Rational Equations 10. ODZ
We have already learned how to solve quadratic equations. Let us now extend the studied methods to rational equations.
What is a rational expression? We have already encountered this concept. Rational expressions called expressions made up of numbers, variables, their degrees and signs of mathematical operations.
Accordingly, rational equations are equations of the form: , where 
- rational expressions.
Previously, we considered only those rational equations that reduce to linear ones. Now let's consider those rational equations that can be reduced to quadratic ones.
Example 1
Solve the equation: .
Solution:
A fraction is 0 if and only if its numerator is 0 and its denominator is not 0.
We get the following system:
The first equation of the system is a quadratic equation. Before solving it, we divide all its coefficients by 3. We get:
We get two roots: ; .
Since 2 is never equal to 0, two conditions must be met: 
. Since none of the roots of the equation obtained above matches the invalid values of the variable that were obtained when solving the second inequality, they are both solutions to this equation.
Answer:.
So, let's formulate an algorithm for solving rational equations:
1. Move all terms to the left side so that 0 is obtained on the right side.
2. Transform and simplify the left side, bring all fractions to a common denominator.
3. Equate the resulting fraction to 0, according to the following algorithm: 
.
4. Write down those roots that are obtained in the first equation and satisfy the second inequality in response.
Let's look at another example.
Example 2
Solve the equation: 
.
Solution
At the very beginning, we transfer all the terms to the left side so that 0 remains on the right. We get:
Now we 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.
The coefficients of this equation: . We calculate the discriminant:
We get two roots: ; .
Now we 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 get 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 are reduced to quadratic equations.
In the next lesson, we will consider rational equations as models of real situations, and also consider motion problems.
Bibliography
- Bashmakov M.I. Algebra, 8th grade. - M.: Enlightenment, 2004.
- Dorofeev G.V., Suvorova S.B., Bunimovich E.A. et al. Algebra, 8. 5th ed. - M.: Education, 2010.
- Nikolsky S.M., Potapov M.A., Reshetnikov N.N., Shevkin A.V. Algebra, 8th grade. Textbook for educational institutions. - M.: Education, 2006.
- Festival of pedagogical ideas "Open Lesson" ().
- School.xvatit.com().
- Rudocs.exdat.com().
Homework
"Rational equations with polynomials" is one of the most frequently encountered topics in the USE tests in mathematics. For this reason, their repetition should be given special attention. Many students are faced with the problem of finding the discriminant, transferring indicators from the right side to the left side and bringing the equation to a common denominator, which makes it difficult to complete such tasks. Solving rational equations in preparation for the exam on our website will help you quickly cope with tasks of any complexity and pass the test perfectly.
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For more effective preparation for testing, we recommend that you follow our special method and start by repeating the rules and solving simple problems, gradually moving on to more complex ones. Thus, the graduate will be able to highlight the most difficult topics for himself and focus on their study.
Start preparing for the final testing with Shkolkovo today, and the result will not keep you waiting! Choose the easiest example from those given. If you quickly mastered the expression, move on to a more difficult task. So you can improve your knowledge up to solving USE tasks in mathematics at the profile level.
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Simply put, these are equations in which there is at least one with a variable in the denominator.
For example:
\(\frac(9x^2-1)(3x)\) \(=0\)
\(\frac(1)(2x)+\frac(x)(x+1)=\frac(1)(2)\)
\(\frac(6)(x+1)=\frac(x^2-5x)(x+1)\)
Example Not fractional rational equations:
\(\frac(9x^2-1)(3)\) \(=0\)
\(\frac(x)(2)\) \(+8x^2=6\)
How are fractional rational equations solved?
The main thing to remember about fractional rational equations is that you need to write in them. And after finding the roots, be sure to check them for admissibility. Otherwise, extraneous roots may appear, and the whole solution will be considered incorrect.
Algorithm for solving a fractional rational equation:
Write out and "solve" the ODZ.
Multiply each term in the equation by a common denominator and reduce the resulting fractions. The denominators will disappear.
Write the equation without opening brackets.
Solve the resulting equation.
Check the found roots with ODZ.
Write down in response the roots that passed the test in step 7.
Do not memorize the algorithm, 3-5 solved equations - and it will be remembered by itself.
Example . Solve fractional rational equation \(\frac(x)(x-2) - \frac(7)(x+2)=\frac(8)(x^2-4)\)
Solution:
Answer: \(3\).
