How to solve equations with fractions. Exponential solution of equations with fractions
In this article, I will show you algorithms for solving seven types of rational equations, which are reduced to square by changing variables. In most cases, the transformations that lead to the replacement are very non-trivial, and it is rather difficult to guess about them on your own.
For each type of equation, I will explain how to change a variable in it, and then I will show a detailed solution in the corresponding video tutorial.
You have the opportunity to continue solving the equations yourself, and then check your solution against the video tutorial.
So, let's begin.
1 ... (x-1) (x-7) (x-4) (x + 2) = 40
Note that there is a product of four brackets on the left side of the equation, and a number on the right.
1. Let's group the brackets by two so that the sum of free terms is the same.
2. Let's multiply them.
3. We introduce a change of variable.
In our equation, we group the first bracket with the third, and the second with the fourth, since (-1) + (- 4) = (- 7) +2:
At this point, the variable replacement becomes obvious:
We get the equation 
Answer: 
2 .
An equation of this type is similar to the previous one with one difference: on the right side of the equation is the product of a number by. And it is solved in a completely different way:
1. We group the brackets by two so that the product of free terms is the same.
2. Multiply each pair of parentheses.
3. From each factor we take out x from the bracket.
4. Divide both sides of the equation by.
5. Introduce variable replacement.
In this equation, we group the first bracket with the fourth, and the second with the third, since:
Note that in each parenthesis the coefficient at and the free term are the same. Take a factor out of each parenthesis:
Since x = 0 is not a root of the original equation, we divide both sides of the equation by. We get:
We get the equation: 
Answer: 
3
. 
Note that the denominators of both fractions contain square trinomials with the same leading coefficient and free term. Let us take out, as in the equation of the second type, x outside the parenthesis. We get:
Divide the numerator and denominator of each fraction by x:
Now we can introduce variable replacement:
We get the equation for the variable t:
4 .
Note that the coefficients of the equation are symmetric with respect to the central one. Such an equation is called returnable .
To solve it,
1. Divide both sides of the equation by (We can do this, since x = 0 is not the root of the equation.) We get:
2. Let us group the terms in this way:
3. In each group, take the common factor out of the bracket:
4. Let's introduce a replacement:
5. Let us express the expression in terms of t:
From here 
We get the equation for t:
Answer:
5. Homogeneous equations.
Equations that have the structure of a homogeneous one can be encountered when solving exponential, logarithmic and trigonometric equations so you need to be able to recognize it.
Homogeneous equations have the following structure:
In this equality, A, B and C are numbers, and the same expressions are denoted by a square and a circle. That is, on the left side of the homogeneous equation there is a sum of monomials with the same degree (in this case, the degree of monomials is 2), and there is no free term.
To solve the homogeneous equation, we divide both sides by
Attention! When dividing the right and left sides of the equation by an expression containing the unknown, you can lose roots. Therefore, it is necessary to check if the roots of the expression by which we divide both sides of the equation are not the roots of the original equation.
Let's go the first way. We get the equation:
Now we introduce variable replacement:
Let's simplify the expression and get a biquadratic equation for t:
Answer: or
7
. 
This equation has the following structure:
To solve it, you need to select a full square on the left side of the equation.
To select a complete square, you need to add or subtract a satisfactory work. Then we get the square of the sum or the difference. This is crucial for a successful variable replacement.
Let's start by finding the doubled product. It will be the key to replace the variable. In our equation, the doubled product is
Now let's estimate what is more convenient for us to have - the square of the sum or the difference. Consider, first, the sum of the expressions:
Excellent! this expression is exactly equal to twice the product. Then, in order to get the square of the sum in brackets, you need to add and subtract the doubled product:
First of all, in order to learn how to work with rational fractions without errors, you need to learn the abbreviated multiplication formulas. And it's not easy to learn - they need to be recognized even when sines, logarithms and roots act as terms.
However, the main tool remains the factorization of the numerator and denominator of a rational fraction. This can be accomplished in three different ways:
- Actually, according to the formula of abbreviated multiplication: they allow you to fold a polynomial into one or more factors;
- Using the factorization of a square trinomial in terms of the discriminant. The same method allows us to make sure that any trinomial does not decompose into factors at all;
- The grouping method is the most difficult tool, but it is the only method that works if the previous two did not work.
As you probably already guessed from the title of this video, we'll talk again about rational fractions. Just a few minutes ago, I finished a lesson with one tenth grader, and there we analyzed these very expressions. Therefore, this lesson will be intended specifically for high school students.
Surely many will now have a question: "Why should students in grades 10-11 learn such simple things as rational fractions, because this is done in grade 8?" But the trouble is that most people just "pass" this topic. In grades 10-11, they no longer remember how multiplication, division, subtraction and addition of rational fractions from the 8th grade is done, and it is on this simple knowledge that further, more complex constructions are built, such as the solution of logarithmic, trigonometric equations and many others complex expressions, so there is practically nothing to do in high school without rational fractions.
Formulas for solving problems
Let's get down to business. First of all, we need two facts - two sets of formulas. First of all, you need to know the abbreviated multiplication formulas:
- $ ((a) ^ (2)) - ((b) ^ (2)) = \ left (a-b \ right) \ left (a + b \ right) $ - difference of squares;
- $ ((a) ^ (2)) \ pm 2ab + ((b) ^ (2)) = ((\ left (a \ pm b \ right)) ^ (2)) $ - the square of the sum or difference;
- $ ((a) ^ (3)) + ((b) ^ (3)) = \ left (a + b \ right) \ left (((a) ^ (2)) - ab + ((b) ^ ( 2)) \ right) $ - the sum of cubes;
- $ ((a) ^ (3)) - ((b) ^ (3)) = \ left (ab \ right) \ left (((a) ^ (2)) + ab + ((b) ^ (2) ) \ right) $ - difference of cubes.
