Test No. 6 General level.

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Lecture: “Methods for solving exponential equations.”

1 . Exponential equations.

Equations containing unknowns in exponents are called exponential equations. The simplest of them is the equation ax = b, where a > 0, a ≠ 1.

1) At b< 0 и b = 0 это уравнение, согласно свойству 1 показательной функции, не имеет решения.

2) For b > 0, using the monotonicity of the function and the root theorem, the equation has a unique root. In order to find it, b must be represented in the form b = aс, аx = bс ó x = c or x = logab.

Exponential equations by algebraic transformations lead to standard equations, which are solved using the following methods:

1) method of reduction to one base;

2) assessment method;

3) graphic method;

4) method of introducing new variables;

5) factorization method;

6) exponential – power equations;

7) demonstrative with a parameter.

2 . Method of reduction to one base.

The method is based on the following property of degrees: if two degrees are equal and their bases are equal, then their exponents are equal, i.e., one must try to reduce the equation to the form

Examples. Solve the equation:

1 . 3x = 81;

Let's represent the right side of the equation in the form 81 = 34 and write the equation equivalent to the original 3 x = 34; x = 4. Answer: 4.

2. https://pandia.ru/text/80/142/images/image004_8.png" width="52" height="49">and let's move on to the equation for exponents 3x+1 = 3 – 5x; 8x = 4; x = 0.5 Answer: 0.5.

3. https://pandia.ru/text/80/142/images/image006_8.png" width="105" height="47">

Note that the numbers 0.2, 0.04, √5 and 25 represent powers of 5. Let's take advantage of this and transform the original equation as follows:

, whence 5-x-1 = 5-2x-2 ó - x – 1 = - 2x – 2, from which we find the solution x = -1. Answer: -1.

5. 3x = 5. By definition of logarithm, x = log35. Answer: log35.

6. 62x+4 = 33x. 2x+8.

Let's rewrite the equation in the form 32x+4.22x+4 = 32x.2x+8, i.e..png" width="181" height="49 src="> Hence x – 4 =0, x = 4. Answer: 4.

7 . 2∙3x+1 - 6∙3x-2 - 3x = 9. Using the properties of powers, we write the equation in the form 6∙3x - 2∙3x – 3x = 9 then 3∙3x = 9, 3x+1 = 32, i.e. i.e. x+1 = 2, x =1. Answer: 1.

Problem bank No. 1.

Solve the equation:

Test No. 1.

What is an exponential equation and how to solve it. Methods for solving exponential equations

At the stage of preparation for the final test, high school students need to improve their knowledge on the topic “Exponential Equations.” The experience of past years indicates that such tasks cause certain difficulties for schoolchildren. Therefore, high school students, regardless of their level of preparation, need to thoroughly master the theory, remember the formulas and understand the principle of solving such equations. Having learned to cope with this type of problem, graduates can count on high scores when passing the Unified State Exam in mathematics.

Get ready for exam testing with Shkolkovo!

When reviewing the materials they have covered, many students are faced with the problem of finding the formulas needed to solve equations. A school textbook is not always at hand, and selecting the necessary information on a topic on the Internet takes a long time.

The Shkolkovo educational portal invites students to use our knowledge base. We are implementing a completely new method of preparing for the final test. By studying on our website, you will be able to identify gaps in knowledge and pay attention to those tasks that cause the most difficulty.

Shkolkovo teachers collected, systematized and presented all the material necessary for successfully passing the Unified State Exam in the simplest and most accessible form.

Basic definitions and formulas are presented in the “Theoretical background” section.

To better understand the material, we recommend that you practice completing the assignments. Carefully review the examples of exponential equations with solutions presented on this page to understand the calculation algorithm. After that, proceed to perform tasks in the “Directories” section. You can start with the easiest tasks or go straight to solving complex exponential equations with several unknowns or . The database of exercises on our website is constantly supplemented and updated.

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Exponential equations are those in which the unknown is contained in the exponent. The simplest exponential equation has the form: a x = a b, where a> 0, a 1, x is unknown.

