Rules for solving equations with variable values. Solving linear equations in one variable
Equation is an equality in which one or more variables are present.
We will consider the case when the equation has one variable, that is, one unknown number. Essentially, an equation is a type of mathematical model. Therefore, first of all, we need equations to solve problems.
Let us recall how a mathematical model is compiled to solve a problem.
For example, in the new academic year the number of students at school No. 5 doubled. After 20 students moved to another school, a total of 720 students began to study at school No. 5. How many students were there last year?
We need to express what is said in the condition in mathematical language. Let the number of students last year be X. Then, according to the conditions of the problem,
2X – 20 = 720. We have a mathematical model that represents equation with one variable. More precisely, it is an equation of the first degree with one variable. All that remains is to find its root.
What is the root of an equation?
The value of the variable at which our equation turns into a true equality is called the root of the equation. There are equations that have many roots. For example, in the equation 2*X = (5-3)*X, any value of X is a root. And the equation X = X +5 has no roots at all, since no matter what value we substitute for X, we will not get the correct equality. Solving an equation means finding all its roots, or determining that it has no roots. So to answer our question, we need to solve the equation 2X – 20 = 720.
How to solve equations with one variable?
First, let's write down some basic definitions. Every equation has a right and left side. In our case, (2X – 20) is the left side of the equation (it is to the left of the equal sign), and 720 is the right side of the equation. The terms on the right and left sides of the equation are called terms of the equation. Our equation terms are 2X, -20 and 720.
Let’s immediately talk about 2 properties of equations:
- Any term of the equation can be transferred from the right side of the equation to the left, and vice versa. In this case, it is necessary to change the sign of this term of the equation to the opposite. That is, records of the form 2X – 20 = 720, 2X – 20 – 720 = 0, 2X = 720 + 20, -20 = 720 – 2X are equivalent.
- Both sides of the equation can be multiplied or divided by the same number. This number must not be zero. That is, records of the form 2X – 20 = 720, 5*(2X – 20) = 720*5, (2X – 20):2 = 720:2 are also equivalent.
Let's move -20 to the right side with opposite sign. We get:
2X = 720 + 20. Let's add what we have on the right side. We get that 2X = 740.
Now divide the left and right sides of the equation by 2.
2X:2 = 740:2 or X = 370. We found the root of our equation and at the same time found the answer to the question of our problem. Last year there were 370 students at school No. 5.
Let's check whether our root really turns the equation into a true equality. Let's substitute the number 370 instead of X into the equation 2X – 20 = 720.
2*370-20 = 720.
That's right.
So, to solve an equation with one variable, it needs to be reduced to a so-called linear equation of the form ax = b, where a and b are some numbers. Then divide the left and right sides by the number a. We get that x = b:a.
What does it mean to reduce an equation to a linear equation?
Consider this equation:
5X - 2X + 10 = 59 - 7X +3X.
This is also an equation with one unknown variable X. Our task is to reduce this equation to the form ax = b.
To do this, we first collect all the terms that have X as a factor on the left side of the equation, and the remaining terms on the right side. Terms that have the same letter as a factor are called similar terms.
5X - 2X + 7X – 3X = 59 – 10.
According to the distributive property of multiplication, we can take the same factor out of brackets and add the coefficients (multipliers for the variable x). This process is also called reduction of like terms.
X(5-2+7-3) = 49.
7X = 49. We have reduced the equation to the form ax = b, where a = 7, b = 49.
And as we wrote above, the root of an equation of the form ax = b is x = b:a.
That is, X = 49:7 = 7.
Algorithm for finding the roots of an equation with one variable.
- Collect similar terms on the left side of the equation, and the remaining terms on the right side of the equation.
- Give similar terms.
- Reduce the equation to the form ax = b.
- Find the roots using the formula x = b:a.
Let's take two expressions with a variable: 4x and 5x + 2. Connecting them with an equal sign, we get the sentence 4x = 5x + 2. It contains a variable and, when substituting the values of the variable, turns into a statement.