In their pure form, they are not found in any examples and in real serious expressions. Therefore, our task is to learn to see much more complex constructions under the letters $ a $ and $ b $, for example, logarithms, roots, sines, etc. You can learn to see this only through constant practice. That is why solving rational fractions is absolutely necessary.
The second, completely obvious formula is the factorization of a square trinomial:
$ ((x) _ (1)) $; $ ((x) _ (2)) $ - roots.
We have dealt with the theoretical part. But how to solve real rational fractions, which are considered in grade 8? Now we are going to practice.
Problem number 1
\ [\ frac (27 ((a) ^ (3)) - 64 ((b) ^ (3))) (((b) ^ (3)) - 4): \ frac (9 ((a) ^ (2)) + 12ab + 16 ((b) ^ (2))) (((b) ^ (2)) + 4b + 4) \]
Let's try to apply the above formulas to solving rational fractions. First of all, I want to explain why factoring is needed at all. The fact is that at the first glance at the first part of the assignment, you want to reduce a cube with a square, but this is absolutely impossible, because they are terms in the numerator and in the denominator, but in no case are factors.
In general, what is a reduction? Abbreviation is a general rule of thumb for dealing with such expressions. The main property of a fraction is that we can multiply the numerator and denominator by the same number, other than zero. In this case, when we reduce, then, on the contrary, we divide by the same number, different from "zero". However, we must divide all the terms in the denominator by the same number. You can't do that. And we have the right to reduce the numerator with the denominator only when both of them are factorized. Let's do it.
Now you need to see how many terms are in a particular element, in accordance with this, find out which formula you need to use.
Let's convert each expression to an exact cube:
Let's rewrite the numerator:
\ [((\ left (3a \ right)) ^ (3)) - ((\ left (4b \ right)) ^ (3)) = \ left (3a-4b \ right) \ left (((\ left (3a \ right)) ^ (2)) + 3a \ cdot 4b + ((\ left (4b \ right)) ^ (2)) \ right) \]
Let's take a look at the denominator. Let's expand it according to the formula of the difference of squares:
\ [((b) ^ (2)) - 4 = ((b) ^ (2)) - ((2) ^ (2)) = \ left (b-2 \ right) \ left (b + 2 \ right) \]
Now let's look at the second part of the expression:
Numerator:
It remains to figure out the denominator:
\ [((b) ^ (2)) + 2 \ cdot 2b + ((2) ^ (2)) = ((\ left (b + 2 \ right)) ^ (2)) \]
Let's rewrite the whole construction taking into account the above facts:
\ [\ frac (\ left (3a-4b \ right) \ left (((\ left (3a \ right)) ^ (2)) + 3a \ cdot 4b + ((\ left (4b \ right)) ^ (2 )) \ right)) (\ left (b-2 \ right) \ left (b + 2 \ right)) \ cdot \ frac (((\ left (b + 2 \ right)) ^ (2))) ( ((\ left (3a \ right)) ^ (2)) + 3a \ cdot 4b + ((\ left (4b \ right)) ^ (2))) = \]
\ [= \ frac (\ left (3a-4b \ right) \ left (b + 2 \ right)) (\ left (b-2 \ right)) \]
Nuances of multiplying rational fractions
The key takeaway from these constructions is as follows:
- Not every polynomial is factorized.
- Even if it unfolds, it is necessary to carefully look at which exact formula for abbreviated multiplication.
To do this, firstly, you need to estimate how many terms in total (if there are two of them, then all we can do is expand them either by the sum of the difference of squares, or by the sum or difference of cubes; and if there are three of them, then this is , uniquely, either the square of the sum or the square of the difference). It often happens that either the numerator or the denominator does not require factorization at all, it can be linear, or its discriminant will be negative.
Problem number 2
\ [\ frac (3-6x) (2 ((x) ^ (2)) + 4x + 8) \ cdot \ frac (2x + 1) (((x) ^ (2)) + 4-4x) \ cdot \ frac (8 - ((x) ^ (3))) (4 ((x) ^ (2)) - 1) \]
In general, the scheme for solving this problem is no different from the previous one - there will simply be more actions, and they will become more diverse.
Let's start with the first fraction: let's look at its numerator and make possible transformations:
Now let's look at the denominator:
With the second fraction: nothing can be done in the numerator at all, because this is a linear expression, and you cannot take out any factor from it. Let's look at the denominator:
\ [((x) ^ (2)) - 4x + 4 = ((x) ^ (2)) - 2 \ cdot 2x + ((2) ^ (2)) = ((\ left (x-2 \ right )) ^ (2)) \]
We go to the third fraction. Numerator:
Let's deal with the denominator of the last fraction:
Let's rewrite the expression taking into account the above facts:
\ [\ frac (3 \ left (1-2x \ right)) (2 \ left (((x) ^ (2)) + 2x + 4 \ right)) \ cdot \ frac (2x + 1) ((( \ left (x-2 \ right)) ^ (2))) \ cdot \ frac (\ left (2-x \ right) \ left (((2) ^ (2)) + 2x + ((x) ^ ( 2)) \ right)) (\ left (2x-1 \ right) \ left (2x + 1 \ right)) = \]
\ [= \ frac (-3) (2 \ left (2-x \ right)) = - \ frac (3) (2 \ left (2-x \ right)) = \ frac (3) (2 \ left (x-2 \ right)) \]
Solution nuances
As you can see, not everything and not always rests on the abbreviated multiplication formulas - sometimes it is just enough to exclude a constant or a variable from the brackets. However, there is also the opposite situation, when there are so many terms or they are constructed in such a way that the formulas for abbreviated multiplication to them are generally impossible. In this case, a universal tool comes to our aid, namely, the grouping method. This is what we will now apply in the next task.
