Coordinates on the coordinate plane. Video tutorial “Coordinate plane

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, you want to work with others, in particular, with equations of the second degree, which are otherwise called quadratic.

Quadratic equations are equations like ax² + bx + c = 0, where the variable is x, the numbers are 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 be classified as an incomplete quadratic equation.

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

a) If coefficient c is equal to 0, and 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 the incomplete quadratic equation, which consists in factoring the left side of it and later using the condition that the product is equal to zero.

For example, 5x² - 20x = 0. We factor the left side of the equation, while doing the usual mathematical operation: moving the total 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 = 0, and the free term is not equal to zero, then the equation ax ² + 0x + c = 0 is reduced to an equation of the form ax ² + c = 0. The equations are solved in two ways: a) by factoring the polynomial of the equation on the left side ; b) using the properties of arithmetic square root. Such an equation can be solved using one of the methods, for example:

x = ± √ 25/4

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

c) If b is equal to 0 and c is equal to 0, then ax ² + 0 + 0 = 0 is reduced to an equation of the form ax ² = 0. In such an equation x will be equal to 0.

As you can see, incomplete quadratic equations can have no more than two roots.

Quadratic equation - easy to solve! *Hereinafter referred to as “KU”. Friends, it would seem that there could be nothing simpler in mathematics than solving such an equation. But something told me that many people have problems with him. I decided to see how many on-demand impressions Yandex gives out per month. Here's what happened, look:

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

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

A quadratic equation is an equation of the form:

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

IN school course material is given in the following form– the equations are divided into three classes:

1. They have two roots.

2. *Have only one root.

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

How are roots calculated? Just!

We calculate the discriminant. Underneath this “terrible” word lies a very simple formula:

The root formulas are as follows:

*You need to know these formulas by heart.

You can immediately write down and solve:

Example:

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

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

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

Let's look at the equation:

In this regard, when the discriminant is equal to zero, the school course says that one root is obtained, here it is equal to nine. Everything is correct, it is so, but...

This idea is somewhat incorrect. In fact, there are two roots. Yes, yes, don’t be surprised, it turns out two equal roots, and to be mathematically precise, the answer should contain two roots:

x 1 = 3 x 2 = 3

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

Now the next example:

As we know, the root of a negative number cannot be taken, so the solutions in in this case No.

That's the whole decision process.

Quadratic function.

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

This is a function of the form:

where x and y are variables

a, b, c – given numbers, with a ≠ 0

The graph is a parabola:

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

Let's look at examples:

Example 1: Solve 2x 2 +8 x–192=0

a=2 b=8 c= –192

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

Answer: x 1 = 8 x 2 = –12

*It was possible to immediately left and right side divide the equation by 2, that is, simplify it. The calculations will be easier.

Example 2: Decide x 2–22 x+121 = 0

a=1 b=–22 c=121

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

We found that x 1 = 11 and x 2 = 11

It is permissible to write x = 11 in the answer.

Answer: x = 11

Example 3: Decide x 2 –8x+72 = 0

a=1 b= –8 c=72

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

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

Answer: no solution

The discriminant is negative. There is a solution!

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

The concept of a complex number.

A little theory.

A complex number z is a number of the form

z = a + bi

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

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

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

Now consider the equation:

We get two conjugate roots.

Incomplete quadratic equation.

Let's consider special cases, this is when the coefficient “b” or “c” is equal to zero (or both are equal to zero). They can be solved easily without any discriminants.

Case 1. Coefficient b = 0.

The equation becomes:

Let's transform:

Example:

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

Case 2. Coefficient c = 0.

The equation becomes:

Let's transform and factorize:

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

Example:

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

x 1 = 0 x 2 = 5

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

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

Useful properties and patterns of coefficients.

There are properties that allow you to solve equations with large coefficients.

Ax 2 + bx+ c=0 equality holds

a + b+ c = 0, That

- if for the coefficients of the equation Ax 2 + bx+ c=0 equality holds

a+ s =b, That

These properties help to decide a certain type equations

Example 1: 5001 x 2 –4995 x – 6=0

The sum of the odds is 5001+( – 4995)+(– 6) = 0, which means

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

Equality holds a+ s =b, Means

Regularities of coefficients.