For example, when x = -2, the sentence 4x = 5x + 2 turns into a true numerical equality 4-(-2) = 5-(-2) + 2, and when x = 1 - into a false 4-1 = 5-1+2. Therefore, the sentence 4x = 5x + 2 is an expressive form. They call her equation with one variable.
In general, an equation with one variable can be defined as follows:
Definition.Let f(x) and q(x) be two expressions with a variable x and a domain of definition X. Then an expressive form of the form f(x) =q(x) is called an equation with one variable.
Variable value X from many X, at which the equation turns into a true numerical equality is called root of the equation (or his decision). Solving an equation means finding its many roots. .
Thus, the root of the equation 4x = 5x + 2, if we consider it on the set R real numbers, is the number -2. This equation has no other roots. This means that the set of its roots is (-2).
Let the equation (x-1)(x+2)=0 be given on the set of real numbers. It has two roots - the numbers 1 and -2. Therefore, the set of roots of this equation is: (-2,- 1).
The equation (3x + 1) × 2 = 6x + 2, given on the set of real numbers, turns into a true numerical equality for all real values of the variable x: if we open the parentheses on the left side, we get 6x + 2 = 6 X+ 2. In this case, we say that its root is any real number, and the set of roots is the set of all real numbers.
The equation (3x + 1)-2 = 6x + 1, given on the set of real numbers, does not turn into a true numerical equality for any real value of x: after opening the parentheses on the left side, we obtain that 6x + 2 = 6x + 1, which impossible for any x. In this case, we say that the given equation has no roots and that the set of its roots is empty.
To solve any equation, it is first transformed, replacing it with another, simpler one; the resulting equation is again transformed, replacing it with a simpler one, etc. This process is continued until an equation is obtained whose roots can be found in a known way. But for these roots to be the roots of a given given equation, it is necessary that the transformation process produces equations whose sets of roots coincide. Such equations are called equivalent.
Definition.Two equations f 1 (x) =q 1 (x) and f 2 (x) =q 2 (x) are called equivalent if the sets of their roots coincide.
For example, equations x 2 - 9 = 0 and (2x + 6)(x - 3) = 0 are equivalent since both have their roots in the numbers 3 and -3. The equations (3x + 1)-2 = 6x + 1 and x 2 + 1 are also equivalent = 0, since both have no roots, i.e. the sets of their roots coincide.
Definition. Replacing an equation with an equivalent equation is called an equivalent transformation.
Let us now find out what transformations allow us to obtain equivalent equations.
Theorem 1. Let the equation f(x) = q(x) be defined on a set and h(x) be an expression defined on the same set. Then the equation f(x) = q(x) (1) and f(x) + h(x) = q(x) + h(x) (2) are equivalent.
Proof. Let us denote by T 1 the set of solutions to equation (1), and by T 2 the set of solutions to equation (2). Then equations (1) and (2) will be equivalent if T 1 = T 2. To verify this, it is necessary to show that any root of T 1 is a root of equation (2) and, conversely, any root of T 2 is a root of equation (1).
Let the number a be the root of equation (1). Then a О Т 1, and when substituting into equation (1) turns it into a true numerical equality f(a) = q(a), and turns the expression h(x) into a numerical expression h(a) that makes sense on the set X. Let us add to both sides of the true equality f(a) = q(a) the numerical expression h(a). We obtain, according to the properties of true numerical equalities, the true numerical equality f(a) + h(a) = q(a) + h(a), which indicates that the number a is the root of equation (2).
So, it has been proven that every root of equation (1) is also a root of equation (2), i.e. Т 1 М Т 2.
Let now a be the root of equation (2). Then a О Т 2, and when substituting into equation (2) turns it into a true numerical equality f(a) + h(a) = q(a) + h(a). Let's add to both sides of this equality the numerical expression - h(a). We obtain the true numerical equality f(a) = q(a), that the number a is the root of equation (1).
So, it has been proven that every root of equation (2) is also a root of equation (1), i.e. Т 2 М Т 1 .
