incomplete
			x 2 - 1/5x = 0
Given	x 2 + bx + c = 0		x 2 - 3x + 5 = 0

A reduced quadratic equation is called, in which the leading coefficient is equal to one. Such an equation can be obtained by dividing the entire expression by the leading coefficient a:

x 2 + px + q =0, p = b/a, q = c/a

A quadratic equation is said to be complete if all of its coefficients are nonzero.

Such a quadratic equation is called incomplete if at least one of the coefficients, except for the highest one (either the second coefficient or the free term), is equal to zero.

Ways to solve quadratic equations

I way. General formula for calculating roots

To find the roots of a quadratic equation ax 2 + b + c = 0 In general, the following algorithm should be used:

Calculate the value of the discriminant of the quadratic equation: this is the expression for it D= b 2 - 4ac

Derivation of the formula:

Note: it is obvious that the formula for a root of multiplicity 2 is a special case of the general formula, it is obtained by substituting the equality D=0 into it, and the conclusion about the absence of real roots at D0, and (displaystyle (sqrt (-1))=i) = i.

The described method is universal, but it is far from the only one. The solution of one equation can be approached in different ways, the preferences usually depend on the solver itself. In addition, often for this some of the methods turns out to be much more elegant, simpler, less time-consuming than the standard one.

II way. The roots of a quadratic equation with an even coefficient b III way. Solving incomplete quadratic equations

IV way. Using partial ratios of coefficients

There are special cases of quadratic equations in which the coefficients are in proportion to each other, which makes it much easier to solve them.

The roots of a quadratic equation in which the sum of the leading coefficient and the free term is equal to the second coefficient

If in a quadratic equation ax 2 + bx + c = 0 the sum of the first coefficient and the free term is equal to the second coefficient: a+b=c, then its roots are -1 and the number opposite to the ratio of the free term to the leading coefficient ( -c/a).

Hence, before solving any quadratic equation, one should check the possibility of applying this theorem to it: compare the sum of the leading coefficient and the free term with the second coefficient.

The roots of a quadratic equation whose sum of all coefficients is zero

If in a quadratic equation the sum of all its coefficients is equal to zero, then the roots of such an equation are 1 and the ratio of the free term to the leading coefficient ( c/a).

Hence, before solving the equation standard methods, you should check the applicability of this theorem to it: add up all the coefficients of this equation and see if this sum is not equal to zero.

V way. Decomposition of a square trinomial into linear factors

If a trinomial of the form (displaystyle ax^(2)+bx+c(anot =0))ax 2 + bx + c(a ≠ 0) can somehow be represented as a product of linear factors (displaystyle (kx+m)(lx+n)=0)(kx + m)(lx + n), then we can find the roots of the equation ax 2 + bx + c = 0- they will be -m / k and n / l, indeed, because (displaystyle (kx+m)(lx+n)=0Longleftrightarrow kx+m=0cup lx+n=0)(kx + m)(lx + n) = 0 kx + mUlx + n, and by solving the indicated linear equations, we obtain the above. Note that a square trinomial is not always decomposed into linear factors with real coefficients: this is possible if the equation corresponding to it has real roots.

Consider some special cases

Using the formula for the square of the sum (difference)

If a square trinomial has the form (displaystyle (ax)^(2)+2abx+b^(2))ax 2 + 2abx + b 2 , then applying the above formula to it, we can factor it into linear factors and, therefore, find roots:

(ax) 2 + 2abx + b 2 = (ax + b) 2

Selection of the full square of the sum (difference)

Also, the named formula is used using the method called "selection of the full square of the sum (difference)". In relation to the given quadratic equation with the notation introduced earlier, this means the following:

Note: if you notice, this formula coincides with the one proposed in the “Roots of the reduced quadratic equation” section, which, in turn, can be obtained from the general formula (1) by substituting the equality a=1. This fact is not just a coincidence: by the described method, after some additional reasoning, it is possible to deduce and general formula, as well as to prove the properties of the discriminant.

VI way. Using direct and inverse Vieta theorem

Vieta's direct theorem (see below in the section of the same name) and its inverse theorem allow us to solve the reduced quadratic equations orally without resorting to rather cumbersome calculations using formula (1).

According to the inverse theorem, any pair of numbers (number) (displaystyle x_(1),x_(2)) x 1 , x 2 being the solution of the system of equations below, are the roots of the equation

In the general case, that is, for a non-reduced quadratic equation ax 2 + bx + c = 0

x 1 + x 2 \u003d -b / a, x 1 * x 2 \u003d c / a

A direct theorem will help you verbally select numbers that satisfy these equations. With its help, you can determine the signs of the roots without knowing the roots themselves. To do this, follow the rule:

1) if the free term is negative, then the roots have a different sign, and the largest absolute value of the roots is the sign opposite to the sign of the second coefficient of the equation;

2) if the free term is positive, then both roots have the same sign, and this is the opposite sign of the second coefficient.

