Formulas for the roots of a quadratic equation. The cases of real, multiple and complex roots are considered. Factorization 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 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)
.
Solution
.
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)
.
Solution
We write the quadratic equation in general view:
.
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)
.
Solution
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.
You can find 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:
;
;
.
Kopyevskaya rural secondary school
10 Ways to Solve Quadratic Equations
Head: Patrikeeva Galina Anatolyevna,
mathematic teacher
s.Kopyevo, 2007
1. History of the development of quadratic equations
1.1 Quadratic equations in ancient Babylon
1.2 How Diophantus compiled and solved quadratic equations
1.3 Quadratic equations in India
1.4 Quadratic equations in al-Khwarizmi
1.5 Quadratic equations in Europe XIII - XVII centuries
1.6 About Vieta's theorem
2. Methods for solving quadratic equations
Conclusion
Literature
1. History of the development of quadratic equations
1.1 Quadratic equations in ancient Babylon
The need to solve equations not only of the first, but also of the second degree in ancient times was caused by the need to solve problems related to finding the areas of land and earthworks of a military nature, as well as the development of astronomy and mathematics itself. Quadratic equations were able to solve about 2000 BC. e. Babylonians.
Applying modern algebraic notation, we can say that in their cuneiform texts there are, in addition to incomplete ones, such, for example, complete quadratic equations:
X 2 + X = ¾; X 2 - X = 14,5
The rule for solving these equations, stated in the Babylonian texts, coincides essentially with the modern one, but it is not known how the Babylonians came to this rule. Almost all the cuneiform texts found so far give only problems with solutions stated in the form of recipes, with no indication of how they were found.
In spite of high level development of algebra in Babylon, in cuneiform texts there is no concept of a negative number and general methods for solving quadratic equations.
1.2 How Diophantus compiled and solved quadratic equations.
Diophantus' Arithmetic does not contain a systematic exposition of algebra, but it contains a systematic series of problems, accompanied by explanations and solved by drawing up equations of various degrees.
When compiling equations, Diophantus skillfully chooses unknowns to simplify the solution.
Here, for example, is one of his tasks.
Task 11."Find two numbers knowing that their sum is 20 and their product is 96"
Diophantus argues as follows: it follows from the condition of the problem that the desired numbers are not equal, since if they were equal, then their product would be equal not to 96, but to 100. Thus, one of them will be more than half of their sum, i.e. . 10+x, the other is smaller, i.e. 10's. The difference between them 2x .
Hence the equation:
(10 + x)(10 - x) = 96
100 - x 2 = 96
x 2 - 4 = 0 (1)
From here x = 2. One of the desired numbers is 12 , other 8 . Solution x = -2 for Diophantus does not exist, since Greek mathematics knew only positive numbers.
If we solve this problem by choosing one of the desired numbers as the unknown, then we will come to the solution of the equation
y(20 - y) = 96,
y 2 - 20y + 96 = 0. (2)
It is clear that Diophantus simplifies the solution by choosing the half-difference of the desired numbers as the unknown; he manages to reduce the problem to solving an incomplete quadratic equation (1).
1.3 Quadratic equations in India
Problems for quadratic equations are already found in the astronomical tract "Aryabhattam", compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scientist, Brahmagupta (7th century), outlined the general rule for solving quadratic equations reduced to a single canonical form:
ah 2+ b x = c, a > 0. (1)
In equation (1), the coefficients, except for a, can also be negative. Brahmagupta's rule essentially coincides with ours.
AT ancient india public competitions in solving difficult problems were common. In one of the old Indian books, the following is said about such competitions: “As the sun outshines the stars with its brilliance, so scientist man eclipse the glory of another in public meetings, proposing and solving algebraic problems. Tasks were often dressed in poetic form.
Here is one of the problems of the famous Indian mathematician of the XII century. Bhaskara.
Task 13.
“A frisky flock of monkeys And twelve in vines ...
Having eaten power, had fun. They began to jump, hanging ...
Part eight of them in a square How many monkeys were there,
Having fun in the meadow. You tell me, in this flock?
Bhaskara's solution indicates that he knew about the two-valuedness of the roots of quadratic equations (Fig. 3).
