Bibliographic description: Gasanov A. R., Kuramshin A. A., Elkov A. A., Shilnenkov N. V., Ulanov D. D., Shmeleva O. V. Methods for solving quadratic equations // Young scientist. - 2016. - No. 6.1. - S. 17-20..02.2019).



Our project is dedicated to the ways of solving quadratic equations. The purpose of the project: to learn how to solve quadratic equations in ways that are not included in the school curriculum. Task: find everything possible ways solve quadratic equations and learn how to use them yourself and introduce classmates to these methods.

What are "quadratic equations"?

Quadratic equation- equation of the form ax2 + bx + c = 0, where a, b, c- some numbers ( a ≠ 0), x- unknown.

The numbers a, b, c are called the coefficients of the quadratic equation.

• a is called the first coefficient;
• b is called the second coefficient;
• c - free member.

And who was the first to "invent" quadratic equations?

Some algebraic techniques for solving linear and quadratic equations were known as early as 4000 years ago in Ancient Babylon. The found ancient Babylonian clay tablets, dated somewhere between 1800 and 1600 BC, are the earliest evidence of the study of quadratic equations. The same tablets contain methods for solving certain types of quadratic equations.

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 areas land plots and with earthworks of a military nature, as well as with the development of astronomy and mathematics itself.

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. In spite of high level development of algebra in Babylon, the cuneiform texts lack the concept negative number and general methods for solving quadratic equations.

Babylonian mathematicians from about the 4th century B.C. used the square complement method to solve equations with positive roots. Around 300 B.C. Euclid came up with a more general geometric solution method. The first mathematician to find solutions to an equation with negative roots in the form algebraic formula, was an Indian scientist Brahmagupta(India, 7th century AD).

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

ax2 + bx = c, a>0

In this equation, the coefficients can 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.

In an algebraic treatise Al-Khwarizmi a classification of linear and quadratic equations is given. The author lists 6 types of equations, expressing them as follows:

1) “Squares are equal to roots”, i.e. ax2 = bx.

2) “Squares are equal to number”, i.e. ax2 = c.

3) "The roots are equal to the number", i.e. ax2 = c.

4) “Squares and numbers are equal to roots”, i.e. ax2 + c = bx.

5) “Squares and roots are equal to number”, i.e. ax2 + bx = c.

6) “Roots and numbers are equal to squares”, i.e. bx + c == ax2.

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 decision, of course, does 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-Khwarizmi, like all mathematicians before the 17th century, does not take into account the zero solution, probably because in specific practical tasks, it does not matter. When solving complete quadratic equations, Al-Khwarizmi sets out the rules for solving them using particular numerical examples, and then their geometric proofs.

Forms for solving quadratic equations on the model of Al-Khwarizmi in Europe were first described in the "Book of the Abacus", written in 1202. Italian mathematician Leonard Fibonacci. The author independently developed some new algebraic examples problem solving and was the first in Europe to approach the introduction of negative numbers.

This book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many tasks from this book were transferred to almost all European textbooks of the 14th-17th centuries. General rule solutions of quadratic equations reduced to a single canonical form x2 + bx = c with all possible combinations of signs and coefficients b, c, was formulated in Europe in 1544. M. Stiefel.

Vieta has a general derivation of the formula for solving a quadratic equation, but Vieta recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli 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 Girard, Descartes, Newton and other scientists, the way of solving quadratic equations takes on a modern form.

Consider several ways to solve quadratic equations.

Standard ways to solve quadratic equations from the school curriculum:

1. Factorization of the left side of the equation.
2. Full square selection method.
3. Solution of quadratic equations by formula.
4. Graphical solution of a quadratic equation.
5. Solution of equations using Vieta's theorem.

Let us dwell in more detail on the solution of reduced and non-reduced quadratic equations using the Vieta theorem.

Recall that to solve the given quadratic equations, it is enough to find two numbers such that the product of which is equal to the free term, and the sum is equal to the second coefficient with the opposite sign.

