Solution fractional rational equations
Reference Guide
Rational equations are equations in which both the left and right sides are rational expressions.
(Remember: rational expressions are integer and fractional expressions without radicals, including the operations of addition, subtraction, multiplication or division - for example: 6x; (m – n)2; x/3y, etc.)
Fractional rational equations are usually reduced to the form:
Where P(x) And Q(x) are polynomials.
To solve such equations, multiply both sides of the equation by Q(x), which can lead to the appearance extraneous roots. Therefore, when solving fractional rational equations, it is necessary to check the found roots.
A rational equation is called whole, or algebraic, if it does not divide by an expression containing a variable.
Examples of a whole rational equation:
5x – 10 = 3(10 – x)
3x
- = 2x – 10
4
If in a rational equation there is a division by an expression containing a variable (x), then the equation is called fractional rational.
Example of a fractional rational equation:
15
x + - = 5x – 17
x
Fractional rational equations are usually solved as follows:
1) find common denominator fractions and multiply both sides of the equation by it;
2) solve the resulting whole equation;
3) exclude from its roots those that reduce the common denominator of the fractions to zero.
Examples of solving integer and fractional rational equations.
Example 1. Let's solve the whole equation
x – 1 2x 5x
-- + -- = --.
2 3 6
Solution:
Finding the lowest common denominator. This is 6. Divide 6 by the denominator and multiply the resulting result by the numerator of each fraction. We obtain an equation equivalent to this:
3(x – 1) + 4x 5x
------ = --
6 6
Since in the left and right parts same denominator, it can be omitted. Then we get a simpler equation:
3(x – 1) + 4x = 5x.
We solve it by opening the brackets and bringing together similar members:
3x – 3 + 4x = 5x
3x + 4x – 5x = 3
The example is solved.
Example 2. Solve a fractional rational equation
x – 3 1 x + 5
-- + - = ---.
x – 5 x x(x – 5)
Finding a common denominator. This is x(x – 5). So:
x 2 – 3x x – 5 x + 5
--- + --- = ---
x(x – 5) x(x – 5) x(x – 5)
Now we get rid of the denominator again, since it is the same for all expressions. We reduce similar terms, equate the equation to zero and get quadratic equation:
x 2 – 3x + x – 5 = x + 5
x 2 – 3x + x – 5 – x – 5 = 0
x 2 – 3x – 10 = 0.
Having solved the quadratic equation, we find its roots: –2 and 5.
Let's check whether these numbers are the roots of the original equation.
At x = –2, the common denominator x(x – 5) does not vanish. This means –2 is the root of the original equation.
At x = 5, the common denominator goes to zero, and two out of three expressions become meaningless. This means that the number 5 is not the root of the original equation.
Answer: x = –2
More examples
Example 1.
x 1 =6, x 2 = - 2.2.
Answer: -2,2;6.
Example 2.
We have already learned how to solve quadratic equations. Now let's extend the studied methods to rational equations.
What is a rational expression? We have already encountered this concept. Rational expressions are expressions made up of numbers, variables, their powers and symbols of mathematical operations.
Accordingly, rational equations are equations of the form: , where 
- rational expressions.
Previously, we considered only those rational equations that can be reduced to linear ones. Now let's look at those rational equations that can be reduced to quadratic equations.
Example 1
Solve the equation: .
Solution:
A fraction is equal to 0 if and only if its numerator is equal to 0 and its denominator is not equal to 0.
We get the following system:
The first equation of the system is a quadratic equation. Before solving it, let's divide all its coefficients by 3. We get:
We get two roots: ; .
Since 2 never equals 0, two conditions must be met: 
. Since none of the roots of the equation obtained above coincide with invalid values variables that were obtained by solving the second inequality, they are both solutions given equation.
Answer:.
So, let's formulate an algorithm for solving rational equations:
1. Transfer all terms to left side, so that the right side turns out to be 0.
2. Transform and simplify the left side, bring all fractions to a common denominator.
3. Equate the resulting fraction to 0, by to the following algorithm: 
.
4. Write down those roots that were obtained in the first equation and satisfy the second inequality in the answer.
Let's look at another example.
Example 2
Solve the equation: 
.
Solution
At the very beginning, let's move all the terms to left side, so that 0 remains on the right. We get:
Now let's bring the left side of the equation to a common denominator:
This equation is equivalent to the system:
The first equation of the system is a quadratic equation.
Coefficients of this equation: . We calculate the discriminant:
We get two roots: ; .
Now let's solve the second inequality: the product of factors is not equal to 0 if and only if none of the factors is equal to 0.
Two conditions must be met: 
. We find that of the two roots of the first equation, only one is suitable - 3.
Answer:.
In this lesson, we remembered what a rational expression is, and also learned how to solve rational equations, which reduce to quadratic equations.
In the next lesson we will look at rational equations as models real situations, and also consider movement tasks.
Bibliography
- Bashmakov M.I. Algebra, 8th grade. - M.: Education, 2004.
- Dorofeev G.V., Suvorova S.B., Bunimovich E.A. and others. Algebra, 8. 5th ed. - M.: Education, 2010.