Problem number 3
\ [\ frac (((a) ^ (2)) + ab) (5a - ((a) ^ (2)) + ((b) ^ (2)) - 5b) \ cdot \ frac (((a ) ^ (2)) - ((b) ^ (2)) + 25-10a) (((a) ^ (2)) - ((b) ^ (2))) \]
Let's take a look at the first part:
\ [((a) ^ (2)) + ab = a \ left (a + b \ right) \]
\ [= 5 \ left (ab \ right) - \ left (ab \ right) \ left (a + b \ right) = \ left (ab \ right) \ left (5-1 \ left (a + b \ right ) \ right) = \]
\ [= \ left (a-b \ right) \ left (5-a-b \ right) \]
Let's rewrite the original expression:
\ [\ frac (a \ left (a + b \ right)) (\ left (ab \ right) \ left (5-ab \ right)) \ cdot \ frac (((a) ^ (2)) - ( (b) ^ (2)) + 25-10a) (((a) ^ (2)) - ((b) ^ (2))) \]
Now let's deal with the second bracket:
\ [((a) ^ (2)) - ((b) ^ (2)) + 25-10a = ((a) ^ (2)) - 10a + 25 - ((b) ^ (2)) = \ left (((a) ^ (2)) - 2 \ cdot 5a + ((5) ^ (2)) \ right) - ((b) ^ (2)) = \]
\ [= ((\ left (a-5 \ right)) ^ (2)) - ((b) ^ (2)) = \ left (a-5-b \ right) \ left (a-5 + b \ right) \]
Since the two elements could not be grouped, we grouped three. It remains to deal only with the denominator of the last fraction:
\ [((a) ^ (2)) - ((b) ^ (2)) = \ left (a-b \ right) \ left (a + b \ right) \]
Now let's rewrite our entire construction:
\ [\ frac (a \ left (a + b \ right)) (\ left (ab \ right) \ left (5-ab \ right)) \ cdot \ frac (\ left (a-5-b \ right) \ left (a-5 + b \ right)) (\ left (ab \ right) \ left (a + b \ right)) = \ frac (a \ left (b-a + 5 \ right)) ((( \ left (ab \ right)) ^ (2))) \]
The problem is solved, and nothing more can be simplified here.
Solution nuances
We figured out the grouping and got another very powerful tool that expands the possibilities for factoring. But the problem is that in real life no one will give us such refined examples, where there are several fractions, in which you only need to factor the numerator and denominator into a factor, and then reduce them if possible. Real expressions will be much more complex.
Most likely, in addition to multiplication and division, there will be subtractions and additions, all kinds of brackets - in general, the order of actions will have to be taken into account. But the worst thing is that when subtracting and adding fractions with different denominators they will have to be reduced to one thing in common. To do this, each of them will need to be factorized, and then these fractions will need to be transformed: to bring similar ones and much more. How to do it correctly, quickly, and at the same time get an unambiguously correct answer? This is exactly what we will talk about now with the example of the following construction.
Problem number 4
\ [\ left (((x) ^ (2)) + \ frac (27) (x) \ right) \ cdot \ left (\ frac (1) (x + 3) + \ frac (1) ((( x) ^ (2)) - 3x + 9) \ right) \]
Let's write out the first fraction and try to deal with it separately:
\ [((x) ^ (2)) + \ frac (27) (x) = \ frac (((x) ^ (2))) (1) + \ frac (27) (x) = \ frac ( ((x) ^ (3))) (x) + \ frac (27) (x) = \ frac (((x) ^ (3)) + 27) (x) = \ frac (((x) ^ (3)) + ((3) ^ (3))) (x) = \]
\ [= \ frac (\ left (x + 3 \ right) \ left (((x) ^ (2)) - 3x + 9 \ right)) (x) \]
Let's move on to the second. Let's immediately calculate the discriminant of the denominator:
It cannot be multiplied, so we write the following:
\ [\ frac (1) (x + 3) + \ frac (1) (((x) ^ (2)) - 3x + 9) = \ frac (((x) ^ (2)) - 3x + 9 + x + 3) (\ left (x + 3 \ right) \ left (((x) ^ (2)) - 3x + 9 \ right)) = \]
\ [= \ frac (((x) ^ (2)) - 2x + 12) (\ left (x + 3 \ right) \ left (((x) ^ (2)) - 3x + 9 \ right)) \]
Let's write out the numerator separately:
\ [((x) ^ (2)) - 2x + 12 = 0 \]
Therefore, this polynomial is not factorized.
The maximum that we could do and decompose, we have already done.
So, we rewrite our original construction and get:
\ [\ frac (\ left (x + 3 \ right) \ left (((x) ^ (2)) - 3x + 9 \ right)) (x) \ cdot \ frac (((x) ^ (2) ) -2x + 12) (\ left (x + 3 \ right) \ left (((x) ^ (2)) - 3x + 9 \ right)) = \ frac (((x) ^ (2)) - 2x + 12) (x) \]
That's it, the problem is solved.
To be honest, it was not such a difficult task: everything was easily decomposed into factors, such terms were quickly given, and everything was beautifully reduced. So now let's try to solve a more serious problem.