Example.x 2 -5x+6=0

You need to find numbers whose product is 6 and the sum is 5. These numbers will be 3 and 2.

Answer: x 1 =2, x 2 =3.

But you can use this method for equations with the first coefficient not equal to one.

Example.3x 2 +2x-5=0

We take the first coefficient and multiply it by the free term: x 2 +2x-15=0

The roots of this equation will be numbers whose product is - 15, and the sum is - 2. These numbers are 5 and 3. To find the roots of the original equation, we divide the obtained roots by the first coefficient.

Answer: x 1 =-5/3, x 2 =1

6. Solution of equations by the method of "transfer".

Consider quadratic equation ax 2 + bx + c = 0, where a≠0.

Multiplying both its parts by a, we get the equation a 2 x 2 + abx + ac = 0.

Let ax = y, whence x = y/a; then we arrive at the equation y 2 + by + ac = 0, which is equivalent to the given one. We find its roots at 1 and at 2 using the Vieta theorem.

Finally we get x 1 = y 1 /a and x 2 = y 2 /a.

With this method, the coefficient a is multiplied by the free term, as if "transferred" to it, therefore it is called the "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.

Example.2x 2 - 11x + 15 = 0.

Let's "transfer" the coefficient 2 to the free term and making the replacement we get the equation y 2 - 11y + 30 = 0.

According to Vieta's inverse theorem

y 1 = 5, x 1 = 5/2, x 1 = 2.5; y 2 ​​= 6, x 2 = 6/2, x 2 = 3.

Answer: x 1 =2.5; X 2 = 3.

7. Properties of the coefficients of a quadratic equation.

Let the quadratic equation ax 2 + bx + c \u003d 0, a ≠ 0 be given.

1. If a + b + c \u003d 0 (i.e., the sum of the coefficients of the equation is zero), then x 1 \u003d 1.

2. If a - b + c \u003d 0, or b \u003d a + c, then x 1 \u003d - 1.

Example.345x 2 - 137x - 208 = 0.

Since a + b + c \u003d 0 (345 - 137 - 208 \u003d 0), then x 1 \u003d 1, x 2 \u003d -208/345.

Answer: x 1 =1; X 2 = -208/345 .

Example.132x 2 + 247x + 115 = 0

Because a-b + c \u003d 0 (132 - 247 + 115 \u003d 0), then x 1 \u003d - 1, x 2 \u003d - 115/132

Answer: x 1 = - 1; X 2 =- 115/132

There are other properties of the coefficients of a quadratic equation. but their usage is more complicated.

8. Solving quadratic equations using a nomogram.

Fig 1. Nomogram

This is an old and currently forgotten method for solving quadratic equations, placed on p. 83 of the collection: Bradis V.M. Four-digit mathematical tables. - M., Education, 1990.

Table XXII. Nomogram for Equation Solving z2 + pz + q = 0. This nomogram allows, without solving the quadratic equation, to determine the roots of the equation by its coefficients.

The curvilinear scale of the nomogram is built according to the formulas (Fig. 1):

Assuming OS = p, ED = q, OE = a(all in cm), from Fig. 1 similarity of triangles SAN and CDF we get the proportion

whence, after substitutions and simplifications, the equation follows z 2 + pz + q = 0, and the letter z means the label of any point on the curved scale.

Rice. 2 Solving a quadratic equation using a nomogram

Examples.

1) For the equation z 2 - 9z + 8 = 0 the nomogram gives the roots z 1 = 8.0 and z 2 = 1.0

Answer: 8.0; 1.0.

2) Solve the equation using the nomogram

2z 2 - 9z + 2 = 0.

Divide the coefficients of this equation by 2, we get the equation z 2 - 4.5z + 1 = 0.

The nomogram gives the roots z 1 = 4 and z 2 = 0.5.

Answer: 4; 0.5.

9. Geometric method for solving quadratic equations.

Example.X 2 + 10x = 39.

In the original, this problem is formulated as follows: "The square and ten roots are equal to 39."

