Irrational equations and ways to solve them. Irrational equations

Answer: x = 3.

8 method. Application of the derivative in solving irrational equations

Most often, when solving equations using the derivative method, the estimation method is used.

EXAMPLE 15:

Solve the equation: (1)

Solution: Since https://pandia.ru/text/78/021/images/image122_1.gif" width="371" height="29">, or (2). Consider the function ..gif" width="400" height="23 src=">.gif" width="215" height="49"> at all and therefore increasing. Therefore, the equation is equivalent to an equation that has a root that is the root of the original equation.

Answer:

EXAMPLE 16:

Solve the irrational equation:

The domain of definition of the function is a segment. Find the largest and smallest value the values ​​of this function on the interval . To do this, we find the derivative of the function f(x): https://pandia.ru/text/78/021/images/image136_1.gif" width="37 height=19" height="19">. Let's find the values ​​of the function f(x) at the ends of the segment and at the point : So, But and, therefore, equality is possible only under the condition https://pandia.ru/text/78/021/images/image136_1.gif" width="37" height="19 src=" > Verification shows that the number 3 is the root of this equation.

Answer: x = 3.

9 method. Functional

In exams, they sometimes offer to solve equations that can be written in the form , where is a certain function.

For example, some equations: 1) 2) . Indeed, in the first case , in the second case . Therefore, solve irrational equations using the following statement: if a function is strictly increasing on the set X and for any , then the equations, etc., are equivalent on the set X .

Solve the irrational equation: https://pandia.ru/text/78/021/images/image145_1.gif" width="103" height="25"> strictly increasing on the set R, and https://pandia.ru/text/78/021/images/image153_1.gif" width="45" height="24 src=">..gif" width="104" height="24 src=" > which has a unique root Therefore, the equivalent equation (1) also has a unique root

Answer: x = 3.

EXAMPLE 18:

Solve the irrational equation: (1)

By virtue of the definition of the square root, we obtain that if equation (1) has roots, then they belong to the set DIV_ADBLOCK166">

. (2)

Consider the function https://pandia.ru/text/78/021/images/image147_1.gif" width="35" height="21"> strictly increasing on this set for any ..gif" width="100" height ="41"> which has a single root Therefore, and equivalent to it on the set X equation (1) has a single root

Answer: https://pandia.ru/text/78/021/images/image165_0.gif" width="145" height="27 src=">

Solution: This equation is equivalent to a mixed system

If the equation contains a variable under the square root sign, then the equation is called irrational.
Consider the irrational equation

This equality, by definition of the square root, means that 2x + 1 = 32. In fact, we went from the given irrational equation to the rational equation 2x + 1 = 9 by squaring both sides of the irrational equation. The method of squaring both sides of an equation is the main method for solving irrational equations. However, this is understandable: how else to get rid of the sign of the square root? From the equation 2x + 1 = 9 we find x = 4.
This is both the root of the equation 2x + 1 = 9 and the given irrational equation.
The squaring method is technically simple, but sometimes leads to trouble. Consider, for example, the irrational equation

By squaring both sides, we get

Next we have:
2x-4x = -7 +5; -2x = -2; x = 1.
But the value x - 1, being the root of the rational equation 2x - 5 = 4x - 7, is not the root of the given irrational equation. Why? Substituting 1 instead of x in the given irrational equation, we get . How can we talk about the fulfillment of numerical equality, if both its left and right parts contain expressions that do not make sense? In such cases, they say: x \u003d 1 is an extraneous root for a given irrational equation. It turns out that the given irrational equation has no roots.
Let's solve the irrational equation

-
The roots of this equation can be found orally, as we did at the end of the previous paragraph: their product is - 38, and the sum is - 17; it is easy to guess that these are the numbers 2
and - 19. So, x 1 \u003d 2, x 2 \u003d - 19.
Substituting the value 2 instead of x in the given irrational equation, we get

This is not true.
Substituting the value - 19 instead of x in the given irrational equation, we get

This is also incorrect.
What is the conclusion? Both found values ​​are extraneous roots. In other words, the given irrational equation, like the previous one, has no roots.
An extraneous root is not a new concept for you, extraneous roots have already been encountered when solving rational equations, checking helps to detect them. For irrational equations, checking is a mandatory step in solving the equation, which will help to detect extraneous roots, if any, and discard them (usually they say “weed out”).

So, an irrational equation is solved by squaring both its parts; having solved the resulting rational equation, it is necessary to make a check, eliminating possible extraneous roots.

Using this derivation, let's look at a few examples.

