Elective course “Methods for solving irrational equations. Equation and its roots: definitions, examples
After we have studied the concept of equalities, namely one of their types - numerical equalities, we can move on to another important type - equations. Within the framework of this material, we will explain what an equation and its root are, formulate basic definitions and give various examples of equations and finding their roots.
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Concept of equation
Typically, the concept of an equation is taught at the very beginning of a school algebra course. Then it is defined like this:
Definition 1
Equation called an equality with an unknown number that needs to be found.
It is customary to denote unknowns in small Latin letters, for example, t, r, m, etc., but x, y, z are most often used. In other words, the equation is determined by the form of its recording, that is, equality will be an equation only when it is reduced to a certain form - it must contain a letter, the value that must be found.
Let us give some examples of the simplest equations. These can be equalities of the form x = 5, y = 6, etc., as well as those that include arithmetic operations, for example, x + 7 = 38, z − 4 = 2, 8 t = 4, 6: x = 3.
After the concept of brackets is learned, the concept of equations with brackets appears. These include 7 · (x − 1) = 19, x + 6 · (x + 6 · (x − 8)) = 3, etc. The letter that needs to be found can appear more than once, but several times, like, for example, in the equation x + 2 + 4 · x − 2 − x = 10 . Also, unknowns can be located not only on the left, but also on the right or in both parts at the same time, for example, x (8 + 1) − 7 = 8, 3 − 3 = z + 3 or 8 x − 9 = 2 (x + 17) .
Further, after students become familiar with the concepts of integers, reals, rationals, natural numbers, as well as logarithms, roots and powers, new equations appear that include all these objects. We have devoted a separate article to examples of such expressions.
In the 7th grade curriculum, the concept of variables appears for the first time. These are letters that can take different meanings(For more information, see the article on numeric, literal, and variable expressions). Based on this concept, we can redefine the equation:
Definition 2
The equation is an equality involving a variable whose value needs to be calculated.
That is, for example, the expression x + 3 = 6 x + 7 is an equation with the variable x, and 3 y − 1 + y = 0 is an equation with the variable y.
One equation can have more than one variable, but two or more. They are called, respectively, equations with two, three variables, etc. Let us write down the definition:
Definition 3
Equations with two (three, four or more) variables are equations that include a corresponding number of unknowns.
For example, an equality of the form 3, 7 · x + 0, 6 = 1 is an equation with one variable x, and x − z = 5 is an equation with two variables x and z. An example of an equation with three variables would be x 2 + (y − 6) 2 + (z + 0, 6) 2 = 26.
Root of the equation
When we talk about an equation, the need immediately arises to define the concept of its root. Let's try to explain what it means.
Example 1
We are given a certain equation that includes one variable. If we substitute a number for the unknown letter, the equation becomes a numerical equality - true or false. So, if in the equation a + 1 = 5 we replace the letter with the number 2, then the equality will become false, and if 4, then the correct equality will be 4 + 1 = 5.
We are more interested in precisely those values with which the variable will turn into a true equality. They are called roots or solutions. Let's write down the definition.
Definition 4
Root of the equation They call the value of a variable that turns a given equation into a true equality.
The root can also be called a solution, or vice versa - both of these concepts mean the same thing.
Example 2
Let's take an example to clarify this definition. Above we gave the equation a + 1 = 5. According to the definition, the root in this case will be 4, because when substituted instead of a letter it gives the correct numerical equality, and two will not be a solution, since it corresponds to the incorrect equality 2 + 1 = 5.
How many roots can one equation have? Does every equation have a root? Let's answer these questions.
Equations that do not have a single root also exist. An example would be 0 x = 5. We can substitute an infinite number of different numbers into it, but none of them will turn it into a true equality, since multiplying by 0 always gives 0.
There are also equations that have several roots. They can be either finite or infinite a large number of roots
Example 3
So, in the equation x − 2 = 4 there is only one root - six, in x 2 = 9 two roots - three and minus three, in x · (x − 1) · (x − 2) = 0 three roots - zero, one and two, there are infinitely many roots in the equation x=x.
Now let us explain how to correctly write the roots of the equation. If there are none, then we write: “the equation has no roots.” In this case, you can also indicate the sign of the empty set ∅. If there are roots, then we write them separated by commas or indicate them as elements of a set, enclosing them in curly braces. So, if any equation has three roots - 2, 1 and 5, then we write - 2, 1, 5 or (- 2, 1, 5).
