Equations
How to solve equations?
In this section, we will recall (or study - as anyone else) the most elementary equations. So what is an equation? In human terms, this is some kind of mathematical expression, where there is an equal sign and an unknown. Which is usually denoted by the letter "NS". Solve the equation is to find such x values that, when substituted in initial expression, will give us the correct identity. Let me remind you that identity is an expression that does not raise doubts even in a person who is absolutely not burdened with mathematical knowledge. Like 2 = 2, 0 = 0, ab = ab, etc. So how do you solve equations? Let's figure it out.
There are all sorts of equations (I was surprised, right?). But all their infinite variety can be divided into only four types.
4. Other.)
All the rest, of course, most of all, yes ...) This includes cubic, and exponential, and logarithmic, and trigonometric and all sorts of others. We will work closely with them in the relevant sections.
I must say right away that sometimes the equations of the first three types will wind up so that you don'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 are solved in one way, square others, fractional rational - third, a rest do not dare at all! Well, it’s not that they don’t make up their minds at all, I shouldn’t have offended the mathematics.) It’s just that they have their own special techniques and methods.
But for any (I repeat - for any!) equations have a reliable and trouble-free basis for solving. Works anywhere and anytime. This foundation - Sounds scary, but the thing is very simple. And very (very!) important.
Actually, the solution to the equation consists of these very transformations. 99%. The answer to the question: " How to solve equations?"lies, just in these transformations. Is the hint clear?)
Identical transformations of equations.
V any equations to find the unknown, it is necessary to transform and simplify the original example. And so that when changing appearance the essence of the equation did not change. Such transformations are called identical or equivalent.
Note that these transformations are precisely to the equations. There are still identical transformations in mathematics expressions. This is a different 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 the same thing!) a number or an expression (including an expression with an unknown!). This does not change the essence of the equation.
By the way, you constantly used this transformation, just thought that you were transferring some terms from one side of the equation to another with a change in sign. Type:
The matter is familiar, we transfer the two to the right, and we get:
In fact you taken away from both sides of the equation two. The result is the same:
x + 2 - 2 = 3 - 2
The transfer of terms to the left-to the right with a change of sign is simply an abbreviated version of the first identical transformation. And why do we need such deep knowledge? - you ask. The equations are low. Move, for God's sake. Just don't forget to change the sign. But in inequalities, the habit of transference can be confusing….
Second Identity Transformation: both sides of the equation can be multiplied (divided) by the same nonzero number or expression. An understandable limitation already appears here: multiplying by zero is stupid, but dividing is not at all possible. You use this transformation when you're doing something cool like
It's clear business NS= 2. How did you find it? By selection? Or did it just light up? In order not to pick up and not wait for insight, you need to understand that you just divided both sides of the equation by 5. When dividing the left side (5x), the five was reduced, leaving a pure x. Which is what we needed. And when dividing the right side (10) by five, it turned out, obviously, a two.
That's all.
It's funny, but these two (only two!) Identical transformations underlie the solution all equations of mathematics. How! It makes sense to look at examples of what and how, right?)
Examples of identical transformations of equations. Main problems.
Let's start with the first identical transformation. Move left-right.
An example for the youngest.)
Let's say you need to solve the following equation:
3-2x = 5-3x
Remember the spell: "with x - to the left, without x - to the right!" This spell is an instruction on how to apply the first identical transformation.) What expression with an x do we have on the right? 3x? The answer is wrong! On our right - 3x! Minus three x! Therefore, when moving to the left, the sign will change to a plus. It will turn out:
3-2x + 3x = 5
So, the X's were gathered in a pile. Let's get down to numbers. There is a three on the left. What is your sign? The answer "with no" is not accepted!) In front of the three, really, nothing is drawn. And this means that in front of the three is a plus. So the mathematicians agreed. Nothing is written, so a plus. Therefore, the triplet 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 is immediately obtained:
In this example, one identical transformation was enough. The second was not needed. Well, okay.)
An example for the older ones.)
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. Instant validation testing. Learning - with interest!)
you can get acquainted with functions and derivatives.
Let us consider two types of solutions to systems of equations:
1. Solution of the system by the substitution method.
2. Solution of the system by term-by-term addition (subtraction) of the equations of the system.
In order to solve the system of equations substitution method you need to follow a simple algorithm:
1. We express. Express one variable from any equation.
2. Substitute. We substitute the obtained value into another equation instead of the expressed variable.
3. We solve the resulting equation with one variable. We find a solution to the system.
To solve system by term-by-term addition (subtraction) necessary:
1.Choose a variable for which we will make the same coefficients.
2. We add or subtract equations, in the end we get an equation with one variable.
3. Solve the resulting linear equation. We find a solution to the system.
The solution of the system is the intersection points of the graphs of the function.
Let's consider in detail the solution of systems using examples.