Consider a square with side x, rectangles are built on its sides so that the other side of each of them is 2.5, therefore, the area of ​​\u200b\u200beach is 2.5x. The resulting figure is then supplemented to a new square ABCD, completing four equal squares in the corners, the side of each of them is 2.5, and the area is 6.25

Rice. 3 Graphical way to solve the equation x 2 + 10x = 39

The area S of square ABCD can be represented as the sum of the areas: the original square x 2, four rectangles (4 ∙ 2.5x = 10x) and four attached squares (6.25 ∙ 4 = 25), i.e. S \u003d x 2 + 10x \u003d 25. Replacing x 2 + 10x with the number 39, we get that S \u003d 39 + 25 \u003d 64, which implies that the side of the square ABCD, i.e. segment AB \u003d 8. For the desired side x of the original square, we get

10. Solution of equations using Bezout's theorem.

Bezout's theorem. The remainder after dividing the polynomial P(x) by the binomial x - α is equal to P(α) (that is, the value of P(x) at x = α).

If the number α is the root of the polynomial P(x), then this polynomial is divisible by x -α without remainder.

Example.x²-4x+3=0

Р(x)= x²-4x+3, α: ±1,±3, α=1, 1-4+3=0. Divide P(x) by (x-1): (x²-4x+3)/(x-1)=x-3

x²-4x+3=(x-1)(x-3), (x-1)(x-3)=0

x-1=0; x=1, or x-3=0, x=3; Answer: x1 =2, x2 =3.

Conclusion: The ability to quickly and rationally solve quadratic equations is simply necessary for solving more complex equations, for example, fractional rational equations, equations of higher powers, biquadratic equations, and in high school trigonometric, exponential and logarithmic equations. Having studied all the found ways to solve quadratic equations, we can advise classmates, except standard ways, the solution by the transfer method (6) and the solution of equations by the property of the coefficients (7), since they are more accessible for understanding.

Literature:

1. Bradis V.M. Four-digit mathematical tables. - M., Education, 1990.
2. Algebra grade 8: textbook for grade 8. general education institutions Makarychev Yu. N., Mindyuk N. G., Neshkov K. I., Suvorova S. B. ed. S. A. Telyakovsky 15th ed., revised. - M.: Enlightenment, 2015
3. https://en.wikipedia.org/wiki/%D0%9A%D0%B2%D0%B0%D0%B4%D1%80%D0%B0%D1%82%D0%BD%D0%BE%D0 %B5_%D1%83%D1%80%D0%B0%D0%B2%D0%BD%D0%B5%D0%BD%D0%B8%D0%B5
4. Glazer G.I. History of mathematics at school. A guide for teachers. / Ed. V.N. Younger. - M.: Enlightenment, 1964.

Quadratic equations often appear during solution various tasks physics and mathematics. In this article, we will look at how to solve these equalities. universal way"through the discriminant". Examples of using the acquired knowledge are also given in the article.

What equations are we talking about?

The figure below shows a formula in which x is an unknown variable, and the Latin characters a, b, c represent some known numbers.

Each of these symbols is called a coefficient. As you can see, the number "a" is in front of the squared variable x. This is the maximum power of the represented expression, which is why it is called a quadratic equation. Another name is often used: a second-order equation. The value a itself is a square coefficient (squaring the variable), b is a linear coefficient (it is next to the variable raised to the first power), and finally the number c is a free term.

Note that the form of the equation shown in the figure above is a general classical quadratic expression. In addition to it, there are other second-order equations in which the coefficients b, c can be zero.

When the task is set to solve the equality under consideration, this means that such values ​​of the variable x must be found that would satisfy it. The first thing to remember here next thing: since the maximum power of x is 2, then given type expressions cannot have more than 2 solutions. This means that if, when solving the equation, 2 x values ​​\u200b\u200bthat satisfy it were found, then you can be sure that there is no 3rd number, substituting which instead of x, the equality would also be true. Solutions to an equation in mathematics are called its roots.