The main properties of powers by which exponential equations are transformed: a>0, b>0.

When solving exponential equations, the following properties of the exponential function are also used: y = a x, a > 0, a1:

To represent a number as a power, use the basic logarithmic identity: b = , a > 0, a1, b > 0.

Problems and tests on the topic "Exponential Equations"

  • Exponential equations

    Lessons: 4 Assignments: 21 Tests: 1

  • Exponential equations - Important topics for reviewing the Unified State Examination in mathematics

    Tasks: 14

  • Systems of exponential and logarithmic equations - Exponential and logarithmic functions grade 11

    Lessons: 1 Assignments: 15 Tests: 1

  • §2.1. Solving exponential equations

    Lessons: 1 Tasks: 27

  • §7 Exponential and logarithmic equations and inequalities - Section 5. Exponential and logarithmic functions, grade 10

    Lessons: 1 Tasks: 17

To successfully solve exponential equations, you must know the basic properties of powers, properties of the exponential function, and the basic logarithmic identity.

When solving exponential equations, two main methods are used:

  1. transition from the equation a f(x) = a g(x) to the equation f(x) = g(x);
  2. introduction of new lines.

Examples.

1. Equations reduced to the simplest. They are solved by reducing both sides of the equation to a power with the same base.

3 x = 9 x – 2.

Solution:

3 x = (3 2) x – 2 ;
3 x = 3 2x – 4 ;
x = 2x –4;
x = 4.

Answer: 4.

2. Equations solved by taking the common factor out of brackets.

Solution:

3 x – 3 x – 2 = 24
3 x – 2 (3 2 – 1) = 24
3 x – 2 × 8 = 24
3 x – 2 = 3
x – 2 = 1
x = 3.

Answer: 3.

3. Equations solved using a change of variable.

Solution:

2 2x + 2 x – 12 = 0
We denote 2 x = y.
y 2 + y – 12 = 0
y 1 = - 4; y2 = 3.
a) 2 x = - 4. The equation has no solutions, because 2 x > 0.
b) 2 x = 3; 2 x = 2 log 2 3 ; x = log 2 3.

Answer: log 2 3.

4. Equations containing powers with two different (not reducible to each other) bases.

3 × 2 x + 1 - 2 × 5 x – 2 = 5 x + 2 x – 2.

3× 2 x + 1 – 2 x – 2 = 5 x – 2 × 5 x – 2
2 x – 2 ×23 = 5 x – 2
×23
2 x – 2 = 5 x – 2
(5/2) x– 2 = 1
x – 2 = 0
x = 2.

Answer: 2.

5. Equations that are homogeneous with respect to a x and b x.

General form: .

9 x + 4 x = 2.5 × 6 x.

Solution:

3 2x – 2.5 × 2 x × 3 x +2 2x = 0 |: 2 2x > 0
(3/2) 2x – 2.5 × (3/2) x + 1 = 0.
Let us denote (3/2) x = y.
y 2 – 2.5y + 1 = 0,
y 1 = 2; y 2 = ½.

Answer: log 3/2 2; - log 3/2 2.

Solving exponential equations. Examples.

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

What's happened exponential equation? This is an equation in which the unknowns (x's) and expressions with them are in indicators some degrees. And only there! It is important.

There you are examples of exponential equations:

3 x 2 x = 8 x+3

Note! In the bases of degrees (below) - only numbers. IN indicators degrees (above) - a wide variety of expressions with an X. If, suddenly, an X appears in the equation somewhere other than an indicator, for example:

this will already be an equation of mixed type. Such equations do not have clear rules for solving them. We will not consider them for now. Here we will deal with solving exponential equations in its purest form.

In fact, even pure exponential equations are not always solved clearly. But there are certain types of exponential equations that can and should be solved. These are the types we will consider.

Solving simple exponential equations.

First, let's solve something very basic. For example:

Even without any theories, by simple selection it is clear that x = 2. Nothing more, right!? No other value of X works. Now let's look at the solution to this tricky exponential equation:

What have we done? We, in fact, simply threw out the same bases (triples). Completely thrown out. And, the good news is, we hit the nail on the head!