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

ax 2 + (a 2 +1)∙x+ a= 0 = > x 1 = –a x 2 = –1/a.

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

x 1 = –6 x 2 = –1/6.

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

ax 2 – (a 2 +1)∙x+ a= 0 = > x 1 = a x 2 = 1/a.

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

x 1 = 15 x 2 = 1/15.

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

ax 2 + (a 2 –1)∙x – a= 0 = > x 1 = – a x 2 = 1/a.

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

x 1 = – 17 x 2 = 1/17.

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

ax 2 – (a 2 –1)∙x – a= 0 = > x 1 = a x 2 = – 1/a.

Example. Consider the equation 10x 2 – 99x –10 = 0.

x 1 = 10 x 2 = – 1/10

Vieta's theorem.

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

45 = 1∙45 45 = 3∙15 45 = 5∙9.

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

Vieta's theorem, in addition. It is convenient in that after solving a quadratic equation in the usual way (through a discriminant), the resulting roots can be checked. I recommend doing this always.

TRANSPORTATION METHOD

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

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

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

Using Vieta's theorem in equation (2), it is easy to determine that x 1 = 10 x 2 = 1

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

x 1 = 5 x 2 = 0.5.

What is the rationale? Look what's happening.

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

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

The second (modified) one has roots that are 2 times larger.

Therefore, we divide the result by 2.

*If we reroll the three, we will divide the result by 3, etc.

Answer: x 1 = 5 x 2 = 0.5

Sq. ur-ie and Unified State Examination.

I’ll tell you briefly about its importance - YOU MUST BE ABLE TO DECIDE quickly and without thinking, you need to know the formulas of roots and discriminants by heart. Many of the problems included in the Unified State Examination tasks come down to solving a quadratic equation (geometric ones included).

Something worth noting!

1. The form of writing an equation can be “implicit”. For example, the following entry is possible:

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

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

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

More in a simple way. To do this, put z out of brackets. You will get: z(аz + b) = 0. The factors can be written: z=0 and аz + b = 0, since both can result in zero. In the notation az + b = 0, we move the second one to the right with a different sign. From here we get z1 = 0 and z2 = -b/a. These are the roots of the original.

If there is incomplete equation of the form аz² + с = 0, in this case are found by simply moving the free term to the right side of the equation. Also change its sign. The result will be az² = -с. Express z² = -c/a. Take the root and write down two solutions - positive and negative meaning square root.

note

When present in Eq. fractional odds multiply the entire equation by the appropriate factor to eliminate fractions.

Knowledge of how to solve quadratic equations is necessary for both schoolchildren and students; sometimes this can also help an adult in ordinary life. There are several certain methods decisions.

Solving Quadratic Equations

Quadratic equation of the form a*x^2+b*x+c=0. Coefficient x is the desired variable, a, b, c are numerical coefficients. Remember that the “+” sign can change to a “-” sign.

In order to solve this equation, it is necessary to use Vieta’s theorem or find the discriminant. The most common method is to find the discriminant, since for some values ​​of a, b, c it is not possible to use Vieta’s theorem.

To find the discriminant (D), you need to write the formula D=b^2 - 4*a*c. The D value can be greater than, less than, or equal to zero. If D is greater or less than zero, then there will be two roots; if D = 0, then only one root remains; more precisely, we can say that D in this case has two equivalent roots. Substitute the known coefficients a, b, c into the formula and calculate the value.

After you have found the discriminant, use the formulas to find x: x(1) = (- b+sqrt(D))/2*a; x(2) = (- b-sqrt(D))/2*a, where sqrt is a function that means taking the square root of given number. After calculating these expressions, you will find two roots of your equation, after which the equation is considered solved.

If D is less than zero, then it still has roots. At school this section practically not studied. University students should be aware of what is emerging a negative number under the root. They get rid of it by highlighting the imaginary part, that is, -1 under the root is always equal to the imaginary element “i”, which is multiplied by the root with the same positive number. For example, if D=sqrt(-20), after the transformation we get D=sqrt(20)*i. After this transformation, solving the equation is reduced to the same finding of roots as described above.