Since T 1 Ì T 2 and T 2 Ì T 1, then by definition of equal sets T 1 = T 2, which means that equations (1) and (2) are equivalent.
This Theorem 1 can be formulated differently: If we add the same expression with a variable, defined on the same set, to both sides of the equation with the domain of definition X, we obtain a new equation equivalent to the given one.
Corollaries follow from this theorem and are used in solving equations:
1. If we add the same number to both sides of the equation, we get an equation equivalent to the given one.
2. If any term (numerical expression or expression with a variable) is transferred from one part of the equation to another, changing the sign of the term to the opposite, then we obtain an equation equivalent to the given one.
Theorem 2.Let the equation f(x) = q(x) be defined on the set X and let h(x) be an expression that is defined on the same set and does not vanish for any values of x from the set X. Then the equations f(x) = q(x) and f(x) × h(x) = q(x) × h(x) are equivalent.
The proof of this theorem is similar to the proof of Theorem 1.
Theorem 2 can be formulated differently: If both sides of the equation with the domain of definition X are multiplied by the same expression, which is defined on the same set and does not vanish on it, then we obtain a new equation equivalent to the given one.
A corollary follows from this theorem: If both sides of the equation are multiplied (or divided) by the same number other than zero, we obtain an equation equivalent to the given one.
Let's solve the equation , x О R, and justify all the transformations that we will perform in the solution process.
Etc., it is logical to get acquainted with equations of other types. Next in line are linear equations, the targeted study of which begins in algebra lessons in the 7th grade.
It is clear that first you need to explain what a linear equation is, give a definition of a linear equation, its coefficients, show it general view. Then you can figure out how many solutions a linear equation has depending on the values of the coefficients, and how the roots are found. This will allow you to move on to solving examples, and thereby consolidate the learned theory. In this article we will do this: we will dwell in detail on all the theoretical and practical points relating to linear equations and their decisions.
Let’s say right away that here we will consider only linear equations with one variable, and in a separate article we will study the principles of solution linear equations with two variables.
Page navigation.
What is a linear equation?
The definition of a linear equation is given by the way it is written. Moreover, in different mathematics and algebra textbooks, the formulations of the definitions of linear equations have some differences that do not affect the essence of the issue.
For example, in the algebra textbook for grade 7 by Yu. N. Makarychev et al., a linear equation is defined as follows:
Definition.
Equation of the form a x=b, where x is a variable, a and b are some numbers, is called linear equation with one variable.
Let us give examples of linear equations that meet the stated definition. For example, 5 x = 10 is a linear equation with one variable x, here the coefficient a is 5, and the number b is 10. Another example: −2.3·y=0 is also a linear equation, but with a variable y, in which a=−2.3 and b=0. And in linear equations x=−2 and −x=3.33 a are not present explicitly and are equal to 1 and −1, respectively, while in the first equation b=−2, and in the second - b=3.33.
And a year earlier, in the textbook of mathematics by N. Ya. Vilenkin, linear equations with one unknown, in addition to equations of the form a x = b, also considered equations that can be brought to this form by transferring terms from one part of the equation to another with the opposite sign, as well as by reducing similar terms. According to this definition, equations of the form 5 x = 2 x + 6, etc. also linear.
In turn, in the algebra textbook for grade 7 by A. G. Mordkovich the following definition is given:
Definition.
Linear equation with one variable x is an equation of the form a·x+b=0, where a and b are some numbers called coefficients of the linear equation.
For example, linear equations of this type are 2 x−12=0, here the coefficient a is 2, and b is equal to −12, and 0.2 y+4.6=0 with coefficients a=0.2 and b =4.6. But at the same time, there are examples of linear equations that have the form not a·x+b=0, but a·x=b, for example, 3·x=12.