Methods for solving second-order equations

Solving equations of this type requires knowledge of some theory about them. AT school course algebras consider 4 different method solutions. Let's list them:

• using factorization;
• using the formula for the perfect square;
• applying the graph of the corresponding quadratic function;
• using the discriminant equation.

The advantage of the first method is its simplicity, however, it can not be applied to all equations. The second method is universal, but somewhat cumbersome. The third method is distinguished by its clarity, but it is not always convenient and applicable. And finally, using the discriminant equation is a universal and fairly simple way to find the roots of absolutely any second-order equation. Therefore, in the article we will consider only it.

Formula for obtaining the roots of the equation

Let's turn to general view quadratic equation. Let's write it down: a*x²+ b*x + c =0. Before using the method of solving it "through the discriminant", equality should always be reduced to the written form. That is, it must consist of three terms (or less if b or c is 0).

For example, if there is an expression: x²-9*x+8 = -5*x+7*x², then first you should transfer all its members to one side of equality and add the terms containing the variable x in the same powers.

AT this case this operation will lead to the following expression: -6*x²-4*x+8=0, which is equivalent to the equation 6*x²+4*x-8=0 (here we have multiplied the left and right sides of the equation by -1).

In the example above, a = 6, b=4, c=-8. Note that all terms of the considered equality are always summed among themselves, therefore, if the "-" sign appears, this means that the corresponding coefficient is negative, like the number c in this case.

Having analyzed this point, we now turn to the formula itself, which makes it possible to obtain the roots of a quadratic equation. It looks like the photo below.

As can be seen from this expression, it allows you to get two roots (you should pay attention to the "±" sign). To do this, it is enough to substitute the coefficients b, c, and a into it.

The concept of discriminant

In the previous paragraph, a formula was given that allows you to quickly solve any second-order equation. In it, the radical expression is called the discriminant, that is, D \u003d b²-4 * a * c.

Why is this part of the formula singled out, and does it even have its own name? The point is that the discriminant links single expression all three coefficients of the equation. The last fact means that it completely carries information about the roots, which can be expressed by the following list:

1. D>0: equality has 2 various solutions, both of which are real numbers.
2. D=0: The equation has only one root, and it is a real number.

The task of determining the discriminant

Here is a simple example of how to find the discriminant. Let the following equality be given: 2*x² - 4+5*x-9*x² = 3*x-5*x²+7.

Let's bring it to the standard form, we get: (2*x²-9*x²+5*x²) + (5*x-3*x) + (- 4-7) = 0, from which we come to equality: -2*x² +2*x-11 = 0. Here a=-2, b=2, c=-11.

Now you can use the named formula for the discriminant: D \u003d 2² - 4 * (-2) * (-11) \u003d -84. The resulting number is the answer to the task. Since the discriminant in the example is less than zero, we can say that this quadratic equation has no real roots. Its solution will be only numbers of complex type.

An example of inequality through the discriminant

Let's solve problems of a slightly different type: the equality -3*x²-6*x+c = 0 is given. It is necessary to find such values ​​of c for which D>0.

In this case, only 2 out of 3 coefficients are known, so it will not be possible to calculate the exact value of the discriminant, but it is known that it is positive. We use the last fact when compiling the inequality: D= (-6)²-4*(-3)*c>0 => 36+12*c>0. The solution of the obtained inequality leads to the result: c>-3.

Let's check the resulting number. To do this, we calculate D for 2 cases: c=-2 and c=-4. The number -2 satisfies the result (-2>-3), the corresponding discriminant will have the value: D = 12>0. In turn, the number -4 does not satisfy the inequality (-4Thus, any numbers c that are greater than -3 will satisfy the condition.

An example of solving an equation

Here is a problem that consists not only in finding the discriminant, but also in solving the equation. It is necessary to find the roots for the equality -2*x²+7-9*x = 0.

