How is the system of equations solved? Methods for solving systems of equations. Online calculator. Solving exponential equations How to solve equation 2a 4
Recall the basic properties of a degree. Let a > 0, b > 0, n, m be any real numbers. Then
1) a n a m = a n+m
2) \(\frac(a^n)(a^m) = a^(n-m) \)
3) (a n) m = a nm
4) (ab) n = a n b n
5) \(\left(\frac(a)(b) \right)^n = \frac(a^n)(b^n) \)
7) a n > 1 if a > 1, n > 0
8) a n 1, n
9) a n > a m , if 0
In practice, functions of the form y = a x are often used, where a is a given positive number, x is a variable. Such functions are called demonstrative. This name is explained by the fact that the argument of the exponential function is the exponent, and the base of the degree is a given number.
Definition. An exponential function is a function of the form y = a x , where a is a given number, a > 0, \(a \neq 1\)
An exponential function has the following properties
1) The domain of the exponential function is the set of all real numbers.
This property follows from the fact that the degree a x where a > 0 is defined for all real numbers x.
2) The set of values of the exponential function is the set of all positive numbers.
To verify this, we need to show that the equation a x = b, where a > 0, \(a \neq 1\), has no roots if \(b \leq 0\), and has a root for any b > 0 .
3) The exponential function y \u003d a x is increasing on the set of all real numbers if a > 1, and decreasing if 0 This follows from the properties of the degree (8) and (9)
We construct graphs of exponential functions y \u003d a x for a > 0 and for 0 Using the considered properties, we note that the graph of the function y \u003d a x for a > 0 passes through the point (0; 1) and is located above the Ox axis.
If x is 0.
If x > 0 and |x| increases, the graph quickly rises.
Graph of the function y \u003d a x at 0 If x\u003e 0 and increases, then the graph quickly approaches the Ox axis (without crossing it). Thus, the x-axis is the horizontal asymptote of the graph.
If x 
exponential equations
Consider several examples of exponential equations, i.e. equations in which the unknown is contained in the exponent. Solving exponential equations often comes down to solving the equation a x = a b where a > 0, \(a\neq 1\), x is the unknown. This equation is solved using the power property: powers with the same base a > 0, \(a \neq 1\) are equal if and only if their exponents are equal.
Solve equation 2 3x 3 x = 576
Since 2 3x \u003d (2 3) x \u003d 8 x, 576 \u003d 24 2, the equation can be written in the form 8 x 3 x \u003d 24 2, or in the form 24 x \u003d 24 2, from where x \u003d 2.
Answer x = 2
Solve the equation 3 x + 1 - 2 3 x - 2 = 25
Bracketing the common factor 3 x - 2 on the left side, we get 3 x - 2 (3 3 - 2) \u003d 25, 3 x - 2 25 \u003d 25,
whence 3 x - 2 = 1, x - 2 = 0, x = 2
Answer x = 2
Solve the equation 3 x = 7 x
Since \(7^x \neq 0 \) , the equation can be written as \(\frac(3^x)(7^x) = 1 \), whence \(\left(\frac(3)( 7) \right) ^x = 1 \), x = 0
Answer x = 0
Solve the equation 9 x - 4 3 x - 45 = 0
By replacing 3 x \u003d t, this equation is reduced to a quadratic equation t 2 - 4t - 45 \u003d 0. Solving this equation, we find its roots: t 1 \u003d 9, t 2 \u003d -5, from which 3 x \u003d 9, 3 x \u003d -5 .
The equation 3 x = 9 has a root x = 2, and the equation 3 x = -5 has no roots, since the exponential function cannot take negative values.
Answer x = 2
Solve equation 3 2 x + 1 + 2 5 x - 2 = 5 x + 2 x - 2
We write the equation in the form
3 2 x + 1 - 2 x - 2 = 5 x - 2 5 x - 2, whence
2 x - 2 (3 2 3 - 1) = 5 x - 2 (5 2 - 2)
2 x - 2 23 = 5 x - 2 23
\(\left(\frac(2)(5) \right) ^(x-2) = 1 \)
x - 2 = 0
Answer x = 2
Solve equation 3 |x - 1| = 3 |x + 3|
Since 3 > 0, \(3 \neq 1\), the original equation is equivalent to the equation |x-1| = |x+3|
Squaring this equation, we obtain its corollary (x - 1) 2 = (x + 3) 2, whence
x 2 - 2x + 1 = x 2 + 6x + 9, 8x = -8, x = -1
The check shows that x = -1 is the root of the original equation.