Indeed, if in an exponential equation there are left and right the same numbers in any powers, these numbers can be removed and the exponents can be equalized. Mathematics allows. It remains to solve a much simpler equation. Great, right?)

However, let us remember firmly: You can remove bases only when the base numbers on the left and right are in splendid isolation! Without any neighbors and coefficients. Let's say in the equations:

2 x +2 x+1 = 2 3, or

twos cannot be removed!

Well, we have mastered the most important thing. How to move from evil exponential expressions to simpler equations.

"Those are the times!" - you say. “Who would give such a primitive lesson on tests and exams!?”

I have to agree. Nobody will. But now you know where to aim when solving tricky examples. It must be brought to the form where the same base number is on the left and right. Then everything will be easier. Actually, this is a classic of mathematics. We take the original example and transform it to the desired one us mind. According to the rules of mathematics, of course.

Let's look at examples that require some additional effort to reduce them to the simplest. Let's call them simple exponential equations.

Solving simple exponential equations. Examples.

When solving exponential equations, the main rules are actions with degrees. Without knowledge of these actions nothing will work.

To actions with degrees, one must add personal observation and ingenuity. Do we need the same base numbers? So we look for them in the example in explicit or encrypted form.

Let's see how this is done in practice?

Let us be given an example:

2 2x - 8 x+1 = 0

The first keen glance is at grounds. They... They are different! Two and eight. But it’s too early to become discouraged. It's time to remember that

Two and eight are relatives in degree.) It is quite possible to write:

8 x+1 = (2 3) x+1

If we recall the formula from operations with degrees:

(a n) m = a nm ,

this works out great:

8 x+1 = (2 3) x+1 = 2 3(x+1)

The original example began to look like this:

2 2x - 2 3(x+1) = 0

We transfer 2 3 (x+1) to the right (no one has canceled the elementary operations of mathematics!), we get:

2 2x = 2 3(x+1)

That's practically all. Removing the bases:

We solve this monster and get

This is the correct answer.

In this example, knowing the powers of two helped us out. We identified in eight there is an encrypted two. This technique (encoding common bases under different numbers) is a very popular technique in exponential equations! Yes, and in logarithms too. You must be able to recognize powers of other numbers in numbers. This is extremely important for solving exponential equations.

The fact is that raising any number to any power is not a problem. Multiply, even on paper, and that’s it. For example, anyone can raise 3 to the fifth power. 243 will work out if you know the multiplication table.) But in exponential equations, much more often it is not necessary to raise to a power, but vice versa... Find out what number to what degree is hidden behind the number 243, or, say, 343... No calculator will help you here.

You need to know the powers of some numbers by sight, right... Let's practice?

Determine what powers and what numbers the numbers are:

2; 8; 16; 27; 32; 64; 81; 100; 125; 128; 216; 243; 256; 343; 512; 625; 729, 1024.

Answers (in a mess, of course!):

5 4 ; 2 10 ; 7 3 ; 3 5 ; 2 7 ; 10 2 ; 2 6 ; 3 3 ; 2 3 ; 2 1 ; 3 6 ; 2 9 ; 2 8 ; 6 3 ; 5 3 ; 3 4 ; 2 5 ; 4 4 ; 4 2 ; 2 3 ; 9 3 ; 4 5 ; 8 2 ; 4 3 ; 8 3 .

If you look closely, you can see a strange fact. There are significantly more answers than tasks! Well, it happens... For example, 2 6, 4 3, 8 2 - that's all 64.

Let us assume that you have taken note of the information about familiarity with numbers.) Let me also remind you that to solve exponential equations we use all stock of mathematical knowledge. Including those from junior and middle classes. You didn’t go straight to high school, right?)