It is allowed to write roots in the form of simple equalities. So, if the unknown in the equation is denoted by the letter y, and the roots are 2 and 7, then we write y = 2 and y = 7. Sometimes subscripts are added to letters, for example, x 1 = 3, x 2 = 5. In this way we point to the numbers of the roots. If the equation has an infinite number of solutions, then we write the answer as a numerical interval or use generally accepted notation: the set of natural numbers is denoted N, integers - Z, real numbers - R. Let's say, if we need to write that the solution to the equation will be any integer, then we write that x ∈ Z, and if any real number from one to nine, then y ∈ 1, 9.
When an equation has two, three roots or more, then, as a rule, we talk not about roots, but about solutions to the equation. Let us formulate the definition of a solution to an equation with several variables.
Definition 5
The solution to an equation with two, three or more variables is two, three or more values of the variables that turn the given equation into a correct numerical equality.
Let us explain the definition with examples.
Example 4
Let's say we have the expression x + y = 7, which is an equation with two variables. Let's substitute one instead of the first, and two instead of the second. We will get an incorrect equality, which means that this pair of values will not be a solution to this equation. If we take the pair 3 and 4, then the equality becomes true, which means we have found a solution.
Such equations may also have no roots or an infinite number of them. If we need to write down two, three, four or more values, then we write them separated by commas in parentheses. That is, in the example above, the answer will look like (3, 4).
In practice, you most often have to deal with equations containing one variable. We will consider the algorithm for solving them in detail in the article devoted to solving equations.
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Solving irrational equations.
In this article we will talk about solutions simplest irrational equations.
Irrational equation is an equation that contains an unknown under the root sign.
Let's look at two types irrational equations, which are very similar at first glance, but in essence are very different from each other.
(1)
(2)
In the first equation we see that the unknown is under the sign of the root of the third degree. We can take the odd root of a negative number, so in this equation there are no restrictions on either the expression under the root sign or the expression on the right side of the equation. We can raise both sides of the equation to the third power to get rid of the root. We get an equivalent equation:
When raising the right and left sides of the equation to an odd power, we can not be afraid of getting extraneous roots.
Example 1. Let's solve the equation 
Let's raise both sides of the equation to the third power. We get an equivalent equation:
Let's move all the terms to one side and put x out of brackets:
Equating each factor to zero, we get:
Answer: (0;1;2)
Let's look closely at the second equation: . On the left side of the equation is the square root, which takes only non-negative values. Therefore, for the equation to have solutions, the right-hand side must also be non-negative. Therefore, the condition is imposed on the right side of the equation:
Title="g(x)>=0"> - это !} condition for the existence of roots.
To solve an equation of this type, you need to square both sides of the equation:
(3)
Squaring can lead to the appearance of extraneous roots, so we need the equations:
Title="f(x)>=0"> (4)!}
However, inequality (4) follows from condition (3): if the right side of the equality contains the square of some expression, and the square of any expression can only take non-negative values, therefore the left side must also be non-negative. Therefore, condition (4) automatically follows from condition (3) and our the equation is equivalent to the system:
Title="delim(lbrace)(matrix(2)(1)((f(x)=g^2((x))) (g(x)>=0) ))( )">!}
Example 2. Let's solve the equation:
.
Let's move on to an equivalent system:
Title="delim(lbrace)(matrix(2)(1)((2x^2-7x+5=((1-x))^2) (1-x>=0) ))( )">!}
Let's solve the first equation of the system and check which roots satisfy the inequality.
Inequality title="1-x>=0">удовлетворяет только корень !}
Answer: x=1
Attention! If in the process of solving we square both sides of the equation, then we must remember that extraneous roots may appear. Therefore, you either need to move on to an equivalent system, or at the end of the solution, DO A CHECK: find the roots and substitute them into the original equation.
Example 3. Let's solve the equation:
To solve this equation, we also need to square both sides. Let's not bother with the ODZ and the condition for the existence of roots in this equation, but simply do a check at the end of the solution.
Let's square both sides of the equation:
Let's move the term containing the root to the left, and all other terms to the right:
Let's square both sides of the equation again:
On Vieta's theme:
Let's do a check. To do this, we substitute the found roots into the original equation. Obviously, at , the right-hand side of the original equation is negative, and the left-hand side is positive.