Answer x = -1
Service for solving equations online will help you solve any equation. Using our site, you will not only get 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 for high school students and their parents. Students will be able to prepare for tests, exams, test their knowledge, and parents will be able to control the solution of mathematical equations by their children. The ability to solve equations is a mandatory requirement for students. The service will help you self-learn and improve your knowledge in the field of mathematical equations. With it, you can solve any equation: quadratic, cubic, irrational, trigonometric, etc. The benefit of the online service is invaluable, because in addition to the correct answer, you get a detailed solution to each equation. Benefits of solving equations online. You can solve any equation online on our website absolutely free of charge. The service is fully automatic, you do not have to install anything on your computer, you just need to enter the data and the program will issue a solution. Any calculation errors or typographical errors are excluded. It is very easy to solve any equation online with us, 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 seconds. The program works independently, without human intervention, and you get 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 desired roots in a general form. Our service allows you to solve even the most complex algebraic equation online. You can get both the general solution of the equation and the private one for the numerical values of the coefficients you specified. To solve an algebraic equation on the site, it is enough to correctly fill in only two fields: the left and right parts of the given equation. Algebraic equations with variable coefficients have an infinite number of solutions, and by setting certain conditions, particular ones are selected from the set of solutions. Quadratic equation. The quadratic equation has the form ax^2+bx+c=0 for a>0. The solution of equations of a square form implies finding the values of x, at which the equality ax ^ 2 + bx + c \u003d 0 is satisfied. To do this, the value of the discriminant is found by 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 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 \u003d -b + -sqrt / 2a. To solve a quadratic equation online, you just need to enter the coefficients of such an equation (whole numbers, fractions or decimal values). If there are subtraction signs in the equation, you must put a minus in front of the corresponding terms of the equation. You can also solve a quadratic equation online depending on the parameter, that is, the variables in the coefficients of the equation. Our online service for finding common solutions perfectly copes with this task. Linear equations. To solve linear equations (or systems of equations), four main methods are used in practice. Let's describe each method in detail. Substitution method. Solving equations using the substitution method requires expressing one variable in terms of the others. After that, 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 through the rest of the variables is substituted. In practice, the method requires complex calculations, although it is easy to understand, so solving such an equation online will save time and make calculations easier. You just need to specify the number of unknowns in the equation and fill in the data from 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. The unknowns are determined one by one from it. In practice, you need to solve such an equation online with a detailed description, thanks to which you will learn the Gauss method well 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 correctly solve the system. Cramer's method. This method solves systems of equations in cases where the system has a unique solution. The main mathematical operation here is the calculation of matrix determinants. The solution of equations by the Cramer method is carried out online, you get the result instantly with a complete and detailed description. It is enough just to fill the system with coefficients and choose the number of unknown variables. matrix method. This method consists in 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 the matrix A is non-zero, otherwise the system has no solutions, or an infinite number of solutions. The solution of equations by the matrix method is to find the inverse matrix A.
2x 4 + 5x 3 - 11x 2 - 20x + 12 = 0
First you need to use the selection method to find one root. It is usually the divisor of the free term. In this case, the divisors of the number 12 are ±1, ±2, ±3, ±4, ±6, ±12. Let's start substituting them in turn:
1: 2 + 5 - 11 - 20 + 12 = -12 ⇒ number 1
-1: 2 - 5 - 11 + 20 + 12 = 18 ⇒ number -1 is not a root of a polynomial
2: 2 ∙ 16 + 5 ∙ 8 - 11 ∙ 4 - 20 ∙ 2 + 12 = 0 ⇒ number 2 is the root of the polynomial
We have found 1 of the roots of the polynomial. The root of the polynomial is 2, which means that the original polynomial must be divisible by x - 2. In order to perform the division of polynomials, we use Horner's scheme:
| 2 | 5 | -11 | -20 | 12 | |
| 2 |
The top line contains the coefficients of the original polynomial. In the first cell of the second row, we put the root we found 2. The second line contains the coefficients of the polynomial, which will be obtained as a result of division. They count like this:
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In the second cell of the second row, write the number 2, simply by moving it from the corresponding cell of the first row. | ||||||||||||
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2 ∙ 2 + 5 = 9 | ||||||||||||
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2 ∙ 9 - 11 = 7 | ||||||||||||
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2 ∙ 7 - 20 = -6 | ||||||||||||
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2 ∙ (-6) + 12 = 0 |
The last number is the remainder of the division. If it is equal to 0, then we counted everything correctly.