In this example, the discriminant is equal to the following value: D = 81-4*(-2)*7= 137. Then the roots of the equation are determined as follows: x = (9±√137)/(-4). This is exact values roots, if you calculate the approximate root, then you get the numbers: x \u003d -5.176 and x \u003d 0.676.

geometric problem

Let's solve a problem that will require not only the ability to calculate the discriminant, but also the use of abstract thinking skills and knowledge of how to write quadratic equations.

Bob had duvet size 5 x 4 meters. The boy wanted to sew a continuous strip of beautiful fabric around the entire perimeter. How thick will this strip be if it is known that Bob has 10 m² of fabric.

Let the strip have a thickness of x m, then the area of ​​​​the fabric along the long side of the blanket will be (5 + 2 * x) * x, and since there are 2 long sides, we have: 2 * x * (5 + 2 * x). On the short side, the area of ​​the sewn fabric will be 4*x, since there are 2 of these sides, we get the value 8*x. Note that 2*x has been added to the long side because the length of the quilt has increased by that number. The total area of ​​fabric sewn to the blanket is 10 m². Therefore, we get the equality: 2*x*(5+2*x) + 8*x = 10 => 4*x²+18*x-10 = 0.

For this example, the discriminant is: D = 18²-4*4*(-10) = 484. Its root is 22. Using the formula, we find the desired roots: x = (-18±22)/(2*4) = (- 5; 0.5). Obviously, of the two roots, only the number 0.5 is suitable for the condition of the problem.

Thus, the strip of fabric that Bob sews to his blanket will be 50 cm wide.

Formulas for the roots of a quadratic equation. The cases of real, multiple and complex roots are considered. Factorization of a square trinomial. Geometric interpretation. Examples of determining roots and factorization.

Basic formulas

Consider the quadratic equation:
(1) .
The roots of a quadratic equation(1) are determined by the formulas:
; .
These formulas can be combined like this:
.
When the roots of the quadratic equation are known, then the polynomial of the second degree can be represented as a product of factors (factored):
.

Further, we assume that are real numbers.
Consider discriminant of a quadratic equation:
.
If the discriminant is positive, then the quadratic equation (1) has two different real roots:
; .
Then the factorization of the square trinomial has the form:
.
If the discriminant is zero, then the quadratic equation (1) has two multiple (equal) real roots:
.
Factorization:
.
If the discriminant is negative, then the quadratic equation (1) has two complex conjugate roots:
;
.
Here is the imaginary unit, ;
and are the real and imaginary parts of the roots:
; .
Then

.

Graphic interpretation

If build function graph
,
which is a parabola, then the points of intersection of the graph with the axis will be the roots of the equation
.
When , the graph intersects the abscissa axis (axis) at two points.
When , the graph touches the x-axis at one point.
When , the graph does not cross the x-axis.

Below are examples of such graphs.

Useful Formulas Related to Quadratic Equation

(f.1) ;
(f.2) ;
(f.3) .

Derivation of the formula for the roots of a quadratic equation

We perform transformations and apply formulas (f.1) and (f.3):




,
where
; .

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

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

Examples of determining the roots of a quadratic equation

Example 1

(1.1) .

Decision

.
Comparing with our equation (1.1), we find the values ​​of the coefficients:
.
Finding the discriminant:
.
Since the discriminant is positive, the equation has two real roots:
;
;
.

From here we obtain the decomposition of the square trinomial into factors:

.

Graph of the function y = 2 x 2 + 7 x + 3 crosses the x-axis at two points.

Let's plot the function
.
The graph of this function is a parabola. It crosses the x-axis (axis) at two points:
and .
These points are the roots of the original equation (1.1).

Answer

;
;
.

Example 2

Find the roots of a quadratic equation:
(2.1) .

Decision

We write the quadratic equation in general form:
.
Comparing with the original equation (2.1), we find the values ​​of the coefficients:
.
Finding the discriminant:
.
Since the discriminant is zero, the equation has two multiple (equal) roots:
;
.

Then the factorization of the trinomial has the form:
.

Graph of the function y = x 2 - 4 x + 4 touches the x-axis at one point.

