T. Kosyakova,
School No. 80, Krasnodar

Solving quadratic and fractional rational equations containing parameters

Lesson 4

Lesson topic:

The purpose of the lesson: develop the ability to solve fractional rational equations containing parameters.

Lesson type: introduction of new material.

1. (Orally) Solve the equations:

Example 1. Solve the equation

Solution.

Let's find invalid values a:

Answer. If If a = – 19 , then there are no roots.

Example 2. Solve the equation

Solution.

Let's find invalid parameter values a :

10 – a = 5, a = 5;

10 – a = a, a = 5.

Answer. If a = 5 a № 5 , That x=10– a .

Example 3. At what parameter values b the equation It has:

a) two roots; b) the only root?

Solution.

1) Find invalid parameter values b :

x = b, b 2 (b 2 – 1) – 2b 3 + b 2 = 0, b 4 – 2b 3 = 0,
b= 0 or b = 2;
x = 2, 4( b 2 – 1) – 4b 2 + b 2 = 0, b 2 – 4 = 0, (b – 2)(b + 2) = 0,
b= 2 or b = – 2.

2) Solve the equation x 2 ( b 2 – 1) – 2b 2x+ b 2 = 0:

D=4 b 4 – 4b 2 (b 2 – 1), D = 4 b 2 .

A)

Excluding invalid parameter values b , we find that the equation has two roots if b № – 2, b № – 1, b № 0, b № 1, b № 2 .

b) 4b 2 = 0, b = 0, but this is an invalid parameter value b ; If b 2 –1=0 , i.e. b=1 or.

Answer: a) if b № –2 , b № –1, b № 0, b № 1, b № 2 , then two roots; b) if b=1 or b=–1 , then the only root.

Independent work

Option 1

Solve the equations:

Option 2

Solve the equations:

Answers

IN 1. and if a=3 , then there are no roots; If b) if if a № 2 , then there are no roots.

AT 2. If a=2 , then there are no roots; If a=0 , then there are no roots; If
b) if a=– 1 , then the equation becomes meaningless; if there are no roots;
If

Homework assignment.

Solve the equations:

Answers: a) If a № –2 , That x= a ; If a=–2 , then there are no solutions; b) if a № –2 , That x=2; If a=–2 , then there are no solutions; c) if a=–2 , That x– any number except 3 ; If a № –2 , That x=2; d) if a=–8 , then there are no roots; If a=2 , then there are no roots; If

Lesson 5

Lesson topic:"Solving fractional rational equations containing parameters."

Lesson objectives:

training in solving equations with non-standard conditions;
conscious assimilation by students of algebraic concepts and connections between them.

Lesson type: systematization and generalization.

Checking homework.

Example 1. Solve the equation

a) relative to x; b) relative to y.

Solution.

a) Find invalid values y: y=0, x=y, y 2 =y 2 –2y,

y=0– invalid parameter value y.

If y№ 0 , That x=y–2; If y=0, then the equation becomes meaningless.

b) Find invalid parameter values x: y=x, 2x–x 2 +x 2 =0, x=0– invalid parameter value x; y(2+x–y)=0, y=0 or y=2+x;

y=0 does not satisfy the condition y(y–x)№ 0 .

Answer: a) if y=0, then the equation becomes meaningless; If y№ 0 , That x=y–2; b) if x=0 x№ 0 , That y=2+x .

Example 2. For what integer values ​​of the parameter a are the roots of the equation belong to the interval

D = (3 a + 2) 2 – 4a(a+ 1) 2 = 9 a 2 + 12a + 4 – 8a 2 – 8a,

D = ( a + 2) 2 .

If a № 0 or a № – 1 , That

Answer: 5 .

Example 3. Find relatively x integer solutions to the equation

Answer. If y=0, then the equation does not make sense; If y=–1, That x– any integer except zero; If y№ 0, y№ – 1, then there are no solutions.

Example 4. Solve the equation with parameters a And b .