The equation corresponding to problem 13 is:
( x /8) 2 + 12 = x
Bhaskara writes under the guise of:
x 2 - 64x = -768
and, to complete the left side of this equation to a square, he adds to both sides 32 2 , getting then:
x 2 - 64x + 32 2 = -768 + 1024,
(x - 32) 2 = 256,
x - 32 = ± 16,
x 1 = 16, x 2 = 48.
1.4 Quadratic equations in al-Khorezmi
Al-Khorezmi's algebraic treatise gives a classification of linear and quadratic equations. The author lists 6 types of equations, expressing them as follows:
1) "Squares are equal to roots", i.e. ax 2 + c = b X.
2) "Squares are equal to number", i.e. ax 2 = s.
3) "The roots are equal to the number", i.e. ah = s.
4) "Squares and numbers are equal to roots", i.e. ax 2 + c = b X.
5) "Squares and roots are equal to the number", i.e. ah 2+ bx = s.
6) "Roots and numbers are equal to squares", i.e. bx + c \u003d ax 2.
For al-Khwarizmi, who avoided the use of negative numbers, the terms of each of these equations are addends, not subtractions. In this case, equations that do not have positive solutions are obviously not taken into account. The author outlines the methods for solving these equations, using the methods of al-jabr and al-muqabala. His decisions, of course, do not completely coincide with ours. Not to mention the fact that it is purely rhetorical, it should be noted, for example, that when solving an incomplete quadratic equation of the first type
al-Khorezmi, like all mathematicians before the 17th century, does not take into account the zero solution, probably because it does not matter in specific practical problems. When solving complete quadratic equations, al-Khorezmi sets out the rules for solving, and then geometric proofs, using particular numerical examples.
Task 14.“The square and the number 21 are equal to 10 roots. Find the root" (assuming the root of the equation x 2 + 21 = 10x).
The author's solution goes something like this: divide the number of roots in half, you get 5, multiply 5 by itself, subtract 21 from the product, 4 remains. Take the root of 4, you get 2. Subtract 2 from 5, you get 3, this will be the desired root. Or add 2 to 5, which will give 7, this is also a root.
Treatise al - Khorezmi is the first book that has come down to us, in which the classification of quadratic equations is systematically stated and formulas for their solution are given.
1.5 Quadratic equations in Europe XIII - XVII centuries
Formulas for solving quadratic equations on the model of al - Khorezmi in Europe were first set forth in the "Book of the Abacus", written in 1202 by the Italian mathematician Leonardo Fibonacci. This voluminous work, which reflects the influence of mathematics, both the countries of Islam and Ancient Greece, differs in both completeness and clarity of presentation. The author independently developed some new algebraic examples of problem solving and was the first in Europe to approach the introduction of negative numbers. His book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many tasks from the "Book of the Abacus" passed into almost all European textbooks of the 16th - 17th centuries. and partly XVIII.
The general rule for solving quadratic equations reduced to a single canonical form:
x 2+ bx = with,
for all possible combinations of signs of the coefficients b , With was formulated in Europe only in 1544 by M. Stiefel.
Vieta has a general derivation of the formula for solving a quadratic equation, but Vieta recognized only positive roots. The Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 16th century. Take into account, in addition to positive, and negative roots. Only in the XVII century. Thanks to the work of Girard, Descartes, Newton and other scientists, the way to solve quadratic equations takes on a modern look.
1.6 About Vieta's theorem
The theorem expressing the relationship between the coefficients of a quadratic equation and its roots, bearing the name of Vieta, was formulated by him for the first time in 1591 as follows: “If B + D multiplied by A - A 2 , equals BD, then A equals AT and equal D ».
To understand Vieta, one must remember that BUT, like any vowel, meant for him the unknown (our X), the vowels AT, D- coefficients for the unknown. In the language of modern algebra, Vieta's formulation above means: if
(a + b )x - x 2 = ab ,
x 2 - (a + b )x + a b = 0,
x 1 = a, x 2 = b .
Expressing the relationship between the roots and coefficients of the equations general formulas, written using symbols, Viet established uniformity in the methods of solving equations. However, the symbolism of Vieta is still far from its modern form. He did not recognize negative numbers, and therefore, when solving equations, he considered only cases where all roots are positive.
2. Methods for solving quadratic equations
Quadratic equations are the foundation on which the majestic edifice of algebra rests. Quadratic equations are widely used in solving trigonometric, exponential, logarithmic, irrational and transcendental equations and inequalities. We all know how to solve quadratic equations from school (grade 8) until graduation.