At we obtain the correct equality.
Methodological developments for the elective course
"Methods for solving irrational equations""
INTRODUCTION
The proposed elective course “Methods for solving irrational equations” is intended for students in the 11th grade of a general education school and is subject-oriented, aimed at expanding the theoretical and practical knowledge of students. The elective course is built on the knowledge and skills that students acquire when studying mathematics in high school.
The specificity of this course is that it is intended primarily for students who want to expand, deepen, systematize, generalize their mathematical knowledge, and learn common methods and techniques for solving irrational equations. The program includes questions that partially go beyond the current mathematics programs and non-standard methods that allow you to more effectively solve various problems.
Most USE tasks require graduates to master various methods solutions of various kinds of equations and their systems. Material related to equations and systems of equations makes up a significant part of the school mathematics course. The relevance of choosing a topic for an elective course is determined by the significance of the topic “Irrational Equations” in school course mathematics and, at the same time, a lack of time to consider non-standard methods and approaches to solving irrational equations that are found in tasks of group “C” of the Unified State Examination.
Along with the basic task of teaching mathematics - ensuring students’ strong and conscious mastery of the system of mathematical knowledge and skills - this elective course provides for the formation of a sustainable interest in the subject, the development of mathematical abilities, increasing the level of mathematical culture of students, creating the basis for successfully passing the Unified State Exam and continuing education at universities .
Purpose of the course:
Increase the level of understanding and practical training in solving irrational equations;
Study techniques and methods for solving irrational equations;
Develop the ability to analyze, highlight the main thing, form elements of creative search based on generalization techniques;
Expand students’ knowledge on this topic, improve their skills in solving various problems in order to successfully pass the Unified State Exam.
Course objectives:
Expanding knowledge about methods and techniques for solving algebraic equations;
Generalization and systematization of knowledge when studying in grades 10-11 and preparing for the Unified State Exam;
Development of the ability to independently acquire and apply knowledge;
Introducing students to work with mathematical literature;
Development of students’ logical thinking, their algorithmic culture and mathematical intuition;
Improving the student’s mathematical culture.
The elective course program involves studying various methods and approaches to solving irrational equations and developing practical skills on the issues under consideration. The course lasts 17 hours.
The program is complicated, exceeds the usual course of study, promotes the development abstract thinking, expands the student’s field of knowledge. At the same time, it maintains continuity with existing programs, being their logical continuation.
Educational and thematic plan
№p/p
Lesson topic
Number of hours
Solving equations taking into account the range of acceptable values
Solving irrational equations by raising to natural powers
Solving equations by introducing auxiliary variables (replacement method)
Solving an equation with a radical of the third degree.
Identical transformations when solving irrational equations
Unconventional tasks. Problems of group “C” of the Unified State Exam
Forms of control: home tests, independent work, essays and research papers.
As a result of studying this elective course, students should be able to solve various irrational equations using standard and non-standard methods and techniques;
master the algorithm for solving standard irrational equations;
be able to use the properties of equations to solve non-standard problems;
be able to perform identity transformations when solving equations;
have a clear understanding of the topics of the unified state exam, the main methods of solving them;
gain experience in choosing methods for solving non-standard problems.
MAIN PART.
Equations in which the unknown quantity is under the radical sign are called irrational.
The simplest irrational equations include equations of the form:
The main idea of the solution of an irrational equation consists in reducing it to a rational algebraic equation, which is either equivalent to the original irrational equation or is its consequence. When solving irrational equations, we are always talking about finding real roots.
Let's look at some ways to solve irrational equations.
1. Solving irrational equations taking into account the range of permissible values (APV).
The region of permissible values of an irrational equation consists of those values of the unknowns for which all expressions under the sign of a radical of even degree are non-negative.
Sometimes knowledge of the ODZ allows you to prove that the equation has no solutions, and sometimes allows you to find solutions to the equation by directly substituting numbers from the ODZ.
Example1 . Solve the equation.
Solution . Having found the ODZ of this equation, we come to the conclusion that the ODZ of the original equation is a single-element set. Substitutingx=2into this equation, we come to the conclusion thatx=2is the root of the original equation.