7th way. Transfer method

The so-called "transfer" method makes it possible to reduce the solution of non-reduced and non-transformable equations to the form of equations reduced with integer coefficients by dividing them by the leading coefficient of equations to the solution of equations reduced with integer coefficients. It is as follows:

Next, the equation is solved orally in the manner described above, then they return to the original variable and find the roots of the equations (displaystyle y_(1)=ax_(1)) y 1 = ax 1 and y 2 = ax 2 .(displaystyle y_(2)=ax_(2))

geometric sense

The graph of a quadratic function is a parabola. The solutions (roots) of a quadratic equation are the abscissas of the points of intersection of the parabola with the abscissa axis. If the parabola described quadratic function, does not intersect with the x-axis, the equation has no real roots. If the parabola intersects the x-axis at one point (at the vertex of the parabola), the equation has one real root (the equation is also said to have two coinciding roots). If the parabola intersects the x-axis at two points, the equation has two real roots (see image on the right.)

If coefficient (displaystyle a) a positive, the branches of the parabola are directed upwards and vice versa. If the coefficient (display style b) bpositive (when positive (displaystyle a) a, if negative, vice versa), then the vertex of the parabola lies in the left half-plane and vice versa.

Application of quadratic equations in life

The quadratic equation is widespread. It is used in many calculations, structures, sports, and also around us.

Consider and give some examples of the application of the quadratic equation.

Sport. High jumps: during the run-up of the jumper for the most accurate hit on the repulsion bar and high flight use the calculations associated with the parabola.

Also, similar calculations are needed in throwing. The flight range of an object depends on a quadratic equation.

Astronomy. The trajectory of the planets can be found using a quadratic equation.

Airplane flight. The takeoff of an aircraft is the main component of the flight. Here the calculation is taken for a small resistance and takeoff acceleration.

Also, quadratic equations are used in various economic disciplines, in programs for processing sound, video, vector and raster graphics.

Conclusion

As a result of the work done, it turned out that quadratic equations attracted scientists in ancient times, they already encountered them when solving some problems and tried to solve them. Considering various ways solving quadratic equations, I came to the conclusion that not all of them are simple. In my opinion the most the best way solving quadratic equations is a solution by formulas. Formulas are easy to remember, this method is universal. The hypothesis that equations are widely used in life and mathematics was confirmed. Having studied the topic, I learned a lot interesting facts about quadratic equations, their use, application, types, solutions. And I will continue to study them with pleasure. I hope this will help me do well in my exams.

List of used literature

Site materials:

Wikipedia

Open lesson.rf

Reference book elementary mathematics Vygodsky M. Ya.

AT modern society the ability to operate with equations containing a squared variable can be useful in many areas of activity and is widely used in practice in scientific and technical developments. This can be evidenced by the design of sea and river vessels, aircraft and missiles. With the help of such calculations, the trajectories of movement of the most different bodies, including space objects. Examples with the solution of quadratic equations are used not only in economic forecasting, in the design and construction of buildings, but also in the most ordinary everyday circumstances. They may be needed on camping trips, at sports events, in stores when shopping and in other very common situations.

Let's break the expression into component factors

The degree of an equation is determined by the maximum value of the degree of the variable that the given expression contains. If it is equal to 2, then such an equation is called a quadratic equation.

If we speak in the language of formulas, then these expressions, no matter how they look, can always be brought to the form when the left side of the expression consists of three terms. Among them: ax 2 (that is, a variable squared with its coefficient), bx (an unknown without a square with its coefficient) and c (free component, that is, an ordinary number). All this on the right side is equal to 0. In the case when such a polynomial does not have one of its constituent terms, with the exception of ax 2, it is called an incomplete quadratic equation. Examples with the solution of such problems, in which the value of the variables is not difficult to find, should be considered first.

If the expression looks like it has two terms on the right side of the expression, more precisely ax 2 and bx, it is easiest to find x by bracketing the variable. Now our equation will look like this: x(ax+b). Further, it becomes obvious that either x=0, or the problem is reduced to finding a variable from the following expression: ax+b=0. This is dictated by one of the properties of multiplication. The rule says that the product of two factors results in 0 only if one of them is zero.

Example

x=0 or 8x - 3 = 0

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

Equations of this kind can describe the movement of bodies under the action of gravity, which began to move from a certain point, taken as the origin. Here the mathematical notation takes the following form: y = v 0 t + gt 2 /2. Substituting the necessary values, equating right side 0 and finding possible unknowns, one can find out the time elapsed from the moment the body rises to the moment it falls, as well as many other quantities. But we will talk about this later.

Factoring an Expression

The rule described above makes it possible to solve these problems and in more difficult cases. Consider examples with the solution of quadratic equations of this type.

X2 - 33x + 200 = 0

This square trinomial is complete. First, we transform the expression and decompose it into factors. There are two of them: (x-8) and (x-25) = 0. As a result, we have two roots 8 and 25.

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

For example: 2x 3 + 2x 2 - 18x - 18 = 0. When factoring the right side into factors with a variable, there are three of them, that is, (x + 1), (x-3) and (x + 3).

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

Extracting the square root

Another case incomplete equation the second order is an expression expressed in the language of letters in such a way that the right side is built from the components ax 2 and c. Here, to obtain the value of the variable, the free term is transferred to the right side, and after that, the square root is extracted from both sides of the equality. It should be noted that in this case There are usually two roots of an equation. The only exceptions are equalities that do not contain the term c at all, where the variable is equal to zero, as well as variants of expressions when the right side turns out to be negative. In the latter case, there are no solutions at all, since the above actions cannot be performed with roots. Examples of solutions to quadratic equations of this type should be considered.