Vieta's theorem consists of selecting the values ​​of x(1) and x(2). Two are used identical equations: x(1) + x(2)= -b; x(1)*x(2)=с. And very important point is the sign in front of the coefficient b, remember that this sign is opposite to the one in the equation. At first glance, it seems that calculating x(1) and x(2) is very simple, but when solving, you will be faced with the fact that you will have to select the numbers.

Elements of solving quadratic equations

According to the rules of mathematics, some can be factorized: (a+x(1))*(b-x(2))=0, if you managed to transform this quadratic equation in a similar way using mathematical formulas, then feel free to write down the answer. x(1) and x(2) will be equal to the adjacent coefficients in brackets, but with the opposite sign.

Also, do not forget about incomplete quadratic equations. You may be missing some of the terms; if so, then all its coefficients are simply equal to zero. If there is nothing in front of x^2 or x, then the coefficients a and b are equal to 1.

Yakupova M.I. 1

Smirnova Yu.V. 1

1 Municipal budget educational institution average comprehensive school № 11

The text of the work is posted without images and formulas.
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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 could be solved around 2000 BC. e. Babylonians. The rules for solving these equations, set out in the Babylonian texts, essentially coincide with modern ones, but in these texts there is no concept of a negative number and general methods solving quadratic equations.

Ancient Greece

Solving quadratic equations was also done in Ancient Greece such scientists as Diophantus, Euclid and Heron. Diophantus Diophantus of Alexandria is 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 first in Greece in the 1st century AD. gives a purely algebraic way to solve a quadratic equation

India

Problems on quadratic equations are found already in the astronomical treatise “Aryabhattiam”, compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scientist, Brahmagupta (7th century), outlined general rule solving quadratic equations reduced to a unified canonical form: ax2 + bх = с, а> 0. (1) In equation (1) the coefficients can be negative. Brahmagupta's rule is essentially the same as ours. Public competitions in solving difficult problems were common in India. One of the old Indian books says the following about such competitions: “As the sun eclipses the stars with its brilliance, so learned man will eclipse his glory in public assemblies by proposing and solving algebraic problems.” Problems were often presented in poetic form.

This is one of the problems of the famous Indian mathematician of the 12th century. Bhaskars.

“A flock of frisky monkeys

And twelve along the vines, having eaten to my heart’s content, had fun

They began to jump, hanging

Part eight of them squared

How many monkeys were there?

I was having fun in the clearing

Tell me, in this pack?

Bhaskara's solution indicates that the author knew that the roots of quadratic equations are two-valued. Bhaskar writes the equation corresponding to the problem as x2 - 64x = - 768 and, in order to complete the left side of this equation to a square, adds 322 to both sides, then obtaining: x2 - b4x + 322 = -768 + 1024, (x - 32)2 = 256, x - 32= ±16, x1 = 16, x2 = 48.

Quadratic equations in Europe XVII century

Formulas for solving quadratic equations modeled after Al-Khorezmi in Europe were first set forth in the Book of Abacus, written in 1202 by the Italian mathematician Leonardo Fibonacci. This voluminous work, which reflects the influence of mathematics, both from the countries of Islam and from ancient Greece, is distinguished by its completeness and clarity of presentation. The author independently developed some new algebraic examples solving problems and was the first in Europe to introduce 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 problems from the Book of Abacus were used in 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 it, but Viet only recognized positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 16th century. They take into account, in addition to the positive, and negative roots. Only in the 17th century. Thanks to the work of Girard, Descartes, Newton and others scientists way solving quadratic equations takes a modern form.

Definition of a quadratic equation

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

Quadratic equation coefficients

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

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

Types of quadratic equations

Name	General form of the equation	Feature (what are the coefficients)	Examples of equations
	ax 2 + 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

Reduced is a quadratic equation 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 called complete if all its coefficients are nonzero.

A quadratic equation is called incomplete in which at least one of the coefficients, except the leading one (either the second coefficient or the free term), is equal to zero.