- Nikolsky S.M., Potapov M.A., Reshetnikov N.N., Shevkin A.V. Algebra, 8th grade. Tutorial for educational institutions. - M.: Education, 2006.
- Festival pedagogical ideas "Public lesson" ().
- School.xvatit.com ().
- Rudocs.exdat.com ().
Homework
The whole expression is mathematical expression, composed of numbers and alphabetic variables using the operations of addition, subtraction and multiplication. Integers also include expressions that involve division by any number other than zero.
The concept of a fractional rational expression
A fractional expression is a mathematical expression that, in addition to the operations of addition, subtraction and multiplication performed with numbers and letter variables, as well as division by a number not equal to zero, also contains division into expressions with letter 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 integer expressions, then such a rational equation is called an integer.
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))
Scheme for solving a fractional rational equation
1. Find the common denominator of all fractions that are included 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 that make the common denominator vanish.
Since we are solving fractional rational equations, there will be variables in the denominators of the fractions. This means that they will be a 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 multiplying by it will be meaningless. Therefore, at the end it is necessary to check the obtained roots.
Let's look at an example:
Solve the fractional rational equation: (x-3)/(x-5) + 1/x = (x+5)/(x*(x-5)).
We will stick to general scheme: Let's 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 get a simple reduced quadratic equation. We solve it with any of known methods, 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. At x=-2 the common denominator x*(x-5) does not vanish, -2*(-2-5)=14. This means that the number -2 will be the root of the original fractional rational equation.
When x=5 the common denominator x*(x-5) becomes equal to zero. Therefore, this number is not the root of the original fractional rational equation, since there will be a division by zero.
We introduced the equation above in § 7. First, let us recall what a rational expression is. This - algebraic expression, composed of numbers and the variable x using the operations of addition, subtraction, multiplication, division and exponentiation with a natural exponent.
If r(x) is a rational expression, then the equation r(x) = 0 is called a rational equation.
However, in practice it is more convenient to use a slightly broader interpretation of the term “rational equation”: this is an equation of the form h(x) = q(x), where h(x) and q(x) are rational expressions.
Until now, we could not solve any rational equation, but only one that, as a result of various transformations and reasoning, was reduced to linear equation. Now our capabilities are much greater: we will be able to solve a rational equation that reduces not only to linear
mu, but also to the quadratic equation.
Let us recall how we solved rational equations before and try to formulate a solution algorithm.
Example 1. Solve the equation
Solution. Let us rewrite the equation in the form
In this case, as usual, we take advantage of the fact that the equalities A = B and A - B = 0 express the same relationship between A and B. This allowed us to move the term to the left side of the equation with the opposite sign.
Let's transform the left side of the equation. We have
Let us recall the conditions of equality fractions zero: if and only if two relations are simultaneously satisfied:
1) the numerator of the fraction is zero (a = 0); 2) the denominator of the fraction is different from zero).
Equating the numerator of the fraction on the left side of equation (1) to zero, we obtain
It remains to check the fulfillment of the second condition indicated above. The relation means for equation (1) that . The values x 1 = 2 and x 2 = 0.6 satisfy the indicated relationships and therefore serve as the roots of equation (1), and at the same time the roots of the given equation.
1) Let's transform the equation to the form
2) Let us transform the left side of this equation:
(simultaneously changed the signs in the numerator and
fractions).
Thus, given equation takes the form
3) Solve the equation x 2 - 6x + 8 = 0. Find
4) For the found values, check the fulfillment of the condition 
. The number 4 satisfies this condition, but the number 2 does not. This means that 4 is the root of the given equation, and 2 is an extraneous root.
ANSWER: 4.
2. Solving rational equations by introducing a new variable
The method of introducing a new variable is familiar to you; we have used it more than once. Let us show with examples how it is used in solving rational equations.
Example 3. Solve the equation x 4 + x 2 - 20 = 0.
Solution. Let's introduce a new variable y = x 2 . Since x 4 = (x 2) 2 = y 2, the given equation can be rewritten as
y 2 + y - 20 = 0.
This is a quadratic equation, the roots of which can be found using known formulas; we get y 1 = 4, y 2 = - 5.
But y = x 2, which means the problem has been reduced to solving two equations:
x 2 =4; x 2 = -5.
From the first equation we find that the second equation has no roots.
Answer: .
An equation of the form ax 4 + bx 2 + c = 0 is called a biquadratic equation (“bi” is two, i.e., a kind of “double quadratic” equation). The equation just solved was precisely biquadratic. Any biquadratic equation is solved in the same way as the equation from example 3: introduce a new variable y = x 2, solve the resulting quadratic equation with respect to the variable y, and then return to the variable x.
Example 4. Solve the equation
Solution. Note that the same expression x 2 + 3x appears twice here. This means that it makes sense to introduce a new variable y = x 2 + 3x. This will allow us to rewrite the equation in a simpler and more pleasant form (which, in fact, is the purpose of introducing a new variable- and simplifying the recording
becomes clearer, and the structure of the equation becomes clearer):
Now let’s use the algorithm for solving a rational equation.
