Fractionally irrational. The simplest rational equations
Today we will figure out how to solve fractional rational equations.
Let's see: from the equations
(1) 2x + 5 = 3(8 - x),
(3)
(4)
fractional rational equations are only (2) and (4), while (1) and (3) are entire equations.
I propose to solve equation (4), and then formulate the rule.
Since the equation is fractional, we need to find common denominator. In this equation, this expression is 6 (x - 12) (x - 6). Then we multiply both sides of the equation by a common denominator:
After reduction, we get the whole equation:
6 (x - 6) 2 - 6 (x - 12) 2 \u003d 5 (x - 12) (x - 6).
Having solved this equation, it is necessary to check whether the obtained roots turn the denominators of the fractions in the original equation to zero.
Expanding the brackets:
6x 2 - 72x + 216 - 6x 2 + 144x - 864 = 5x 2 - 90x + 360, we simplify the equation: 5x 2 - 162x + 1008 = 0.
Finding the roots of the equation
D=6084, √D=78,
x 1 = (162 - 78) / 10 = 84/10 = 8.4 and x 2 = (162 + 78) / 10 = 240/10 = 24.
At x = 8.4 and 24, the common denominator is 6(x - 12)(x - 6) ≠ 0, which means that these numbers are the roots of equation (4).
Answer: 8,4; 24.
Solving the proposed equation, we arrive at the following provisions:
1) We find a common denominator.
2) Multiply both sides of the equation by a common denominator.
3) We solve the resulting whole equation.
4) We check which of the roots turn the common denominator to zero and exclude them from the solution.
Let us now look at an example of how the resulting positions work.
Solve the equation:
1) Common denominator: x 2 - 1
2) We multiply both parts of the equation by a common denominator, we get the whole equation: 6 - 2 (x + 1) \u003d 2 (x 2 - 1) - (x + 4) (x - 1)
3) We solve the equation: 6 - 2x - 2 \u003d 2x 2 - 2 - x 2 - 4x + x + 4
x 2 - x - 2 = 0
x 1 = - 1 and x 2 = 2
4) When x \u003d -1, the common denominator x 2 - 1 \u003d 0. The number -1 is not a root.
For x \u003d 2, the common denominator is x 2 - 1 ≠ 0. The number 2 is the root of the equation.
Answer: 2.
As you can see, our provisions work. Don't be afraid, you will succeed! The most important find the common denominator correctly and do the transformations carefully. We hope that when solving fractional rational equations, you will always get the correct answers. If you have any questions or want to practice solving such equations, sign up for lessons with the author of this article, tutor Valentina Galinevskaya.
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We introduced the equation above in § 7. First, we recall what a rational expression is. It - 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 somewhat 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 possibilities are much greater: we will be able to solve a rational equation, which reduces not only to linear
mu, but also to the quadratic equation.
Recall how we solved rational equations earlier and try to formulate a solution algorithm.
Example 1 solve the equation
Solution. We rewrite the equation in the form
In this case, as usual, we use the fact that the equalities A \u003d B and A - B \u003d 0 express the same relationship between A and B. This allowed us to transfer the term to the left side of the equation with the opposite sign.
Let's perform transformations of the left side of the equation. We have
Recall the equality conditions 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 to zero the numerator of the fraction on the left side of equation (1), we obtain
It remains to check the fulfillment of the second condition mentioned above. The ratio means for equation (1) that . The values x 1 \u003d 2 and x 2 \u003d 0.6 satisfy the indicated relationships and therefore serve as the roots of equation (1), and at the same time the roots for given equation.
1) Let's transform the equation to the form
2) Let's perform the transformations of the left side of this equation:
(simultaneously changed the signs in the numerator and
fractions).
Thus, the given equation takes the form
3) Solve the equation x 2 - 6x + 8 = 0. Find
4) For the found values, check the condition . The number 4 satisfies this condition, but the number 2 does not. So 4 is the root of the given equation, and 2 is an extraneous root.