Let's plot the function
.
The graph of this function is a parabola. It touches the x-axis (axis) at one point:
.
This point is the root of the original equation (2.1). Since this root is factored twice:
,
then such a root is called a multiple. That is, they consider that there are two equal roots:
.

Answer

;
.

Example 3

Find the roots of a quadratic equation:
(3.1) .

Decision

We write the quadratic equation in general form:
(1) .
Let us rewrite the original equation (3.1):
.
Comparing with (1), we find the values ​​of the coefficients:
.
Finding the discriminant:
.
The discriminant is negative, . Therefore, there are no real roots.

Can be found complex roots:
;
;
.

Then


.

The graph of the function does not cross the x-axis. There are no real roots.

Let's plot the function
.
The graph of this function is a parabola. It does not cross the abscissa (axis). Therefore, there are no real roots.

Answer

There are no real roots. Complex roots:
;
;
.

Quadratic equations. Discriminant. Solution, examples.

Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")

Types of quadratic equations

What is a quadratic equation? What does it look like? In term quadratic equation keyword is "square". It means that in the equation necessarily there must be an x ​​squared. In addition to it, in the equation there may be (or may not be!) Just x (to the first degree) and just a number (free member). And there should not be x's in a degree greater than two.

In mathematical terms, a quadratic equation is an equation of the form:

Here a, b and c- some numbers. b and c- absolutely any, but a- anything but zero. For example:

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

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

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

Well, you get the idea...

In these quadratic equations, on the left, there is full set members. x squared with coefficient a, x to the first power with coefficient b and free member of

Such quadratic equations are called complete.

And if b= 0, what will we get? We have X will disappear in the first degree. This happens from multiplying by zero.) It turns out, for example:

5x 2 -25 = 0,

2x 2 -6x=0,

-x 2 +4x=0

Etc. And if both coefficients b and c are equal to zero, then it is even simpler:

2x 2 \u003d 0,

-0.3x 2 \u003d 0

Such equations, where something is missing, are called incomplete quadratic equations. Which is quite logical.) Please note that x squared is present in all equations.

By the way why a can't be zero? And you substitute instead a zero.) The X in the square will disappear! The equation will become linear. And it's done differently...

That's all the main types of quadratic equations. Complete and incomplete.

Solution of quadratic equations.

Solution of complete quadratic equations.

Quadratic equations are easy to solve. According to formulas and clear simple rules. At the first stage, it is necessary to bring the given equation to the standard form, i.e. to the view:

If the equation is already given to you in this form, you do not need to do the first stage.) The main thing is to correctly determine all the coefficients, a, b and c.

The formula for finding the roots of a quadratic equation looks like this:

The expression under the root sign is called discriminant. But more about him below. As you can see, to find x, we use only a, b and c. Those. coefficients from the quadratic equation. Just carefully substitute the values a, b and c into this formula and count. Substitute with your signs! For example, in the equation:

a =1; b = 3; c= -4. Here we write:

Example almost solved:

This is the answer.

Everything is very simple. And what do you think, you can't go wrong? Well, yes, how...

The most common mistakes are confusion with the signs of values a, b and c. Or rather, not with their signs (where is there to be confused?), But with the substitution of negative values ​​​​into the formula for calculating the roots. Here, a detailed record of the formula with specific numbers saves. If there are problems with calculations, so do it!

Suppose we need to solve the following example:

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

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

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

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

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

Did you know?) Yes! This is incomplete quadratic equations.

Solution of incomplete quadratic equations.

They can also be solved by the general formula. You just need to correctly figure out what is equal here a, b and c.

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

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

And what from this? And the fact that the product is equal to zero if, and only if any of the factors is equal to zero! Don't believe? Well, then come up with two non-zero numbers that, when multiplied, will give zero!
Does not work? Something...
Therefore, we can confidently write: x 1 = 0, x 2 = 4.

Everything. These will be the roots of our equation. Both fit. When substituting any of them into the original equation, we get the correct identity 0 = 0. As you can see, the solution is much simpler than the general formula. I note, by the way, which X will be the first, and which the second - it is absolutely indifferent. Easy to write in order x 1- whichever is less x 2- that which is more.

