We introduced the equation above in § 7. First, let us recall what a rational expression is. This is an algebraic expression made up 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's 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, the 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, then 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 with a 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.
Solving equations with fractions Let's look at examples. The examples are simple and illustrative. With their help, you will be able to understand in the most understandable way.
For example, you need to solve the simple equation x/b + c = d.
An equation of this type is 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 side cancels.
For example, how to solve a fractional equation:
x/5+4=9
We multiply both sides 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 quadratic equation, which is solved in the usual way. You just need to 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.
This is where the concept of the region of permissible values (ADV) comes into force - these are the values of the roots of the equation for which the equation makes sense.
Thus, when 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 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.
When solving this equation, we will not move everything to one side and bring the fractions to a common denominator. We will immediately multiply both sides of the equation by an expression that will cancel out 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. This means that 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 slightly differently
The left side is reduced by (x+2), and the right by 2. After the reduction, we obtain 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 how to solve equations with fractions, then unsubscribe in the comments.
Maintaining your privacy is important to us. For this reason, we have developed a Privacy Policy that describes how we use and store your information. Please review our privacy practices and let us know if you have any questions.
Collection and use of personal information
Personal information refers to data that can be used to identify or contact a specific person.
You may be asked to provide your personal information at any time when you contact us.
Below are some examples of the types of personal information we may collect and how we may use such information.
What personal information do we collect:
- When you submit an application on the site, we may collect various information, including your name, phone number, email address, etc.
How we use your personal information:
- The personal information we collect allows us to contact you with unique offers, promotions and other events and upcoming events.
- From time to time, we may use your personal information to send important notices and communications.
- We may also use personal information for internal purposes, such as conducting audits, data analysis and various research in order to improve the services we provide and provide you with recommendations regarding our services.
- If you participate in a prize draw, contest or similar promotion, we may use the information you provide to administer such programs.
Disclosure of information to third parties
We do not disclose the information received from you to third parties.
Exceptions:
- If necessary - in accordance with the law, judicial procedure, in legal proceedings, and/or on the basis of public requests or requests from government authorities in the territory of the Russian Federation - to disclose your personal information. We may also disclose information about you if we determine that such disclosure is necessary or appropriate for security, law enforcement, or other public importance purposes.
- In the event of a reorganization, merger, or sale, we may transfer the personal information we collect to the applicable successor third party.
Protection of personal information
We take precautions - including administrative, technical and physical - to protect your personal information from loss, theft, and misuse, as well as unauthorized access, disclosure, alteration and destruction.
Respecting your privacy at the company level
To ensure that your personal information is secure, we communicate privacy and security standards to our employees and strictly enforce privacy practices.
Presentation and lesson on the topic: "Rational equations. Algorithm and examples of solving rational equations"
Additional materials
Dear users, do not forget to leave your comments, reviews, wishes! All materials have been checked by an anti-virus program.
Educational aids and simulators in the Integral online store for grade 8
A manual for the textbook by Makarychev Yu.N. A manual for the textbook by Mordkovich A.G.
Introduction to Irrational Equations
Guys, we learned how to solve quadratic equations. But mathematics is not limited to them only. Today we will learn how to solve rational equations. The concept of rational equations is in many ways similar to the concept of rational numbers. Only in addition to numbers, now we have introduced some variable $x$. And thus we get an expression in which the operations of addition, subtraction, multiplication, division and raising to an integer power are present.Let $r(x)$ be rational expression. Such an expression can be a simple polynomial in the variable $x$ or a ratio of polynomials (a division operation is introduced, as for rational numbers).
The equation $r(x)=0$ is called rational equation.
Any equation of the form $p(x)=q(x)$, where $p(x)$ and $q(x)$ are rational expressions, will also be rational equation.
Let's look at examples of solving rational equations.
Example 1.Solve the equation: $\frac(5x-3)(x-3)=\frac(2x-3)(x)$.
Solution.
Let's move all the expressions to the left side: $\frac(5x-3)(x-3)-\frac(2x-3)(x)=0$.
If the left side of the equation were represented by ordinary numbers, then we would reduce the two fractions to a common denominator.
Let's do this: $\frac((5x-3)*x)((x-3)*x)-\frac((2x-3)*(x-3))((x-3)*x )=\frac(5x^2-3x-(2x^2-6x-3x+9))((x-3)*x)=\frac(3x^2+6x-9)((x-3) *x)=\frac(3(x^2+2x-3))((x-3)*x)$.
We got the equation: $\frac(3(x^2+2x-3))((x-3)*x)=0$.
A fraction is equal to zero if and only if the numerator of the fraction is zero and the denominator is non-zero. Then we separately equate the numerator to zero and find the roots of the numerator.
$3(x^2+2x-3)=0$ or $x^2+2x-3=0$.
$x_(1,2)=\frac(-2±\sqrt(4-4*(-3)))(2)=\frac(-2±4)(2)=1;-3$.
Now let's check the denominator of the fraction: $(x-3)*x≠0$.
The product of two numbers is equal to zero when at least one of these numbers is equal to zero. Then: $x≠0$ or $x-3≠0$.
$x≠0$ or $x≠3$.
The roots obtained in the numerator and denominator do not coincide. So we write down both roots of the numerator in the answer.
Answer: $x=1$ or $x=-3$.