2x 4 + 5x 3 - 11x 2 - 20x + 12 = (x - 2)(2x 3 + 9x 2 + 7x - 6)
But this is not the end. You can try to expand the polynomial in the same way 2x 3 + 9x 2 + 7x - 6.
Again we are looking for the root among the divisors of the free term. Number Divisors -6 are ±1, ±2, ±3, ±6.
1: 2 + 9 + 7 - 6 = 12 ⇒ number 1 is not a root of a polynomial
-1: -2 + 9 - 7 - 6 = -6 ⇒ number -1 is not a root of a polynomial
2: 2 ∙ 8 + 9 ∙ 4 + 7 ∙ 2 - 6 = 60 ⇒ number 2 is not a root of a polynomial
-2: 2 ∙ (-8) + 9 ∙ 4 + 7 ∙ (-2) - 6 = 0 ⇒ number -2 is the root of the polynomial
Let's write the found root into our Horner scheme and start filling in the empty cells:
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In the second cell of the third row, write the number 2, simply by moving it from the corresponding cell of the second row. | ||||||||||||||||||
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-2 ∙ 2 + 9 = 5 | ||||||||||||||||||
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-2 ∙ 5 + 7 = -3 | ||||||||||||||||||
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-2 ∙ (-3) - 6 = 0 |
Thus, we factored the original polynomial:
2x 4 + 5x 3 - 11x 2 - 20x + 12 = (x - 2)(x + 2)(2x 2 + 5x - 3)
Polynomial 2x 2 + 5x - 3 can also be factored. To do this, you can solve the quadratic equation through the discriminant, or you can look for the root among the divisors of the number -3. One way or another, we will come to the conclusion that the root of this polynomial is the number -3
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In the second cell of the fourth row, write the number 2, simply by transferring it from the corresponding cell of the third row. | ||||||||||||||||||||||||
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-3 ∙ 2 + 5 = -1 | ||||||||||||||||||||||||
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-3 ∙ (-1) - 3 = 0 |
Thus, we decomposed the original polynomial into linear factors:
2x 4 + 5x 3 - 11x 2 - 20x + 12 = (x - 2)(x + 2)(x + 3)(2x - 1)
And the roots of the equation are.
An equation with one unknown, which, after opening the brackets and reducing like terms, takes the form
ax + b = 0, where a and b are arbitrary numbers, is called linear equation with one unknown. Today we will figure out how to solve these linear equations.
For example, all equations:
2x + 3 \u003d 7 - 0.5x; 0.3x = 0; x / 2 + 3 \u003d 1/2 (x - 2) - linear.
The value of the unknown that turns the equation into a true equality is called decision or the root of the equation .
For example, if in the equation 3x + 7 \u003d 13 we substitute the number 2 instead of the unknown x, then we get the correct equality 3 2 + 7 \u003d 13. Hence, the value x \u003d 2 is the solution or the root of the equation.
And the value x \u003d 3 does not turn the equation 3x + 7 \u003d 13 into a true equality, since 3 2 + 7 ≠ 13. Therefore, the value x \u003d 3 is not a solution or a root of the equation.
The solution of any linear equations is reduced to the solution of equations of the form
ax + b = 0.
We transfer the free term from the left side of the equation to the right, while changing the sign in front of b to the opposite, we get
If a ≠ 0, then x = – b/a .
Example 1 Solve the equation 3x + 2 =11.
We transfer 2 from the left side of the equation to the right, while changing the sign in front of 2 to the opposite, we get
3x \u003d 11 - 2.
Let's do the subtraction, then
3x = 9.
To find x, you need to divide the product by a known factor, that is,
x = 9:3.
So the value x = 3 is the solution or the root of the equation.
Answer: x = 3.
If a = 0 and b = 0, then we get the equation 0x \u003d 0. This equation has infinitely many solutions, since when multiplying any number by 0, we get 0, but b is also 0. The solution to this equation is any number.
Example 2 Solve the equation 5(x - 3) + 2 = 3 (x - 4) + 2x - 1.
Let's expand the brackets:
5x - 15 + 2 \u003d 3x - 12 + 2x - 1.
5x - 3x - 2x \u003d - 12 - 1 + 15 - 2.
Here are similar members:
0x = 0.
Answer: x is any number.
If a = 0 and b ≠ 0, then we get the equation 0x = - b. This equation has no solutions, since when multiplying any number by 0, we get 0, but b ≠ 0.
Example 3 Solve the equation x + 8 = x + 5.