Answer: 4.
2. Solution of 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 by examples how it is used in solving rational equations.
Example 3 Solve the equation x 4 + x 2 - 20 = 0.
Solution. We introduce a new variable y \u003d x 2. Since x 4 \u003d (x 2) 2 \u003d y 2, then the given equation can be rewritten in the form
y 2 + y - 20 = 0.
This is a quadratic equation, the roots of which we will find using the known formulas; we get y 1 = 4, y 2 = - 5.
But y \u003d x 2, which means that the problem has been reduced to solving two equations:
x2=4; x 2 \u003d -5.
From the first equation we find the second equation has no roots.
Answer: .
An equation of the form ax 4 + bx 2 + c \u003d 0 is called a biquadratic equation (“bi” - two, i.e., as it were, a “twice square” equation). The equation just solved was exactly biquadratic. Any biquadratic equation is solved in the same way as the equation from example 3: a new variable y \u003d x 2 is introduced, the resulting quadratic equation is solved with respect to the variable y, and then returned to the variable x.
Example 4 solve the equation
Solution. Note that the same expression x 2 + 3x occurs twice here. Hence, it makes sense to introduce a new variable y = x 2 + Zx. 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 recording is easier
, and the structure of the equation becomes clearer):
And now we will use the algorithm for solving a rational equation.
1) Let's move all the terms of the equation into one part:
= 0
2) Let's transform the left side of the equation
So, we have transformed the given equation into the form
3) From the equation - 7y 2 + 29y -4 = 0 we find (we 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 the condition 5 (y - 3) (y + 1). Both roots satisfy this condition.
So, the quadratic equation for the new variable y is solved:
Since y \u003d x 2 + Zx, and y, as we have established, takes two values: 4 and, - we still have to solve two equations: x 2 + Zx \u003d 4; x 2 + Zx \u003d. 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 was clearly encountered in the equation record several times and it was reasonable to designate this expression with a new letter. But this is not always the case, sometimes a new variable "appears" only in the process of transformations. 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) \u003d x 2 - 3x;
(x - 1) (x - 2) \u003d x 2 -3x + 2.
So the given equation can be rewritten as
(x 2 - 3x)(x 2 + 3x + 2) = 24
Now a new variable has "appeared": y = x 2 - Zx.
With its help, the equation can be rewritten in the form y (y + 2) \u003d 24 and then y 2 + 2y - 24 \u003d 0. The roots of this equation are the numbers 4 and -6.
Returning to the original variable x, we obtain two equations x 2 - Zx \u003d 4 and x 2 - Zx \u003d - 6. From the first equation we find x 1 \u003d 4, x 2 \u003d - 1; the second equation has no roots.
Answer: 4, - 1.
As is already known (see § 2 of the previous chapter), an equation of the form
where rational functions, at least one of which is fractionally rational, is called a fractional rational equation with one unknown.
To solve equation (1), we transfer it to the left side, perform the necessary identical transformations and write the given equation in the form
where and are polynomials from
Equation (2) is a consequence of equation (1). Indeed, if there is a solution to equation (1), then Let's perform all the transformations on this equality that we performed on the equation. We get the equality and this means that c is a solution to equation (2).
However, equation (2) is not necessarily equivalent to equation (1). When transforming equation (1), the set of admissible values of the unknown can change, and it cannot narrow, but can expand,
and then equation (2) will have solutions that are extraneous for equation (1). This will happen when, in the process of transforming equation (1), some fractional expressions are mutually canceled or a reduction is made algebraic fractions into factors that include the unknown
For example, performing in the equation
these transformations, we obtain
Equation (4) is not equivalent to equation (3). Indeed, it has roots. The second of them is extraneous for equation (3), because at the expression does not make sense. This happened because when the equation (3) was transformed, the terms canceled each other out
Transforming another equation
Reducing the fraction by we will have
Equation (6) has a root that does not satisfy equation (5), because its left side loses its meaning at Therefore, equation (6) is not equivalent to equation (5). This happened because in the process of transforming the given equation, we reduced the algebraic fraction by
Thus, equation (2) is a consequence of equation (1), but is not necessarily equivalent to it; hence it follows that the solutions of Eq. (1) should be sought among the solutions of Eq. (2). The solutions of equation (2) can only be those values at which it equals zero, i.e., only solutions to the equation, which means that solutions to equation (1) must be sought among the solutions of the equation
Therefore, to solve equation (1), it is sufficient to determine all the roots of the equation and then, by directly substituting them into equation (1), find out which of them are the roots of the given equation (1).