Let us, so that we do not have any discrepancies in the future, by a linear equation with one variable x and coefficients a and b we mean an equation of the form a x + b = 0. This type of linear equation seems to be the most justified, since linear equations are algebraic equations first degree. And all the other equations indicated above, as well as equations that, using equivalent transformations, are reduced to the form a x + b = 0, we will call equations that reduce to linear equations. With this approach, the equation 2 x+6=0 is a linear equation, and 2 x=−6, 4+25 y=6+24 y, 4 (x+5)=12, etc. - These are equations that reduce to linear ones.
How to solve linear equations?
Now it’s time to figure out how linear equations a·x+b=0 are solved. In other words, it's time to find out whether a linear equation has roots, and if so, how many of them and how to find them.
The presence of roots of a linear equation depends on the values of the coefficients a and b. In this case, the linear equation a x+b=0 has
- the only root for a≠0,
- has no roots for a=0 and b≠0,
- has infinitely many roots for a=0 and b=0, in which case any number is a root of a linear equation.
Let us explain how these results were obtained.
We know that to solve equations we can move from the original equation to equivalent equations, that is, to equations with the same roots or, like the original one, without roots. To do this, you can use the following equivalent transformations:
- transferring a term from one part of the equation to another with the opposite sign,
- as well as multiplying or dividing both sides of an equation by the same non-zero number.
So, in a linear equation with one variable of the form a·x+b=0, we can move the term b from the left side to the right side with the opposite sign. In this case, the equation will take the form a·x=−b.
And then it begs the question of dividing both sides of the equation by the number a. But there is one thing: the number a can be equal to zero, in which case such division is impossible. To deal with this problem, we first assume that the number a is nonzero, and the case equal to zero We’ll look at it separately a little later.
So, when a is not equal to zero, then we can divide both sides of the equation a·x=−b by a, after which it will transform to the form x=(−b):a, this result can be written using the fractional slash as.
Thus, for a≠0, the linear equation a·x+b=0 is equivalent to the equation, from which its root is visible.
It is easy to show that this root is unique, that is, the linear equation has no other roots. This allows you to do the opposite method.
Let's denote the root as x 1. Let us assume that there is another root of the linear equation, which we denote as x 2, and x 2 ≠x 1, which, due to determining equal numbers through difference is equivalent to the condition x 1 −x 2 ≠0. Since x 1 and x 2 are roots of the linear equation a·x+b=0, then the numerical equalities a·x 1 +b=0 and a·x 2 +b=0 hold. We can subtract the corresponding parts of these equalities, which the properties of numerical equalities allow us to do, we have a·x 1 +b−(a·x 2 +b)=0−0, from which a·(x 1 −x 2)+( b−b)=0 and then a·(x 1 −x 2)=0 . But this equality is impossible, since both a≠0 and x 1 − x 2 ≠0. So we came to a contradiction, which proves the uniqueness of the root of the linear equation a·x+b=0 for a≠0.
So we solved the linear equation a·x+b=0 for a≠0. The first result given at the beginning of this paragraph is justified. There are two more left that meet the condition a=0.
When a=0, the linear equation a·x+b=0 takes the form 0·x+b=0. From this equation and the property of multiplying numbers by zero it follows that no matter what number we take as x, when it is substituted into the equation 0 x + b=0, the numerical equality b=0 will be obtained. This equality is true when b=0, and in other cases when b≠0 this equality is false.
Therefore, with a=0 and b=0, any number is the root of the linear equation a·x+b=0, since under these conditions, substituting any number for x gives the correct numerical equality 0=0. And when a=0 and b≠0, the linear equation a·x+b=0 has no roots, since under these conditions, substituting any number for x leads to the incorrect numerical equality b=0.
The given justifications allow us to formulate a sequence of actions that allows us to solve any linear equation. So, algorithm for solving linear equation is:
- First, by writing the linear equation, we find the values of the coefficients a and b.
- If a=0 and b=0, then this equation has infinitely many roots, namely, any number is a root of this linear equation.
- If a is nonzero, then
- the coefficient b is transferred to the right side with the opposite sign, and the linear equation is transformed to the form a·x=−b,
- after which both sides of the resulting equation are divided by a nonzero number a, which gives the desired root of the original linear equation.