Let us group the terms containing unknowns on the left side, and the free terms on the right side:
x - x \u003d 5 - 8.
Here are similar members:
0x = - 3.
Answer: no solutions.
On figure 1 the scheme for solving the linear equation is shown
Let us compose a general scheme for solving equations with one variable. Consider the solution of example 4.
Example 4 Let's solve the equation
1) Multiply all terms of the equation by the least common multiple of the denominators, equal to 12.
2) After reduction we get
4 (x - 4) + 3 2 (x + 1) - 12 = 6 5 (x - 3) + 24x - 2 (11x + 43)
3) To separate members containing unknown and free members, open the brackets:
4x - 16 + 6x + 6 - 12 \u003d 30x - 90 + 24x - 22x - 86.
4) We group in one part the terms containing unknowns, and in the other - free terms:
4x + 6x - 30x - 24x + 22x \u003d - 90 - 86 + 16 - 6 + 12.
5) Here are similar members:
- 22x = - 154.
6) Divide by - 22 , We get
x = 7.
As you can see, the root of the equation is seven.
In general, such equations can be solved as follows:
a) bring the equation to an integer form;
b) open brackets;
c) group the terms containing the unknown in one part of the equation, and the free terms in the other;
d) bring similar members;
e) solve an equation of the form aх = b, which was obtained after bringing like terms.
However, this scheme is not required for every equation. When solving many simpler equations, one has to start not from the first, but from the second ( Example. 2), third ( Example. 13) and even from the fifth stage, as in example 5.
Example 5 Solve the equation 2x = 1/4.
We find the unknown x \u003d 1/4: 2,
x = 1/8 .
Consider the solution of some linear equations encountered in the main state exam.
Example 6 Solve equation 2 (x + 3) = 5 - 6x.
2x + 6 = 5 - 6x
2x + 6x = 5 - 6
Answer: - 0.125
Example 7 Solve the equation - 6 (5 - 3x) \u003d 8x - 7.
– 30 + 18x = 8x – 7
18x - 8x = - 7 +30
Answer: 2.3
Example 8 Solve the Equation
3(3x - 4) = 4 7x + 24
9x - 12 = 28x + 24
9x - 28x = 24 + 12
Example 9 Find f(6) if f (x + 2) = 3 7's
Solution
Since we need to find f(6), and we know f (x + 2),
then x + 2 = 6.
We solve the linear equation x + 2 = 6,
we get x \u003d 6 - 2, x \u003d 4.
If x = 4 then
f(6) = 3 7-4 = 3 3 = 27
Answer: 27.
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Quadratic equations are studied in grade 8, so there is nothing complicated here. The ability to solve them is essential.
A quadratic equation is an equation of the form ax 2 + bx + c = 0, where the coefficients a , b and c are arbitrary numbers, and a ≠ 0.
Before studying specific solution methods, we note that all quadratic equations can be divided into three classes:
- Have no roots;
- They have exactly one root;
- They have two different roots.
This is an important difference between quadratic and linear equations, where the root always exists and is unique. How to determine how many roots an equation has? There is a wonderful thing for this - discriminant.
Discriminant
Let the quadratic equation ax 2 + bx + c = 0 be given. Then the discriminant is simply the number D = b 2 − 4ac .
This formula must be known by heart. Where it comes from is not important now. Another thing is important: by the sign of the discriminant, you can determine how many roots a quadratic equation has. Namely:
- If D< 0, корней нет;
- If D = 0, there is exactly one root;
- If D > 0, there will be two roots.
Please note: the discriminant indicates the number of roots, and not at all their signs, as for some reason many people think. Take a look at the examples and you will understand everything yourself:
Task. How many roots do quadratic equations have:
- x 2 - 8x + 12 = 0;
- 5x2 + 3x + 7 = 0;
- x 2 − 6x + 9 = 0.
We write the coefficients for the first equation and find the discriminant:
a = 1, b = −8, c = 12;
D = (−8) 2 − 4 1 12 = 64 − 48 = 16
So, the discriminant is positive, so the equation has two different roots. We analyze the second equation in the same way:
a = 5; b = 3; c = 7;
D \u003d 3 2 - 4 5 7 \u003d 9 - 140 \u003d -131.
The discriminant is negative, there are no roots. The last equation remains:
a = 1; b = -6; c = 9;
D = (−6) 2 − 4 1 9 = 36 − 36 = 0.
The discriminant is equal to zero - the root will be one.