Our reasoning can be briefly formulated as the following rule for solving fractional rational equations.
To solve fractional rational equations
with one unknown you need:
1) move all its members to the left side;
2) perform the necessary identical transformations and write the given equation in the form
where and are polynomials from
3) solve the equation
4) by substituting the solutions of the equation into the original equation, determine which of them satisfy the given equation.
Example. solve the equation
Transferring all the terms to the left side and reducing them to a common denominator, we get:
Equating the numerator of the left side to zero, we will have the equation
The first of these solutions is extraneous to the given equation, while the second satisfies it.
Note that in school practice often, when solving fractional rational equations, both parts of a given equation are multiplied by the common denominator of all algebraic fractions included in the left and right sides of the equation, and then the equation obtained in this way is solved. Obviously, the resulting algebraic equation is a consequence of the given equation, but is not equivalent to it.
Therefore, having found solutions to this algebraic equation, it is necessary to substitute them into the given equation to determine which of them will be solutions to the given equation.
Solution of fractional rational equations
Help Guide
Rational equations are equations in which both the left and right sides are rational expressions.
(Recall: 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, as a rule, are 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 of extraneous roots. Therefore, when solving fractional rational equations, it is necessary to check the found roots.
A rational equation is called an integer, or algebraic, if it does not have a division 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 the variable (x), then the equation is called fractional rational.
An example of a fractional rational equation:
15
x + - = 5x - 17
x
Fractional rational equations are usually solved as follows:
1) find a common denominator of fractions and multiply both parts of the equation by it;
2) solve the resulting whole equation;
3) exclude from its roots those that turn the common denominator of the fractions to zero.
Examples of solving integer and fractional rational equations.
Example 1. 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 result by the numerator of each fraction. We get an equation equivalent to this one:
3(x - 1) + 4x 5x
------ = --
6 6
Since the denominator is the same on the left and right sides, it can be omitted. Then we have a simpler equation:
3(x - 1) + 4x = 5x.
We solve it by opening brackets and reducing like terms:
3x - 3 + 4x = 5x
3x + 4x - 5x = 3
Example solved.
Example 2. Solve a fractional rational equation
x – 3 1 x + 5
-- + - = ---.
x - 5 x x(x - 5)
We find 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 like terms, equate the equation to zero and get a 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 if these numbers are the roots of the original equation.
For x = –2, the common denominator x(x – 5) does not vanish. So -2 is the root of the original equation.
At x = 5, the common denominator vanishes, and two of the three expressions lose their meaning. So the number 5 is not the root of the original equation.
Answer: x = -2
More examples
Example 1
x 1 \u003d 6, x 2 \u003d - 2.2.
Answer: -2.2; 6.
Example 2
Rational equations are equations containing rational expressions.
Definition 1
In this case, rational expressions are expressions that can be written in the form common fraction of the form $\frac(m)(n)$, while $m$ and $n$ are integers and $n$ cannot be equal to zero. Rational expressions include not only expressions containing fractions of the form $\frac(2)(3)$, but also expressions containing only integers, since any integer can be represented as an improper fraction.
Now let's take a closer look at what rational equations are.
As we mentioned above, rational equations are equations containing rational expressions and variables.
According to the position of the variable in a rational equation, it can be either a fractional rational equation or an entire rational equation.