Methods for solving quadratic equations

Method I General formula for calculating roots

To find the roots of a quadratic equation ax 2 + b + c = 0 V general case you should use the algorithm below:

Calculate the value of the discriminant of a 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, 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 presented method is universal, but it is far from the only one. Solving a single equation can be approached in a variety of ways, with preferences usually depending on the solver. In addition, often for this purpose some of the methods turn out to be much more elegant, simple, and less labor-intensive than the standard one.

Method II. Roots of a quadratic equation with an even coefficient b III method. Solving incomplete quadratic equations

IV method. Using partial ratios of coefficients

There are special cases of quadratic equations in which the coefficients are in relationships with each other, making them much easier to solve.

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 attitude free term to the leading coefficient ( -c/a).

Hence, before solving any quadratic equation, you 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.

Roots of a quadratic equation whose sum of all coefficients is zero

If in a quadratic equation the sum of all its coefficients is 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 using standard methods, you should check the applicability of this theorem to it: add all the coefficients given equation and see if this amount is equal to zero.

V method. Factoring a quadratic trinomial into linear factors

If the trinomial is 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, after all (displaystyle (kx+m)(lx+n)=0Longleftrightarrow kx+m=0cup lx+n=0)(kx + m)(lx + n) = 0 kx + mUlx + n, and having solved the indicated linear equations, we get the above. Note that quadratic trinomial does not always decompose into linear factors with real coefficients: this is possible if the corresponding equation has real roots.

Let's consider some special cases

Using the squared sum (difference) formula

If the quadratic trinomial has the form (displaystyle (ax)^(2)+2abx+b^(2))ax 2 + 2abx + b 2 , then by 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 full square amounts (differences)

The above formula is also used using a method called “selecting the full square of the sum (difference).” In relation to the above quadratic equation with the previously introduced notation, this means the following:

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

VI method. Using the direct and inverse Vieta theorem

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

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

In the general case, that is, for an unreduced quadratic equation ax 2 + bx + c = 0

x 1 + x 2 = -b/a, x 1 * x 2 = c/a

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

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

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

VII method. Transfer method

The so-called “transfer” method allows you to reduce the solution of unreduced and irreducible equations to the form of reduced equations with integer coefficients by dividing them by the leading coefficient to the solution of reduced equations 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 meaning

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 a 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 upward and vice versa. If the coefficient (displaystyle b) bpositive (if 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 widely used. It is used in many calculations, structures, sports, and also around us.

Let us consider and give some examples of the application of the quadratic equation.

Sport. High jumps: during the jumper’s run-up to hit the take-off bar as clearly as possible and flying high use calculations involving parabolas.

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

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

Airplane flight. Airplane takeoff is the main component of flight. Here we take the calculation for low resistance and acceleration of takeoff.

Quadratic equations are also used in various economic disciplines, in programs for processing audio, video, vector and raster graphics.

Conclusion

As a result of the work done, it turned out that quadratic equations attracted scientists back in ancient times; they had 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 solving by formulas. The formulas are easy to remember, this method is universal. The hypothesis that equations are widely used in life and mathematics was confirmed. After studying the topic, I learned a lot interesting facts about quadratic equations, their use, application, types, solutions. And I will be happy to continue studying them. I hope this will help me do well in my exams.

List of used literature

Site materials:

Wikipedia

Open lesson.rf

Handbook of Elementary Mathematics Vygodsky M. Ya.

I hope, having studied this article, you will learn to find the roots of a complete quadratic equation.

Using the discriminant, only complete quadratic equations are solved; to solve incomplete quadratic equations, other methods are used, which you will find in the article “Solving incomplete quadratic equations.”

What quadratic equations are called complete? This equations of the form ax 2 + b x + c = 0, where coefficients a, b and c are not equal to zero. So, to solve a complete quadratic equation, we need to calculate the discriminant D.

D = b 2 – 4ac.

Depending on the value of the discriminant, we will write down the answer.

If the discriminant is a negative number (D< 0),то корней нет.