If a№ – b , That

Answer. If a= 0 or b= 0 , then the equation becomes meaningless; If a№ 0, b№ 0, a=–b , That x– any number except zero; If a№ 0, b№ 0, a№ –b, That x=–a, x=–b .

Example 5. Prove that for any value of the parameter n other than zero, the equation has a single root equal to – n .

Solution.

i.e. x=–n, which was what needed to be proven.

Homework assignment.

1. Find integer solutions to the equation

2. At what parameter values c the equation It has:
a) two roots; b) the only root?

3. Find all the integer roots of the equation If a ABOUT N .

4. Solve the equation 3xy – 5x + 5y = 7: a) relatively y; b) relatively x .

1. The equation is satisfied by any integer equal values ​​of x and y other than zero.
2. a) When
b) at or
3. – 12; – 9; 0 .
4. a) If then there are no roots; If
b) if then there are no roots; If

Test

Option 1

1. Determine the type of equation 7c(c + 3)x 2 +(c–2)x–8=0 when: a) c=–3; b) c=2 ; V) c=4 .

2. Solve the equations: a) x 2 –bx=0 ; b) cx 2 –6x+1=0; V)

3. Solve the equation 3x–xy–2y=1:

a) relatively x ;
b) relatively y .

nx 2 – 26x + n = 0, knowing that the parameter n accepts only integer values.

5. For what values ​​of b does the equation It has:

a) two roots;
b) the only root?

Option 2

1. Determine the type of equation 5c(c + 4)x 2 +(c–7)x+7=0 when: a) c=–4 ; b) c=7 ; V) c=1 .

2. Solve the equations: a) y 2 +cy=0 ; b) ny 2 –8y+2=0 ; V)

3. Solve the equation 6x–xy+2y=5:

a) relatively x ;
b) relatively y .

4. Find the integer roots of the equation nx 2 –22x+2n=0 , knowing that the parameter n accepts only integer values.

5. For what values ​​of the parameter a does the equation It has:

a) two roots;
b) the only root?

Answers

IN 1. 1. a) Linear equation;
b) incomplete quadratic equation; c) quadratic equation.
2. a) If b=0, That x=0; If b№ 0, That x=0, x=b;
b) If cО (9;+Ґ ), then there are no roots;
c) if a=–4 , then the equation becomes meaningless; If a№ –4 , That x=– a .
3. a) If y=3, then there are no roots; If);
b) a=–3, a=1.

Additional tasks

Solve the equations:

Literature

1. Golubev V.I., Goldman A.M., Dorofeev G.V. About the parameters from the very beginning. – Tutor, No. 2/1991, p. 3–13.
2. Gronshtein P.I., Polonsky V.B., Yakir M.S. The necessary conditions in problems with parameters. – Kvant, No. 11/1991, p. 44–49.
3. Dorofeev G.V., Zatakavay V.V. Problem solving containing parameters. Part 2. – M., Perspective, 1990, p. 2–38.
4. Tynyakin S.A. Five hundred and fourteen problems with parameters. – Volgograd, 1991.
5. Yastrebinetsky G.A. Problems with parameters. – M., Education, 1986.

Fractional equations. ODZ.

Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)

We continue to master the equations. We already know how to work with linear and quadratic equations. The last view left - fractional equations . Or they are also called much more respectably - fractional rational equations. It is the same.

Fractional equations.

As the name implies, these equations necessarily contain fractions. But not just fractions, but fractions that have unknown in denominator. At least in one. For example:

Let me remind you that if the denominators are only numbers, these are linear equations.

How to decide fractional equations? First of all, get rid of fractions! After this, the equation most often turns into linear or quadratic. And then we know what to do... In some cases it can turn into an identity, such as 5=5 or an incorrect expression, such as 7=2. But this rarely happens. I will mention this below.

But how to get rid of fractions!? Very simple. Applying the same identical transformations.