For example, when solving exponential equations, putting the common factor out of brackets often helps (hello to 7th grade!). Let's look at an example:

3 2x+4 -11 9 x = 210

And again, the first glance is at the foundations! The bases of the degrees are different... Three and nine. But we want them to be the same. Well, in this case the desire is completely fulfilled!) Because:

9 x = (3 2) x = 3 2x

Using the same rules for dealing with degrees:

3 2x+4 = 3 2x ·3 4

That’s great, you can write it down:

3 2x 3 4 - 11 3 2x = 210

We gave an example for the same reasons. So, what is next!? You can't throw out threes... Dead end?

Not at all. Remember the most universal and powerful decision rule everyone math tasks:

If you don’t know what you need, do what you can!

Look, everything will work out).

What's in this exponential equation Can do? Yes, on the left side it just begs to be taken out of brackets! The overall multiplier of 3 2x clearly hints at this. Let's try, and then we'll see:

3 2x (3 4 - 11) = 210

3 4 - 11 = 81 - 11 = 70

The example keeps getting better and better!

We remember that to eliminate grounds we need a pure degree, without any coefficients. The number 70 bothers us. So we divide both sides of the equation by 70, we get:

Oops! Everything got better!

This is the final answer.

It happens, however, that taxiing on the same basis is achieved, but their elimination is not possible. This happens in other types of exponential equations. Let's master this type.

Replacing a variable in solving exponential equations. Examples.

Let's solve the equation:

4 x - 3 2 x +2 = 0

First - as usual. Let's move on to one base. To a deuce.

4 x = (2 2) x = 2 2x

We get the equation:

2 2x - 3 2 x +2 = 0

And this is where we hang out. The previous techniques will not work, no matter how you look at it. We'll have to pull out another powerful and universal method from our arsenal. It's called variable replacement.

The essence of the method is surprisingly simple. Instead of one complex icon (in our case - 2 x) we write another, simpler one (for example - t). Such a seemingly meaningless replacement leads to amazing results!) Everything just becomes clear and understandable!

So let

Then 2 2x = 2 x2 = (2 x) 2 = t 2

In our equation we replace all powers with x's by t:

Well, does it dawn on you?) Have you forgotten the quadratic equations yet? Solving through the discriminant, we get:

The main thing here is not to stop, as happens... This is not the answer yet, we need x, not t. Let's return to the X's, i.e. we make a reverse replacement. First for t 1:

That is,

One root was found. We are looking for the second one from t 2:

Hm... 2 x on the left, 1 on the right... Problem? Not at all! It is enough to remember (from operations with powers, yes...) that a unit is any number to the zero power. Any. Whatever is needed, we will install it. We need a two. Means:

That's it now. We got 2 roots:

This is the answer.

At solving exponential equations at the end sometimes you end up with some kind of awkward expression. Type:

Seven cannot be converted to two through a simple power. They are not relatives... How can we be? Someone may be confused... But the person who read on this site the topic “What is a logarithm?” , just smiles sparingly and writes down with a firm hand the absolutely correct answer:

There cannot be such an answer in tasks “B” on the Unified State Examination. There a specific number is required. But in tasks “C” it’s easy.

This lesson provides examples of solving the most common exponential equations. Let's highlight the main points.

Practical tips:

1. First of all, we look at grounds degrees. We are wondering if it is possible to make them identical. Let's try to do this by actively using actions with degrees. Don't forget that numbers without x's can also be converted to powers!

2. We try to bring the exponential equation to the form when on the left and on the right there are the same numbers in any powers. We use actions with degrees And factorization. What can be counted in numbers, we count.

3. If the second tip doesn’t work, try using variable replacement. The result may be an equation that can be easily solved. Most often - square. Or fractional, which also reduces to square.

4. To successfully solve exponential equations, you need to know the powers of some numbers by sight.

As usual, at the end of the lesson you are invited to decide a little.) On your own. From simple to complex.

Solve exponential equations:

More difficult:

2 x+3 - 2 x+2 - 2 x = 48

9 x - 8 3 x = 9

2 x - 2 0.5x+1 - 8 = 0

Find the product of roots:

2 3's + 2 x = 9

Happened?