Example . Find the roots of the fractional rational equation \(=0\)
Solution:
|
\(\frac(x)(x+2) + \frac(x+1)(x+5)-\frac(7-x)(x^2+7x+10)\)\(=0\) ODZ: \(x+2≠0⇔x≠-2\) |
We write down and "solve" ODZ. Expand \(x^2+7x+10\) into the formula: \(ax^2+bx+c=a(x-x_1)(x-x_2)\). |
|
|
\(\frac(x)(x+2) + \frac(x+1)(x+5)-\frac(7-x)((x+2)(x+5))\)\(=0\) |
Obviously, the common denominator of fractions: \((x+2)(x+5)\). We multiply the whole equation by it. |
|
|
\(\frac(x(x+2)(x+5))(x+2) + \frac((x+1)(x+2)(x+5))(x+5)-\) |
We reduce fractions |
|
|
\(x(x+5)+(x+1)(x+2)-7+x=0\) |
Expanding the brackets |
|
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\(x^2+5x+x^2+3x+2-7+x=0\) |
|
We give like terms |
|
\(2x^2+9x-5=0\) |
|
Finding the roots of the equation |
|
\(x_1=-5;\) \(x_2=\frac(1)(2).\) |
|
One of the roots does not fit under the ODZ, so in response we write down only the second root. |
Answer: \(\frac(1)(2)\).
We have already learned how to solve quadratic equations. Let us now extend the studied methods to rational equations.
What is a rational expression? We have already encountered this concept. Rational expressions called expressions made up of numbers, variables, their degrees and signs of mathematical operations.
Accordingly, rational equations are equations of the form: , where - rational expressions.
Previously, we considered only those rational equations that reduce to linear ones. Now let's consider those rational equations that can be reduced to quadratic ones.
Example 1
Solve the equation: .
Solution:
A fraction is 0 if and only if its numerator is 0 and its denominator is not 0.
We get the following system:
The first equation of the system is a quadratic equation. Before solving it, we divide all its coefficients by 3. We get:
We get two roots: ; .
Since 2 is never equal to 0, two conditions must be met: . Since none of the roots of the equation obtained above matches the invalid values of the variable that were obtained when solving the second inequality, they are both solutions to this equation.
Answer:.
So, let's formulate an algorithm for solving rational equations:
1. Move all terms to the left side so that 0 is obtained on the right side.
2. Transform and simplify the left side, bring all fractions to a common denominator.
3. Equate the resulting fraction to 0, according to the following algorithm: .
4. Write down those roots that are obtained in the first equation and satisfy the second inequality in response.
Let's look at another example.
Example 2
Solve the equation: .
Solution
At the very beginning, we transfer all the terms to the left side so that 0 remains on the right. We get:
Now we 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.
The coefficients of this equation: . We calculate the discriminant:
We get two roots: ; .
Now we 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 get 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 are reduced to quadratic equations.
In the next lesson, we will consider rational equations as models of real situations, and also consider motion problems.
Bibliography
- Bashmakov M.I. Algebra, 8th grade. - M.: Enlightenment, 2004.
- Dorofeev G.V., Suvorova S.B., Bunimovich E.A. et al. Algebra, 8. 5th ed. - M.: Education, 2010.
- Nikolsky S.M., Potapov M.A., Reshetnikov N.N., Shevkin A.V. Algebra, 8th grade. Textbook for educational institutions. - M.: Education, 2006.
- Festival of pedagogical ideas "Open Lesson" ().
- School.xvatit.com().
- Rudocs.exdat.com().
Homework
\(\bullet\) A rational equation is an equation expressed as \[\dfrac(P(x))(Q(x))=0\] where \(P(x), \ Q(x)\) - polynomials (the sum of "xes" in various degrees, multiplied by various numbers).
The expression on the left side of the equation is called the rational expression.
The ODV (range of acceptable values) of a rational equation is all values \(x\) for which the denominator does NOT vanish, i.e. \(Q(x)\ne 0\) .
\(\bullet\) For example, equations \[\dfrac(x+2)(x-3)=0,\qquad \dfrac 2(x^2-1)=3, \qquad x^5-3x=2\] are rational equations.