The written algorithm is a comprehensive answer to the question of how to solve linear equations.
In conclusion of this point, it is worth saying that a similar algorithm is used to solve equations of the form a·x=b. Its difference is that when a≠0, both sides of the equation are immediately divided by this number; here b is already in the required part of the equation and there is no need to transfer it.
To solve equations of the form a x = b, the following algorithm is used:
- If a=0 and b=0, then the equation has infinitely many roots, which are any numbers.
- If a=0 and b≠0, then the original equation has no roots.
- If a is non-zero, then both sides of the equation are divided by a non-zero number a, from which the only root of the equation is found, equal to b/a.
Examples of solving linear equations
Let's move on to practice. Let's look at how the algorithm for solving linear equations is used. Let us give solutions to typical examples corresponding to different meanings coefficients of linear equations.
Example.
Solve the linear equation 0·x−0=0.
Solution.
In this linear equation, a=0 and b=−0 , which is the same as b=0 . Therefore, this equation has infinitely many roots; any number is a root of this equation.
Answer:
x – any number.
Example.
Does the linear equation 0 x + 2.7 = 0 have solutions?
Solution.
In this case, coefficient a is equal to zero, and coefficient b of this linear equation is equal to 2.7, that is, different from zero. Therefore, a linear equation has no roots.
When studying Russian at school, many wondered: why the word plain written through A , because the test word smooth written through O ? Actually the answer is simple. After all, the plain is so called because all its points are located on equal terms distance (from sea level) and a test word for it - equals.
ABOUT definition: An equation with a variable x is an equality of the form A(x)=B(x), where A(x) and B(x) are expressions for x. Many T values of x, when substituted into the equation to obtain a true numerical equality, are called truth set given equation or decision of this equation, and each such value variable - root of the equation.
Thus it becomes clear that the basis of any equation is equalitiesO its two parts. And when, when solving equations, one works on its parts, this equality must always be observed.
Methods for solving equations with one variable
Exists huge amount a wide variety of types of equations for solving which are used different ways. But in order to easily solve equations you need to know three basic methods:
Identical transformation of equations
Factoring an Expression
Introduction of a new variable
Identical transformations of equations
The simplest and at the same time one of the most common ways to solve equations is the method of identity transformations. IN any equations To find the unknown, you need to transform and simplify the original example. And so that when changing appearance the essence of the equation has not changed. Such transformations are called identical or equivalent. Let's consider the main methods of identical transformations of algebraic expressions.
Examples and formulas of identity transformations:
First identity transformation: you can add (subtract) to both sides of any equation any(but one and the same!) number or expression (including an expression with an unknown!). This does not change the essence of the equation.
Example: 9 x 2 + 12x+ 10 = 15x+ 10 → subtract ten from both sides → 9 x 2 + 12x = 15x
Second identity transformation: transferring terms of an equation from one side to the other with opposite signs.
Example: 9 x 2 + 12x = 15x→ move 15x to the left → 9 x 2 + 12x — 15x =0. After simplification we get: 9 x 2 - 3x =0
Third identity transformation: both sides of the equation can be multiplied (divided) by the same thing non-zero number or expression. Here an understandable limitation already appears: multiplying by zero is stupid, and dividing is completely impossible.
Example: 9 x 2 - 3x =0 → divide both sides of the equation by three → 3x 2 - x =0
Fourth identity transformation: Can raise both sides of the equation to an odd power or extractthe odd root of both sides of the equation. It must be remembered that:
a) construction in even may lead to purchaseextraneous roots;
b) wrong extraction even root may lead to loss of roots.
Example: 49 x 2 = 1225 → take the square root of both parts → | 7 x| = 35
Factoring an Expression
Let us now list some of the most common techniques for factoring polynomials, as the simplest algebraic ones.
Taking the common factor out of brackets
In the case when all terms of a polynomial have the same common factor, it can be taken out of brackets, thereby obtaining an expansion of the polynomial.