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

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

also two roots . x 1 = -3, x 2 = 3.

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

Discriminant. Discriminant formula.

Magic word discriminant ! A rare high school student has not heard this word! The phrase “decide through the discriminant” is reassuring and reassuring. Because there is no need to wait for tricks from the discriminant! It is simple and trouble-free in handling.) I remind you of the most general formula for solutions any quadratic equations:

The expression under the root sign is called the discriminant. The discriminant is usually denoted by the letter D. Discriminant formula:

D = b 2 - 4ac

And what is so special about this expression? Why does it deserve a special name? What meaning of the discriminant? After all -b, or 2a in this formula they don’t specifically name ... Letters and letters.

The point is this. When solving a quadratic equation using this formula, it is possible only three cases.

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

2. The discriminant is zero. Then you have one solution. Since adding or subtracting zero in the numerator does not change anything. Strictly speaking, this is not a single root, but two identical. But, in a simplified version, it is customary to talk about one solution.

3. The discriminant is negative. A negative number does not take the square root. Well, okay. This means there are no solutions.

To be honest, at simple solution quadratic equations, the concept of discriminant is not particularly required. We substitute the values ​​​​of the coefficients in the formula, and we consider. There everything turns out by itself, and two roots, and one, and not a single one. However, when solving more complex tasks, without knowledge meaning and discriminant formula not enough. Especially - in equations with parameters. Such equations are aerobatics for the GIA and the Unified State Examination!)

So, how to solve quadratic equations through the discriminant you remembered. Or learned, which is also not bad.) You know how to correctly identify a, b and c. Do you know how attentively substitute them into the root formula and attentively count the result. Did you understand that the key word here is - attentively?

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

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

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

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

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

Second reception. Check your roots! According to Vieta's theorem. Don't worry, I'll explain everything! Checking last thing the equation. Those. the one by which we wrote down the formula of the roots. If (as in this example) the coefficient a = 1, check the roots easily. It is enough to multiply them. You should get a free term, i.e. in our case -2. Pay attention, not 2, but -2! free member with your sign . If it didn’t work out, it means they already messed up somewhere. Look for an error.

If it worked out, you need to fold the roots. Last and final check. Should be a ratio b with opposite sign. In our case -1+2 = +1. A coefficient b, which is before the x, is equal to -1. So, everything is correct!
It is a pity that it is so simple only for examples where x squared is pure, with a coefficient a = 1. But at least check in such equations! There will be fewer mistakes.

Reception third . If your equation has fractional coefficients, get rid of the fractions! Multiply the equation by the common denominator as described in the lesson "How to solve equations? Identity transformations". When working with fractions, errors, for some reason, climb ...

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

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

That's all! Deciding is fun!

So let's recap the topic.

Practical Tips:

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

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

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

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

Now you can decide.)

Solve Equations:

8x 2 - 6x + 1 = 0

x 2 + 3x + 8 = 0

x 2 - 4x + 4 = 0

(x+1) 2 + x + 1 = (x+1)(x+2)

Answers (in disarray):

x 1 = 0
x 2 = 5

x 1.2 =2

x 1 = 2
x 2 \u003d -0.5

x - any number

x 1 = -3
x 2 = 3

no solutions

x 1 = 0.25
x 2 \u003d 0.5

Does everything fit? Fine! Quadratic equations are not your headache. The first three turned out, but the rest did not? Then the problem is not in quadratic equations. The problem is in identical transformations of equations. Take a look at the link, it's helpful.

Doesn't quite work? Or does it not work at all? Then Section 555 will help you. There, all these examples are sorted by bones. Showing main errors in the solution. Of course, the application of identical transformations in solving various equations is also described. Helps a lot!

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

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 short form records of the form 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 case study will allow us to visually 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 Square root from 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 when it is 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 \u003d 0, it is possible to immediately determine that the sum of its roots is 7 3, and the product of the roots is 22 3.

You can also find a number of other 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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