Problem number 5
\ [\ left (\ frac (x) (((x) ^ (2)) + 2x + 4) + \ frac (((x) ^ (2)) + 8) (((x) ^ (3) ) -8) - \ frac (1) (x-2) \ right) \ cdot \ left (\ frac (((x) ^ (2))) (((x) ^ (2)) - 4) - \ frac (2) (2-x) \ right) \]
First, let's deal with the first parenthesis. From the very beginning, factor the denominator of the second fraction separately:
\ [((x) ^ (3)) - 8 = ((x) ^ (3)) - ((2) ^ (3)) = \ left (x-2 \ right) \ left (((x) ^ (2)) + 2x + 4 \ right) \]
\ [\ frac (x) (((x) ^ (2)) + 2x + 4) + \ frac (((x) ^ (2)) + 8) (((x) ^ (3)) - 8 ) - \ frac (1) (((x) ^ (2))) = \]
\ [= \ frac (x) (((x) ^ (2)) + 2x + 4) + \ frac (((x) ^ (2)) + 8) (\ left (x-2 \ right) \ left (((x) ^ (2)) + 2x + 4 \ right)) - \ frac (1) (x-2) = \]
\ [= \ frac (x \ left (x-2 \ right) + ((x) ^ (2)) + 8- \ left (((x) ^ (2)) + 2x + 4 \ right)) ( \ left (x-2 \ right) \ left (((x) ^ (2)) + 2x + 4 \ right)) = \]
\ [= \ frac (((x) ^ (2)) - 2x + ((x) ^ (2)) + 8 - ((x) ^ (2)) - 2x-4) (\ left (x-2 \ right) \ left (((x) ^ (2)) + 2x + 4 \ right)) = \]
\ [= \ frac (((x) ^ (2)) - 4x + 4) (\ left (x-2 \ right) \ left (((x) ^ (2)) + 2x + 4 \ right)) = \ frac (((\ left (x-2 \ right)) ^ (2))) (\ left (x-2 \ right) \ left (((x) ^ (2)) + 2x + 4 \ right )) = \ frac (x-2) (((x) ^ (2)) + 2x + 4) \]
Now let's work with the second fraction:
\ [\ frac (((x) ^ (2))) (((x) ^ (2)) - 4) - \ frac (2) (2-x) = \ frac (((x) ^ (2 ))) (\ left (x-2 \ right) \ left (x + 2 \ right)) - \ frac (2) (2-x) = \ frac (((x) ^ (2)) + 2 \ left (x-2 \ right)) (\ left (x-2 \ right) \ left (x + 2 \ right)) = \]
\ [= \ frac (((x) ^ (2)) + 2x + 4) (\ left (x-2 \ right) \ left (x + 2 \ right)) \]
Go back to our original construction and write:
\ [\ frac (x-2) (((x) ^ (2)) + 2x + 4) \ cdot \ frac (((x) ^ (2)) + 2x + 4) (\ left (x-2 \ right) \ left (x + 2 \ right)) = \ frac (1) (x + 2) \]
Key points
Once again, the key facts of today's video tutorial:
- It is necessary to know "by heart" the formulas of abbreviated multiplication - and not just know, but be able to see in those expressions that you will meet in real tasks... A wonderful rule can help us in this: if there are two terms, then this is either the difference of squares, or the difference or sum of cubes; if three, it can only be the square of the sum or difference.
- If any construction cannot be decomposed using the abbreviated multiplication formulas, then either the standard formula for factoring trinomials into factors, or the grouping method comes to our aid.
- If something does not work out, take a close look at the original expression - and whether any conversions are required with it at all. Perhaps it will be enough just to put the factor outside the parenthesis, and this is very often just a constant.
- In complex expressions, where you need to perform several actions in a row, do not forget to lead to common denominator, and only after that, when all the fractions are reduced to it, be sure to bring something like that in a new numerator, and then factor the new numerator into factors again - perhaps something will be canceled.
That's all I wanted to tell you today about rational fractions. If something is not clear - the site still has a bunch of video tutorials, as well as a bunch of tasks for independent decision... Therefore, stay with us!
Solving equations with fractions Let's look at examples. The examples are simple and illustrative. With their help, you will be able to learn in the most understandable way,.
For example, you want to solve a simple equation x / b + c = d.
Equations of this type are called linear, because the denominator contains only numbers.
The solution is performed by multiplying both sides of the equation by b, then the equation takes the form x = b * (d - c), i.e. the denominator of the fraction on the left is canceled.
For example, how to solve fractional equation:
x / 5 + 4 = 9
We multiply both parts by 5. We get:
x + 20 = 45
x = 45 - 20 = 25
Another example, when the unknown is in the denominator:
Equations of this type are called fractional rational or simply fractional.
We would solve a fractional equation by getting rid of fractions, after which this equation, most often, turns into a linear or square one, which is solved in the usual way. You should only consider the following points:
- the value of a variable that turns the denominator to 0 cannot be a root;
- you cannot divide or multiply an equation by the expression = 0.
Here comes into force such a concept as the range of permissible values (ODV) - these are the values of the roots of the equation for which the equation makes sense.
Thus, solving the equation, it is necessary to find the roots, and then check them for compliance with the ODZ. Those roots that do not correspond to our ODZ are excluded from the answer.
For example, you need to solve a fractional equation:
Based on the above rule, x cannot be = 0, i.e. ODZ in this case: x - any value other than zero.
We get rid of the denominator by multiplying all the terms of the equation by x
And we solve the usual equation
5x - 2x = 1
3x = 1
x = 1/3
Answer: x = 1/3
Let's solve a more complicated equation:
ODZ is also present here: x -2.
Solving this equation, we will not transfer everything to one side and reduce fractions to a common denominator. We will immediately multiply both sides of the equation by an expression that will cancel all the denominators at once.
To reduce the denominators, you need to multiply the left side by x + 2, and the right side by 2. Hence, both sides of the equation must be multiplied by 2 (x + 2):
This is the most common multiplication of fractions, which we have already discussed above.
Let's write the same equation, but in a slightly different way
The left side is canceled by (x + 2), and the right by 2. After the cancellation, we get the usual linear equation:
x = 4 - 2 = 2, which corresponds to our ODZ
Answer: x = 2.
Solving equations with fractions not as difficult as it might seem. In this article, we have shown this with examples. If you have any difficulties with that, how to solve equations with fractions, then unsubscribe in the comments.
An integer expression is a mathematical expression made up of numbers and literal variables using addition, subtraction, and multiplication. Also, integers include expressions that include division by any number other than zero.
The concept of fractional rational expression
A fractional expression is a mathematical expression that, in addition to the operations of addition, subtraction and multiplication performed on numbers and alphabetic variables, as well as division by a number not equal to zero, also contains division by expressions with alphabetic variables.