Quadratic equation- easy to solve! *Further in the text "KU". Friends, it would seem that in mathematics it can be easier than solving such an equation. But something told me that many people have problems with him. I decided to see how many impressions Yandex gives per request per month. Here's what happened, take a look:
What does it mean? This means that about 70,000 people a month are looking for this information, what does this summer have to do with it, and what will happen among school year- requests will be twice as large. This is not surprising, because those guys and girls who have long graduated from school and are preparing for the exam are looking for this information, and schoolchildren are also trying to refresh their memory.
Despite the fact that there are a lot of sites that tell how to solve this equation, I decided to also contribute and publish the material. Firstly, I want visitors to come to my site on this request; secondly, in other articles, when the speech “KU” comes up, I will give a link to this article; thirdly, I will tell you a little more about his solution than is usually stated on other sites. Let's get started! The content of the article:
A quadratic equation is an equation of the form:
where coefficients a,band with arbitrary numbers, with a≠0.
In the school course, the material is given in the following form - the division of equations into three classes is conditionally done:
1. Have two roots.
2. * Have only one root.
3. Have no roots. It is worth noting here that they do not have real roots
How are roots calculated? Just!
We calculate the discriminant. Under this "terrible" word lies a very simple formula:
The root formulas are as follows:
*These formulas must be known by heart.
You can immediately write down and solve:
Example:
1. If D > 0, then the equation has two roots.
2. If D = 0, then the equation has one root.
3. If D< 0, то уравнение не имеет действительных корней.
Let's look at the equation:
By this occasion, when the discriminant is zero, the school course says that one root is obtained, here it is equal to nine. That's right, it is, but...
This representation is somewhat incorrect. In fact, there are two roots. Yes, yes, do not be surprised, it turns out two equal roots, and to be mathematically accurate, then two roots should be written in the answer:
x 1 = 3 x 2 = 3
But this is so - a small digression. At school, you can write down and say that there is only one root.
Now the following example:
As we know, the root of a negative number is not extracted, so the solutions in this case no.
That's the whole decision process.
Quadratic function.
Here is how the solution looks geometrically. This is extremely important to understand (in the future, in one of the articles, we will analyze in detail the solution of a quadratic inequality).
This is a function of the form:
where x and y are variables
a, b, c - given numbers, where a ≠ 0
The graph is a parabola:
That is, it turns out that by solving a quadratic equation with "y" equal to zero, we find the points of intersection of the parabola with the x-axis. There can be two of these points (the discriminant is positive), one (the discriminant is zero) or none (the discriminant is negative). More about the quadratic function You can view article by Inna Feldman.
Consider examples:
Example 1: Decide 2x 2 +8 x–192=0
a=2 b=8 c= -192
D = b 2 –4ac = 8 2 –4∙2∙(–192) = 64+1536 = 1600
Answer: x 1 = 8 x 2 = -12
* You could immediately left and right side divide the equation by 2, that is, simplify it. The calculations will be easier.
Example 2: Decide x2–22 x+121 = 0
a=1 b=-22 c=121
D = b 2 –4ac =(–22) 2 –4∙1∙121 = 484–484 = 0
We got that x 1 \u003d 11 and x 2 \u003d 11
In the answer, it is permissible to write x = 11.
Answer: x = 11
Example 3: Decide x 2 –8x+72 = 0
a=1 b= -8 c=72
D = b 2 –4ac =(–8) 2 –4∙1∙72 = 64–288 = –224
The discriminant is negative, there is no solution in real numbers.
Answer: no solution
The discriminant is negative. There is a solution!
Here we will talk about solving the equation in the case when a negative discriminant is obtained. Do you know anything about complex numbers? I will not go into detail here about why and where they arose and what their specific role and necessity in mathematics is, this is a topic for a large separate article.
The concept of a complex number.
A bit of theory.
A complex number z is a number of the form
z = a + bi
where a and b are real numbers, i is the so-called imaginary unit.
a+bi is a SINGLE NUMBER, not an addition.
The imaginary unit is equal to the root of minus one:
Now consider the equation:
Get two conjugate roots.
Incomplete quadratic equation.
Consider special cases, this is when the coefficient "b" or "c" is equal to zero (or both are equal to zero). They are solved easily without any discriminants.
Case 1. Coefficient b = 0.
