Solving exponential equations online with detailed solutions. How is a system of equations solved? Methods for solving systems of 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.

Those examples with indicators that caused you difficulties can be added to “Favorites”. This way you can quickly find them and discuss the solution with your teacher.

To successfully pass the Unified State Exam, study on the Shkolkovo portal every day!

I. ax 2 =0 – incomplete quadratic equation (b=0, c=0 ). Solution: x=0. Answer: 0.

Solve equations.

2x·(x+3)=6x-x 2 .

Solution. Let's open the brackets by multiplying 2x for each term in brackets:

2x 2 +6x=6x-x 2 ; We move the terms from the right side to the left:

2x 2 +6x-6x+x 2 =0; Here are similar terms:

3x 2 =0, hence x=0.

Answer: 0.

II. ax 2 +bx=0 –incomplete quadratic equation (c=0 ). Solution: x (ax+b)=0 → x 1 =0 or ax+b=0 → x 2 =-b/a. Answer: 0; -b/a.

5x 2 -26x=0.

Solution. Let's take out the common factor X outside of brackets:

x(5x-26)=0; each factor can be equal to zero:

x=0 or 5x-26=0→ 5x=26, divide both sides of the equality by 5 and we get: x=5.2.

Answer: 0; 5,2.

Example 3. 64x+4x 2 =0.

Solution. Let's take out the common factor 4x outside of brackets:

4x(16+x)=0. We have three factors, 4≠0, therefore, or x=0 or 16+x=0. From the last equality we get x=-16.

Answer: -16; 0.

Example 4.(x-3) 2 +5x=9.

Solution. Applying the formula for the square of the difference of two expressions, we will open the brackets:

x 2 -6x+9+5x=9; transform to the form: x 2 -6x+9+5x-9=0; Let us present similar terms:

x 2 -x=0; we'll take it out X outside the brackets, we get: x (x-1)=0. From here or x=0 or x-1=0→ x=1.

Answer: 0; 1.

III. ax 2 +c=0 –incomplete quadratic equation (b=0 ); Solution: ax 2 =-c → x 2 =-c/a.

If (-c/a)<0 , then there are no real roots. If (-с/а)>0

Example 5. x 2 -49=0.

Solution.

x 2 =49, from here x=±7. Answer:-7; 7.

Example 6. 9x 2 -4=0.

Solution.

Often you need to find the sum of squares (x 1 2 +x 2 2) or the sum of cubes (x 1 3 +x 2 3) of the roots of a quadratic equation, less often - the sum of the reciprocal values ​​of the squares of the roots or the sum of arithmetic square roots of the roots of a quadratic equation:

Vieta's theorem can help with this:

x 2 +px+q=0

x 1 + x 2 = -p; x 1 ∙x 2 =q.

Let's express through p And q:

1) sum of squares of the roots of the equation x 2 +px+q=0;

2) sum of cubes of the roots of the equation x 2 +px+q=0.

Solution.

1) Expression x 1 2 +x 2 2 obtained by squaring both sides of the equation x 1 + x 2 = -p;

(x 1 +x 2) 2 =(-p) 2 ; open the brackets: x 1 2 +2x 1 x 2 + x 2 2 =p 2 ; we express the required amount: x 1 2 +x 2 2 =p 2 -2x 1 x 2 =p 2 -2q. We got a useful equality: x 1 2 +x 2 2 =p 2 -2q.

2) Expression x 1 3 +x 2 3 Let us represent the sum of cubes using the formula:

(x 1 3 +x 2 3)=(x 1 +x 2)(x 1 2 -x 1 x 2 +x 2 2)=-p·(p 2 -2q-q)=-p·(p 2 -3q).

Another useful equation: x 1 3 +x 2 3 = -p·(p 2 -3q).

Examples.

3) x 2 -3x-4=0. Without solving the equation, calculate the value of the expression x 1 2 +x 2 2.