Rational expressions are all whole and fractional expressions. Rational equations are equations in which the left and right sides are rational expressions. If in a rational equation the left and right sides are whole expressions, then such a rational equation is called whole.
If in a rational equation the left or right sides are fractional expressions, then such a rational equation is called fractional.
Examples of fractional rational expressions
1.x-3 / x = -6 * x + 19
2. (x-4) / (2 * x + 5) = (x + 7) / (x-2)
3. (x-3) / (x-5) + 1 / x = (x + 5) / (x * (x-5))
A scheme for solving a fractional rational equation
1. Find the common denominator of all fractions in the equation.
2. Multiply both sides of the equation by a common denominator.
3. Solve the resulting whole equation.
4. Check the roots, and exclude those from them that vanish the common denominator.
Since we are solving fractional rational equations, there will be variables in the denominators of the fractions. This means that they will be in the common denominator. And in the second point of the algorithm, we multiply by a common denominator, then extraneous roots may appear. At which the common denominator will be equal to zero, which means that multiplying by it will be meaningless. Therefore, at the end, be sure to check the obtained roots.
Let's consider an example:
Solve the rational fractional equation: (x-3) / (x-5) + 1 / x = (x + 5) / (x * (x-5)).
Let's stick to general scheme: first find the common denominator of all fractions. We get x * (x-5).
Multiply each fraction by a common denominator and write the resulting whole equation.
(x-3) / (x-5) * (x * (x-5)) = x * (x + 3);
1 / x * (x * (x-5)) = (x-5);
(x + 5) / (x * (x-5)) * (x * (x-5)) = (x + 5);
x * (x + 3) + (x-5) = (x + 5);
Let us simplify the resulting equation. We get:
x ^ 2 + 3 * x + x-5 - x - 5 = 0;
x ^ 2 + 3 * x-10 = 0;
We got a simple reduced quadratic equation. We solve it in any of the known ways, we get the roots x = -2 and x = 5.
Now we check the obtained solutions:
Substitute the numbers -2 and 5 into the common denominator. When x = -2, the common denominator x * (x-5) does not vanish, -2 * (- 2-5) = 14. Hence, the number -2 will be the root of the original fractional rational equation.
When x = 5, the common denominator x * (x-5) becomes zero. Therefore, this number is not the root of the original fractional rational equation, since there will be division by zero.
Let's get acquainted with rational and fractional rational equations, give their definition, give examples, and also analyze the most common types of problems.
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Rational equation: definition and examples
Acquaintance with rational expressions begins in the 8th grade of the school. At this time, in algebra lessons, students increasingly begin to meet tasks with equations that contain rational expressions in their notes. Let's brush up on what it is.
Definition 1
Rational equation Is an equation in which both sides contain rational expressions.
One more formulation can be found in various manuals.
Definition 2
Rational equation- this is such an equation, the record of the left side of which contains a rational expression, and the right side contains zero.
The definitions that we gave for rational equations are equivalent, since they say the same thing. What confirms the correctness of our words is the fact that for any rational expression P and Q equations P = Q and P - Q = 0 are equivalent expressions.
Now let's turn to some examples.
Example 1
Rational equations:
x = 1, 2 x - 12 x 2 y z 3 = 0, xx 2 + 3 x - 1 = 2 + 2 7 x - a (x + 2), 1 2 + 3 4 - 12 x - 1 = 3.
Rational equations, just like equations of other types, can contain any number of variables from 1 to several. We'll start by looking at simple examples in which equations will only contain one variable. And then we will begin to gradually complicate the task.
Rational equations are divided into two large groups: whole and fractional. Let's see which equations will apply to each of the groups.
Definition 3
A rational equation will be whole if the record of the left and right parts of it contains whole rational expressions.
Definition 4
A rational equation will be fractional if one or both of its parts contain a fraction.
Fractionally rational equations necessarily contain division by a variable, or the variable is in the denominator. There is no such division in the recording of entire equations.
Example 2
3 x + 2 = 0 and (x + y) (3 x 2 - 1) + x = - y + 0.5- whole rational equations. Here, both sides of the equation are represented by whole expressions.
1 x - 1 = x 3 and x: (5 x 3 + y 2) = 3: (x - 1): 5 Are fractionally rational equations.
The number of whole rational equations includes linear and quadratic equations.
Solving entire equations
The solution of such equations is usually reduced to transforming them into equivalent algebraic equations. This can be achieved by carrying out equivalent transformations of the equations in accordance with the following algorithm:
- first we get zero on the right side of the equation, for this you need to transfer the expression that is on the right side of the equation to its left side and change the sign;
- then we transform the expression on the left side of the equation into a standard polynomial.
We have to get an algebraic equation. This equation will be the same as the original equation. Easy cases allow us to reduce the whole equation to a linear or quadratic one to solve the problem. In general, we solve the algebraic equation of degree n.
Example 3
It is necessary to find the roots of the whole equation 3 (x + 1) (x - 3) = x (2 x - 1) - 3.
Solution
Let's transform the original expression in order to obtain an algebraic equation equivalent to it. To do this, we will carry out the transfer of the expression contained in the right side of the equation to the left side and replace the sign with the opposite. As a result, we get: 3 (x + 1) (x - 3) - x (2 x - 1) + 3 = 0.