Example: Factor the polynomial x 5 – 2x 3 + x 2.
Solution: Each term of this polynomial contains a factor x 2. Let's take it out of the bracket and get the answer:
x 5 – 2x 3 + x 2 = x 2 (x 3 – 2x + 1).
Application of abbreviated multiplication formulas
Abbreviations are used quite effectively when factoring a polynomial. It is useful to remember the following formulas:
1. The square of the sum of two quantities is equal to the square of the first plus twice the product of the first and the second plus the square of the second.
(a+b) 2 =a 2 +2ab+b 2
2. The square of the difference between two quantities is equal to the square of the first minus twice the product of the first and the second plus the square of the second.
(a-b) 2 =a 2 -2ab+b 2
3. The product of the sum of two quantities and their difference is equal to the difference of their squares.
(a+b)(a-b)=a 2 -b 2
4. The cube of the sum of two quantities is equal to the cube of the first plus triple the product of the square of the first and the second plus the triple product of the first by the square of the second plus the cube of the second.
(a+b) 3 =a 3 +3a 2 b+3ab 2 +b 3
5. The cube of the difference between two quantities is equal to the cube of the first minus the triple product of the square of the first and the second plus the triple product of the first by the square of the second minus the cube of the second.
(a-b) 3 =a 3 -3a 2 b+3ab 2 -b 3
6. The product of the sum of two quantities and the partial square of the difference is equal to the sum of their cubes.
(a+b)(a 2 -ab+b 2)=a 3 +b 3
7. The product of the difference of two quantities by the partial square of the sum is equal to the difference of their cubes.
(a-b)(a 2 +ab+b 2)=a 3 -b 3
Example: (3x+5) 2 =9x 2 +30x+25=0
Solution: using formula (1) 9x 2 +30x+25=(3x+5) 2
Applying a full square selection
Without exaggeration, we can say that the method of isolating a complete square is one of the most effective methods factorization used when passing and
Equation is an equality containing a variable denoted by a letter.
Root of the equation(or solution to an equation) is the value of a variable at which the equation turns into a true equality.
Example: solve the equation (that is, find the root of the equation): 4 x - 15 = x + 15
So:
4x - x= 15 + 15
3X = 30
X= 30: 3
X = 10
Result: the equation has one root - the number 10.
An equation can have two, three, four or more roots.
For example, the equation ( X- 4)(X- 5)(X- 6) = 0 has three roots :
4, 5 and 6.
The equation may have no roots at all.
For example, the equation X + 2 = X has no roots, because at any value X equality is impossible.
Equivalence of equations.
Two equations are equivalent if they have the same roots or if both equations have no roots.
Example1:
Equations X+ 3 = 5 and 3 X- 1 = 5 are equivalent, since in both equations X= 2.
Example 2:
Equations X 4 + 2 = 1 and X 2 + 5 = 0 are equivalent, since both equations have no roots.
Whole equation with one variable is an equation whose left and right sides are integer expressions (for integer expressions, see the section “Rational Expressions”).
An equation with one variable can be written as P(x) = 0, where P(x) is a polynomial of standard form.
For example:
y 2 + 3y - 6 = 0
(Here P(x) represented as a polynomial y 2 + 3y - 6).
In such an equation, the degree of the polynomial is called degree of equation.
Our example represents an equation of the second degree (since it contains a polynomial of the second degree).
Equation of the first degree.
The first degree equation can be reduced to the form:
ax+b = 0,
Where x- variable, a And b- some numbers, and a ≠ 0.
From here it is easy to derive the value x:
b
x = - —
a
This is the meaning x is the root of the equation.
Equations of the first degree have one root.
Equation of the second degree.
The second degree equation can be reduced to the form:
ax 2 + bx + c = 0,
Where x- variable, a,b,c- some numbers, and a ≠ 0.
The number of roots of the second degree equation depends on the discriminant:
If D > 0, then the equation has two roots;
If D = 0, then the equation has one root;
If D< 0, то уравнение корней не имеет.