Solution.

x 1 +x 2 =-p=3, and the work x 1 ∙x 2 =q=in example 1) equality:

x 1 2 +x 2 2 =p 2 -2q. We have -p=x 1 +x 2 = 3 → p 2 =3 2 =9; q= x 1 x 2 = -4. Then x 1 2 +x 2 2 =9-2·(-4)=9+8=17.

Answer: x 1 2 +x 2 2 =17.

4) x 2 -2x-4=0. Calculate: x 1 3 +x 2 3 .

Solution.

By Vieta's theorem, the sum of the roots of this reduced quadratic equation is x 1 +x 2 =-p=2, and the work x 1 ∙x 2 =q=-4. Let's apply what we have received ( in example 2) equality: x 1 3 +x 2 3 =-p·(p 2 -3q)= 2·(2 2 -3·(-4))=2·(4+12)=2·16=32.

Answer: x 1 3 +x 2 3 =32.

Question: what if we are given an unreduced quadratic equation? Answer: it can always be “reduced” by dividing term by term by the first coefficient.

5) 2x 2 -5x-7=0. Without deciding, calculate: x 1 2 +x 2 2.

Solution. We are given a complete quadratic equation. Divide both sides of the equality by 2 (the first coefficient) and obtain the following quadratic equation: x 2 -2.5x-3.5=0.

According to Vieta's theorem, the sum of the roots is equal to 2,5 ; the product of the roots is equal -3,5 .

We solve it in the same way as the example 3) using the equality: x 1 2 +x 2 2 =p 2 -2q.

x 1 2 +x 2 2 =p 2 -2q= 2,5 2 -2∙(-3,5)=6,25+7=13,25.

Answer: x 1 2 + x 2 2 = 13,25.

6) x 2 -5x-2=0. Find:

Let us transform this equality and, using Vieta’s theorem, replace the sum of roots through -p, and the product of the roots through q, we get another useful formula. When deriving the formula, we used equality 1): x 1 2 +x 2 2 =p 2 -2q.

In our example x 1 +x 2 =-p=5; x 1 ∙x 2 =q=-2. We substitute these values ​​into the resulting formula:

7) x 2 -13x+36=0. Find:

Let's transform this sum and get a formula that can be used to find the sum of arithmetic square roots from the roots of a quadratic equation.

We have x 1 +x 2 =-p=13; x 1 ∙x 2 =q=36. We substitute these values ​​into the resulting formula:

Advice : always check the possibility of finding the roots of a quadratic equation using a suitable method, because 4 reviewed useful formulas allow you to quickly complete a task, especially in cases where the discriminant is an “inconvenient” number. In all simple cases, find the roots and operate on them. For example, in the last example we select the roots using Vieta’s theorem: the sum of the roots should be equal to 13 , and the product of the roots 36 . What are these numbers? Certainly, 4 and 9. Now calculate the sum of the square roots of these numbers: 2+3=5. That's it!

I. Vieta's theorem for the reduced quadratic equation.

Sum of roots of the reduced quadratic equation x 2 +px+q=0 is equal to the second coefficient taken with the opposite sign, and the product of the roots is equal to the free term:

x 1 + x 2 = -p; x 1 ∙x 2 =q.

Find the roots of the given quadratic equation using Vieta's theorem.

Example 1) x 2 -x-30=0. This is the reduced quadratic equation ( x 2 +px+q=0), second coefficient p=-1, and the free member q=-30. First, let's make sure that this equation has roots, and that the roots (if any) will be expressed in integers. To do this, it is enough that the discriminant be a perfect square of an integer.

Finding the discriminant D=b 2 — 4ac=(-1) 2 -4∙1∙(-30)=1+120=121= 11 2 .

Now, according to Vieta’s theorem, the sum of the roots must be equal to the second coefficient taken with the opposite sign, i.e. ( -p), and the product is equal to the free term, i.e. ( q). Then:

x 1 +x 2 =1; x 1 ∙x 2 =-30. We need to choose two numbers such that their product is equal to -30 , and the amount is unit. These are numbers -5 And 6 . Answer: -5; 6.