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

Calculation of the area of ​​land

The need for this kind of calculations appeared in ancient times, because the development of mathematics in those distant times was largely due to the need to determine the areas and perimeters of land plots with the greatest accuracy.

We should also consider examples with the solution of quadratic equations compiled on the basis of problems of this kind.

So, let's say there is a rectangular piece of land, the length of which is 16 meters more than the width. You should find the length, width and perimeter of the site, if it is known that its area is 612 m 2.

Getting down to business, at first we will make the necessary equation. Let's denote the width of the section as x, then its length will be (x + 16). It follows from what has been written that the area is determined by the expression x (x + 16), which, according to the condition of our problem, is 612. This means that x (x + 16) \u003d 612.

The solution of complete quadratic equations, and this expression is just that, cannot be done in the same way. Why? Although the left side of it still contains two factors, the product of them is not equal to 0 at all, so other methods are used here.

Discriminant

First of all, we make the necessary transformations, then appearance this expression will look like this: x 2 + 16x - 612 = 0. This means that we have received an expression in the form corresponding to the previously specified standard, where a=1, b=16, c=-612.

This can be an example of solving quadratic equations through the discriminant. Here necessary calculations produced according to the scheme: D = b 2 - 4ac. This auxiliary value not only makes it possible to find the desired values ​​in the second-order equation, it determines the number options. In case D>0, there are two of them; for D=0 there is one root. In case D<0, никаких шансов для решения у уравнения вообще не имеется.

About roots and their formula

In our case, the discriminant is: 256 - 4(-612) = 2704. This indicates that our problem has an answer. If you know, to, the solution of quadratic equations must be continued using the formula below. It allows you to calculate the roots.

This means that in the presented case: x 1 =18, x 2 =-34. The second option in this dilemma cannot be a solution, because the size of the land plot cannot be measured in negative values, which means that x (that is, the width of the plot) is 18 m. From here we calculate the length: 18+16=34, and the perimeter 2(34+ 18) = 104 (m 2).

Examples and tasks

We continue the study of quadratic equations. Examples and a detailed solution of several of them will be given below.

1) 15x2 + 20x + 5 = 12x2 + 27x + 1

Let's transfer everything to the left side of the equality, make a transformation, that is, we get the form of the equation, which is usually called the standard one, and equate it to zero.

15x 2 + 20x + 5 - 12x 2 - 27x - 1 = 0

Having added similar ones, we determine the discriminant: D \u003d 49 - 48 \u003d 1. So our equation will have two roots. We calculate them according to the above formula, which means that the first of them will be equal to 4/3, and the second 1.

2) Now we will reveal riddles of a different kind.

Let's find out if there are roots x 2 - 4x + 5 = 1 here at all? To obtain an exhaustive answer, we bring the polynomial to the corresponding familiar form and calculate the discriminant. In this example, it is not necessary to solve the quadratic equation, because the essence of the problem is not at all in this. In this case, D \u003d 16 - 20 \u003d -4, which means that there really are no roots.

Vieta's theorem

It is convenient to solve quadratic equations through the above formulas and the discriminant, when the square root is extracted from the value of the latter. But this does not always happen. However, there are many ways to get the values ​​of variables in this case. Example: solving quadratic equations using Vieta's theorem. It is named after a man who lived in 16th-century France and had a brilliant career thanks to his mathematical talent and connections at court. His portrait can be seen in the article.

The pattern that the famous Frenchman noticed was as follows. He proved that the sum of the roots of the equation is equal to -p=b/a, and their product corresponds to q=c/a.

Now let's look at specific tasks.

3x2 + 21x - 54 = 0

For simplicity, let's transform the expression:

x 2 + 7x - 18 = 0

Using the Vieta theorem, this will give us the following: the sum of the roots is -7, and their product is -18. From here we get that the roots of the equation are the numbers -9 and 2. Having made a check, we will make sure that these values ​​of the variables really fit into the expression.

Graph and Equation of a Parabola

The concepts of a quadratic function and quadratic equations are closely related. Examples of this have already been given previously. Now let's look at some mathematical puzzles in a little more detail. Any equation of the described type can be represented visually. Such a dependence, drawn in the form of a graph, is called a parabola. Its various types are shown in the figure below.

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

Visual representations of functions help to solve any equations, including quadratic ones. This method is called graphic. And the value of the x variable is the abscissa coordinate at the points where the graph line intersects with 0x. The coordinates of the vertex can be found by the formula just given x 0 = -b / 2a. And, substituting the resulting value into the original equation of the function, you can find out y 0, that is, the second coordinate of the parabola vertex belonging to the y-axis.

The intersection of the branches of the parabola with the abscissa axis

There are a lot of examples with the solution of quadratic equations, but there are also general patterns. Let's consider them. It is clear that the intersection of the graph with the 0x axis for a>0 is possible only if y 0 takes negative values. And for a<0 координата у 0 должна быть положительна. Для указанных вариантов D>0. Otherwise D<0. А когда D=0, вершина параболы расположена непосредственно на оси 0х.

From the graph of a parabola, you can also determine the roots. The reverse is also true. That is, if it is not easy to get a visual representation of a quadratic function, you can equate the right side of the expression to 0 and solve the resulting equation. And knowing the points of intersection with the 0x axis, it is easier to plot.