Now we will transform the expression on the left side into a polynomial of the standard form and perform the necessary actions with this polynomial:
3 (x + 1) (x - 3) - x (2 x - 1) + 3 = (3 x + 3) (x - 3) - 2 x 2 + x + 3 = = 3 x 2 - 9 x + 3 x - 9 - 2 x 2 + x + 3 = x 2 - 5 x - 6
We managed to reduce the solution of the original equation to the solution quadratic equation kind x 2 - 5 x - 6 = 0... The discriminant of this equation is positive: D = (- 5) 2 - 4 1 (- 6) = 25 + 24 = 49. This means there will be two real roots. We find them using the formula for the roots of the quadratic equation:
x = - - 5 ± 49 2 1,
x 1 = 5 + 7 2 or x 2 = 5 - 7 2,
x 1 = 6 or x 2 = - 1
Let's check the correctness of the roots of the equation that we found during the solution. For this, the numbers we got will be substituted into the original equation: 3 (6 + 1) (6 - 3) = 6 (2 6 - 1) - 3 and 3 (- 1 + 1) (- 1 - 3) = (- 1) (2 (- 1) - 1) - 3... In the first case 63 = 63 , in the second 0 = 0 ... Roots x = 6 and x = - 1 are indeed the roots of the equation given in the example condition.
Answer: 6 , − 1 .
Let's take a look at what "degree of the whole equation" means. We will often come across this term in those cases when we need to represent the whole equation in the form of an algebraic one. Let's give a definition to the concept.
Definition 5
The degree of the whole equation Is the degree of an algebraic equation equivalent to the original whole equation.
If you look at the equations from the example above, you can establish: the degree of this whole equation is second.
If our course was limited to solving equations of the second degree, then the consideration of the topic could have ended there. But it’s not that simple. Solving equations of the third degree is fraught with difficulties. And for equations higher than the fourth degree, it does not exist at all general formulas roots. In this regard, the solution of entire equations of the third, fourth and other degrees requires us to use a number of other techniques and methods.
The most commonly used approach to solving entire rational equations, which is based on the method of factorization. The algorithm of actions in this case is as follows:
- transfer the expression from the right side to the left so that zero remains on the right side of the record;
- we represent the expression on the left as a product of factors, and then move on to a collection of several simpler equations.
Find the solution to the equation (x 2 - 1) (x 2 - 10 x + 13) = 2 x (x 2 - 10 x + 13).
Solution
We transfer the expression from the right side of the record to the left with the opposite sign: (x 2 - 1) (x 2 - 10 x + 13) - 2 x (x 2 - 10 x + 13) = 0... Converting the left-hand side to a polynomial of the standard form is impractical due to the fact that this will give us an algebraic equation of the fourth degree: x 4 - 12 x 3 + 32 x 2 - 16 x - 13 = 0... The ease of conversion does not justify all the difficulties with solving such an equation.
It is much easier to go the other way: take the common factor out of the brackets x 2 - 10 x + 13. So we come to an equation of the form (x 2 - 10 x + 13) (x 2 - 2 x - 1) = 0... Now we replace the resulting equation with a set of two quadratic equations x 2 - 10 x + 13 = 0 and x 2 - 2 x - 1 = 0 and find their roots through the discriminant: 5 + 2 3, 5 - 2 3, 1 + 2, 1 - 2.
Answer: 5 + 2 3, 5 - 2 3, 1 + 2, 1 - 2.
Similarly, we can use the method of introducing a new variable. This method allows us to go to equivalent equations with degrees lower than the degrees in the original whole equation were.
Example 5
Does the equation have roots (x 2 + 3 x + 1) 2 + 10 = - 2 (x 2 + 3 x - 4)?
Solution
If we now try to reduce the whole rational equation to an algebraic one, we get an equation of degree 4, which has no rational roots. Therefore, it will be easier for us to go the other way: introduce a new variable y, which will replace the expression in the equation x 2 + 3 x.
We will now work with the whole equation (y + 1) 2 + 10 = - 2 (y - 4)... Move the right side of the equation to the left with the opposite sign and carry out the necessary transformations. We get: y 2 + 4 y + 3 = 0... Find the roots of the quadratic equation: y = - 1 and y = - 3.
Now let's do the reverse replacement. We get two equations x 2 + 3 x = - 1 and x 2 + 3 x = - 3. Rewrite them as x 2 + 3 x + 1 = 0 and x 2 + 3 x + 3 = 0... We use the formula for the roots of the quadratic equation in order to find the roots of the first equation from the obtained ones: - 3 ± 5 2. The discriminant of the second equation is negative. This means that the second equation has no real roots.
Answer:- 3 ± 5 2
Whole equations high degrees come across in tasks quite often. You don't need to be afraid of them. You need to be ready to apply a non-standard method of solving them, including a number of artificial transformations.
Solving fractionally rational equations
We begin our consideration of this subtopic with an algorithm for solving fractionally rational equations of the form p (x) q (x) = 0, where p (x) and q (x)- whole rational expressions. The solution of the remaining fractionally rational equations can always be reduced to the solution of equations of the indicated form.
The most commonly used method for solving the equations p (x) q (x) = 0 is based on the following statement: the number fraction u v, where v Is a number that is different from zero, equal to zero only in those cases when the numerator of the fraction is equal to zero. Following the logic of the above statement, we can assert that the solution to the equation p (x) q (x) = 0 can be reduced in the fulfillment of two conditions: p (x) = 0 and q (x) ≠ 0... This is used to construct an algorithm for solving fractional rational equations of the form p (x) q (x) = 0:
- find the solution of the whole rational equation p (x) = 0;
- we check whether the condition is satisfied for the roots found during the solution q (x) ≠ 0.
If this condition is met, then the found root. If not, then the root is not a solution to the problem.
Example 6
Find the roots of the equation 3 x - 2 5 x 2 - 2 = 0.
Solution
We are dealing with a fractional rational equation of the form p (x) q (x) = 0, in which p (x) = 3 x - 2, q (x) = 5 x 2 - 2 = 0. Let's start solving the linear equation 3 x - 2 = 0... The root of this equation will be x = 2 3.
Let's check the found root, whether it satisfies the condition 5 x 2 - 2 ≠ 0... To do this, we substitute a numeric value into the expression. We get: 5 2 3 2 - 2 = 5 4 9 - 2 = 20 9 - 2 = 2 9 ≠ 0.