Example 2) x 2 +6x+8=0. We have the reduced quadratic equation with the second coefficient p=6 and free member q=8. Let's make sure that there are integer roots. Let's find the discriminant D 1 D 1=3 2 -1∙8=9-8=1=1 2 . The discriminant D 1 is the perfect square of the number 1 , which means that the roots of this equation are integers. Let us select the roots using Vieta’s theorem: the sum of the roots is equal to –р=-6, and the product of the roots is equal to q=8. These are numbers -4 And -2 .

In fact: -4-2=-6=-р; -4∙(-2)=8=q. Answer: -4; -2.

Example 3) x 2 +2x-4=0. In this reduced quadratic equation, the second coefficient p=2, and the free member q=-4. Let's find the discriminant D 1, since the second coefficient is an even number. D 1=1 2 -1∙(-4)=1+4=5. The discriminant is not a perfect square of the number, so we do conclusion: The roots of this equation are not integers and cannot be found using Vieta’s theorem. This means that we solve this equation, as usual, using the formulas (in this case, using the formulas). We get:

Example 4). Write a quadratic equation using its roots if x 1 =-7, x 2 =4.

Solution. The required equation will be written in the form: x 2 +px+q=0, and, based on Vieta’s theorem –p=x 1 +x 2=-7+4=-3 → p=3; q=x 1 ∙x 2=-7∙4=-28 . Then the equation will take the form: x 2 +3x-28=0.

Example 5). Write a quadratic equation using its roots if:

II. Vieta's theorem for a complete quadratic equation ax 2 +bx+c=0.

The sum of the roots is minus b, divided by A, the product of the roots is equal to With, divided by A:

x 1 + x 2 = -b/a; x 1 ∙x 2 =c/a.

Example 6). Find the sum of the roots of a quadratic equation 2x 2 -7x-11=0.

Solution.

We make sure that this equation will have roots. To do this, it is enough to create an expression for the discriminant, and, without calculating it, just make sure that the discriminant is greater than zero. D=7 2 -4∙2∙(-11)>0 . Now let's use theorem Vieta for complete quadratic equations.

x 1 +x 2 =-b:a=- (-7):2=3,5.

Example 7). Find the product of the roots of a quadratic equation 3x 2 +8x-21=0.

Solution.

Let's find the discriminant D 1, since the second coefficient ( 8 ) is an even number. D 1=4 2 -3∙(-21)=16+63=79>0 . The quadratic equation has 2 root, according to Vieta’s theorem, the product of roots x 1 ∙x 2 =c:a=-21:3=-7.

I. ax 2 +bx+c=0– general quadratic equation

Discriminant D=b 2 - 4ac.

If D>0, then we have two real roots:

If D=0, then we have a single root (or two equal roots) x=-b/(2a).

If D<0, то действительных корней нет.

Example 1) 2x 2 +5x-3=0.

Solution. a=2; b=5; c=-3.

D=b 2 - 4ac=5 2 -4∙2∙(-3)=25+24=49=7 2 >0; 2 real roots.

4x 2 +21x+5=0.

Solution. a=4; b=21; c=5.

D=b 2 - 4ac=21 2 - 4∙4∙5=441-80=361=19 2 >0; 2 real roots.

II. ax 2 +bx+c=0 – quadratic equation of particular form with even second

coefficient b

Example 3) 3x 2 -10x+3=0.

Solution. a=3; b=-10 (even number); c=3.

Example 4) 5x 2 -14x-3=0.

Solution. a=5; b= -14 (even number); c=-3.

Example 5) 71x 2 +144x+4=0.

Solution. a=71; b=144 (even number); c=4.

Example 6) 9x 2 -30x+25=0.

Solution. a=9; b=-30 (even number); c=25.

III. ax 2 +bx+c=0 – quadratic equation private type provided: a-b+c=0.

The first root is always equal to minus one, and the second root is always equal to minus With, divided by A:

x 1 =-1, x 2 =-c/a.

Example 7) 2x 2 +9x+7=0.

Solution. a=2; b=9; c=7. Let's check the equality: a-b+c=0. We get: 2-9+7=0 .

Then x 1 =-1, x 2 =-c/a=-7/2=-3.5. Answer: -1; -3,5.