From the history

With the help of equations containing a squared variable, in the old days, not only did mathematical calculations and determined the area of ​​\u200b\u200bgeometric shapes. The ancients needed such calculations for grandiose discoveries in the field of physics and astronomy, as well as for making astrological forecasts.

As modern scientists suggest, the inhabitants of Babylon were among the first to solve quadratic equations. It happened four centuries before the advent of our era. Of course, their calculations were fundamentally different from those currently accepted and turned out to be much more primitive. For example, Mesopotamian mathematicians had no idea about the existence of negative numbers. They were also unfamiliar with other subtleties of those known to any student of our time.

Perhaps even earlier than the scientists of Babylon, the sage from India, Baudhayama, took up the solution of quadratic equations. This happened about eight centuries before the advent of the era of Christ. True, the second-order equations, the methods for solving which he gave, were the simplest. In addition to him, Chinese mathematicians were also interested in similar questions in the old days. In Europe, quadratic equations began to be solved only at the beginning of the 13th century, but later they were used in their work by such great scientists as Newton, Descartes and many others.

5x (x - 4) = 0

5 x = 0 or x - 4 = 0

x = ± √ 25/4

Having learned to solve equations of the first degree, of course, I want to work with others, in particular, with equations of the second degree, which are otherwise called quadratic.

Quadratic equations are equations of the type ax² + bx + c = 0, where the variable is x, the numbers will be - a, b, c, where a is not equal to zero.

If in a quadratic equation one or the other coefficient (c or b) is equal to zero, then this equation will refer to an incomplete quadratic equation.

How to solve an incomplete quadratic equation if students have only been able to solve equations of the first degree so far? Consider incomplete quadratic equations of various types and simple ways to solve them.

a) If the coefficient c is equal to 0, and the coefficient b is not equal to zero, then ax ² + bx + 0 = 0 is reduced to an equation of the form ax ² + bx = 0.

To solve such an equation, you need to know the formula for solving an incomplete quadratic equation, which consists in decomposing the left side of it into factors and later using the condition that the product is equal to zero.

For example, 5x ² - 20x \u003d 0. We factor out the left side of the equation, while performing the usual mathematical operation: taking the common factor out of brackets

5x (x - 4) = 0

We use the condition that the products are equal to zero.

5 x = 0 or x - 4 = 0

The answer will be: the first root is 0; the second root is 4.

b) If b \u003d 0, and the free term is not equal to zero, then the equation ax ² + 0x + c \u003d 0 is reduced to an equation of the form ax ² + c \u003d 0. Solve equations in two ways: a) decomposing the polynomial of the equation on the left side into factors ; b) using the properties of the arithmetic square root. Such an equation is solved by one of the methods, for example:

x = ± √ 25/4

x = ± 5/2. The answer is: the first root is 5/2; the second root is - 5/2.

c) If b is equal to 0 and c is equal to 0, then ax² + 0 + 0 = 0 reduces to an equation of the form ax² = 0. In such an equation, x will be equal to 0.

As you can see, incomplete quadratic equations can have at most two roots.

In continuation of the topic “Solving Equations”, the material in this article will introduce you to quadratic equations.

Let's consider everything in detail: the essence and notation of a quadratic equation, set the accompanying terms, analyze the scheme for solving incomplete and complete equations, get acquainted with the formula of roots and the discriminant, establish connections between roots and coefficients, and of course we will give a visual solution of practical examples.

Yandex.RTB R-A-339285-1

Quadratic equation, its types

Definition 1

Quadratic equation is the equation written as a x 2 + b x + c = 0, where x– variable, a , b and c are some numbers, while a is not zero.

Often, quadratic equations are also called equations of the second degree, since in fact a quadratic equation is an algebraic equation of the second degree.

Let's give an example to illustrate the given definition: 9 x 2 + 16 x + 2 = 0 ; 7, 5 x 2 + 3, 1 x + 0, 11 = 0, etc. are quadratic equations.

Definition 2

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

For example, in the quadratic equation 6 x 2 - 2 x - 11 = 0 the highest coefficient is 6 , the second coefficient is − 2 , and the free term is equal to − 11 . Let us pay attention to the fact that when the coefficients b and/or c are negative, then the shorthand form is used 6 x 2 - 2 x - 11 = 0, but not 6 x 2 + (− 2) x + (− 11) = 0.

Let us also clarify this aspect: if the coefficients a and/or b equal 1 or − 1 , then they may not take an explicit part in writing the quadratic equation, which is explained by the peculiarities of writing the indicated numerical coefficients. For example, in the quadratic equation y 2 − y + 7 = 0 the senior coefficient is 1 and the second coefficient is − 1 .

Reduced and non-reduced quadratic equations

According to the value of the first coefficient, quadratic equations are divided into reduced and non-reduced.

Definition 3

Reduced quadratic equation is a quadratic equation where the leading coefficient is 1 . For other values ​​of the leading coefficient, the quadratic equation is unreduced.

Here are some examples: quadratic equations x 2 − 4 · x + 3 = 0 , x 2 − x − 4 5 = 0 are reduced, in each of which the leading coefficient is 1 .

9 x 2 - x - 2 = 0- unreduced quadratic equation, where the first coefficient is different from 1 .

Any unreduced quadratic equation can be converted into a reduced equation by dividing both its parts by the first coefficient (equivalent transformation). The transformed equation will have the same roots as the given non-reduced equation or will also have no roots at all.