The condition is met. It means that x = 2 3 is the root of the original equation.
Answer: 2 3 .
There is another option for solving fractional rational equations p (x) q (x) = 0. Recall that this equation is equivalent to the whole equation p (x) = 0 on the range of admissible values of the variable x of the original equation. This allows us to use the following algorithm in solving the equations p (x) q (x) = 0:
- we solve the equation p (x) = 0;
- find the range of valid values of the variable x;
- we take the roots that lie in the range of admissible values of the variable x as the desired roots of the original fractional rational equation.
Solve the equation x 2 - 2 x - 11 x 2 + 3 x = 0.
Solution
First, let's solve the quadratic equation x 2 - 2 x - 11 = 0... To calculate its roots, we use the root formula for an even second coefficient. We get D 1 = (- 1) 2 - 1 (- 11) = 12, and x = 1 ± 2 3.
Now we can find the ODV of the variable x for the original equation. These are all numbers for which x 2 + 3 x ≠ 0... This is the same as x (x + 3) ≠ 0, whence x ≠ 0, x ≠ - 3.
Now let us check whether the roots x = 1 ± 2 3 obtained at the first stage are included in the range of admissible values of the variable x. We see what comes in. This means that the original fractional rational equation has two roots x = 1 ± 2 3.
Answer: x = 1 ± 2 3
The second described solution method is simpler than the first in cases when the range of admissible values of the variable x is easily found, and the roots of the equation p (x) = 0 irrational. For example, 7 ± 4 26 9. Roots can be rational, but with a large numerator or denominator. For example, 127 1101 and − 31 59 ... This saves time on checking the condition. q (x) ≠ 0: it is much easier to exclude roots that do not fit according to the DHS.
In cases where the roots of the equation p (x) = 0 integers, it is more expedient to use the first of the described algorithms for solving equations of the form p (x) q (x) = 0. Find the roots of an entire equation faster p (x) = 0 and then check whether the condition q (x) ≠ 0, and not find the ODZ, and then solve the equation p (x) = 0 on this ODZ. This is due to the fact that in such cases it is usually easier to make a check than to find an LDZ.
Example 8
Find the roots of the equation (2 x - 1) (x - 6) (x 2 - 5 x + 14) (x + 1) x 5 - 15 x 4 + 57 x 3 - 13 x 2 + 26 x + 112 = 0.
Solution
Let's start by looking at the whole equation (2 x - 1) (x - 6) (x 2 - 5 x + 14) (x + 1) = 0 and finding its roots. To do this, we will apply the method of solving equations through factorization. It turns out that the original equation is equivalent to a set of four equations 2 x - 1 = 0, x - 6 = 0, x 2 - 5 x + 14 = 0, x + 1 = 0, of which three are linear and one is square. Find the roots: from the first equation x = 1 2, from the second - x = 6, from the third - x = 7, x = - 2, from the fourth - x = - 1.
Let's check the obtained roots. It is difficult for us to determine the ODZ in this case, since for this we will have to solve the algebraic equation of the fifth degree. It will be easier to check the condition that the denominator of the fraction on the left side of the equation should not vanish.
In turn, substitute the roots in place of the variable x in the expression x 5 - 15 x 4 + 57 x 3 - 13 x 2 + 26 x + 112 and calculate its value:
1 2 5 - 15 1 2 4 + 57 1 2 3 - 13 1 2 2 + 26 1 2 + 112 = = 1 32 - 15 16 + 57 8 - 13 4 + 13 + 112 = 122 + 1 32 ≠ 0;
6 5 - 15 6 4 + 57 6 3 - 13 6 2 + 26 6 + 112 = 448 ≠ 0;
7 5 - 15 7 4 + 57 7 3 - 13 7 2 + 26 7 + 112 = 0;
(- 2) 5 - 15 (- 2) 4 + 57 (- 2) 3 - 13 (- 2) 2 + 26 (- 2) + 112 = - 720 ≠ 0;
(- 1) 5 - 15 (- 1) 4 + 57 (- 1) 3 - 13 (- 1) 2 + 26 (- 1) + 112 = 0.
The performed check allows us to establish that the roots of the original fractional rational equation are 1 2, 6 and − 2 .
Answer: 1 2 , 6 , - 2
Example 9
Find the roots of the fractional rational equation 5 x 2 - 7 x - 1 x - 2 x 2 + 5 x - 14 = 0.
Solution
Let's get started with the equation (5 x 2 - 7 x - 1) (x - 2) = 0... Let's find its roots. It is easier for us to represent this equation as a combination of quadratic and linear equations 5 x 2 - 7 x - 1 = 0 and x - 2 = 0.
We use the formula for the roots of the quadratic equation to find the roots. We obtain from the first equation two roots x = 7 ± 69 10, and from the second x = 2.
It will be quite difficult for us to substitute the value of the roots into the original equation to test the conditions. It will be easier to determine the ODV of the variable x. In this case, the ODZ of the variable x is all numbers, except for those for which the condition x 2 + 5 x - 14 = 0... We get: x ∈ - ∞, - 7 ∪ - 7, 2 ∪ 2, + ∞.
Now let's check if the roots we found belong to the range of valid values of the variable x.
The roots x = 7 ± 69 10 - belong, therefore, they are the roots of the original equation, and x = 2- does not belong, therefore, it is an extraneous root.
Answer: x = 7 ± 69 10.
Let us consider separately the cases when a number is found in the numerator of a fractional rational equation of the form p (x) q (x) = 0. In such cases, if the numerator contains a nonzero number, then the equation will have no roots. If this number is equal to zero, then the root of the equation will be any number from the ODZ.
Example 10
Solve the fractional rational equation - 3, 2 x 3 + 27 = 0.
Solution
This equation will not have roots, since there is a nonzero number in the numerator of the fraction on the left side of the equation. This means that at no value of x will the value of the fraction given in the problem statement equal zero.