IV. ax 2 +bx+c=0 – quadratic equation of a particular form subject to : a+b+c=0.

The first root is always equal to one, and the second root is equal to With, divided by A:

x 1 =1, x 2 =c/a.

Example 8) 2x 2 -9x+7=0.

Solution. a=2; b=-9; c=7. Let's check the equality: a+b+c=0. We get: 2-9+7=0 .

Then x 1 =1, x 2 =c/a=7/2=3.5. Answer: 1; 3,5.

Page 1 of 1 1

In the 7th grade mathematics course, we encounter for the first time equations with two variables, but they are studied only in the context of systems of equations with two unknowns. That is why a whole series of problems in which certain conditions are introduced on the coefficients of the equation that limit them fall out of sight. In addition, methods for solving problems like “Solve an equation in natural or integer numbers” are also ignored, although problems of this kind are found more and more often in the Unified State Examination materials and in entrance exams.

Which equation will be called an equation with two variables?

So, for example, the equations 5x + 2y = 10, x 2 + y 2 = 20, or xy = 12 are equations in two variables.

Consider the equation 2x – y = 1. It becomes true when x = 2 and y = 3, so this pair of variable values ​​is a solution to the equation in question.

Thus, the solution to any equation with two variables is a set of ordered pairs (x; y), values ​​of the variables that turn this equation into a true numerical equality.

An equation with two unknowns can:

A) have one solution. For example, the equation x 2 + 5y 2 = 0 has a unique solution (0; 0);

b) have multiple solutions. For example, (5 -|x|) 2 + (|y| – 2) 2 = 0 has 4 solutions: (5; 2), (-5; 2), (5; -2), (-5; - 2);

V) have no solutions. For example, the equation x 2 + y 2 + 1 = 0 has no solutions;

G) have infinitely many solutions. For example, x + y = 3. The solutions to this equation will be numbers whose sum is equal to 3. The set of solutions to this equation can be written in the form (k; 3 – k), where k is any real number.

The main methods for solving equations with two variables are methods based on factoring expressions, isolating a complete square, using the properties of a quadratic equation, limited expressions, and estimation methods. The equation is usually transformed into a form from which a system for finding the unknowns can be obtained.

Factorization

Example 1.

Solve the equation: xy – 2 = 2x – y.

Solution.

We group the terms for the purpose of factorization:

(xy + y) – (2x + 2) = 0. From each bracket we take out a common factor:

y(x + 1) – 2(x + 1) = 0;

(x + 1)(y – 2) = 0. We have:

y = 2, x – any real number or x = -1, y – any real number.

Thus, the answer is all pairs of the form (x; 2), x € R and (-1; y), y € R.

Equality of non-negative numbers to zero

Example 2.

Solve the equation: 9x 2 + 4y 2 + 13 = 12(x + y).

Solution.

Grouping:

(9x 2 – 12x + 4) + (4y 2 – 12y + 9) = 0. Now each bracket can be folded using the squared difference formula.

(3x – 2) 2 + (2y – 3) 2 = 0.

The sum of two non-negative expressions is zero only if 3x – 2 = 0 and 2y – 3 = 0.

This means x = 2/3 and y = 3/2.

Answer: (2/3; 3/2).

Estimation method

Example 3.

Solve the equation: (x 2 + 2x + 2)(y 2 – 4y + 6) = 2.

Solution.

In each bracket we select a complete square:

((x + 1) 2 + 1)((y – 2) 2 + 2) = 2. Let’s estimate the meaning of the expressions in parentheses.

(x + 1) 2 + 1 ≥ 1 and (y – 2) 2 + 2 ≥ 2, then the left side of the equation is always at least 2. Equality is possible if:

(x + 1) 2 + 1 = 1 and (y – 2) 2 + 2 = 2, which means x = -1, y = 2.

Answer: (-1; 2).

Let's get acquainted with another method for solving equations with two variables of the second degree. This method consists of treating the equation as square with respect to some variable.

Example 4.

Solve the equation: x 2 – 6x + y – 4√y + 13 = 0.