Note that coefficients have been written out for each equation. Yes, it's long, yes, it's tedious - but you won't mix up the odds and don't make stupid mistakes. Choose for yourself: speed or quality.
By the way, if you “fill your hand”, after a while you will no longer need to write out all the coefficients. You will perform such operations in your head. Most people start doing this somewhere after 50-70 solved equations - in general, not so much.
The roots of a quadratic equation
Now let's move on to the solution. If the discriminant D > 0, the roots can be found using the formulas:
The basic formula for the roots of a quadratic equation
When D = 0, you can use any of these formulas - you get the same number, which will be the answer. Finally, if D< 0, корней нет — ничего считать не надо.
- x 2 - 2x - 3 = 0;
- 15 - 2x - x2 = 0;
- x2 + 12x + 36 = 0.
First equation:
x 2 - 2x - 3 = 0 ⇒ a = 1; b = −2; c = -3;
D = (−2) 2 − 4 1 (−3) = 16.
D > 0 ⇒ the equation has two roots. Let's find them:
Second equation:
15 − 2x − x 2 = 0 ⇒ a = −1; b = −2; c = 15;
D = (−2) 2 − 4 (−1) 15 = 64.
D > 0 ⇒ the equation again has two roots. Let's find them
\[\begin(align) & ((x)_(1))=\frac(2+\sqrt(64))(2\cdot \left(-1 \right))=-5; \\ & ((x)_(2))=\frac(2-\sqrt(64))(2\cdot \left(-1 \right))=3. \\ \end(align)\]
Finally, the third equation:
x 2 + 12x + 36 = 0 ⇒ a = 1; b = 12; c = 36;
D = 12 2 − 4 1 36 = 0.
D = 0 ⇒ the equation has one root. Any formula can be used. For example, the first one:
As you can see from the examples, everything is very simple. If you know the formulas and be able to count, there will be no problems. Most often, errors occur when negative coefficients are substituted into the formula. Here, again, the technique described above will help: look at the formula literally, paint each step - and get rid of mistakes very soon.
Incomplete quadratic equations
It happens that the quadratic equation is somewhat different from what is given in the definition. For example:
- x2 + 9x = 0;
- x2 − 16 = 0.
It is easy to see that one of the terms is missing in these equations. Such quadratic equations are even easier to solve than standard ones: they do not even need to calculate the discriminant. So let's introduce a new concept:
The equation ax 2 + bx + c = 0 is called an incomplete quadratic equation if b = 0 or c = 0, i.e. the coefficient of the variable x or the free element is equal to zero.
Of course, a very difficult case is possible when both of these coefficients are equal to zero: b \u003d c \u003d 0. In this case, the equation takes the form ax 2 \u003d 0. Obviously, such an equation has a single root: x \u003d 0.
Let's consider other cases. Let b \u003d 0, then we get an incomplete quadratic equation of the form ax 2 + c \u003d 0. Let's slightly transform it:
Since the arithmetic square root exists only from a non-negative number, the last equality only makes sense when (−c / a ) ≥ 0. Conclusion:
- If an incomplete quadratic equation of the form ax 2 + c = 0 satisfies the inequality (−c / a ) ≥ 0, there will be two roots. The formula is given above;
- If (−c / a )< 0, корней нет.
As you can see, the discriminant was not required - there are no complex calculations at all in incomplete quadratic equations. In fact, it is not even necessary to remember the inequality (−c / a ) ≥ 0. It is enough to express the value of x 2 and see what is on the other side of the equal sign. If there is a positive number, there will be two roots. If negative, there will be no roots at all.
Now let's deal with equations of the form ax 2 + bx = 0, in which the free element is equal to zero. Everything is simple here: there will always be two roots. It is enough to factorize the polynomial:
Taking the common factor out of the bracketThe product is equal to zero when at least one of the factors is equal to zero. This is where the roots come from. In conclusion, we will analyze several of these equations:
Task. Solve quadratic equations:
- x2 − 7x = 0;
- 5x2 + 30 = 0;
- 4x2 − 9 = 0.
x 2 − 7x = 0 ⇒ x (x − 7) = 0 ⇒ x 1 = 0; x2 = −(−7)/1 = 7.
5x2 + 30 = 0 ⇒ 5x2 = -30 ⇒ x2 = -6. There are no roots, because the square cannot be equal to a negative number.
4x 2 − 9 = 0 ⇒ 4x 2 = 9 ⇒ x 2 = 9/4 ⇒ x 1 = 3/2 = 1.5; x 2 \u003d -1.5.
