The online equation solving service will help you solve any equation. Using our site, you will not only receive the answer to the equation, but also see 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 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 given equation. U algebraic equations with variable odds infinite number solutions, and by setting certain conditions, private ones are selected from a set of solutions. Quadratic equation. The quadratic equation has the form ax^2+bx+c=0 for a>0. Solving equations square look implies 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 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 quadratic equation online you just need to enter the coefficients of such an equation (integers, fractions or decimal values). 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. To solve 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 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 matrix method is to find inverse matrix A.
Equations
How to solve equations?
In this section we will recall (or study, depending on who you choose) the most elementary equations. So what is the equation? In human language, this is some kind of mathematical expression where there is an equal sign and an unknown. Which is usually denoted by the letter "X". Solve the equation- this is to find such values of x that, when substituted into original expression will give us the correct identity. Let me remind you that identity is an expression that is beyond doubt even for a person who is absolutely not burdened with mathematical knowledge. Like 2=2, 0=0, ab=ab, etc. So how to solve equations? Let's figure it out.
There are all sorts of equations (I’m surprised, right?). But all their infinite variety can be divided into only four types.
4. Everyone else.)
All the rest, of course, most of all, yes...) This includes cubic, exponential, logarithmic, trigonometric and all sorts of others. We will work closely with them in the appropriate sections.
I’ll say right away that sometimes the equations of the first three types they will cheat you so much that you won’t even recognize them... Nothing. We will learn how to unwind them.
And why do we need these four types? And then what linear equations solved in one way square others, fractional rationals - third, A rest They don’t dare at all! Well, it’s not that they can’t decide at all, it’s that I was wrong with mathematics.) It’s just that they have their own special techniques and methods.
But for any (I repeat - for any!) equations provide a reliable and fail-safe basis for solving. Works everywhere and always. This foundation - Sounds scary, but it's very simple. And very (Very!) important.
Actually, the solution to the equation consists of these very transformations. 99% Answer to the question: " How to solve equations?" lies precisely in these transformations. Is the hint clear?)
Identical transformations of equations.
IN any equations To find the unknown, you need to transform and simplify the original example. And so that when changing appearance the essence of the equation has not changed. Such transformations are called identical or equivalent.
Note that these transformations apply specifically to the equations. There are also identity transformations in mathematics expressions. This is another topic.
Now we will repeat all, all, all basic identical transformations of equations.
Basic because they can be applied to any equations - linear, quadratic, fractional, trigonometric, exponential, logarithmic, etc. etc.
First identity transformation: you can add (subtract) to both sides of any equation any(but one and the same!) number or expression (including an expression with an unknown!). This does not change the essence of the equation.
By the way, you constantly used this transformation, you just thought that you were transferring some terms from one part of the equation to another with a change of sign. Type:
The case is familiar, we move the two to the right, and we get:
Actually you taken away from both sides of the equation is two. The result is the same:
x+2 - 2 = 3 - 2
Moving terms left and right with a change of sign is simply a shortened version of the first identity transformation. And why do we need such deep knowledge? – you ask. Nothing in the equations. For God's sake, bear it. Just don’t forget to change the sign. But in inequalities, the habit of transference can lead to a dead end...
Second identity transformation: both sides of the equation can be multiplied (divided) by the same thing non-zero number or expression. Here an understandable limitation already appears: multiplying by zero is stupid, and dividing is completely impossible. This is the transformation you use when you solve something cool like
It's clear X= 2. How did you find it? By selection? Or did it just dawn on you? In order not to select and not wait for insight, you need to understand that you are just divided both sides of the equation by 5. When dividing the left side (5x), the five was reduced, leaving pure X. Which is exactly what we needed. And when dividing the right side of (10) by five, the result is, of course, two.
That's it.
It's funny, but these two (only two!) identical transformations are the basis of the solution all equations of mathematics. Wow! It makes sense to look at examples of what and how, right?)
Examples of identical transformations of equations. Main problems.
Let's start with first identity transformation. Transfer left-right.
An example for the younger ones.)
Let's say we need to solve the following equation:
3-2x=5-3x
Let's remember the spell: "with X's - to the left, without X's - to the right!" This spell is instructions for using the first identity transformation.) What expression with an X is on the right? 3x? The answer is incorrect! On our right - 3x! Minus three x! Therefore, when moving to the left, the sign will change to plus. It will turn out:
3-2x+3x=5
So, the X’s were collected in a pile. Let's get into the numbers. There is a three on the left. With what sign? The answer “with none” is not accepted!) In front of the three, indeed, nothing is drawn. And this means that before the three there is plus. So the mathematicians agreed. Nothing is written, which means plus. Therefore, the triple will be transferred to the right side with a minus. We get:
-2x+3x=5-3
There are mere trifles left. On the left - bring similar ones, on the right - count. The answer comes straight away:
In this example, one identity transformation was enough. The second one was not needed. Well, okay.)