Solution.

Let's solve the equation as a quadratic equation for x. Let's find the discriminant:

D = 36 – 4(y – 4√y + 13) = -4y + 16√y – 16 = -4(√y – 2) 2 . The equation will have a solution only when D = 0, that is, if y = 4. We substitute the value of y into the original equation and find that x = 3.

Answer: (3; 4).

Often in equations with two unknowns they indicate restrictions on variables.

Example 5.

Solve the equation in whole numbers: x 2 + 5y 2 = 20x + 2.

Solution.

Let's rewrite the equation in the form x 2 = -5y 2 + 20x + 2. The right side of the resulting equation when divided by 5 gives a remainder of 2. Therefore, x 2 is not divisible by 5. But the square of a number not divisible by 5 gives a remainder of 1 or 4. Thus, equality is impossible and there are no solutions.

Answer: no roots.

Example 6.

Solve the equation: (x 2 – 4|x| + 5)(y 2 + 6y + 12) = 3.

Solution.

Let's highlight the complete squares in each bracket:

((|x| – 2) 2 + 1)((y + 3) 2 + 3) = 3. The left side of the equation is always greater than or equal to 3. Equality is possible provided |x| – 2 = 0 and y + 3 = 0. Thus, x = ± 2, y = -3.

Answer: (2; -3) and (-2; -3).

Example 7.

For every pair of negative integers (x;y) satisfying the equation
x 2 – 2xy + 2y 2 + 4y = 33, calculate the sum (x + y). Please indicate the smallest amount in your answer.

Solution.

Let's select complete squares:

(x 2 – 2xy + y 2) + (y 2 + 4y + 4) = 37;

(x – y) 2 + (y + 2) 2 = 37. Since x and y are integers, their squares are also integers. We get the sum of the squares of two integers equal to 37 if we add 1 + 36. Therefore:

(x – y) 2 = 36 and (y + 2) 2 = 1

(x – y) 2 = 1 and (y + 2) 2 = 36.

Solving these systems and taking into account that x and y are negative, we find solutions: (-7; -1), (-9; -3), (-7; -8), (-9; -8).

Answer: -17.

Don't despair if you have difficulty solving equations with two unknowns. With a little practice, you can handle any equation.

Still have questions? Don't know how to solve equations in two variables?
To get help from a tutor, register.
The first lesson is free!

website, when copying material in full or in part, a link to the source is required.