Answer : 2 .
Example 2.
The equation has no solutions, because for each permissible value of the variable, the sum of two not negative numbers cannot be negative.
Example 3. 
+ 3 = 
.
ODZ:
The ODZ equation is an empty set.
Answer: the equation has no roots.
Example 4. 3
−4
−
=−(2+
).
ODZ:
ODZ: 
. By checking we are convinced that x=1 is the root of the equation.
Answer: 1.
Prove that the equation does not have
roots
1. 
= 0.
2. 
=1.
3. 5
.
4.
+ 
=2.
5.=
.
Solve the equation.
1. .
2. = 0.
3. 
= 92.
4. = 0.
5. 
+
+(x+3)(2005−x)=0.
2. B raising both sides of the equation to the natural power , that is, the transition from the equation
(1)
to the equation
. (2)
The following statements are true:
1) for any equation (2) is a consequence of equation (1);
2) if ( n is an odd number), then equations (1) and (2 ) are equivalent;
3) if ( n is an even number), then equation (2) is equivalent to the equation
, (3)
and equation (3) is equivalent to the set of equations
. (4)
In particular, the equation
(5)
is equivalent to the set of equations (4).
Example 1. Solve the equation
.
The equation is equivalent to the system
whence it follows that x=1, and the root does not satisfy the second inequality. At the same time, a competent solution does not require verification.
Answer:x=1.
Example 2. Solve the equation.
Solving the first equation of this system, which is equivalent to the equation 
, we get the roots and . However, at these values x the inequality does not hold, and therefore this equation has no roots.
Answer: no roots.
Example 3. Solve the equation
Isolating the first radical, we obtain the equation
equivalent to the original one.
By squaring both sides of this equation, since they are both positive, we get the equation
,
which is a consequence of the original equation. By squaring both sides of this equation under the condition that , we arrive at the equation
.
This equation has roots , . The first root satisfies the initial condition, but the second does not.
Answer: x=2.
If the equation contains two or more radicals, then they are first isolated and then squared.
Example 1.
Isolating the first radical, we obtain an equation equivalent to the given one. Let's square both sides of the equation:
Having performed the necessary transformations, we square the resulting equation 
After checking, we notice that
is not within the range of acceptable values.
Answer: 8.
Answer: 2
Answer: 3; 1.4.
3. Many irrational equations are solved by introducing auxiliary variables.
A convenient means of solving irrational equations is sometimes the method of introducing a new variable, or "replacement method" The method is usually applied when in Eq. some expression appears repeatedly, depending on an unknown quantity. Then it makes sense to denote this expression with some new letter and try to solve the equation first with respect to the introduced unknown, and then find the original unknown.
The successful choice of a new variable makes the structure of the equation more transparent. The new variable is sometimes obvious, sometimes somewhat veiled, but “felt”, and sometimes “manifests” only in the process of transformation.
Example 1. 
Let 
t>0, then
t = 
,
t 2 +5t-14=0,
t 1 =-7, t 2 =2. t=-7 does not satisfy the condition t>0, then
,
x 2 -2x-5=0,
x 1 =1- 
, x 2 =1+ .
Answer: 1- ; 1+.
Example 2. Solve an irrational equation 
Replacement:
Reverse replacement: /
Answer:
Example 3. Solve the equation 
.
Let's make replacements: , . The original equation will be rewritten in the form , from which we find that A = 4b And . Next, raising both sides of the equation 
squared, we get: From here X= 15. All that remains is to check:
- right!
Answer: 15.
Example 4. Solve the equation
Putting , we obtain a significantly simpler irrational equation. Let's square both sides of the equation: .
; 
;
; 
; , .
Checking the found values and substituting them into the equation shows that is the root of the equation, and is an extraneous root.
Returning to the original variable x, we obtain an equation, that is, a quadratic equation, solving which we find two roots: ,. Both roots satisfy the original equation.
Answer: , .
Replacement is especially useful if a new quality is achieved as a result, for example, an irrational equation turns into a rational one.
Example 6. Solve the equation.
Let's rewrite the equation like this: .
It can be seen that if we introduce a new variable 
, then the equation takes the form 
, where is the extraneous root and .
From the equation we obtain , .
Answer: , .
Example 7. Solve the equation 
.
Let's introduce a new variable, .