Answer: no roots.
Example 11
Solve the equation 0 x 4 + 5 x 3 = 0.
Solution
Since there is zero in the numerator of the fraction, the solution to the equation will be any value of x from the ODZ of the variable x.
Now let's define the ODZ. It will include all values of x for which x 4 + 5 x 3 ≠ 0... Equation solutions x 4 + 5 x 3 = 0 are 0 and − 5 , since this equation is equivalent to the equation x 3 (x + 5) = 0, and it, in turn, is equivalent to the combination of two equations x 3 = 0 and x + 5 = 0 from where these roots are visible. We come to the conclusion that any x, except x = 0 and x = - 5.
It turns out that the fractional rational equation 0 x 4 + 5 x 3 = 0 has an infinite set of solutions, which are any numbers other than zero and - 5.
Answer: - ∞ , - 5 ∪ (- 5 , 0 ∪ 0 , + ∞
Now let's talk about fractional rational equations of an arbitrary form and methods for solving them. They can be written as r (x) = s (x), where r (x) and s (x)- rational expressions, and at least one of them is fractional. The solution of such equations is reduced to the solution of equations of the form p (x) q (x) = 0.
We already know that we can get an equivalent equation by transferring the expression from the right side of the equation to the left with the opposite sign. This means that the equation r (x) = s (x) is equivalent to the equation r (x) - s (x) = 0... Also, we have already analyzed the ways of converting a rational expression into a rational fraction. Thanks to this, we can easily transform the equation r (x) - s (x) = 0 into its identical rational fraction of the form p (x) q (x).
So we pass from the original fractional rational equation r (x) = s (x) to an equation of the form p (x) q (x) = 0, which we have already learned how to solve.
It should be borne in mind that when making transitions from r (x) - s (x) = 0 to p (x) q (x) = 0 and then to p (x) = 0 we can ignore the expansion of the range of admissible values of the variable x.
The situation is quite real when the original equation r (x) = s (x) and the equation p (x) = 0 as a result of the transformations, they will cease to be equivalent. Then the solution to the equation p (x) = 0 can give us roots that will be strangers to r (x) = s (x)... In this regard, in each case, it is necessary to check by any of the methods described above.
To make it easier for you to study the topic, we generalized all the information into an algorithm for solving a fractional rational equation of the form r (x) = s (x):
- we transfer the expression from the right side with the opposite sign and get zero on the right;
- transform the original expression into a rational fraction p (x) q (x), sequentially performing actions with fractions and polynomials;
- we solve the equation p (x) = 0;
- we identify extraneous roots by checking their belonging to the ODZ or by substituting them into the original equation.
Visually, the chain of actions will look like this:
r (x) = s (x) → r (x) - s (x) = 0 → p (x) q (x) = 0 → p (x) = 0 → after
Example 12
Solve the rational fractional equation x x + 1 = 1 x + 1.
Solution
Let's go to the equation x x + 1 - 1 x + 1 = 0. We transform the fractional rational expression on the left side of the equation to the form p (x) q (x).
To do this, we will have to bring rational fractions to a common denominator and simplify the expression:
xx + 1 - 1 x - 1 = x x - 1 (x + 1) - 1 x (x + 1) x (x + 1) = = x 2 - x - 1 - x 2 - xx (X + 1) = - 2 x - 1 x (x + 1)
In order to find the roots of the equation - 2 x - 1 x (x + 1) = 0, we need to solve the equation - 2 x - 1 = 0... We get one root x = - 1 2.
It remains for us to check with any of the methods. Let's consider both of them.
Substitute this value into the original equation. We get - 1 2 - 1 2 + 1 = 1 - 1 2 + 1. We have arrived at the correct numerical equality − 1 = − 1 ... It means that x = - 1 2 is the root of the original equation.
Now let's check through the ODZ. Let us define the range of admissible values of the variable x. This will be the entire set of numbers, except for - 1 and 0 (for x = - 1 and x = 0, the denominators of the fractions vanish). The root we got x = - 1 2 belongs to ODZ. This means that it is the root of the original equation.
Answer: − 1 2 .
Example 13
Find the roots of the equation x 1 x + 3 - 1 x = - 2 3 x.
Solution
We are dealing with a fractional rational equation. Therefore, we will act according to the algorithm.
Move the expression from the right side to the left with the opposite sign: x 1 x + 3 - 1 x + 2 3 x = 0
Let's carry out the necessary transformations: x 1 x + 3 - 1 x + 2 3 x = x 3 + 2 x 3 = 3 x 3 = x.
We arrive at the equation x = 0... The root of this equation is zero.
Let us check if this root is not extraneous to the original equation. Substitute the value in the original equation: 0 1 0 + 3 - 1 0 = - 2 3 0. As you can see, the resulting equation does not make sense. This means that 0 is an extraneous root, and the original fractional rational equation has no roots.
Answer: no roots.
If we have not included other equivalent transformations in the algorithm, this does not mean at all that they cannot be used. The algorithm is universal, but it is designed to help, not limit.
Example 14
Solve the equation 7 + 1 3 + 1 2 + 1 5 - x 2 = 7 7 24
Solution
The easiest way is to solve the given fractional rational equation according to the algorithm. But there is also another way. Let's consider it.
Subtract 7 from the right and left parts, we get: 1 3 + 1 2 + 1 5 - x 2 = 7 24.
From this we can conclude that the expression in the denominator of the left side should be equal to the reciprocal of the number from the right side, that is, 3 + 1 2 + 1 5 - x 2 = 24 7.
Subtract 3 from both parts: 1 2 + 1 5 - x 2 = 3 7. By analogy 2 + 1 5 - x 2 = 7 3, whence 1 5 - x 2 = 1 3, and further 5 - x 2 = 3, x 2 = 2, x = ± 2
Let's check in order to establish whether the found roots are the roots of the original equation.
Answer: x = ± 2
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