Consideration of a specific example will allow us to clearly demonstrate the transition from an unreduced quadratic equation to a reduced one.

Example 1

Given the equation 6 x 2 + 18 x − 7 = 0 . It is necessary to convert the original equation into the reduced form.

Decision

According to the above scheme, we divide both parts of the original equation by the leading coefficient 6 . Then we get: (6 x 2 + 18 x - 7) : 3 = 0: 3, and this is the same as: (6 x 2) : 3 + (18 x) : 3 − 7: 3 = 0 and further: (6: 6) x 2 + (18: 6) x − 7: 6 = 0 . From here: x 2 + 3 x - 1 1 6 = 0 . Thus, an equation equivalent to the given one is obtained.

Answer: x 2 + 3 x - 1 1 6 = 0 .

Complete and incomplete quadratic equations

Let us turn to the definition of a quadratic equation. In it, we specified that a ≠ 0. A similar condition is necessary for the equation a x 2 + b x + c = 0 was exactly square, since a = 0 it essentially transforms into linear equation b x + c = 0.

In the case where the coefficients b and c are equal to zero (which is possible, both individually and jointly), the quadratic equation is called incomplete.

Definition 4

Incomplete quadratic equation is a quadratic equation a x 2 + b x + c \u003d 0, where at least one of the coefficients b and c(or both) is zero.

Complete quadratic equation is a quadratic equation in which all numerical coefficients are not equal to zero.

Let's discuss why the types of quadratic equations are given precisely such names.

For b = 0, the quadratic equation takes the form a x 2 + 0 x + c = 0, which is the same as a x 2 + c = 0. At c = 0 the quadratic equation is written as a x 2 + b x + 0 = 0, which is equivalent a x 2 + b x = 0. At b = 0 and c = 0 the equation will take the form a x 2 = 0. The equations that we have obtained 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 at once. Actually, this fact gave the name to this type of equations - incomplete.

For example, x 2 + 3 x + 4 = 0 and − 7 x 2 − 2 x + 1, 3 = 0 are complete quadratic equations; x 2 \u003d 0, − 5 x 2 \u003d 0; 11 x 2 + 2 = 0 , − x 2 − 6 x = 0 are incomplete quadratic equations.

Solving incomplete quadratic equations

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

• a x 2 = 0, coefficients correspond to such an equation b = 0 and c = 0 ;
• a x 2 + c \u003d 0 for b \u003d 0;
• a x 2 + b x = 0 for c = 0 .

Consider successively the solution of each type of incomplete quadratic equation.

Solution of the equation a x 2 \u003d 0

As already mentioned above, such an equation corresponds to the coefficients b and c, equal to zero. The equation a x 2 = 0 can be converted into an equivalent equation x2 = 0, which we get by dividing both sides of the original equation by the number a, not equal to zero. The obvious fact is that the root of the equation x2 = 0 is zero because 0 2 = 0 . This equation has no other roots, which is explained by the properties of the degree: for any number p , not equal to zero, the inequality is true p2 > 0, from which it follows that when p ≠ 0 equality p2 = 0 will never be reached.

Definition 5

Thus, for the incomplete quadratic equation a x 2 = 0, there is a single root x=0.

Example 2

For example, let's solve an incomplete quadratic equation − 3 x 2 = 0. It is equivalent to the equation x2 = 0, its only root is x=0, then the original equation has a single root - zero.

The solution is summarized as follows:

− 3 x 2 \u003d 0, x 2 \u003d 0, x \u003d 0.

Solution of the equation a x 2 + c \u003d 0

Next in line is the solution of incomplete quadratic equations, where b \u003d 0, c ≠ 0, that is, equations of the form a x 2 + c = 0. Let's transform this equation by transferring the term from one side of the equation to the other, changing the sign to the opposite and dividing both sides of the equation by a number that is not equal to zero:

• endure c to the right side, which gives the equation a x 2 = − c;
• divide both sides of the equation by a, we get as a result x = - c a .

Our transformations are equivalent, respectively, the resulting equation is also equivalent to the original one, and this fact makes it possible to draw a conclusion about the roots of the equation. From what are the values a and c depends on the value of the expression - c a: it can have a minus sign (for example, if a = 1 and c = 2, then - c a = - 2 1 = - 2) or a plus sign (for example, if a = -2 and c=6, then - c a = - 6 - 2 = 3); it is not equal to zero because c ≠ 0. Let us dwell in more detail on situations when - c a< 0 и - c a > 0 .

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

Everything is different when - c a > 0: remember the square root, and it will become obvious that the root of the equation x 2 \u003d - c a will be the number - c a, since - c a 2 \u003d - c a. It is easy to understand that the number - - c a - is also the root of the equation x 2 = - c a: indeed, - - c a 2 = - c a .

The equation will have no other roots. We can demonstrate this using the opposite method. First, let's set the notation of the roots found above as x 1 and − x 1. Let's assume that the equation x 2 = - c a also has a root x2, which is different from the roots x 1 and − x 1. We know that by substituting into the equation instead of x its roots, we transform the equation into a fair numerical equality.