1) Let’s move all the terms of the equation into one part:
= 0
2) Transform the left side of the equation
So, we have transformed the given equation to the form
3) From the equation - 7y 2 + 29y -4 = 0 we find (you and I have already solved quite a lot of quadratic equations, so it’s probably not worth always giving detailed calculations in the textbook).
4) Let's check the found roots using condition 5 (y - 3) (y + 1). Both roots satisfy this condition.
So, the quadratic equation for the new variable y is solved: 
Since y = x 2 + 3x, and y, as we have established, takes two values: 4 and , we still have to solve two equations: x 2 + 3x = 4; x 2 + Zx = . The roots of the first equation are the numbers 1 and - 4, the roots of the second equation are the numbers
In the examples considered, the method of introducing a new variable was, as mathematicians like to say, adequate to the situation, that is, it corresponded well to it. Why? Yes, because the same expression clearly appeared in the equation several times and there was a reason to designate this expression new letter. But this does not always happen; sometimes a new variable “appears” only during the transformation process. This is exactly what will happen in the next example.
Example 5. Solve the equation
x(x-1)(x-2)(x-3) = 24.
Solution. We have
x(x - 3) = x 2 - 3x;
(x - 1)(x - 2) = x 2 -Зx+2.
This means that the given equation can be rewritten in the form
(x 2 - 3x)(x 2 + 3x + 2) = 24
Now a new variable has “appeared”: y = x 2 - 3x.
With its help, the equation can be rewritten in the form y (y + 2) = 24 and then y 2 + 2y - 24 = 0. The roots of this equation are the numbers 4 and -6.
Returning to the original variable x, we obtain two equations x 2 - 3x = 4 and x 2 - 3x = - 6. From the first equation we find x 1 = 4, x 2 = - 1; the second equation has no roots.
ANSWER: 4, - 1.
A proof of the derivative formula is given complex function. Cases when a complex function depends on one or two variables are considered in detail. A generalization is made to the case any number variables.
Here we present the conclusion following formulas for the derivative of a complex function.
If , then
.
If , then
.
If , then
.
Derivative of a complex function from one variable
Let a function of variable x be represented as a complex function in the following form:
,
where there are some functions. The function is differentiable for some value of the variable x. The function is differentiable at the value of the variable.
Then the complex (composite) function is differentiable at point x and its derivative is determined by the formula:
(1)
.
Formula (1) can also be written as follows:
;
.
Proof
Let us introduce the following notation.
;
.
Here there is a function of the variables and , there is a function of the variables and . But we will omit the arguments of these functions so as not to clutter the calculations.
Since the functions and are differentiable at points x and , respectively, then at these points there are derivatives of these functions, which are the following limits:
;
.
Consider the following function:
.
For a fixed value of the variable u, is a function of . It's obvious that
.
Then
.
Since the function is a differentiable function at the point, it is continuous at that point. That's why
.
Then
.
Now we find the derivative.
.
The formula is proven.
Consequence
If a function of a variable x can be represented as a complex function of a complex function
,
then its derivative is determined by the formula
.
Here , and there are some differentiable functions.
To prove this formula, we sequentially calculate the derivative using the rule for differentiating a complex function.
Consider the complex function
.
Its derivative
.
Consider the original function
.
Its derivative
.
Derivative of a complex function from two variables
Now let the complex function depend on several variables. First let's look at case of a complex function of two variables.
Let a function depending on the variable x be represented as a complex function of two variables in the following form:
,
Where
and there are differentiable functions for some value of the variable x;
- a function of two variables, differentiable at the point , . Then the complex function is defined in a certain neighborhood of the point and has a derivative, which is determined by the formula:
(2)
.
Proof
Since the functions and are differentiable at the point, they are defined in a certain neighborhood of this point, are continuous at the point, and their derivatives exist at the point, which are the following limits:
;
.
Here
;
.
Due to the continuity of these functions at a point, we have:
;
.
Since the function is differentiable at the point, it is defined in a certain neighborhood of this point, is continuous at this point, and its increment can be written in the following form:
(3)
.
Here
- increment of a function when its arguments are incremented by values and ;
;
- partial derivatives of the function with respect to the variables and .
For fixed values of and , and are functions of the variables and . They tend to zero at and:
;
.
Since and , then
;
.
Function increment:
.
:
.
Let's substitute (3):
.
The formula is proven.
Derivative of a complex function from several variables
The above conclusion can easily be generalized to the case when the number of variables of a complex function is more than two.
For example, if f is function of three variables, That
,
Where
, and there are differentiable functions for some value of the variable x;
- differentiable function of three variables at point , , .
Then, from the definition of differentiability of the function, we have:
(4)
.
Because, due to continuity,
;
;
,
That
;
;
.
Dividing (4) by and passing to the limit, we obtain:
.
And finally, let's consider most general case
.
Let a function of variable x be represented as a complex function of n variables in the following form:
,
Where
there are differentiable functions for some value of the variable x;
- differentiable function of n variables at a point
,
,
... , .
Then
.