We need to multiply the entire equation by the same expression. So that all denominators are reduced! Everything will immediately become easier. Let me explain with an example. Let us need to solve the equation:

How were you taught in elementary school? We move everything to one side, bring it to a common denominator, etc. Forget how horrible dream! This is what you need to do when you add or subtract. fractional expressions. Or you work with inequalities. And in equations, we immediately multiply both sides by an expression that will give us the opportunity to reduce all the denominators (i.e., in essence, by a common denominator). And what is this expression?

On the left side, reducing the denominator requires multiplying by x+2. And on the right, multiplication by 2 is required. This means that the equation must be multiplied by 2(x+2). Multiply:

This is a common multiplication of fractions, but I’ll describe it in detail:

Please note that I am not opening the bracket yet (x + 2)! So, in its entirety, I write it:

On the left side it contracts entirely (x+2), and on the right 2. Which is what was required! After reduction we get linear the equation:

And everyone can solve this equation! x = 2.

Let's solve another example, a little more complicated:

If we remember that 3 = 3/1, and 2x = 2x/ 1, we can write:

And again we get rid of what we don’t really like - fractions.

We see that to reduce the denominator with X, we need to multiply the fraction by (x – 2). And a few are not a hindrance to us. Well, let's multiply. All left side and all right side:

Parentheses again (x – 2) I'm not revealing. I work with the bracket as a whole as if it were one number! This must always be done, otherwise nothing will be reduced.

With a feeling of deep satisfaction we reduce (x – 2) and we get an equation without any fractions, with a ruler!

Now let’s open the brackets:

We bring similar ones, move everything to the left side and get:

But before that we will learn to solve other problems. On interest. That's a rake, by the way!

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

Equations with fractions themselves are not difficult and are very interesting. Let's look at the types of fractional equations and how to solve them.

How to solve equations with fractions - x in the numerator

If a fractional equation is given, where the unknown is in the numerator, the solution does not require additional conditions and is solved without unnecessary hassle. General form such an equation is x/a + b = c, where x is the unknown, a, b and c are ordinary numbers.

Find x: x/5 + 10 = 70.

In order to solve the equation, you need to get rid of fractions. Multiply each term in the equation by 5: 5x/5 + 5x10 = 70x5. 5x and 5 are cancelled, 10 and 70 are multiplied by 5 and we get: x + 50 = 350 => x = 350 – 50 = 300.

Find x: x/5 + x/10 = 90.

This example is a slightly more complicated version of the first one. There are two possible solutions here.

• Option 1: We get rid of fractions by multiplying all terms of the equation by a larger denominator, that is, by 10: 10x/5 + 10x/10 = 90×10 => 2x + x = 900 => 3x = 900 => x=300.
• Option 2: Add the left side of the equation. x/5 + x/10 = 90. The common denominator is 10. Divide 10 by 5, multiply by x, we get 2x. Divide 10 by 10, multiply by x, we get x: 2x+x/10 = 90. Hence 2x+x = 90×10 = 900 => 3x = 900 => x = 300.

We often encounter fractional equations in which the x's are on opposite sides of the equal sign. In such situations, it is necessary to move all the fractions with X's to one side, and the numbers to the other.

• Find x: 3x/5 = 130 – 2x/5.
• Move 2x/5 to the right with the opposite sign: 3x/5 + 2x/5 = 130 => 5x/5 = 130.
• We reduce 5x/5 and get: x = 130.

How to solve an equation with fractions - x in the denominator

This type of fractional equations requires writing additional conditions. The indication of these conditions is a mandatory and integral part of the right decision. By not adding them, you run the risk, since the answer (even if it is correct) may simply not be counted.

The general form of fractional equations, where x is in the denominator, is: a/x + b = c, where x is the unknown, a, b, c are ordinary numbers. Please note that x may not be any number. For example, x cannot equal zero, since it cannot be divided by 0. This is exactly what it is additional condition, which we must specify. This is called an area acceptable values, abbreviated as ODZ.

Find x: 15/x + 18 = 21.

We immediately write the ODZ for x: x ≠ 0. Now that the ODZ is indicated, we solve the equation using standard scheme, getting rid of fractions. Multiply all terms of the equation by x. 15x/x+18x = 21x => 15+18x = 21x => 15 = 3x => x = 15/3 = 5.