For x 1 and − x 1 write: x 1 2 = - c a , and for x2- x 2 2 \u003d - c a. Based on the properties of numerical equalities, we subtract one true equality from another term by term, which will give us: x 1 2 − x 2 2 = 0. Use the properties of number operations to rewrite the last equality as (x 1 - x 2) (x 1 + x 2) = 0. It is known that the product of two numbers is zero if and only if at least one of the numbers is zero. From what has been said, it follows that x1 − x2 = 0 and/or x1 + x2 = 0, which is the same x2 = x1 and/or x 2 = − x 1. An obvious contradiction arose, because at first it was agreed that the root of the equation x2 differs from x 1 and − x 1. So, we have proved that the equation has no other roots than x = - c a and x = - - c a .

We summarize all the arguments above.

Definition 6

Incomplete quadratic equation a x 2 + c = 0 is equivalent to the equation x 2 = - c a , which:

• will not have roots at - c a< 0 ;
• will have two roots x = - c a and x = - - c a when - c a > 0 .

Let us give examples of solving equations a x 2 + c = 0.

Example 3

Given a quadratic equation 9 x 2 + 7 = 0 . It is necessary to find its solution.

Decision

We transfer the free term to the right side of the equation, then the equation will take the form 9 x 2 \u003d - 7.
We divide both sides of the resulting equation by 9 , we come to x 2 = - 7 9 . On the right side we see a number with a minus sign, which means: the given equation has no roots. Then the original incomplete quadratic equation 9 x 2 + 7 = 0 will not have roots.

Answer: the equation 9 x 2 + 7 = 0 has no roots.

Example 4

It is necessary to solve the equation − x2 + 36 = 0.

Decision

Let's move 36 to the right side: − x 2 = − 36.
Let's divide both parts into − 1 , we get x2 = 36. On the right side is a positive number, from which we can conclude that x = 36 or x = - 36 .
We extract the root and write the final result: an incomplete quadratic equation − x2 + 36 = 0 has two roots x=6 or x = -6.

Answer: x=6 or x = -6.

Solution of the equation a x 2 +b x=0

Let us analyze the third kind of incomplete quadratic equations, when c = 0. To find a solution to an incomplete quadratic equation a x 2 + b x = 0, we use the factorization method. Let us factorize the polynomial, which is on the left side of the equation, taking the common factor out of brackets x. This step will make it possible to transform the original incomplete quadratic equation into its equivalent x (a x + b) = 0. And this equation, in turn, is equivalent to the set of equations x=0 and a x + b = 0. The equation a x + b = 0 linear, and its root: x = − b a.

Definition 7

Thus, the incomplete quadratic equation a x 2 + b x = 0 will have two roots x=0 and x = − b a.

Let's consolidate the material with an example.

Example 5

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

Decision

Let's take out x outside the brackets and get the equation x · 2 3 · x - 2 2 7 = 0 . This equation is equivalent to the equations x=0 and 2 3 x - 2 2 7 = 0 . Now you should solve the resulting linear equation: 2 3 · x = 2 2 7 , x = 2 2 7 2 3 .

Briefly, we write the solution of the equation as follows:

2 3 x 2 - 2 2 7 x = 0 x 2 3 x - 2 2 7 = 0

x = 0 or 2 3 x - 2 2 7 = 0

x = 0 or x = 3 3 7

Answer: x = 0 , x = 3 3 7 .

Discriminant, formula of the roots of a quadratic equation

To find a solution to quadratic equations, there is a root formula:

Definition 8

x = - b ± D 2 a, where D = b 2 − 4 a c is the so-called discriminant of a quadratic equation.

Writing x \u003d - b ± D 2 a essentially means that x 1 \u003d - b + D 2 a, x 2 \u003d - b - D 2 a.

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

Derivation of the formula of the roots of a quadratic equation

Suppose we are faced with the task of solving a quadratic equation a x 2 + b x + c = 0. Let's carry out a number of equivalent transformations:

• divide both sides of the equation by the number a, different from zero, we obtain the reduced quadratic equation: x 2 + b a x + c a \u003d 0;
• select the full square on the left side of the resulting equation:
  x 2 + b a x + c a = x 2 + 2 b 2 a x + b 2 a 2 - b 2 a 2 + c a = = x + b 2 a 2 - b 2 a 2 + c a
  After this, the equation will take the form: x + b 2 a 2 - b 2 a 2 + c a \u003d 0;
• now it is possible to transfer the last two terms to the right side, changing the sign to the opposite, after which we get: x + b 2 · a 2 = b 2 · a 2 - c a ;
• finally, we transform the expression written on the right side of the last equality:
  b 2 a 2 - c a \u003d b 2 4 a 2 - c a \u003d b 2 4 a 2 - 4 a c 4 a 2 \u003d b 2 - 4 a c 4 a 2.

Thus, we have come to the equation x + b 2 a 2 = b 2 - 4 a c 4 a 2 , which is equivalent to the original equation a x 2 + b x + c = 0.

We discussed the solution of such equations in the previous paragraphs (the solution of incomplete quadratic equations). The experience already gained makes it possible to draw a conclusion regarding the roots of the equation x + b 2 a 2 = b 2 - 4 a c 4 a 2:

• for b 2 - 4 a c 4 a 2< 0 уравнение не имеет действительных решений;
• for b 2 - 4 · a · c 4 · a 2 = 0, the equation has the form x + b 2 · a 2 = 0, then x + b 2 · a = 0.