Well, then a very complex example (though it can be solved in the mind...):

7 0.13x + 13 0.7x+1 + 2 0.5x+1 = -3

What's more interesting? Then here's a bad example for you. Quite tempting for increased difficulty. Let me hint that in this example, what saves you is ingenuity and the most universal rule for solving all mathematical problems.)

2 5x-1 3 3x-1 5 2x-1 = 720 x

A simpler example, for relaxation):

9 2 x - 4 3 x = 0

And for dessert. Find the sum of the roots of the equation:

x 3 x - 9x + 7 3 x - 63 = 0

Yes Yes! This is a mixed type equation! Which we did not consider in this lesson. Why consider them, they need to be solved!) This lesson is quite enough to solve the equation. Well, you need ingenuity... And may seventh grade help you (this is a hint!).

Answers (in disarray, separated by semicolons):

1; 2; 3; 4; there are no solutions; 2; -2; -5; 4; 0.

Is everything successful? Great.

There is a problem? No problem! Special Section 555 solves all these exponential equations with detailed explanations. What, why, and why. And, of course, there is additional valuable information on working with all sorts of exponential equations. Not just these ones.)

One last fun question to consider. In this lesson we worked with exponential equations. Why didn’t I say a word about ODZ here? In equations, this is a very important thing, 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.

1) 0 2) 4 3) -2 4) -4

A2 32x-8 = √3.

1)17/4 2) 17 3) 13/2 4) -17/4

A3

1) 3;1 2) -3;-1 3) 0;2 4) no roots

1) 7;1 2) no roots 3) -7;1 4) -1;-7

A5

1) 0;2; 2) 0;2;3 3) 0 4) -2;-3;0

A6

1) -1 2) 0 3) 2 4) 1

Test No. 2

A1

1) 3 2) -1;3 3) -1;-3 4) 3;-1

A2

1) 14/3 2) -14/3 3) -17 4) 11

A3

1) 2;-1 2) no roots 3) 0 4) -2;1

A4

1) -4 2) 2 3) -2 4) -4;2

A5

1) 3 2) -3;1 3) -1 4) -1;3

3 Evaluation method.

Root theorem: if the function f(x) increases (decreases) on the interval I, the number a is any value taken by f on this interval, then the equation f(x) = a has a single root on the interval I.

When solving equations using the estimation method, this theorem and the monotonicity properties of the function are used.

Examples. Solve equations: 1. 4x = 5 – x.

Solution. Let's rewrite the equation as 4x +x = 5.

1. if x = 1, then 41+1 = 5, 5 = 5 is true, which means 1 is the root of the equation.

Function f(x) = 4x – increases on R, and g(x) = x – increases on R => h(x)= f(x)+g(x) increases on R, as the sum of increasing functions, then x = 1 is the only root of the equation 4x = 5 – x. Answer: 1.

2.

Solution. Let's rewrite the equation in the form .

1. if x = -1, then , 3 = 3 is true, which means x = -1 is the root of the equation.

2. prove that he is the only one.

3. Function f(x) = - decreases on R, and g(x) = - x – decreases on R=> h(x) = f(x)+g(x) – decreases on R, as the sum of decreasing functions . This means, according to the root theorem, x = -1 is the only root of the equation. Answer: -1.

Problem bank No. 2. Solve the equation

a) 4x + 1 =6 – x;

b)

c) 2x – 2 =1 – x;

4. Method of introducing new variables.

The method is described in paragraph 2.1. The introduction of a new variable (substitution) is usually carried out after transformations (simplification) of the terms of the equation. Let's look at examples.

Examples. R Solve the equation: 1. .

Let's rewrite the equation differently: https://pandia.ru/text/80/142/images/image030_0.png" width="128" height="48 src="> i.e..png" width="210" height ="45">

Solution. Let's rewrite the equation differently:

Let's designate https://pandia.ru/text/80/142/images/image035_0.png" width="245" height="57"> - not suitable.

t = 4 => https://pandia.ru/text/80/142/images/image037_0.png" width="268" height="51"> - irrational equation. We note that

The solution to the equation is x = 2.5 ≤ 4, which means 2.5 is the root of the equation. Answer: 2.5.