As a result, the original irrational equation takes the form of a quadratic
,
from where, taking into account the limitation, we obtain . Solving the equation, we get the root. Answer: 2,5.
Tasks for independent solution.
1. 
+
=
.
2. 
+
=.
3.
.
5. 
.
4.Method of introducing two auxiliary variables.
Equations of the form 
(Here a
,
b
,
c
,
d
–
some numbers m
,
n
–
natural numbers) and a number of other equations can often be solved by introducing two auxiliary unknowns: and , where and subsequent transition to equivalent system of rational equations.
Example 1. Solve the equation.
Raising both sides of this equation to the fourth power does not promise anything good. If we put , then the original equation is rewritten as follows: . Since we have introduced two new unknowns, we need to find another equation relating y And z. To do this, we raise the equalities to the fourth power and note that . So, we need to solve the system of equations
By squaring we get:
After substitution we have: or . Then the system has two solutions: , ; , , and the system has no solutions.
It remains to solve the system of two equations with one unknown
and the system The first of them gives, the second gives.
Answer: , .
Example 2.
Let 
Answer: 
5.
Equations with a radical of the third degree.
When solving equations containing radicals of the 3rd degree, it can be useful to use addition by identities:
Example 1.
.
Let's raise both sides of this equation to the 3rd power and use the above identity:
Note that the expression in brackets is equal to 1, which follows from the original equation. Taking this into account and bringing similar terms, we get:
Let's open the brackets, add similar terms and solve the quadratic equation. Its rootsAnd. If we assume (by definition) that odd roots can also be extracted from negative numbers, then both obtained numbers are solutions to the original equation.
Answer:.
6.Multiplying both sides of the equation by the conjugate expression of one of them.
Sometimes an irrational equation can be solved quite quickly if both sides are multiplied by a well-chosen function. Of course, when both sides of the equation are multiplied by a certain function, extraneous solutions may appear; they may turn out to be zeros of this function itself. Therefore, the proposed method requires mandatory research of the resulting values.
Example 1. Solve the equation
Solution: Let's select a function
Let's multiply both sides of the equation by the selected function:
Let us bring similar terms and obtain an equivalent equation
Let's add the original equation and the last one, we get
Answer: .
7. Identical transformations when solving irrational equations
When solving irrational equations, it is often necessary to apply identical transformations associated with the use of well-known formulas. Unfortunately, these actions are sometimes just as unsafe as raising to an even power—solutions can be gained or lost.
Let's look at several situations in which these problems occur, and learn how to recognize and prevent them.
I. Example 1. Solve the equation.
Solution. The formula that applies here is 
.
You just need to think about the safety of its use. It is easy to see that its left and right sides have different domains of definition and that this equality is true only under the condition . Therefore, the original equation is equivalent to the system
Solving the equation of this system, we obtain the roots and . The second root does not satisfy the set of inequalities of the system and, therefore, is an extraneous root of the original equation.
Answer: -1 .
II.The next dangerous transformation when solving irrational equations is determined by the formula.
If you use this formula from left to right, the ODZ expands and you can acquire third-party solutions. Indeed, on the left side both functions must be non-negative; and on the right, their product must be non-negative.
Let's look at an example where a problem is implemented using the formula.
Example 2. Solve the equation.
Solution. Let's try to solve this equation by factoring
Note that with this action the solution turned out to be lost, since it fits the original equation and no longer fits the resulting one: it does not make sense for . Therefore, it is better to solve this equation by ordinary squaring
Solving the equation of this system, we obtain the roots and . Both roots satisfy the system inequality.
Answer: , .
III There is an even more dangerous action - reduction by a common factor.
Example 3. Solve the equation 
.
Incorrect reasoning: Reduce both sides of the equation by , we get 
.
There is nothing more dangerous and wrong than this action. First, a suitable solution to the original equation was lost; secondly, two third-party solutions were purchased. It turns out that the new equation has nothing in common with the original one! Let's give the correct solution.
Solution. Let's move all the terms to the left side of the equation and factor it into factors
.
This equation is equivalent to the system
which has a unique solution.
Answer: 3 .
CONCLUSION.
As part of the study of the elective course, non-standard solutions are shown complex tasks who successfully develop logical thinking, the ability to find among many solutions the one that is comfortable and rational for the student. This course requires a lot of independent work from students, helps prepare students for continuing education, and improves the level of mathematical culture.