From here, the only root x = - b 2 · a is obvious;

• for b 2 - 4 a c 4 a 2 > 0, the correct one is: x + b 2 a = b 2 - 4 a c 4 a 2 or x = b 2 a - b 2 - 4 a c 4 a 2 , which is the same as x + - b 2 a = b 2 - 4 a c 4 a 2 or x = - b 2 a - b 2 - 4 a c 4 a 2 , i.e. the equation has two roots.

It is possible to conclude that the presence or absence of the roots of the equation x + b 2 a 2 = b 2 - 4 a c 4 a 2 (and hence the original equation) depends on the sign of the expression b 2 - 4 a c 4 · a 2 written on the right side. And the sign of this expression is given by the sign of the numerator, (the denominator 4 a 2 will always be positive), that is, the sign of the expression b 2 − 4 a c. This expression b 2 − 4 a c a name is given - the discriminant of a quadratic equation and the letter D is defined as its designation. Here you can write down the essence of the discriminant - by its value and sign, they conclude whether the quadratic equation will have real roots, and, if so, how many roots - one or two.

Let's return to the equation x + b 2 a 2 = b 2 - 4 a c 4 a 2 . Let's rewrite it using the discriminant notation: x + b 2 · a 2 = D 4 · a 2 .

Let's recap the conclusions:

Definition 9

• at D< 0 the equation has no real roots;
• at D=0 the equation has a single root x = - b 2 · a ;
• at D > 0 the equation has two roots: x \u003d - b 2 a + D 4 a 2 or x \u003d - b 2 a - D 4 a 2. Based on the properties of radicals, these roots can be written as: x \u003d - b 2 a + D 2 a or - b 2 a - D 2 a. And when we open the modules and reduce the fractions to a common denominator, we get: x \u003d - b + D 2 a, x \u003d - b - D 2 a.

So, the result of our reasoning was the derivation of the formula for the roots of the quadratic equation:

x = - b + D 2 a , x = - b - D 2 a , discriminant D calculated by the formula D = b 2 − 4 a c.

These formulas make it possible, when the discriminant is greater than zero, to determine both real roots. When the discriminant is zero, applying both formulas will give the same root as the only solution to the quadratic equation. In the case when the discriminant is negative, trying to use the quadratic root formula, we will be faced with the need to extract the square root of a negative number, which will take us beyond real numbers. With a negative discriminant, the quadratic equation will not have real roots, but a pair of complex conjugate roots is possible, determined by the same root formulas we obtained.

Algorithm for solving quadratic equations using root formulas

It is possible to solve a quadratic equation by immediately using the root formula, but basically this is done if necessary to find complex roots.

In the bulk of cases, the search is usually meant not for complex, but for real roots of a quadratic equation. Then it is optimal, before using the formulas for the roots of the quadratic equation, first to determine the discriminant and make sure that it is not negative (otherwise we will conclude that the equation has no real roots), and then proceed to calculate the value of the roots.

The reasoning above makes it possible to formulate an algorithm for solving a quadratic equation.

Definition 10

To solve a quadratic equation a x 2 + b x + c = 0, necessary:

• according to the formula D = b 2 − 4 a c find the value of the discriminant;
• at D< 0 сделать вывод об отсутствии у квадратного уравнения действительных корней;
• for D = 0 find the only root of the equation by the formula x = - b 2 · a ;
• for D > 0, determine two real roots of the quadratic equation by the formula x = - b ± D 2 · a.

Note that when the discriminant is zero, you can use the formula x = - b ± D 2 · a , it will give the same result as the formula x = - b 2 · a .

Consider examples.

Examples of solving quadratic equations

We present the solution of examples for various values ​​of the discriminant.

Example 6

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

Decision

We write the numerical coefficients of the quadratic equation: a \u003d 1, b \u003d 2 and c = − 6. Next, we act according to the algorithm, i.e. Let's start calculating the discriminant, for which we substitute the coefficients a , b and c into the discriminant formula: D = b 2 − 4 a c = 2 2 − 4 1 (− 6) = 4 + 24 = 28 .

So, we got D > 0, which means that the original equation will have two real roots.
To find them, we use the root formula x \u003d - b ± D 2 · a and, substituting the appropriate values, we get: x \u003d - 2 ± 28 2 · 1. We simplify the resulting expression by taking the factor out of the sign of the root, followed by reduction of the fraction:

x = - 2 ± 2 7 2

x = - 2 + 2 7 2 or x = - 2 - 2 7 2

x = - 1 + 7 or x = - 1 - 7

Answer: x = - 1 + 7 , x = - 1 - 7 .

Example 7

It is necessary to solve a quadratic equation − 4 x 2 + 28 x − 49 = 0.

Decision

Let's define the discriminant: D = 28 2 − 4 (− 4) (− 49) = 784 − 784 = 0. With this value of the discriminant, the original equation will have only one root, determined by the formula x = - b 2 · a.

x = - 28 2 (- 4) x = 3, 5

Answer: x = 3, 5.

Example 8

It is necessary to solve the equation 5 y 2 + 6 y + 2 = 0

Decision

The numerical coefficients of this equation will be: a = 5 , b = 6 and c = 2 . We use these values ​​to find the discriminant: D = b 2 − 4 · a · c = 6 2 − 4 · 5 · 2 = 36 − 40 = − 4 . The computed discriminant is negative, so the original quadratic equation has no real roots.