The equation takes the form:
Let's transform:
Example:
4x 2 -16 = 0 => 4x 2 =16 => x 2 = 4 => x 1 = 2 x 2 = -2
Case 2. Coefficient c = 0.
The equation takes the form:
Transform, factorize:
*The product is equal to zero when at least one of the factors is equal to zero.
Example:
9x 2 –45x = 0 => 9x (x–5) =0 => x = 0 or x–5 =0
x 1 = 0 x 2 = 5
Case 3. Coefficients b = 0 and c = 0.
Here it is clear that the solution to the equation will always be x = 0.
Useful properties and patterns of coefficients.
There are properties that allow solving equations with large coefficients.
ax 2 + bx+ c=0 equality
a + b+ c = 0, then
— if for the coefficients of the equation ax 2 + bx+ c=0 equality
a+ with =b, then
These properties help solve a certain kind of equation.
Example 1: 5001 x 2 –4995 x – 6=0
The sum of the coefficients is 5001+( – 4995)+(– 6) = 0, so
Example 2: 2501 x 2 +2507 x+6=0
Equality a+ with =b, means
Regularities of coefficients.
1. If in the equation ax 2 + bx + c \u003d 0 the coefficient "b" is (a 2 +1), and the coefficient "c" is numerically equal to the coefficient "a", then its roots are
ax 2 + (a 2 +1) ∙ x + a \u003d 0 \u003d\u003e x 1 \u003d -a x 2 \u003d -1 / a.
Example. Consider the equation 6x 2 +37x+6 = 0.
x 1 \u003d -6 x 2 \u003d -1/6.
2. If in the equation ax 2 - bx + c \u003d 0, the coefficient "b" is (a 2 +1), and the coefficient "c" is numerically equal to the coefficient "a", then its roots are
ax 2 - (a 2 + 1) ∙ x + a \u003d 0 \u003d\u003e x 1 \u003d a x 2 \u003d 1 / a.
Example. Consider the equation 15x 2 –226x +15 = 0.
x 1 = 15 x 2 = 1/15.
3. If in the equation ax 2 + bx - c = 0 coefficient "b" equals (a 2 – 1), and the coefficient “c” numerically equal to the coefficient "a", then its roots are equal
ax 2 + (a 2 -1) ∙ x - a \u003d 0 \u003d\u003e x 1 \u003d - a x 2 \u003d 1 / a.
Example. Consider the equation 17x 2 + 288x - 17 = 0.
x 1 \u003d - 17 x 2 \u003d 1/17.
4. If in the equation ax 2 - bx - c \u003d 0, the coefficient "b" is equal to (a 2 - 1), and the coefficient c is numerically equal to the coefficient "a", then its roots are
ax 2 - (a 2 -1) ∙ x - a \u003d 0 \u003d\u003e x 1 \u003d a x 2 \u003d - 1 / a.
Example. Consider the equation 10x2 - 99x -10 = 0.
x 1 \u003d 10 x 2 \u003d - 1/10
Vieta's theorem.
Vieta's theorem is named after the famous French mathematician Francois Vieta. Using Vieta's theorem, one can express the sum and product of the roots of an arbitrary KU in terms of its coefficients.
45 = 1∙45 45 = 3∙15 45 = 5∙9.
In sum, the number 14 gives only 5 and 9. These are the roots. With a certain skill, using the presented theorem, you can solve many quadratic equations immediately orally.
Vieta's theorem, moreover. convenient because after solving the quadratic equation in the usual way (through the discriminant), the resulting roots can be checked. I recommend doing this all the time.
TRANSFER METHOD
With this method, the coefficient "a" is multiplied by the free term, as if "transferred" to it, which is why it is called transfer method. This method is used when it is easy to find the roots of an equation using Vieta's theorem and, most importantly, when the discriminant is an exact square.
If a a± b+c≠ 0, then the transfer technique is used, for example:
2X 2 – 11x+ 5 = 0 (1) => X 2 – 11x+ 10 = 0 (2)
According to the Vieta theorem in equation (2), it is easy to determine that x 1 \u003d 10 x 2 \u003d 1
The obtained roots of the equation must be divided by 2 (since the two were “thrown” from x 2), we get
x 1 \u003d 5 x 2 \u003d 0.5.
What is the rationale? See what's happening.
The discriminants of equations (1) and (2) are:
If you look at the roots of the equations, then only different denominators are obtained, and the result depends precisely on the coefficient at x 2:
The second (modified) roots are 2 times larger.