Example 1 solve the equation

Solution. Let's square both sides of equation (1):

Next, successively we have

5x - 16 \u003d x 2 - 4x + 4;
x 2 - 4x + 4 - 5x + 16 = 0;
x 2 - 9x + 20 = 0;
x 1 = 5, x 2 = 4.
Examination. Substituting x \u003d 5 into equation (1), we get - the correct equality. Substituting x \u003d 4 into equation (1), we get - the correct equality. Hence, both found values ​​are the roots of equation (1).
O n e t: 4; five.

Example 2 solve the equation
(we encountered this equation in § 22 and we “postponed” its solution until better times.) of an irrational equation, we obtain
2x2 + 8* + 16 = (44 - 2x) 2 .
Then we have
2x 2 + 8x + 16 \u003d 1936 - 176x + 4x 2;
- 2x 2 + 184x - 1920 = 0;
x 2 - 92x + 960 = 0;
x 1 = 80, x 2 = 12.
Examination. Substituting x = 80 into the given irrational equation, we get

This is obviously an incorrect equality, since its right side contains a negative number, and its left side contains a positive number. So x = 80 is an extraneous root for this equation.

Substituting x = 12 into the given irrational equation, we get

i.e. . = 20, is the correct equality. Therefore, x = 12 is the root of this equation.
Answer: 12.

We divide both parts of the last equation term by term by 2:

Examination. Substituting the value x = 14 into equation (2), we get is an incorrect equality, so x = 14 is an extraneous root.
Substituting the value x = -1 into equation (2), we obtain
- true equality. Therefore, x = - 1 is the root of equation (2).
A n t e t : - 1.

Example 4 solve the equation

Solution. Of course, you can solve this equation in the same way that we used in the previous examples: rewrite the equation as

Square both sides of this equation, solve the resulting rational equation and check the found roots by substituting them into
original irrational equation.

But we will use a more elegant way: we introduce a new variable y = . Then we get 2y 2 + y - 3 \u003d 0 - a quadratic equation with respect to the variable y. Let's find its roots: y 1 = 1, y 2 = -. Thus, the task was reduced to solving two

From the first equation we find x \u003d 1, the second equation has no roots (you remember that it takes only non-negative values).
Answer: 1.
We conclude this section with a rather serious theoretical discussion. The point is the following. You have already gained some experience in solving various equations: linear, square, rational, irrational. You know that when solving equations, various transformations are performed,
for example: a member of the equation is transferred from one part of the equation to another with the opposite sign; both sides of the equation are multiplied or divided by the same non-zero number; get rid of the denominator, i.e., replace the equation = 0 with the equation p (x) = 0; Both sides of the equation are squared.

Of course, you noticed that as a result of some transformations extraneous roots could appear, and therefore you had to be vigilant: check all the roots found. So we will now try to comprehend all this from a theoretical point of view.

Definition. Two equations f (x) = g (x) and r (x) = s (x) are called equivalent if they have the same roots (or, in particular, if both equations have no roots).

Usually, when solving an equation, they try to replace this equation with a simpler one, but equivalent to it. Such a change is called an equivalent transformation of the equation.

The following transformations are equivalent transformations of the equation:

1. Transfer of terms of the equation from one part of the equation to another with opposite signs.
For example, replacing the equation 2x + 5 = 7x - 8 with the equation 2x - 7x = - 8 - 5 is an equivalent transformation of the equation. It means that

the equations 2x + 5 = 7x -8 and 2x - 7x = -8 - 5 are equivalent.

2. Multiplying or dividing both sides of an equation by the same non-zero number.
For example, replacing the equation 0.5x 2 - 0.3x \u003d 2 with the equation 5x 2 - Zx \u003d 20
(both parts of the equation were multiplied term by term by 10) is an equivalent transformation of the equation.

Non-equivalent transformations of the equation are the following transformations:

1. Exemption from denominators containing variables.
For example, replacing an equation with the equation x 2 \u003d 4 is a non-equivalent transformation of the equation. The fact is that the equation x 2 \u003d 4 has two roots: 2 and - 2, and given equation the value x = 2 cannot satisfy (the denominator vanishes). In such cases, we said this: x \u003d 2 is an extraneous root.

2. Squaring both sides of the equation.
We will not give examples, since there were quite a lot of them in this paragraph.
If in the process of solving the equation one of the indicated non-equivalent transformations was used, then all the roots found must be checked by substitution into the original equation, since among them there may be extraneous roots.

Topic: “Irrational equations of the form , .»

(Methodological development.)

Basic concepts

Irrational equations called equations in which the variable is contained under the sign of the root (radical) or the sign of raising to a fractional power.

An equation of the form f(x)=g(x), where at least one of the expressions f(x) or g(x) is irrational irrational equation.