An example for older children.)
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.
4x 3 - 19x 2 + 19x + 6 = 0
First you need to find one root using the selection method. It is usually a divisor free member. In this case, the divisors of the number 6 are ±1, ±2, ±3, ±6.
1: 4 - 19 + 19 + 6 = 10 ⇒ number 1
-1: -4 - 19 - 19 + 6 = -36 ⇒ number -1 is not a root of a polynomial
2: 4 ∙ 8 - 19 ∙ 4 + 19 ∙ 2 + 6 = 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 the original polynomial must be divisible by x - 2. In order to perform the division of polynomials, we use Horner’s scheme:
| 4 | -19 | 19 | 6 | |
| 2 |
The coefficients of the original polynomial are displayed in the top line. The root we found is placed in the first cell of the second row 2. The second line contains the coefficients of the polynomial that results from division. They are counted like this:
|
In the second cell of the second row we write the number 1, simply by moving it from the corresponding cell of the first row. | ||||||||||
|
2 ∙ 4 - 19 = -11 | ||||||||||
|
2 ∙ (-11) + 19 = -3 | ||||||||||
|
2 ∙ (-3) + 6 = 0 |
The last number is the remainder of the division. If it is equal to 0, then we have calculated everything correctly.
Thus, we factorized the original polynomial:
4x 3 - 19x 2 + 19x + 6 = (x - 2)(4x 2 - 11x - 3)
And now all that remains is to find the roots of the quadratic equation
4x 2 - 11x - 3 = 0
D = b 2 - 4ac = (-11) 2 - 4 ∙ 4 ∙ (-3) = 169
D > 0 ⇒ the equation has 2 roots
We have found all the roots of the equation.
We offer you a convenient free online calculator for solving quadratic equations. You can quickly get and understand how they are solved using clear examples.
To produce solve quadratic equation online, first bring the equation to its general form:
ax 2 + bx + c = 0
Fill in the form fields accordingly:
How to solve a quadratic equation
| How to solve a quadratic equation: | Types of roots: |
|
1.
Reduce the quadratic equation to its general form: General view Аx 2 +Bx+C=0 Example: 3x - 2x 2 +1=-1 Reduce to -2x 2 +3x+2=0 2.
Find the discriminant D. 3.
Finding the roots of the equation. |
1.
Real roots. Moreover. x1 is not equal to x2 The situation occurs when D>0 and A is not equal to 0. 2.
The real roots are the same. x1 equals x2 3.
Two complex roots. x1=d+ei, x2=d-ei, where i=-(1) 1/2 5.
The equation has countless solutions. 6.
The equation has no solutions. |
To consolidate the algorithm, here are a few more illustrative examples of solutions to quadratic equations.
Example 1. Solving an ordinary quadratic equation with different real roots.
x 2 + 3x -10 = 0
In this equation
A=1, B=3, C=-10
D=B 2 -4*A*C = 9-4*1*(-10) = 9+40 = 49
square root We will denote it as the number 1/2!
x1=(-B+D 1/2)/2A = (-3+7)/2 = 2
x2=(-B-D 1/2)/2A = (-3-7)/2 = -5
To check, let's substitute:
(x-2)*(x+5) = x2 -2x +5x – 10 = x2 + 3x -10
Example 2. Solving a quadratic equation with matching real roots.
x 2 – 8x + 16 = 0
A=1, B = -8, C=16
D = k 2 – AC = 16 – 16 = 0
X = -k/A = 4
Let's substitute
(x-4)*(x-4) = (x-4)2 = X 2 – 8x + 16
Example 3. Solving a quadratic equation with complex roots.
13x 2 – 4x + 1 = 0
A=1, B = -4, C=9
D = b 2 – 4AC = 16 – 4*13*1 = 16 - 52 = -36
The discriminant is negative – the roots are complex.
X1=(-B+D 1/2)/2A = (4+6i)/(2*13) = 2/13+3i/13
x2=(-B-D 1/2)/2A = (4-6i)/(2*13) = 2/13-3i/13
, where I is the square root of -1
Here are actually all the possible cases of solving quadratic equations.
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