Therefore, we divide the result by 2.
*If we roll three of a kind, then we divide the result by 3, and so on.
Answer: x 1 = 5 x 2 = 0.5
sq. ur-ie and the exam.
I will say briefly about its importance - YOU SHOULD BE ABLE TO DECIDE quickly and without thinking, you need to know the formulas of the roots and the discriminant by heart. A lot of the tasks that are part of the USE tasks come down to solving a quadratic equation (including geometric ones).
What is worth noting!
1. The form of the equation can be "implicit". For example, the following entry is possible:
15+ 9x 2 - 45x = 0 or 15x+42+9x 2 - 45x=0 or 15 -5x+10x 2 = 0.
You need to bring it to a standard form (so as not to get confused when solving).
2. Remember that x is an unknown value and it can be denoted by any other letter - t, q, p, h and others.
More in a simple way. To do this, take z out of brackets. You get: z(az + b) = 0. Factors can be written: z=0 and az + 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 az² + c = 0, in this case they are found by simply transferring the free term to the right side of the equation. Also change its sign. You get the record az² \u003d -s. Express z² = -c/a. Take the root and write down two solutions - a positive and a negative value of the square root.
note
If there are fractional coefficients in the equation, multiply the entire equation by the appropriate factor so as to get rid of the fractions.
Knowing how to solve quadratic equations is necessary for both schoolchildren and students, sometimes it can help an adult in everyday life. There are several specific decision methods.
Solving quadratic equations
A quadratic equation of the form a*x^2+b*x+c=0. Coefficient x is the desired variable, a, b, c - numerical coefficients. Remember that the "+" sign can change to the "-" sign.In order to solve this equation, you must use the Vieta theorem or find the discriminant. The most common way is to find the discriminant, since for some values of a, b, c it is not possible to use the Vieta theorem.
To find the discriminant (D), you must write the formula D=b^2 - 4*a*c. The value of D 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, to find x, use the formulas: x(1) = (- b+sqrt(D))/2*a; x(2) = (- b-sqrt(D))/2*a where sqrt is the function to take the square root of the given number. After calculating these expressions, you will find the 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 is practically not studied. University students should be aware of what is emerging a negative number under the root. We get rid of it by separating 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, D=sqrt(20)*i is obtained. After this transformation, the solution of the equation is reduced to the same finding of the roots, as described above.
Vieta's theorem consists in the selection of x(1) and x(2) values. Two identical equations are used: x(1) + x(2)= -b; x(1)*x(2)=s. Moreover, a 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 encounter the fact that the numbers will have to be selected exactly.
Elements for solving quadratic equations
According to the rules of mathematics, some can be factored: (a + x (1)) * (b-x (2)) = 0, if you managed to convert using mathematical formulas In a similar way this quadratic equation, 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 x^2 or x is preceded by nothing, then the coefficients a and b are equal to 1.
First level
Quadratic equations. Comprehensive guide (2019)
In the term "quadratic equation" the key word is "quadratic". This means that the equation must necessarily contain a variable (the same X) in the square, and at the same time there should not be Xs in the third (or greater) degree.
The solution of many equations is reduced to the solution of quadratic equations.
Let's learn to determine that we have a quadratic equation, and not some other.
Example 1
Get rid of the denominator and multiply each term of the equation by
Let's move everything to the left side and arrange the terms in descending order of powers of x
Now we can say with confidence that this equation is quadratic!
Example 2
Multiply the left and right sides by:
This equation, although it was originally in it, is not a square!
Example 3
Let's multiply everything by:
Scary? The fourth and second degrees ... However, if we make a replacement, we will see that we have a simple quadratic equation:
Example 4
It seems to be, but let's take a closer look. Let's move everything to the left side:
You see, it has shrunk - and now it's a simple linear equation!
Now try to determine for yourself which of the following equations are quadratic and which are not:
Examples:
Answers:
- square;
- square;
- not square;
- not square;
- not square;
- square;
- not square;
- square.
Mathematicians conditionally divide all quadratic equations into the following types:
- Complete quadratic equations- equations in which the coefficients and, as well as the free term c, are not equal to zero (as in the example). In addition, among the complete quadratic equations, there are given are equations in which the coefficient (the equation from example one is not only complete, but also reduced!)