Basic properties of radicals:

• All radicals even degree are arithmetic, those. if the radical expression is negative, then the radical does not make sense (does not exist); if the root expression is equal to zero, then the radical is also zero; if the radical expression is positive, then the value of the radical exists and is positive.
• All radicals odd degree are defined for any value of the radical expression. Moreover, the radical is negative if the radical expression is negative; is zero if the root expression is zero; is positive if the subjugated expression is positive.

Methods for solving irrational equations

Solve an irrational equation - means to find all the real values ​​of the variable, when substituting them into the original equation, it turns into the correct numerical equality, or to prove that such values ​​do not exist. Irrational equations are solved on the set of real numbers R.

The range of valid values ​​of the equation consists of those values ​​of the variable for which all expressions under the sign of radicals of even degree are non-negative.

The main methods for solving irrational equations are:

a) the method of raising both parts of the equation to the same power;

b) the method of introducing new variables (method of substitutions);

c) artificial methods for solving irrational equations.

In this article, we will focus on the consideration of equations of the form defined above and present 6 methods for solving such equations.

1 method. Cube.

This method requires the use of abbreviated multiplication formulas and does not contain "pitfalls", i.e. does not lead to the appearance of extraneous roots.

Example 1 solve the equation

Solution:

We rewrite the equation in the form and cube both sides of it. We obtain an equation equivalent to this equation ,

Answer: x=2, x=11.

Example 2. Solve the equation.

Solution:

Let's rewrite the equation in the form and raise both sides of it into a cube. We obtain an equation equivalent to this equation

and consider the resulting equation as a quadratic one with respect to one of the roots

therefore, the discriminant is 0, and the equation can have a solution x=-2.

Examination:

Answer: x=-2.

Comment: The check can be omitted if the quadratic equation is completed.

2 method. Cube using a formula.

We will continue to cube the equation, but at the same time we will use modified formulas for abbreviated multiplication.

Let's use the formulas:

(minor modification known formula), then

Example3. solve the equation .

Solution:

Let's cube the equation using the formulas given above.

But the expression must be equal to the right side. Therefore we have:

.

Now, when cubed, we get the usual quadratic equation:

, and its two roots

Both values, as shown by the test, are correct.

Answer: x=2, x=-33.

But are all transformations here equivalent? Before answering this question, let's solve one more equation.

Example 4. Solve the equation.

Solution:

Raising, as before, both parts to the third power, we have:

From where (considering that the expression in brackets is ), we get:

We get, .Let's do a check and make sure x=0 is an extraneous root.

Answer: .

Let's answer the question: "Why did extraneous roots arise?"

Equality leads to equality . Replacing from with -s, we get:

It is easy to check the identity

So, if , then either , or . The equation can be represented as , .

Replacing from with -s, we get: if , then either , or

Therefore, when using this solution method, it is imperative to check and make sure that there are no extraneous roots.

3 method. System method.

Example 5 solve the equation .

Solution:

Let be , . Then:

How is it obvious that

The second equation of the system is obtained in such a way that the linear combination of radical expressions does not depend on the original variable.

It is easy to see that the system has no solution, and therefore the original equation has no solution.

Answer: No roots.

Example 6 solve the equation .

Solution:

We introduce a replacement, compose and solve a system of equations.

Let be , . Then

Returning to the original variable, we have:

Answer: x=0.

4 method. Using the monotonicity of functions.

Before using this method, let's turn to the theory.

We will need the following properties:

Example 7 solve the equation .

Solution:

The left side of the equation is an increasing function, and the right side is a number, i.e. constant, therefore, the equation has no more than one root, which we select: x \u003d 9. Checking to make sure that the root is suitable.

Equations are called irrational if they contain an unknown quantity under the root sign. These are, for example, the equations

In many cases, by applying once or repeatedly the exponentiation of both parts of the equation, it is possible to reduce the irrational equation to an algebraic equation of one degree or another (which is a consequence of the original equation). Since when raising the equation to a power, extraneous solutions may appear, then, having solved the algebraic equation to which we have reduced this irrational equation, we should check the found roots by substituting into the original equation and save only those that satisfy it, and discard the rest - extraneous.

When solving irrational equations, we restrict ourselves only to their real roots; all roots of even degree in the notation of equations are understood in the arithmetic sense.

Consider some typical examples irrational equations.

A. Equations containing the unknown under the square root sign. If this equation contains only one Square root, under the sign of which there is an unknown, then this root should be isolated, that is, placed in one part of the equation, and all other terms should be transferred to another part. After squaring both sides of the equation, we will already be free from irrationality and obtain an algebraic equation for

Example 1. Solve the equation.