Solution. Let's rewrite the equation in the form and divide both sides by 56x+6 ≠ 0. We get the equation

2x2-6x-7 = 2x2-6x-8 +1 = 2(x2-3x-4)+1, t..png" width="118" height="56">

The roots of the quadratic equation are t1 = 1 and t2<0, т. е..png" width="200" height="24">.

Solution . Let's rewrite the equation in the form

and note that it is a homogeneous equation of the second degree.

Divide the equation by 42x, we get

Let's replace https://pandia.ru/text/80/142/images/image049_0.png" width="16" height="41 src="> .

Answer: 0; 0.5.

Problem bank No. 3. Solve the equation

b)

G)

Test No. 3 with a choice of answers. Minimum level.

A1

1) -0.2;2 2) log52 3) –log52 4) 2

A2 0.52x – 3 0.5x +2 = 0.

1) 2;1 2) -1;0 3) no roots 4) 0

1) 0 2) 1; -1/3 3) 1 4) 5

A4 52x-5x - 600 = 0.

1) -24;25 2) -24,5; 25,5 3) 25 4) 2

1) no roots 2) 2;4 3) 3 4) -1;2

Test No. 4 with a choice of answers. General level.

A1

1) 2;1 2) ½;0 3)2;0 4) 0

A2 2x – (0.5)2x – (0.5)x + 1 = 0

1) -1;1 2) 0 3) -1;0;1 4) 1

1) 64 2) -14 3) 3 4) 8

1)-1 2) 1 3) -1;1 4) 0

A5

1) 0 2) 1 3) 0;1 4) no roots

5. Factorization method.

1. Solve the equation: 5x+1 - 5x-1 = 24.

Solution..png" width="169" height="69"> , from where

2. 6x + 6x+1 = 2x + 2x+1 + 2x+2.

Solution. Let's put 6x out of brackets on the left side of the equation, and 2x on the right side. We get the equation 6x(1+6) = 2x(1+2+4) ó 6x = 2x.

Since 2x >0 for all x, we can divide both sides of this equation by 2x without fear of losing solutions. We get 3x = 1ó x = 0.

3.

Solution. Let's solve the equation using the factorization method.

Let us select the square of the binomial

4. https://pandia.ru/text/80/142/images/image067_0.png" width="500" height="181">

x = -2 is the root of the equation.

Equation x + 1 = 0 " style="border-collapse:collapse;border:none">

A1 5x-1 +5x -5x+1 =-19.

1) 1 2) 95/4 3) 0 4) -1

A2 3x+1 +3x-1 =270.

1) 2 2) -4 3) 0 4) 4

A3 32x + 32x+1 -108 = 0. x=1.5

1) 0,2 2) 1,5 3) -1,5 4) 3

1) 1 2) -3 3) -1 4) 0

A5 2x -2x-4 = 15. x=4

1) -4 2) 4 3) -4;4 4) 2

A1 (22x-1)(24x+22x+1)=7.

1) ½ 2) 2 3) -1;3 4) 0.2

A2

1) 2.5 2) 3;4 3) log43/2 4) 0

A3 2x-1-3x=3x-1-2x+2.

1) 2 2) -1 3) 3 4) -3

A4

1) 1,5 2) 3 3) 1 4) -4

A5

1) 2 2) -2 3) 5 4) 0

6. Exponential – power equations.

Adjacent to exponential equations are the so-called exponential-power equations, i.e., equations of the form (f(x))g(x) = (f(x))h(x).

If it is known that f(x)>0 and f(x) ≠ 1, then the equation, like the exponential one, is solved by equating the exponents g(x) = f(x).

If the condition does not exclude the possibility of f(x)=0 and f(x)=1, then we have to consider these cases when solving an exponential equation.

1..png" width="182" height="116 src=">

2.