- Incomplete quadratic equations- equations in which the coefficient and or free term c are equal to zero:
They are incomplete because some element is missing from them. But the equation must always contain x squared !!! Otherwise, it will no longer be a quadratic, but some other equation.
Why did they come up with such a division? It would seem that there is an X squared, and okay. Such a division is due to the methods of solution. Let's consider each of them in more detail.
Solving incomplete quadratic equations
First, let's focus on solving incomplete quadratic equations - they are much simpler!
Incomplete quadratic equations are of types:
- , in this equation the coefficient is equal.
- , in this equation the free term is equal to.
- , in this equation the coefficient and the free term are equal.
1. i. Since we know how to extract Square root, then let's express from this equation
The expression can be either negative or positive. A squared number cannot be negative, because when multiplying two negative or two positive numbers, the result will always be a positive number, so: if, then the equation has no solutions.
And if, then we get two roots. These formulas do not need to be memorized. The main thing is that you should always know and remember that it cannot be less.
Let's try to solve some examples.
Example 5:
Solve the Equation
Now it remains to extract the root from the left and right parts. After all, do you remember how to extract the roots?
Answer:
Never forget about roots with a negative sign!!!
Example 6:
Solve the Equation
Answer:
Example 7:
Solve the Equation
Ouch! The square of a number cannot be negative, which means that the equation
no roots!
For such equations in which there are no roots, mathematicians came up with a special icon - (empty set). And the answer can be written like this:
Answer:
Thus, this quadratic equation has two roots. There are no restrictions here, since we did not extract the root.
Example 8:
Solve the Equation
Let's take the common factor out of brackets:
In this way,
This equation has two roots.
Answer:
The simplest type of incomplete quadratic equations (although they are all simple, right?). Obviously, this equation always has only one root:
Here we will do without examples.
Solving complete quadratic equations
We remind you that the complete quadratic equation is an equation of the form equation where
Solving full quadratic equations is a bit more complicated (just a little bit) than those given.
Remember, any quadratic equation can be solved using the discriminant! Even incomplete.
The rest of the methods will help you do it faster, but if you have problems with quadratic equations, first master the solution using the discriminant.
1. Solving quadratic equations using the discriminant.
Solving quadratic equations in this way is very simple, the main thing is to remember the sequence of actions and a couple of formulas.
If, then the equation has a root. Special attention should be paid to the step. The discriminant () tells us the number of roots of the equation.
- If, then the formula at the step will be reduced to. Thus, the equation will have only a root.
- If, then we will not be able to extract the root of the discriminant at the step. This indicates that the equation has no roots.
Let's go back to our equations and look at a few examples.
Example 9:
Solve the Equation
Step 1 skip.
Step 2
Finding the discriminant:
So the equation has two roots.
Step 3
Answer:
Example 10:
Solve the Equation
The equation is in standard form, so Step 1 skip.
Step 2
Finding the discriminant:
So the equation has one root.
Answer:
Example 11:
Solve the Equation
The equation is in standard form, so Step 1 skip.
Step 2
Finding the discriminant:
This means that we will not be able to extract the root from the discriminant. There are no roots of the equation.
Now we know how to write down such answers correctly.
Answer: no roots
2. Solution of quadratic equations using the Vieta theorem.
If you remember, then there is such a type of equations that are called reduced (when the coefficient a is equal to):
Such equations are very easy to solve using Vieta's theorem:
The sum of the roots given quadratic equation is equal, and the product of the roots is equal.
Example 12:
Solve the Equation
This equation is suitable for solution using Vieta's theorem, because .
The sum of the roots of the equation is, i.e. we get the first equation:
And the product is:
Let's create and solve the system:
- and. The sum is;
- and. The sum is;
- and. The amount is equal.
and are the solution of the system:
Answer: ; .
Example 13:
Solve the Equation
Answer:
Example 14:
Solve the Equation
The equation is reduced, which means:
Answer:
QUADRATIC EQUATIONS. AVERAGE LEVEL
What is a quadratic equation?
In other words, a quadratic equation is an equation of the form, where - unknown, - some numbers, moreover.
The number is called the highest or first coefficient quadratic equation, - second coefficient, a - free member.
Why? Because if, the equation will immediately become linear, because will disappear.
In this case, and can be equal to zero. In this stool equation is called incomplete. If all the terms are in place, that is, the equation is complete.
Solutions to various types of quadratic equations
Methods for solving incomplete quadratic equations:
To begin with, we will analyze the methods for solving incomplete quadratic equations - they are simpler.