If suddenly one of the roots of the numerator coincides with the root of the denominator, then it should be excluded. Such roots are called extraneous!
Algorithm for solving rational equations:
1. Move all expressions contained in the equation to the left side of the equal sign.2. Convert this part of the equation to an algebraic fraction: $\frac(p(x))(q(x))=0$.
3. Equate the resulting numerator to zero, that is, solve the equation $p(x)=0$.
4. Equate the denominator to zero and solve the resulting equation. If the roots of the denominator coincide with the roots of the numerator, then they should be excluded from the answer.
Example 2.
Solve the equation: $\frac(3x)(x-1)+\frac(4)(x+1)=\frac(6)(x^2-1)$.
Solution.
Let's solve according to the points of the algorithm.
1. $\frac(3x)(x-1)+\frac(4)(x+1)-\frac(6)(x^2-1)=0$.
2. $\frac(3x)(x-1)+\frac(4)(x+1)-\frac(6)(x^2-1)=\frac(3x)(x-1)+\ frac(4)(x+1)-\frac(6)((x-1)(x+1))= \frac(3x(x+1)+4(x-1)-6)((x -1)(x+1))=$ $=\frac(3x^2+3x+4x-4-6)((x-1)(x+1))=\frac(3x^2+7x- 10)((x-1)(x+1))$.
$\frac(3x^2+7x-10)((x-1)(x+1))=0$.
3. Equate the numerator to zero: $3x^2+7x-10=0$.
$x_(1,2)=\frac(-7±\sqrt(49-4*3*(-10)))(6)=\frac(-7±13)(6)=-3\frac( 1)(3);1$.
4. Equate the denominator to zero:
$(x-1)(x+1)=0$.
$x=1$ and $x=-1$.
One of the roots $x=1$ coincides with the root of the numerator, then we do not write it down in the answer.
Answer: $x=-1$.
It is convenient to solve rational equations using the change of variables method. Let's demonstrate this.
Example 3.
Solve the equation: $x^4+12x^2-64=0$.
Solution.
Let's introduce the replacement: $t=x^2$.
Then our equation will take the form:
$t^2+12t-64=0$ - ordinary quadratic equation.
$t_(1,2)=\frac(-12±\sqrt(12^2-4*(-64)))(2)=\frac(-12±20)(2)=-16; $4.
Let's introduce the reverse substitution: $x^2=4$ or $x^2=-16$.
The roots of the first equation are a pair of numbers $x=±2$. The second thing is that it has no roots.
Answer: $x=±2$.
Example 4.
Solve the equation: $x^2+x+1=\frac(15)(x^2+x+3)$.
Solution.
Let's introduce a new variable: $t=x^2+x+1$.
Then the equation will take the form: $t=\frac(15)(t+2)$.
Next we will proceed according to the algorithm.
1. $t-\frac(15)(t+2)=0$.
2. $\frac(t^2+2t-15)(t+2)=0$.
3. $t^2+2t-15=0$.
$t_(1,2)=\frac(-2±\sqrt(4-4*(-15)))(2)=\frac(-2±\sqrt(64))(2)=\frac( -2±8)(2)=-5; $3.
4. $t≠-2$ - the roots do not coincide.
Let's introduce a reverse substitution.
$x^2+x+1=-5$.
$x^2+x+1=3$.
Let's solve each equation separately:
$x^2+x+6=0$.
$x_(1,2)=\frac(-1±\sqrt(1-4*(-6)))(2)=\frac(-1±\sqrt(-23))(2)$ - no roots.
And the second equation: $x^2+x-2=0$.
The roots of this equation will be the numbers $x=-2$ and $x=1$.
Answer: $x=-2$ and $x=1$.
Example 5.
Solve the equation: $x^2+\frac(1)(x^2) +x+\frac(1)(x)=4$.
Solution.
Let's introduce the replacement: $t=x+\frac(1)(x)$.
Then:
$t^2=x^2+2+\frac(1)(x^2)$ or $x^2+\frac(1)(x^2)=t^2-2$.
We got the equation: $t^2-2+t=4$.
$t^2+t-6=0$.
The roots of this equation are the pair:
$t=-3$ and $t=2$.
Let's introduce the reverse substitution:
$x+\frac(1)(x)=-3$.
$x+\frac(1)(x)=2$.
We'll decide separately.
$x+\frac(1)(x)+3=0$.
$\frac(x^2+3x+1)(x)=0$.
$x_(1,2)=\frac(-3±\sqrt(9-4))(2)=\frac(-3±\sqrt(5))(2)$.
Let's solve the second equation:
$x+\frac(1)(x)-2=0$.
$\frac(x^2-2x+1)(x)=0$.
$\frac((x-1)^2)(x)=0$.
The root of this equation is the number $x=1$.
Answer: $x=\frac(-3±\sqrt(5))(2)$, $x=1$.
Problems to solve independently
Solve equations:1. $\frac(3x+2)(x)=\frac(2x+3)(x+2)$.
2. $\frac(5x)(x+2)-\frac(20)(x^2+2x)=\frac(4)(x)$.
3. $x^4-7x^2-18=0$.
4. $2x^2+x+2=\frac(8)(2x^2+x+4)$.
5. $(x+2)(x+3)(x+4)(x+5)=3$.