Solution. x2 +2x-8 – makes sense for any x, since it is a polynomial, which means the equation is equivalent to the totality

https://pandia.ru/text/80/142/images/image078_0.png" width="137" height="35">

b)

7. Exponential equations with parameters.

1. For what values ​​of the parameter p does equation 4 (5 – 3)2 +4p2–3p = 0 (1) have a unique solution?

Solution. Let us introduce the replacement 2x = t, t > 0, then equation (1) will take the form t2 – (5p – 3)t + 4p2 – 3p = 0. (2)

Discriminant of equation (2) D = (5p – 3)2 – 4(4p2 – 3p) = 9(p – 1)2.

Equation (1) has a unique solution if equation (2) has one positive root. This is possible in the following cases.

1. If D = 0, that is, p = 1, then equation (2) will take the form t2 – 2t + 1 = 0, hence t = 1, therefore, equation (1) has a unique solution x = 0.

2. If p1, then 9(p – 1)2 > 0, then equation (2) has two different roots t1 = p, t2 = 4p – 3. The conditions of the problem are satisfied by a set of systems

Substituting t1 and t2 into the systems, we have

https://pandia.ru/text/80/142/images/image084_0.png" alt="no35_11" width="375" height="54"> в зависимости от параметра a?!}

Solution. Let then equation (3) will take the form t2 – 6t – a = 0. (4)

Let us find the values ​​of the parameter a for which at least one root of equation (4) satisfies the condition t > 0.

Let us introduce the function f(t) = t2 – 6t – a. The following cases are possible.

https://pandia.ru/text/80/142/images/image087.png" alt="http://1september.ru/ru/mat/2002/35/no35_14.gif" align="left" width="215" height="73 src=">где t0 - абсцисса вершины параболы и D - дискриминант квадратного трехчлена f(t);!}

https://pandia.ru/text/80/142/images/image089.png" alt="http://1september.ru/ru/mat/2002/35/no35_16.gif" align="left" width="60" height="51 src=">!}

Case 2. Equation (4) has a unique positive solution if

D = 0, if a = – 9, then equation (4) will take the form (t – 3)2 = 0, t = 3, x = – 1.

Case 3. Equation (4) has two roots, but one of them does not satisfy the inequality t > 0. This is possible if

https://pandia.ru/text/80/142/images/image092.png" alt="no35_17" width="267" height="63">!}

Thus, for a 0, equation (4) has a single positive root . Then equation (3) has a unique solution

When a< – 9 уравнение (3) корней не имеет.

if a< – 9, то корней нет; если – 9 < a < 0, то
if a = – 9, then x = – 1;

if a  0, then

Let us compare the methods for solving equations (1) and (3). Note that when solving equation (1) was reduced to a quadratic equation, the discriminant of which is a perfect square; Thus, the roots of equation (2) were immediately calculated using the formula for the roots of a quadratic equation, and then conclusions were drawn regarding these roots. Equation (3) has been reduced to a quadratic equation (4), the discriminant of which is not a perfect square, therefore, when solving equation (3), it is advisable to use theorems on the location of the roots of a quadratic trinomial and a graphical model. Note that equation (4) can be solved using Vieta's theorem.

Let's solve more complex equations.

Problem 3: Solve the equation

Solution. ODZ: x1, x2.

Let's introduce a replacement. Let 2x = t, t > 0, then as a result of transformations the equation will take the form t2 + 2t – 13 – a = 0. (*) Let us find the values ​​of a for which at least one root of the equation (*) satisfies the condition t > 0.

https://pandia.ru/text/80/142/images/image098.png" alt="http://1september.ru/ru/mat/2002/35/no35_23.gif" align="left" width="71" height="68 src=">где t0 - абсцисса вершины f(t) = t2 + 2t – 13 – a, D - дискриминант квадратного трехчлена f(t).!}

https://pandia.ru/text/80/142/images/image100.png" alt="http://1september.ru/ru/mat/2002/35/no35_25.gif" align="left" width="360" height="32 src=">!}

https://pandia.ru/text/80/142/images/image102.png" alt="http://1september.ru/ru/mat/2002/35/no35_27.gif" align="left" width="218" height="42 src=">!}

Answer: if a > – 13, a  11, a  5, then if a – 13,

a = 11, a = 5, then there are no roots.

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