Fractional equations can contain a fraction with a variable in only one part of the equation, while whole equations do not contain fractional expressions with a variable.
Entire rational equations examples: $5x+2= 12$; $3y=-7(-4y + 5)$; $7a-14=256$.
Fractional-rational equations examples: $\frac(3x-2)(x+3)+\frac(1)(2)=\frac(5)(x)$; $\frac(7)(2y-3)=5$;
It is worth noting that only equations containing a fraction in the denominator are called fractional rational equations, since equations containing fractional expressions without variables can easily be reduced to linear integer equations.
How to solve rational equations?
Depending on whether you are dealing with an integer rational equation or a fractional one, slightly different algorithms are used to solve it.
Algorithm for solving entire rational equations
- First, you need to determine the lowest common denominator for the entire equality.
- Then you need to determine the factors by which you need to multiply each term of the equality.
- The next stage is the reduction to a common denominator of all equality.
- Finally, the implementation of the search for the roots of the obtained integer rational equality.
Example 1
Solve the equation: $\frac(5x+9)(2)=\frac(x)(4)$
First, let's find the common factor - in this case, the number $4$. To get rid of the denominator, multiply the left side by $\frac(2)(2)$, we get:
$10x+18=x$ - the resulting equation is linear, its root is $x=-2$.
How to solve fractional rational equations?
In the case of fractional rational equations, the order of solution is similar to the algorithm for solving integer rational ones, that is, points 1-4 are preserved, but after finding the expected roots, in the case of using non-equivalent transformations, the roots must be checked by substituting into the equation.
Example 2
Solve the fractional rational equation: $\frac(x-3)(x-5)+\frac(1)(x)=\frac(x+5)(x \cdot (x-5))$
In order to reduce the fraction to a common denominator, here it is $x \cdot (x-5)$, we multiply each fraction by one, represented as the factor necessary to reduce to a common denominator:
$\frac((x-3) \cdot x)((x-5)\cdot x)+\frac(1 \cdot (x-5))(x \cdot (x-5))=\frac( x+5)(x \cdot (x-5))$
Now that the whole fraction has a common denominator, you can get rid of it:
$(x-3) \cdot x+(x-5)=x+5$
$x^2 - 3x+x-5 = x+5$
Let's use Vieta's theorem to solve the resulting quadratic equation:
$\begin(cases) x_1 + x_2 = 3 \\ x_1 \cdot x_2 = -10 \\ \end(cases)$
$\begin(cases) x_1=5 \\ x_2=-2 \\ \end(cases)$
Since the transformation used to simplify the equation is not equivalent, the obtained roots must be checked in the original equation, for this we substitute them:
$\frac(-2-3)(-2-5) +\frac(1)(-2)=\frac(-2+5)((-2) \cdot (-2-5))$
$\frac(5)(7)-\frac(1)(2)=\frac(3)(14)$
$\frac(3)(14)=\frac(3)(14)$ - hence the root $x_2=-2$ is correct.
$\frac(5-3)(5-5) +\frac(1)(5)=\frac(5+5)((-2) \cdot (5-5))$
Here it is immediately clear that zero is formed in the denominator, therefore, the root $x_1=5$ is an outsider.
It must be remembered that if an equation containing an expression of the form $\frac(m)(n)$ on the left or right side is equal to zero, only the numerator of the fraction can be equal to zero. This is due to the fact that if a zero is formed somewhere in the denominator, the root being checked is not the root of the equation, since the whole equality loses its meaning in this case. Roots that bring the denominator to zero are called extraneous.