Solution. We isolate the root on the left side of the equation;

We square the resulting equation:

We find the roots of this equation:

Verification shows that only satisfies the original equation.

If the equation includes two or more roots containing x, then squaring has to be repeated several times.

Example 2. Solve the following equations:

Solution, a) We square both sides of the equation:

We separate the root:

The resulting equation is again squared:

After transformations, we obtain the following quadratic equation for:

solve it:

By substituting into the original equation, we make sure that there is its root, but it is an extraneous root for it.

b) The example can be solved in the same way that example a) was solved. However, taking advantage of the fact that the right side of this equation does not contain an unknown quantity, we will proceed differently. We multiply the equation by the expression conjugate to its left side; we get

On the right is the product of the sum and the difference, that is, the difference of the squares. From here

On the left side of this equation was the sum of the square roots; on the left side of the equation now obtained is the difference of the same roots. Let's write down the given and received equations:

Taking the sum of these equations, we get

We square the last equation and, after simplifications, we get

From here we find . By checking we are convinced that only the number serves as the root of this equation. Example 3. Solve the equation

Here, already under the radical sign, we have square trinomials.

Solution. We multiply the equation by the expression conjugated with its left side:

Subtract the last equation from the given one:

Let's square this equation:

From the last equation we find . By checking we are convinced that only the number x \u003d 1 serves as the root of this equation.

B. Equations containing roots of the third degree. Systems of irrational equations. We confine ourselves to individual examples of such equations and systems.

Example 4. Solve the equation

Solution. Let us show two ways of solving equation (70.1). First way. Let's cube both sides of this equation (see formula (20.8)):

(here we have replaced the sum cube roots number 4, using the equation).

So we have

i.e., after simplifications,

whence Both roots satisfy the original equation.

The second way. Let's put

Equation (70.1) will be written as . Moreover, it is clear that . From equation (70.1) we have passed to the system

Dividing the first equation of the system term by term by the second, we find

An irrational equation is any equation that contains a function under the root sign. For example:

Such equations are always solved in 3 steps:

1. Separate the root. In other words, if there are other numbers or functions to the left of the equal sign in addition to the root, all this must be moved to the right by changing the sign. At the same time, only the radical should remain on the left - without any coefficients.
2. 2. We square both sides of the equation. At the same time, remember that the range of the root is all non-negative numbers. Hence the function on the right irrational equation must also be non-negative: g (x) ≥ 0.
3. The third step follows logically from the second: you need to perform a check. The fact is that in the second step we could have extra roots. And in order to cut them off, it is necessary to substitute the resulting candidate numbers into the original equation and check: is the correct numerical equality really obtained?

Solving an irrational equation

Let's deal with our irrational equation given at the very beginning of the lesson. Here the root is already secluded: to the left of the equal sign there is nothing but the root. Let's square both sides:

2x 2 - 14x + 13 = (5 - x) 2
2x2 - 14x + 13 = 25 - 10x + x2
x 2 - 4x - 12 = 0

We solve the resulting quadratic equation through the discriminant:

D = b 2 − 4ac = (−4) 2 − 4 1 (−12) = 16 + 48 = 64
x 1 = 6; x 2 \u003d -2

It remains only to substitute these numbers in the original equation, i.e. perform a check. But even here you can do the right thing to simplify the final decision.

How to simplify the decision

Let's think: why do we even check at the end of solving an irrational equation? We want to make sure that when substituting our roots, there will be a non-negative number to the right of the equal sign. After all, we already know for sure that it is a non-negative number on the left, because the arithmetic square root (because of which our equation is called irrational) by definition cannot be less than zero.

Therefore, all we need to check is that the function g ( x ) = 5 − x , which is to the right of the equal sign, is non-negative:

g(x) ≥ 0

We substitute our roots into this function and get:

g (x 1) \u003d g (6) \u003d 5 - 6 \u003d -1< 0
g (x 2) = g (−2) = 5 − (−2) = 5 + 2 = 7 > 0

From the obtained values ​​it follows that the root x 1 = 6 does not suit us, since when substituting into the right side of the original equation, we get a negative number. But the root x 2 \u003d −2 is quite suitable for us, because:

1. This root is the solution to the quadratic equation obtained by raising both sides irrational equation into a square.
2. The right side of the original irrational equation, when the root x 2 = −2 is substituted, turns into a positive number, i.e. range arithmetic root not broken.

That's the whole algorithm! As you can see, solving equations with radicals is not so difficult. The main thing is not to forget to check the received roots, otherwise it is very likely to get extra answers.