In the first equation, the ODZ is all \(x\) such that \(x\ne 3\) (they write \(x\in (-\infty;3)\cup(3;+\infty)\)); in the second equation, these are all \(x\) , such that \(x\ne -1; x\ne 1\) (write \(x\in (-\infty;-1)\cup(-1;1)\cup(1;+\infty)\)); and in the third equation there are no restrictions on the ODZ, that is, the ODZ is all \(x\) (they write \(x\in\mathbb(R)\) ). \(\bullet\) Theorems:
1) The product of two factors is equal to zero if and only if one of them is equal to zero, while the other does not lose its meaning, therefore, the equation \(f(x)\cdot g(x)=0\) is equivalent to the system \[\begin(cases) \left[ \begin(gathered)\begin(aligned) &f(x)=0\\ &g(x)=0 \end(aligned) \end(gathered) \right.\\ \ text(ODV equations) \end(cases)\] 2) The fraction is equal to zero if and only if the numerator is equal to zero and the denominator is not equal to zero, therefore, the equation \(\dfrac(f(x))(g(x))=0\) is equivalent to the system of equations \[\begin(cases) f(x)=0\\ g(x)\ne 0 \end(cases)\]\(\bullet\) Let's look at some examples.
1) Solve the equation \(x+1=\dfrac 2x\) . Let's find the ODZ of this equation - this is \(x\ne 0\) (since \(x\) is in the denominator).
So, the ODZ can be written as follows: .
Let's transfer all the terms into one part and reduce to a common denominator: \[\dfrac((x+1)\cdot x)x-\dfrac 2x=0\quad\Leftrightarrow\quad \dfrac(x^2+x-2)x=0\quad\Leftrightarrow\quad \begin( cases) x^2+x-2=0\\x\ne 0\end(cases)\] The solution to the first equation of the system will be \(x=-2, x=1\) . We see that both roots are non-zero. Therefore, the answer is: \(x\in \(-2;1\)\) .
2) Solve the equation \(\left(\dfrac4x - 2\right)\cdot (x^2-x)=0\). Let us find the ODZ of this equation. We see that the only value \(x\) for which the left side does not make sense is \(x=0\) . So the OD can be written as follows: \(x\in (-\infty;0)\cup(0;+\infty)\).
Thus, this equation is equivalent to the system:
\[\begin(cases) \left[ \begin(gathered)\begin(aligned) &\dfrac 4x-2=0\\ &x^2-x=0 \end(aligned) \end(gathered) \right. \\ x\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) \left[ \begin(gathered)\begin(aligned) &\dfrac 4x=2\\ &x(x-1)= 0 \end(aligned) \end(gathered) \right.\\ x\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) \left[ \begin(gathered)\begin(aligned) &x =2\\ &x=1\\ &x=0 \end(aligned) \end(gathered) \right.\\ x\ne 0 \end(cases) \quad \Leftrightarrow \quad \left[ \begin(gathered) \begin(aligned) &x=2\\ &x=1 \end(aligned) \end(gathered) \right.\] Indeed, despite the fact that \(x=0\) is the root of the second factor, if you substitute \(x=0\) in the original equation, then it will not make sense, because the expression \(\dfrac 40\) is not defined.
So the solution to this equation is \(x\in \(1;2\)\) .
3) Solve the equation \[\dfrac(x^2+4x)(4x^2-1)=\dfrac(3-x-x^2)(4x^2-1)\] In our equation \(4x^2-1\ne 0\) , whence \((2x-1)(2x+1)\ne 0\) , i.e. \(x\ne -\frac12; \frac12\) .
We transfer all the terms to the left side and reduce to a common denominator:
\(\dfrac(x^2+4x)(4x^2-1)=\dfrac(3-x-x^2)(4x^2-1) \quad \Leftrightarrow \quad \dfrac(x^2+4x- 3+x+x^2)(4x^2-1)=0\quad \Leftrightarrow \quad \dfrac(2x^2+5x-3)(4x^2-1)=0 \quad \Leftrightarrow\)
\(\Leftrightarrow \quad \begin(cases) 2x^2+5x-3=0\\ 4x^2-1\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) (2x-1 )(x+3)=0\\ (2x-1)(2x+1)\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) \left[ \begin(gathered) \begin( aligned) &x=\dfrac12\\ &x=-3 \end(aligned)\end(gathered) \right.\\ x\ne \dfrac 12\\ x\ne -\dfrac 12 \end(cases) \quad \ Leftrightarrow \quad x=-3\)
Answer: \(x\in \(-3\)\) .
Comment. If the answer consists of a finite set of numbers, then they can be written through a semicolon in curly braces, as shown in the previous examples.
Tasks that require solving rational equations are encountered every year in the Unified State Examination in mathematics, therefore, in preparation for passing the certification test, graduates should definitely repeat the theory on this topic on their own. To be able to cope with such tasks, graduates who pass both the basic and the profile level of the exam must necessarily. Having mastered the theory and dealt with practical exercises on the topic "Rational Equations", students will be able to solve problems with any number of actions and expect to receive competitive points at the end of the exam.
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