The following types of equations can be distinguished:
I. , in this equation the coefficient and the free term are equal.
II. , in this equation the coefficient is equal.
III. , in this equation the free term is equal to.
Now consider the solution of each of these subtypes.
Obviously, this equation always has only one root:
A number squared cannot be negative, because when multiplying two negative or two positive numbers, the result will always be a positive number. That's why:
if, then the equation has no solutions;
if we have two roots
These formulas do not need to be memorized. The main thing to remember is that it cannot be less.
Examples:
Solutions:
Answer:
Never forget about roots with a negative sign!
The square of a number cannot be negative, which means that the equation
no roots.
To briefly write that the problem has no solutions, we use the empty set icon.
Answer:
So, this equation has two roots: and.
Answer:
Let's take the common factor out of brackets:
The product is equal to zero if at least one of the factors is equal to zero. This means that the equation has a solution when:
So, this quadratic equation has two roots: and.
Example:
Solve the equation.
Solution:
We factorize the left side of the equation and find the roots:
Answer:
Methods for solving complete quadratic equations:
1. Discriminant
Solving quadratic equations in this way is easy, the main thing is to remember the sequence of actions and a couple of formulas. Remember, any quadratic equation can be solved using the discriminant! Even incomplete.
Did you notice the root of the discriminant in the root formula? But the discriminant can be negative. What to do? We need to pay special attention to step 2. The discriminant tells us the number of roots of the equation.
- If, then the equation has a root:
- If, then the equation has the same root, but in fact, one root:
Such roots are called double roots.
- If, then the root of the discriminant is not extracted. This indicates that the equation has no roots.
Why is it possible different amount roots? Let's turn to geometric sense quadratic equation. The graph of the function is a parabola:
In a particular case, which is a quadratic equation, . And this means that the roots of the quadratic equation are the points of intersection with the x-axis (axis). The parabola may not cross the axis at all, or it may intersect it at one (when the top of the parabola lies on the axis) or two points.
In addition, the coefficient is responsible for the direction of the branches of the parabola. If, then the branches of the parabola are directed upwards, and if - then downwards.
Examples:
Solutions:
Answer:
Answer: .
Answer:
This means there are no solutions.
Answer: .
2. Vieta's theorem
Using the Vieta theorem is very easy: you just need to choose a pair of numbers whose product is equal to the free term of the equation, and the sum is equal to the second coefficient, taken with the opposite sign.
It is important to remember that Vieta's theorem can only be applied to given quadratic equations ().
Let's look at a few examples:
Example #1:
Solve the equation.
Solution:
This equation is suitable for solution using Vieta's theorem, because . Other coefficients: ; .
The sum of the roots of the equation is:
And the product is:
Let's select such pairs of numbers, the product of which is equal, and check if their sum is equal:
- and. The sum is;
- and. The sum is;
- and. The amount is equal.
and are the solution of the system:
Thus, and are the roots of our equation.
Answer: ; .
Example #2:
Solution:
We select such pairs of numbers that give in the product, and then check whether their sum is equal:
and: give in total.
and: give in total. To get it, you just need to change the signs of the alleged roots: and, after all, the product.
Answer:
Example #3:
Solution:
The free term of the equation is negative, and hence the product of the roots is a negative number. This is possible only if one of the roots is negative and the other is positive. So the sum of the roots is differences of their modules.
We select such pairs of numbers that give in the product, and the difference of which is equal to:
and: their difference is - not suitable;
and: - not suitable;
and: - not suitable;
and: - suitable. It remains only to remember that one of the roots is negative. Since their sum must be equal, then the root, which is smaller in absolute value, must be negative: . We check:
Answer:
Example #4:
Solve the equation.
Solution:
The equation is reduced, which means:
The free term is negative, and hence the product of the roots is negative. And this is possible only when one root of the equation is negative and the other is positive.
We select such pairs of numbers whose product is equal, and then determine which roots should have a negative sign:
Obviously, only roots and are suitable for the first condition:
Answer:
Example #5:
Solve the equation.
Solution:
The equation is reduced, which means:
The sum of the roots is negative, which means that at least one of the roots is negative. But since their product is positive, it means both roots are minus.
We select such pairs of numbers, the product of which is equal to:
Obviously, the roots are the numbers and.