If the fractional-rational equation has a rather complex form, for its further simplification and solution it is possible to use the replacement of part of the equation with a new variable, you have probably already seen examples of such fractional-rational equations:
Example 3
Solve the equation:
$\frac(1)(x^2+3x-3)+\frac(2)(x^2+3x+1)=\frac(7)(5)$
To simplify the solution, we introduce the variable $t= x^2+3x$:
$\frac(1)(t-3)+\frac(2)(t+1)=\frac(7)(5)$
The common denominator here is $5 \cdot (t-3)(t+1)$, we multiply all parts of the equation by the necessary factors to get rid of it:
$\frac(5(t+1))(5(t-3)(t+1))+\frac(2 \cdot 5(t-3))(5(t+1)(t-3) )=\frac(7(t+1)(t-3))(5(t-3)(t+1))$
$5(t+1)+10(t-3)=7(t+1)(t-3)$
$5t+5+10t-30=7(t^2-3t+t-3)$
$15t-25=7t^2-14t-21$
Through the discriminant, we calculate the roots:
$t_1=4;t_2=\frac(1)(7)$
Since we used non-equivalent transformations, it is necessary to check the obtained roots in the denominator, they must satisfy the condition $5(t-3)(t+1)≠0$. Both roots meet this condition.
Now we substitute the obtained roots for $t$ and get two equations:
$x^2+3x=4$ and $x^2+3x=\frac(1)(7)$.
By Vieta's theorem, the roots of the first equation $x_1=-4; x_2=1$, we calculate the roots of the second one in terms of the discriminant and have $x_(1,2)=\frac(-3±\sqrt(\frac(67)(7)))(2)$.
All roots of the equation will be: $x_1=-4; x_2=1, x_(3,4)=\frac(-3±\sqrt(\frac(67)(7)))(2)$.
Transformations to Simplify the Equation Form
As you can see above, various transformations are used to solve rational equations.
There are two types of equation transformations: equivalent (identical) and unequal.
Transformations are called equivalent if they lead to a new type of equation, the roots of which are the same as those of the original.
Identity transformations that can be used to change the form of the original equation without any further checks are as follows:
- Multiplying or dividing the entire equation by some non-zero number;
- Transferring parts of an equation from the left side to the right side and vice versa.
Non-equivalent transformations are transformations in the course of which extraneous roots may appear. Non-equivalent transformations include:
- Squaring both sides of the equation;
- Getting rid of the denominators containing the variable;
The roots of rational equations solved using non-equivalent transformations must be checked by substitution into the original equation, since extraneous roots may appear during non-equivalent transformations. Non-equivalent transformations do not always lead to the appearance of extraneous roots, but nevertheless it is necessary to take this into account.
Solution of rational equations with powers greater than two
The most commonly used methods for solving equations with powers greater than two are the change of variable method, which we discussed above using the example of a fractional rational equation, as well as the factorization method.
Let us consider the factorization method in more detail.
Let an equation of the form $P(x)= 0$ be given, where $P(x)$ is a polynomial whose degree is greater than two. If this equation can be factorized so that it takes the form $P_1(x)P_2(x)P_3(x)..\cdot P_n(x)=0$, then the solution to this equation is the set of solutions of the equations $P_1(x )=0, P_2(x)=0, P_3(x)=0...P_n(x)=0$.
For those who do not remember: a free term of an equation is a term of equations that does not contain a variable as a factor. At the same time, having found one of the roots of such an equation, it can be used to further factorize the equation.
Example 5
Solve the equation:
The divisors of the free term will be the numbers $±1, ±2, ±3, ±4, ±6, ±8, ±12$ and $±24$. When they were checked, $x=2$ turned out to be a suitable root. This means that this polynomial can be expanded using this root: $(x-2)(x^2+6+12)=0$.
The polynomial in the second pair of root brackets has no roots, which means that the only root of this equation will be $x=2$.
Another type of equations with degree greater than two are bi quadratic equations of the form $ax^4+bx^2+ c=0$. Such equations are solved by replacing $x^2$ with $y$, applying it, we obtain an equation of the form $ay^2+y+c=0$, and then the resulting value of the new variable is used to calculate the original variable.
There is also another type of equation called returnable. Such equations look like this: $ax^4+bx^3+cx^2+bx+a=0$. They have such a name because of the repetition of coefficients at higher degrees and lower ones.