The work discussed the main methods for solving irrational equations, some approaches to solving equations of higher degrees, the use of which is assumed when solving Unified State Examination tasks, as well as when entering universities and continuing mathematical education. The content of basic concepts and statements related to the theory of solving irrational equations was also revealed. Having determined the most common method for solving equations, we identified its use in standard and non-standard situations. In addition, typical errors when performing identity transformations and ways to overcome them were considered.
When completing the course, students will have the opportunity to master various methods and techniques for solving equations, while learning to systematize and generalize theoretical information, independently search for solutions to certain problems and, in connection with this, compose a number of tasks and exercises on these topics. The choice of complex material will help schoolchildren to express themselves in research activities.
On the positive side The course is the possibility of further application by students of the studied material when passing the Unified State Exam and entering universities.
The negative side is that not every student is able to master all the techniques of this course, even if they want to, due to the difficulty of most of the problems solved.
LITERATURE:
Sharygin I.F. “Mathematics for those entering universities.” - 3rd ed., - M.: Bustard, 2000.
Equations and inequalities. Reference manual./ Vavilov V.V., Melnikov I.I., Olehnik S.N., Pasichenko P.I. –M.: Exam, 1998.
Cherkasov O.Yu., Yakushev A.G. "Mathematics: an intensive exam preparation course." – 8th ed., rev. and additional – M.:Iris, 2003. – (Home tutor)
Balayan E.N. Complex exercises and variants of training tasks for the Unified State Exam in mathematics. Rostov-on-Don: Phoenix Publishing House, 2004.
Skanavi M.I. “Collection of problems in mathematics for those entering universities.” - M., “Higher School”, 1998.
Igusman O.S. "Mathematics in the oral exam." - M., Iris, 1999.
Examination materials for preparing for the Unified State Exam – 2008 – 2012.
V.V. Kochagin, M.N. Kochagina “Unified State Examination - 2010. Mathematics. Tutor" Moscow "Enlightenment" 2010
V.A.Gusev, A.G.Mordkovich “Mathematics. Reference materials" Moscow "Enlightenment" 1988
Quadratic equations are studied in 8th grade, so there is nothing complicated here. The ability to solve them is absolutely necessary.
A quadratic equation is an equation of the form ax 2 + bx + c = 0, where the coefficients a, b and c are arbitrary numbers, and a ≠ 0.
Before studying specific solution methods, note that all quadratic equations can be divided into three classes:
- Have no roots;
- Have exactly one root;
- They have two different roots.
This is an important difference between quadratic equations and linear ones, where the root always exists and is unique. How to determine how many roots an equation has? There is a wonderful thing for this - discriminant.
Discriminant
Let the quadratic equation ax 2 + bx + c = 0 be given. Then the discriminant is simply the number D = b 2 − 4ac.
You need to know this formula by heart. Where it comes from is not important now. Another thing is important: by the sign of the discriminant you can determine how many roots a quadratic equation has. Namely:
- If D< 0, корней нет;
- If D = 0, there is exactly one root;
- If D > 0, there will be two roots.
Please note: the discriminant indicates the number of roots, and not at all their signs, as for some reason many people believe. Take a look at the examples and you will understand everything yourself:
Task. How many roots do quadratic equations have:
- x 2 − 8x + 12 = 0;
- 5x 2 + 3x + 7 = 0;
- x 2 − 6x + 9 = 0.
Let's write out the coefficients for the first equation and find the discriminant:
a = 1, b = −8, c = 12;
D = (−8) 2 − 4 1 12 = 64 − 48 = 16
So the discriminant is positive, so the equation has two different roots. We analyze the second equation in a similar way:
a = 5; b = 3; c = 7;
D = 3 2 − 4 5 7 = 9 − 140 = −131.
The discriminant is negative, there are no roots. The last equation left is:
a = 1; b = −6; c = 9;
D = (−6) 2 − 4 1 9 = 36 − 36 = 0.
The discriminant is zero - the root will be one.
Please note that coefficients have been written down for each equation. Yes, it’s long, yes, it’s tedious, but you won’t mix up the odds and make stupid mistakes. Choose for yourself: speed or quality.