Answer:
Agree, it is very convenient - to invent roots orally, instead of counting this nasty discriminant. Try to use Vieta's theorem as often as possible.
But the Vieta theorem is needed in order to facilitate and speed up finding the roots. To make it profitable for you to use it, you must bring the actions to automatism. And for this, solve five more examples. But don't cheat: you can't use the discriminant! Only Vieta's theorem:
Solutions for tasks for independent work:
Task 1. ((x)^(2))-8x+12=0
According to Vieta's theorem:
As usual, we start the selection with the product:
Not suitable because the amount;
: the amount is what you need.
Answer: ; .
Task 2.
And again, our favorite Vieta theorem: the sum should work out, but the product is equal.
But since it should be not, but, we change the signs of the roots: and (in total).
Answer: ; .
Task 3.
Hmm... Where is it?
It is necessary to transfer all the terms into one part:
The sum of the roots is equal to the product.
Yes, stop! The equation is not given. But Vieta's theorem is applicable only in the given equations. So first you need to bring the equation. If you can’t bring it up, drop this idea and solve it in another way (for example, through the discriminant). Let me remind you that to bring a quadratic equation means to make the leading coefficient equal to:
Excellent. Then the sum of the roots is equal, and the product.
It's easier to pick up here: after all - a prime number (sorry for the tautology).
Answer: ; .
Task 4.
The free term is negative. What's so special about it? And the fact that the roots will be of different signs. And now, during the selection, we check not the sum of the roots, but the difference between their modules: this difference is equal, but the product.
So, the roots are equal and, but one of them is with a minus. Vieta's theorem tells us that the sum of the roots is equal to the second coefficient with the opposite sign, that is. This means that the smaller root will have a minus: and, since.
Answer: ; .
Task 5.
What needs to be done first? That's right, give the equation:
Again: we select the factors of the number, and their difference should be equal to:
The roots are equal and, but one of them is minus. Which? Their sum must be equal, which means that with a minus there will be a larger root.
Answer: ; .
Let me summarize:
- Vieta's theorem is used only in the given quadratic equations.
- Using the Vieta theorem, you can find the roots by selection, orally.
- If the equation is not given or no suitable pair of factors of the free term was found, then there are no integer roots, and you need to solve it in another way (for example, through the discriminant).
3. Full square selection method
If all the terms containing the unknown are represented as terms from the formulas of abbreviated multiplication - the square of the sum or difference - then after the change of variables, the equation can be represented as an incomplete quadratic equation of the type.
For example:
Example 1:
Solve the equation: .
Solution:
Answer:
Example 2:
Solve the equation: .
Solution:
Answer:
In general, the transformation will look like this:
This implies: .
Doesn't it remind you of anything? It's the discriminant! That's exactly how the discriminant formula was obtained.
QUADRATIC EQUATIONS. BRIEFLY ABOUT THE MAIN
Quadratic equation is an equation of the form, where is the unknown, are the coefficients of the quadratic equation, is the free term.
Complete quadratic equation- an equation in which the coefficients are not equal to zero.
Reduced quadratic equation- an equation in which the coefficient, that is: .
Incomplete quadratic equation- an equation in which the coefficient and or free term c are equal to zero:
- if the coefficient, the equation has the form: ,
- if a free term, the equation has the form: ,
- if and, the equation has the form: .
1. Algorithm for solving incomplete quadratic equations
1.1. An incomplete quadratic equation of the form, where, :
1) Express the unknown: ,
2) Check the sign of the expression:
- if, then the equation has no solutions,
- if, then the equation has two roots.
1.2. An incomplete quadratic equation of the form, where, :
1) Let's take the common factor out of brackets: ,
2) The product is equal to zero if at least one of the factors is equal to zero. Therefore, the equation has two roots:
1.3. An incomplete quadratic equation of the form, where:
This equation always has only one root: .
2. Algorithm for solving complete quadratic equations of the form where
2.1. Solution using the discriminant
1) Let's bring the equation to the standard form: ,
2) Calculate the discriminant using the formula: , which indicates the number of roots of the equation:
3) Find the roots of the equation:
- if, then the equation has a root, which are found by the formula:
- if, then the equation has a root, which is found by the formula:
- if, then the equation has no roots.
2.2. Solution using Vieta's theorem
The sum of the roots of the reduced quadratic equation (an equation of the form, where) is equal, and the product of the roots is equal, i.e. , a.
2.3. Full square solution