If the discriminant is zero, then x = (-b)/2a. When the discriminant positive number(D > 0),

then x 1 = (-b - √D)/2a, and x 2 = (-b + √D)/2a.

For example. Solve the equation x 2– 4x + 4= 0.

D = 4 2 – 4 4 = 0

x = (- (-4))/2 = 2

Answer: 2.

Solve Equation 2 x 2 + x + 3 = 0.

D = 1 2 – 4 2 3 = – 23

Answer: no roots.

Solve Equation 2 x 2 + 5x – 7 = 0.

D = 5 2 – 4 2 (–7) = 81

x 1 = (-5 - √81)/(2 2)= (-5 - 9)/4= – 3.5

x 2 = (-5 + √81)/(2 2) = (-5 + 9)/4=1

Answer: – 3.5; 1.

So let’s imagine the solution of complete quadratic equations using the diagram in Figure 1.

Using these formulas you can solve any complete quadratic equation. You just need to be careful to the equation was written as a polynomial standard view

A x 2 + bx + c, otherwise you may make a mistake. For example, in writing the equation x + 3 + 2x 2 = 0, you can mistakenly decide that

a = 1, b = 3 and c = 2. Then

D = 3 2 – 4 1 2 = 1 and then the equation has two roots. And this is not true. (See solution to example 2 above).

Therefore, if the equation is not written as a polynomial of the standard form, first the complete quadratic equation must be written as a polynomial of the standard form (the monomial with the largest exponent should come first, that is A x 2 , then with less – bx and then a free member With.

When solving the reduced quadratic equation and a quadratic equation with an even coefficient in the second term, you can use other formulas. Let's get acquainted with these formulas. If in a complete quadratic equation the second term has an even coefficient (b = 2k), then you can solve the equation using the formulas shown in the diagram in Figure 2.

A complete quadratic equation is called reduced if the coefficient at x 2 is equal to one and the equation takes the form x 2 + px + q = 0. Such an equation can be given for solution, or it can be obtained by dividing all coefficients of the equation by the coefficient A, standing at x 2 .

Figure 3 shows a diagram for solving the reduced square
equations. Let's look at an example of the application of the formulas discussed in this article.

Example. Solve the equation

3x 2 + 6x – 6 = 0.

Let's solve this equation using the formulas shown in the diagram in Figure 1.

D = 6 2 – 4 3 (– 6) = 36 + 72 = 108

√D = √108 = √(36 3) = 6√3

x 1 = (-6 - 6√3)/(2 3) = (6 (-1- √(3)))/6 = –1 – √3

x 2 = (-6 + 6√3)/(2 3) = (6 (-1+ √(3)))/6 = –1 + √3

Answer: –1 – √3; –1 + √3

You can notice that the coefficient of x in this equation is an even number, that is, b = 6 or b = 2k, whence k = 3. Then let’s try to solve the equation using the formulas shown in the diagram of the figure D 1 = 3 2 – 3 · (– 6 ) = 9 + 18 = 27

√(D 1) = √27 = √(9 3) = 3√3

x 1 = (-3 - 3√3)/3 = (3 (-1 - √(3)))/3 = – 1 – √3

x 2 = (-3 + 3√3)/3 = (3 (-1 + √(3)))/3 = – 1 + √3

Answer: –1 – √3; –1 + √3. Noticing that all the coefficients in this quadratic equation are divisible by 3 and performing the division, we get the reduced quadratic equation x 2 + 2x – 2 = 0 Solve this equation using the formulas for the reduced quadratic
equations figure 3.

D 2 = 2 2 – 4 (– 2) = 4 + 8 = 12

√(D 2) = √12 = √(4 3) = 2√3

x 1 = (-2 - 2√3)/2 = (2 (-1 - √(3)))/2 = – 1 – √3

x 2 = (-2 + 2√3)/2 = (2 (-1+ √(3)))/2 = – 1 + √3

Answer: –1 – √3; –1 + √3.

As you can see, when solving this equation using different formulas, we received the same answer. Therefore, having thoroughly mastered the formulas shown in the diagram in Figure 1, you will always be able to solve any complete quadratic equation.

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