An equation of the second degree can have no more than two roots.
(about what a discriminant is and how to find the roots of an equation, see the sections “Formula for the roots of a quadratic equation. Discriminant” and “Another way to solve a quadratic equation”).
Equation of the third degree.
The third degree equation can be reduced to the form:
ax 3 + bx 2 + cx + d = 0,
Where x- variable, a,b,c,d- some numbers, and a ≠ 0.
An equation of the third degree can have no more than three roots.
Equation of the fourth degree.
The fourth degree equation can be reduced to the form:
ax 4 + bx 3 + cx 2 + dx +e = 0,
Where x- variable, a,b,c,d,e- some numbers, and a ≠ 0.
An equation of the third degree can have no more than four roots.
Summary:
1) equation of the fifth, sixth, etc. degrees can be easily derived independently by following the above diagram;
2) equation n- degree can have no more n roots
Example 1 : Let's solve the equation
x 3 - 8x 2 - x + 8 = 0.
We see that this is an equation of the third degree. This means that it can have from zero to three roots.
Let's find them and thereby solve the equation.
Let's factorize the left side of the equation:
x 2 (x - 8) - (x - 8) = 0.
Let us apply the rule for decomposing a polynomial by grouping its terms. To do this, put the number 1 in front of the second brackets:
x 2 (x - 8) - 1(x - 8) = 0.
Now let's group the polynomials x 2 and -1, which are factors of the polynomial x-8. We get two groups of polynomials: ( x 2 -1) and ( x- 8). Therefore, our equation will take a new form:
(x - 8)(x 2 - 1) = 0.
Here the expression x 2 - 1 can be represented as x 2 - 1 2 . This means we can apply the abbreviated multiplication formula: x 2 - 1 2 = (x - 1)(x+ 1). Let's substitute this expression into our equation and get:
(x - 8)(x - 1)(x + 1) = 0.
x - 8 = 0
x - 1 = 0
x + 1 = 0
It remains to find the roots of our equation:
x 1 = 0 + 8 = 8
x 2 = 0 + 1 = 1
x 3 = 0 - 1 = -1.
The equation is solved. It has three roots: 8, 1 and -1.
Example 2 : Let's solve the equation
(x 2 - 5x + 4)(x 2 - 5x +6) = 120
This equation is more complicated. But it can be simplified in an original way - by introducing a new variable.
In our equation the expression occurs twice x 2 - 5x.
We can denote it by a variable y. That is, let’s imagine that x 2 - 5x = y.
Then our equation takes on a simpler form:
(y + 4)(y + 6) = 120.
Let's expand the brackets:
y 2 + 4y + 6y + 24 = 120
y 2 + 10y + 24 = 120
Let's set the equation equal to zero:
y 2 + 10y + 24 - 120 = 0
y 2 + 10y - 96 = 0
We got the usual quadratic equation. Let's find its roots. There is no need to make calculations: how to solve such equations is written in detail in the sections “Quadratic Equations” and “Formula for the Roots of a Quadratic Equation.” Discriminant." Here we will immediately display the result. The quadratic equation y 2 + 10y - 96 = 0 has two roots:
y 1 = -16
y 2 = 6
Letter y we replaced the expression x 2 - 5x. This means we can already substitute the values y and find the roots of the given equation, thereby solving the problem:
1) First we apply the value y 1 = -16:
x 2 - 5x = -16
To solve this equation, we turn it into a quadratic equation:
x 2 - 5x + 16 = 0
Having solved it, we find that it has no roots.
2) Now apply the value y 2 = 6:
x 2 - 5x = 6
x 2 - 5x - 6 = 0
Having solved this quadratic equation, we see that it has two roots:
x 1 = -1
x 2 = 6.
The equation is solved. It has two roots: -1 and 6.
The method of introducing a new variable makes it easy to solve equations of the fourth degree that are quadratic in x 2 (such equations are called biquadratic).