The online equation solving service will help you solve any equation. Using our website, you will receive not just the answer to the equation, but also see a detailed solution, that is, a step-by-step display of the process of obtaining the result. Our service will be useful to high school students and their parents. Students will be able to prepare for tests and exams, test their knowledge, and parents will be able to monitor the solution of mathematical equations by their children. The ability to solve equations is a mandatory requirement for schoolchildren. The service will help you educate yourself and improve your knowledge in the field of mathematical equations. With its help, you can solve any equation: quadratic, cubic, irrational, trigonometric, etc. The benefits of the online service are priceless, because in addition to the correct answer, you receive a detailed solution to each equation. Benefits of solving equations online. You can solve any equation online on our website absolutely free. The service is completely automatic, you don’t have to install anything on your computer, you just need to enter the data and the program will give you a solution. Any errors in calculations or typos are excluded. With us, solving any equation online is very easy, so be sure to use our site to solve any kind of equations. You only need to enter the data and the calculation will be completed in a matter of seconds. The program works independently, without human intervention, and you receive an accurate and detailed answer. Solution of the equation in general form. In such an equation, the variable coefficients and the desired roots are interconnected. The highest power of a variable determines the order of such an equation. Based on this, various methods and theorems are used for equations to find solutions. Solving equations of this type means finding the required roots in general form. Our service allows you to solve even the most complex algebraic equation online. You can obtain both a general solution to the equation and a particular one for the numerical values ​​of the coefficients you specify. To solve an algebraic equation on the website, it is enough to correctly fill out only two fields: the left and right sides of the given equation. Algebraic equations with variable coefficients have an infinite number of solutions, and by setting certain conditions, partial ones are selected from the set of solutions. Quadratic equation. The quadratic equation has the form ax^2+bx+c=0 for a>0. Solving quadratic equations involves finding the values ​​of x at which the equality ax^2+bx+c=0 holds. To do this, find the discriminant value using the formula D=b^2-4ac. If the discriminant is less than zero, then the equation has no real roots (the roots are from the field of complex numbers), if it is equal to zero, then the equation has one real root, and if the discriminant is greater than zero, then the equation has two real roots, which are found by the formula: D = -b+-sqrt/2a. To solve a quadratic equation online, you just need to enter the coefficients of the equation (integers, fractions or decimals). If there are subtraction signs in an equation, you must put a minus sign in front of the corresponding terms of the equation. You can solve a quadratic equation online depending on the parameter, that is, the variables in the coefficients of the equation. Our online service for finding general solutions copes well with this task. Linear equations. To solve linear equations (or systems of equations), four main methods are used in practice. We will describe each method in detail. Substitution method. Solving equations using the substitution method requires expressing one variable in terms of the others. After this, the expression is substituted into other equations of the system. Hence the name of the solution method, that is, instead of a variable, its expression is substituted through the remaining variables. In practice, the method requires complex calculations, although it is easy to understand, so solving such an equation online will help save time and make calculations easier. You just need to indicate the number of unknowns in the equation and fill in the data from the linear equations, then the service will make the calculation. Gauss method. The method is based on the simplest transformations of the system in order to arrive at an equivalent triangular system. From it, the unknowns are determined one by one. In practice, you need to solve such an equation online with a detailed description, thanks to which you will have a good understanding of the Gaussian method for solving systems of linear equations. Write down the system of linear equations in the correct format and take into account the number of unknowns in order to accurately solve the system. Cramer's method. This method solves systems of equations in cases where the system has a unique solution. The main mathematical action here is the calculation of matrix determinants. Solving equations using the Cramer method is carried out online, you receive the result instantly with a complete and detailed description. It is enough just to fill the system with coefficients and select the number of unknown variables. Matrix method. This method consists of collecting the coefficients of the unknowns in matrix A, the unknowns in column X, and the free terms in column B. Thus, the system of linear equations is reduced to a matrix equation of the form AxX=B. This equation has a unique solution only if the determinant of matrix A is different from zero, otherwise the system has no solutions, or an infinite number of solutions. Solving equations using the matrix method involves finding the inverse matrix A.

Instructions. To solve online, you need to select the type of equation and set the dimension of the corresponding matrices.

Example No. 1. Exercise. Find the solution to the matrix equation
Solution. Let's denote:
Then the matrix equation will be written in the form: A·X·B = C.
The determinant of matrix A is equal to detA=-1
Since A is a non-singular matrix, there is an inverse matrix A -1 . Multiply both sides of the equation on the left by A -1: Multiply both sides of this equation on the left by A -1 and on the right by B -1: A -1 ·A·X·B·B -1 = A -1 ·C·B -1 . Since A A -1 = B B -1 = E and E X = X E = X, then X = A -1 C B -1

Inverse matrix A -1:
Let's find the inverse matrix B -1.
Transposed matrix B T:
Inverse matrix B -1:
We look for matrix X using the formula: X = A -1 ·C·B -1

Answer:

Example No. 2. Exercise. Solve matrix equation
Solution. Let's denote:
Then the matrix equation will be written in the form: A·X = B.
The determinant of matrix A is detA=0
Since A is a singular matrix (the determinant is 0), therefore the equation has no solution.

Example No. 3. Exercise. Find the solution to the matrix equation
Solution. Let's denote:
Then the matrix equation will be written in the form: X A = B.
The determinant of matrix A is detA=-60
Since A is a non-singular matrix, there is an inverse matrix A -1 . Let's multiply both sides of the equation on the right by A -1: X A A -1 = B A -1, from where we find that X = B A -1
Let's find the inverse matrix A -1 .
Transposed matrix A T:
Inverse matrix A -1:
We look for matrix X using the formula: X = B A -1


Answer: >