In the case when the task is to indicate complex roots, we apply the root formula by performing operations with complex numbers:

x \u003d - 6 ± - 4 2 5,

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

x = - 3 5 + 1 5 i or x = - 3 5 - 1 5 i .

Answer: there are no real roots; the complex roots are: - 3 5 + 1 5 i , - 3 5 - 1 5 i .

AT school curriculum by default, there is no requirement to look for complex roots, therefore, if the discriminant is determined as negative during the solution, the answer is immediately recorded that there are no real roots.

Root formula for even second coefficients

The root formula x = - b ± D 2 a (D = b 2 − 4 a c) makes it possible to obtain another formula, more compact, allowing you to find solutions to quadratic equations with an even coefficient at x (or with a coefficient of the form 2 a n, for example, 2 3 or 14 ln 5 = 2 7 ln 5). Let us show how this formula is derived.

Suppose we are faced with the task of finding a solution to the quadratic equation a · x 2 + 2 · n · x + c = 0. We act according to the algorithm: we determine the discriminant D = (2 n) 2 − 4 a c = 4 n 2 − 4 a c = 4 (n 2 − a c) , and then use the root formula:

x \u003d - 2 n ± D 2 a, x \u003d - 2 n ± 4 n 2 - a c 2 a, x \u003d - 2 n ± 2 n 2 - a c 2 a, x = - n ± n 2 - a · c a .

Let the expression n 2 − a c be denoted 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 will take the form:

x \u003d - n ± D 1 a, where D 1 \u003d 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 a quarter of the discriminant. Obviously, the sign of D 1 is the same as the sign of D, which means that the sign of D 1 can also serve as an indicator of the presence or absence of the roots of a quadratic equation.

Definition 11

Thus, to find a solution to a quadratic equation with a second coefficient of 2 n, it is necessary:

• find D 1 = n 2 − a c ;
• at D 1< 0 сделать вывод, что действительных корней нет;
• for D 1 = 0, determine the only root of the equation by the formula x = - n a ;
• for D 1 > 0, determine two real roots using the formula x = - n ± D 1 a.

Example 9

It is necessary to solve the quadratic equation 5 · x 2 − 6 · x − 32 = 0.

Decision

The second coefficient of the given equation can be represented as 2 · (− 3) . Then we rewrite the given quadratic equation as 5 · x 2 + 2 · (− 3) · x − 32 = 0 , where a = 5 , n = − 3 and c = − 32 .

Let's calculate the fourth part of the discriminant: D 1 = n 2 − a c = (− 3) 2 − 5 (− 32) = 9 + 160 = 169 . The resulting value is positive, which means that the equation has two real roots. We define them by the corresponding formula of the roots:

x = - n ± D 1 a , x = - - 3 ± 169 5 , x = 3 ± 13 5 ,

x = 3 + 13 5 or x = 3 - 13 5

x = 3 1 5 or x = - 2

It would be possible to perform calculations using the usual formula for the roots of a quadratic equation, but in this case the solution would be more cumbersome.

Answer: x = 3 1 5 or x = - 2 .

Simplification of the form of quadratic equations

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

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

More often, the simplification of the form of a quadratic equation is performed by multiplying or dividing its both parts by a certain number. For example, above we showed a simplified representation of the equation 1200 x 2 - 400 x - 700 = 0, obtained by dividing both of its parts by 100.

Such a transformation is possible when the coefficients of the quadratic equation are not mutually prime numbers. Then it is common to divide both sides of the equation by the largest common divisor absolute values ​​of its coefficients.

As an example, we use the quadratic equation 12 x 2 − 42 x + 48 = 0. Let's define the gcd of the absolute values ​​of its coefficients: gcd (12 , 42 , 48) = gcd(gcd (12 , 42) , 48) = gcd (6 , 48) = 6 . Let's divide both parts of the original quadratic equation by 6 and get the equivalent quadratic equation 2 · x 2 − 7 · x + 8 = 0 .

By multiplying both sides of the quadratic equation, fractional coefficients are usually eliminated. In this case, multiply by the least common multiple of the denominators of its coefficients. For example, if each part of the quadratic equation 1 6 x 2 + 2 3 x - 3 \u003d 0 is multiplied with LCM (6, 3, 1) \u003d 6, then it will be written in a simpler form x 2 + 4 x - 18 = 0 .

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

Relationship between roots and coefficients

The already known formula for the roots of quadratic equations x = - b ± D 2 · a expresses the roots of the equation in terms of its numerical coefficients. Based on this formula, we have the opportunity to set other dependencies between the roots and coefficients.

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

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

In particular, for the given quadratic equation, the sum of the roots is the second coefficient with the opposite sign, and the product of the roots is equal to the free term. For example, by the form of the quadratic equation 3 · x 2 − 7 · x + 22 = 0, it is possible to immediately determine that the sum of its roots is 7 3 , and the product of the roots is 22 3 .

You can also find a number of other relationships between the roots and coefficients of a quadratic equation. For example, the sum of the squares of the roots of a quadratic equation can be expressed in terms of coefficients:

x 1 2 + x 2 2 = (x 1 + x 2) 2 - 2 x 1 x 2 = - b a 2 - 2 c a = b 2 a 2 - 2 c a = b 2 - 2 a c a 2.

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