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 the 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 the 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 from the roots of the 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 equal to the second coefficient taken from 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 is p=2, and the free member q=-4. Let's find the discriminant D 1, since the second coefficient is 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 formulas (in this case, using 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.

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Application

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However, in principle, this very solution of equations online helps to improve the understanding and perception of a holistic approach to studying and practical consolidation of the theoretical material covered by students. It is much easier to do this during your study time.

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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 for high school students secondary schools and their parents. Students will be able to prepare for tests and exams, test their knowledge, and parents will be able to control the decision mathematical equations with your 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. Benefit online service and is 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. Solving the equation in general view. 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. U algebraic equations with variable coefficients there is an infinite number of solutions, and by setting certain conditions, private 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 less than zero, then the equation has no real roots (the roots are from the field complex numbers), if 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. Linear equations. For solutions linear equations(or systems of equations) there are four main methods 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 system triangular in appearance. From it, the unknowns are determined one by one. In practice, it is required to solve such an equation online with 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 matrix method is to find inverse matrix A.

The free calculator we bring to your attention has a rich arsenal of possibilities for mathematical calculations. It allows you to use the online calculator in various fields of activity: educational, professional And commercial. Of course, using an online calculator is especially popular among students And schoolchildren, it makes it much easier for them to perform a variety of calculations.

However, the calculator can become a useful tool in some areas of business and for people different professions. Of course, the need to use a calculator in business or labor activity determined primarily by the type of activity itself. If your business and profession are associated with constant calculations and calculations, then it is worth trying out an electronic calculator and assessing the degree of its usefulness for a particular task.

This online calculator can

  • Correctly perform standard mathematical functions written in one line like - 12*3-(7/2) and can process numbers larger than we can count huge numbers in an online calculator. We don’t even know what to call such a number correctly ( there are 34 characters and this is not the limit at all).
  • Except tangent, cosine, sine and other standard functions - the calculator supports calculation operations arctangent, arccotangent and others.
  • Available in Arsenal logarithms, factorials and other interesting features
  • This online calculator knows how to build graphs!!!

To plot graphs, the service uses a special button (the graph is drawn in gray) or a letter representation of this function (Plot). To build a graph in an online calculator, just write the function: plot(tan(x)),x=-360..360.

We took the simplest graph for the tangent, and after the decimal point we indicated the range of the X variable from -360 to 360.

You can build absolutely any function, with any number of variables, for example this: plot(cos(x)/3z, x=-180..360,z=4) or even more complex that you can come up with. Pay attention to the behavior of the variable X - the interval from and to is indicated using two dots.

The only negative (although it is difficult to call it a disadvantage) of this online calculator this is that he does not know how to build spheres and other three-dimensional figures - only a plane.

How to use the Math Calculator

1. The display (calculator screen) displays the entered expression and the result of its calculation in ordinary symbols, as we write on paper. This field is simply for viewing the current transaction. The entry appears on the display as you type a mathematical expression in the input line.

2. The expression input field is intended for recording the expression that needs to be calculated. It should be noted here that the mathematical symbols used in computer programs, do not always coincide with those that we usually use on paper. In the overview of each calculator function, you will find the correct designation for a specific operation and examples of calculations in the calculator. On this page below is a list of all possible operations in the calculator, also indicating their correct spelling.

3. Toolbar - these are calculator buttons that replace manual input of mathematical symbols indicating the corresponding operation. Some calculator buttons (additional functions, unit converter, solving matrices and equations, graphs) supplement the taskbar with new fields where data for a specific calculation is entered. The "History" field contains examples of writing mathematical expressions, as well as your six most recent entries.

Please note that when you press the buttons for calling additional functions, a unit converter, solving matrices and equations, and plotting graphs, the entire calculator panel moves up, covering part of the display. Fill in the required fields and press the "I" key (highlighted in red in the picture) to see the full-size display.

4. The numeric keypad contains numbers and arithmetic symbols. The "C" button deletes the entire entry in the expression entry field. To delete characters one by one, you need to use the arrow to the right of the input line.

Try to always close parentheses at the end of an expression. For most operations this is not critical; the online calculator will calculate everything correctly. However, in some cases errors may occur. For example, when raising to a fractional power, unclosed parentheses will cause the denominator of the fraction in the exponent to go into the denominator of the base. The closing bracket is shown in pale gray on the display and should be closed when recording is complete.

Key Symbol Operation
pi pi Constant pi
e e Euler number
% % Percent
() () Open/Close Brackets
, , Comma
sin sin(?) Sine of angle
cos cos(?) Cosine
tan tan(y) Tangent
sinh sinh() Hyperbolic sine
cosh cosh() Hyperbolic cosine
tanh tanh() Hyperbolic tangent
sin -1 asin() Reverse sine
cos -1 acos() Inverse cosine
tan -1 atan() Reverse tangent
sinh -1 asinh() Inverse hyperbolic sine
cosh -1 acosh() Inverse hyperbolic cosine
tanh -1 atanh() Inverse hyperbolic tangent
x 2 ^2 Squaring
x 3 ^3 Cube
x y ^ Exponentiation
10 x 10^() Exponentiation to base 10
e x exp() Exponentiation of Euler's number
vx sqrt(x) Square root
3 vx sqrt3(x) 3rd root
yvx sqrt(x,y) Root extraction
log 2 x log2(x) Binary logarithm
log log(x) Decimal logarithm
ln ln(x) Natural logarithm
log y x log(x,y) Logarithm
I/II Collapse/Call additional functions
Unit Unit converter
Matrix Matrices
Solve Equations and systems of equations
Graphing
Additional functions (call with key II)
mod mod Division with remainder
! ! Factorial
i/j i/j Imaginary unit
Re Re() Isolating the whole real part
Im Im() Excluding the real part
|x| abs() The absolute value of a number
Arg arg() Function argument
nCr ncr() Binominal coefficient
gcd gcd() GCD
lcm lcm() NOC
sum sum() Total value of all decisions
fac factorize() Prime factorization
diff diff() Differentiation
Deg Degrees
Rad Radians