By the way, if you get the hang of it, after a while you won’t need to write down all the coefficients. You will perform such operations in your head. Most people start doing this somewhere after 50-70 solved equations - in general, not that much.
Roots of a quadratic equation
Now let's move on to the solution itself. If the discriminant D > 0, the roots can be found using the formulas:
Basic formula for the roots of a quadratic equation
When D = 0, you can use any of these formulas - you will get the same number, which will be the answer. Finally, if D< 0, корней нет — ничего считать не надо.
- x 2 − 2x − 3 = 0;
- 15 − 2x − x 2 = 0;
- x 2 + 12x + 36 = 0.
First equation:
x 2 − 2x − 3 = 0 ⇒ a = 1; b = −2; c = −3;
D = (−2) 2 − 4 1 (−3) = 16.
D > 0 ⇒ the equation has two roots. Let's find them:
Second equation:
15 − 2x − x 2 = 0 ⇒ a = −1; b = −2; c = 15;
D = (−2) 2 − 4 · (−1) · 15 = 64.
D > 0 ⇒ the equation again has two roots. Let's find them
\[\begin(align) & ((x)_(1))=\frac(2+\sqrt(64))(2\cdot \left(-1 \right))=-5; \\ & ((x)_(2))=\frac(2-\sqrt(64))(2\cdot \left(-1 \right))=3. \\ \end(align)\]
Finally, the third equation:
x 2 + 12x + 36 = 0 ⇒ a = 1; b = 12; c = 36;
D = 12 2 − 4 1 36 = 0.
D = 0 ⇒ the equation has one root. Any formula can be used. For example, the first one:
As you can see from the examples, everything is very simple. If you know the formulas and can count, there will be no problems. Most often, errors occur when substituting negative coefficients into the formula. Here again, the technique described above will help: look at the formula literally, write down each step - and very soon you will get rid of errors.
Incomplete quadratic equations
It happens that a quadratic equation is slightly different from what is given in the definition. For example:
- x 2 + 9x = 0;
- x 2 − 16 = 0.
It is easy to notice that these equations are missing one of the terms. Such quadratic equations are even easier to solve than standard ones: they don’t even require calculating the discriminant. So, let's introduce a new concept:
The equation ax 2 + bx + c = 0 is called an incomplete quadratic equation if b = 0 or c = 0, i.e. the coefficient of the variable x or the free element is equal to zero.
Of course, a very difficult case is possible when both of these coefficients are equal to zero: b = c = 0. In this case, the equation takes the form ax 2 = 0. Obviously, such an equation has a single root: x = 0.
Let's consider the remaining cases. Let b = 0, then we obtain an incomplete quadratic equation of the form ax 2 + c = 0. Let us transform it a little:
Since the arithmetic square root exists only of a non-negative number, the last equality makes sense only for (−c /a) ≥ 0. Conclusion:
- If in an incomplete quadratic equation of the form ax 2 + c = 0 the inequality (−c /a) ≥ 0 is satisfied, there will be two roots. The formula is given above;
- If (−c /a)< 0, корней нет.
As you can see, a discriminant was not required—there are no complex calculations at all in incomplete quadratic equations. In fact, it is not even necessary to remember the inequality (−c /a) ≥ 0. It is enough to express the value x 2 and see what is on the other side of the equal sign. If there is a positive number, there will be two roots. If it is negative, there will be no roots at all.
Now let's look at equations of the form ax 2 + bx = 0, in which the free element is equal to zero. Everything is simple here: there will always be two roots. It is enough to factor the polynomial:
Taking the common factor out of bracketsThe product is zero when at least one of the factors is zero. This is where the roots come from. In conclusion, let’s look at a few of these equations:
Task. Solve quadratic equations:
- x 2 − 7x = 0;
- 5x 2 + 30 = 0;
- 4x 2 − 9 = 0.
x 2 − 7x = 0 ⇒ x · (x − 7) = 0 ⇒ x 1 = 0; x 2 = −(−7)/1 = 7.
5x 2 + 30 = 0 ⇒ 5x 2 = −30 ⇒ x 2 = −6. There are no roots, because a square cannot be equal to a negative number.
4x 2 − 9 = 0 ⇒ 4x 2 = 9 ⇒ x 2 = 9/4 ⇒ x 1 = 3/2 = 1.5; x 2 = −1.5.