Often there are equations where the denominator contains not only x, but also some other operation with it, for example, addition or subtraction.

Find x: 15/(x-3) + 18 = 21.

We already know that the denominator cannot be equal to zero, which means x-3 ≠ 0. We move -3 to the right side, changing the “-” sign to “+” and we get that x ≠ 3. The ODZ is indicated.

We solve the equation, multiply everything by x-3: 15 + 18×(x – 3) = 21×(x – 3) => 15 + 18x – 54 = 21x – 63.

Move the X's to the right, numbers to the left: 24 = 3x => x = 8.

Presentation and lesson on the topic: "Rational equations. Algorithm and examples of solving rational equations"

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Introduction to Irrational Equations

Guys, we have learned 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 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 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. $(x+2)(x+3)(x+4)(x+5)=3$.

The lowest common denominator is used to simplify this equation. This method is used when you cannot write a given equation with one rational expression on each side of the equation (and use the crisscross method of multiplication). This method is used when you are given a rational equation with 3 or more fractions (in the case of two fractions, it is better to use criss-cross multiplication).

Find the lowest common denominator of the fractions (or least common multiple). NOZ is the smallest number that is evenly divisible by each denominator.

• Sometimes NPD is an obvious number. For example, if given the equation: x/3 + 1/2 = (3x +1)/6, then it is obvious that the least common multiple of the numbers 3, 2 and 6 is 6.
• If the NCD is not obvious, write down the multiples of the largest denominator and find among them one that will be a multiple of the other denominators. Often the NOD can be found by simply multiplying two denominators. For example, if the equation is given x/8 + 2/6 = (x - 3)/9, then NOS = 8*9 = 72.
• If one or more denominators contain a variable, the process becomes somewhat more complicated (but not impossible). In this case, the NOC is an expression (containing a variable) that is divided by each denominator. For example, in the equation 5/(x-1) = 1/x + 2/(3x) NOZ = 3x(x-1), because this expression is divided by each denominator: 3x(x-1)/(x-1 ) = 3x; 3x(x-1)/3x = (x-1); 3x(x-1)/x = 3(x-1).

Multiply both the numerator and denominator of each fraction by a number equal to the result of dividing the NOC by the corresponding denominator of each fraction. Since you are multiplying both the numerator and denominator by the same number, you are effectively multiplying the fraction by 1 (for example, 2/2 = 1 or 3/3 = 1).

• So in our example, multiply x/3 by 2/2 to get 2x/6, and 1/2 multiply by 3/3 to get 3/6 (the fraction 3x +1/6 does not need to be multiplied because it the denominator is 6).
• Proceed similarly when the variable is in the denominator. In our second example, NOZ = 3x(x-1), so multiply 5/(x-1) by (3x)/(3x) to get 5(3x)/(3x)(x-1); 1/x multiplied by 3(x-1)/3(x-1) and you get 3(x-1)/3x(x-1); 2/(3x) multiplied by (x-1)/(x-1) and you get 2(x-1)/3x(x-1).

Find x. Now that you have reduced the fractions to a common denominator, you can get rid of the denominator. To do this, multiply each side of the equation by the common denominator. Then solve the resulting equation, that is, find “x”. To do this, isolate the variable on one side of the equation.

• In our example: 2x/6 + 3/6 = (3x +1)/6. You can add 2 fractions with same denominator, so write the equation as: (2x+3)/6=(3x+1)/6. Multiply both sides of the equation by 6 and get rid of the denominators: 2x+3 = 3x +1. Solve and get x = 2.
• In our second example (with a variable in the denominator), the equation looks like (after reduction to a common denominator): 5(3x)/(3x)(x-1) = 3(x-1)/3x(x-1) + 2 (x-1)/3x(x-1). By multiplying both sides of the equation by N3, you get rid of the denominator and get: 5(3x) = 3(x-1) + 2(x-1), or 15x = 3x - 3 + 2x -2, or 15x = x - 5 Solve and get: x = -5/14.

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