Example # 1:
Let's solve by substitution method
Solving a System of Equations by the Substitution Method2x + 5y = 1 (1 equation)
x-10y = 3 (2 equation)
1. We express
It can be seen that in the second equation there is a variable x with a coefficient of 1, from which it turns out that it is easiest to express the variable x from the second equation.
x = 3 + 10y
2. After we have expressed, we substitute 3 + 10y in the first equation instead of the variable x.
2 (3 + 10y) + 5y = 1
3. Solve the resulting equation in one variable.
2 (3 + 10y) + 5y = 1 (expand the brackets)
6 + 20y + 5y = 1
25y = 1-6
25y = -5 |: (25)
y = -5: 25
y = -0.2
The solution to the system of equations is the intersection points of the graphs, therefore we need to find x and y, because the intersection point consists of x and y. Find x, in the first paragraph where we expressed there we substitute y.
x = 3 + 10y
x = 3 + 10 * (- 0.2) = 1
It is customary to write dots in the first place we write the variable x, and in the second the variable y.
Answer: (1; -0.2)
Example # 2:
Let's solve by the method of term-by-term addition (subtraction).
Solving a system of equations by the addition method3x-2y = 1 (1 equation)
2x-3y = -10 (2 equation)
1.Choose a variable, say, choose x. In the first equation, the variable x has a coefficient of 3, in the second 2. It is necessary to make the coefficients the same, for this we have the right to multiply the equations or divide by any number. The first equation is multiplied by 2, and the second by 3, and we get a total factor of 6.
3x-2y = 1 | * 2
6x-4y = 2
2x-3y = -10 | * 3
6x-9y = -30
2. Subtract the second from the first equation to get rid of the variable x. Solve the linear equation.
__6x-4y = 2
5y = 32 | :5
y = 6.4
3. Find x. Substitute the found y into any of the equations, let's say in the first equation.
3x-2y = 1
3x-2 * 6.4 = 1
3x-12.8 = 1
3x = 1 + 12.8
3x = 13.8 |: 3
x = 4.6
The intersection point will be x = 4.6; y = 6.4
Answer: (4.6; 6.4)
Do you want to study for exams for free? Online Tutor is free... No kidding.
Goals:
- To systematize and generalize knowledge and skills on the topic: Solutions of equations of the third and fourth degree.
- Deepen your knowledge by completing a series of tasks, some of which are not familiar either by their type or solution.
- Formation of interest in mathematics through the study of new chapters of mathematics, education of a graphic culture through the construction of graphs of equations.
Lesson type: combined.
Equipment: overhead projector.
Visibility: table "Vieta's theorem".
During the classes
1. Verbal counting
a) What is the remainder of dividing the polynomial p n (x) = a n x n + a n-1 x n-1 + ... + a 1 x 1 + a 0 by the binomial x-a?
b) How many roots can a cubic equation have?
c) How do we solve the equation of the third and fourth degree?
d) If b is an even number in a quadratic equation, then what is D and x 1; x 2
2. Independent work (in groups)
Make an equation if the roots are known (answers to the tasks are coded) "Vieta's theorem" is used
1st group
Roots: x 1 = 1; x 2 = -2; x 3 = -3; x 4 = 6
Make an equation:
B = 1-2-3 + 6 = 2; b = -2
c = -2-3 + 6 + 6-12-18 = -23; c = -23
d = 6-12 + 36-18 = 12; d = -12
e = 1 (-2) (- 3) 6 = 36
x 4 -2 x 3 - 23x 2 - 12 x + 36 = 0(this equation is then solved by group 2 on the board)
Solution ... We are looking for integer roots among the divisors of the number 36.
p = ± 1; ± 2; ± 3; ± 4; ± 6 ...
p 4 (1) = 1-2-23-12 + 36 = 0 The number 1 satisfies the equation, therefore, = 1 root of the equation. According to Horner's scheme
p 3 (x) = x 3 -x 2 -24x -36
p 3 (-2) = -8 -4 +48 -36 = 0, x 2 = -2
p 2 (x) = x 2 -3x -18 = 0
x 3 = -3, x 4 = 6
Answer: 1; -2; -3; 6 sum of roots 2 (P)
2nd group
Roots: x 1 = -1; x 2 = x 3 = 2; x 4 = 5
Make an equation:
B = -1 + 2 + 2 + 5-8; b = -8
c = 2 (-1) + 4 + 10-2-5 + 10 = 15; c = 15
D = -4-10 + 20-10 = -4; d = 4
e = 2 (-1) 2 * 5 = -20; e = -20
8 + 15 + 4x-20 = 0 (group 3 solves this equation on the board)
p = ± 1; ± 2; ± 4; ± 5; ± 10; ± 20.
p 4 (1) = 1-8 + 15 + 4-20 = -8
p 4 (-1) = 1 + 8 + 15-4-20 = 0
p 3 (x) = x 3 -9x 2 + 24x -20
p 3 (2) = 8 -36 + 48 -20 = 0
p 2 (x) = x 2 -7x + 10 = 0 x 1 = 2; x 2 = 5
Answer: -1; 2; 2; 5 sum of roots 8 (P)
Group 3
Roots: x 1 = -1; x 2 = 1; x 3 = -2; x 4 = 3
Make an equation:
B = -1 + 1-2 + 3 = 1; B = -1
c = -1 + 2-3-2 + 3-6 = -7; c = -7
D = 2 + 6-3-6 = -1; d = 1
e = -1 * 1 * (- 2) * 3 = 6
x 4 - x 3- 7x 2 + x + 6 = 0(this equation is then solved on the board by group 4)
Solution. We look for integer roots among the divisors of the number 6.
p = ± 1; ± 2; ± 3; ± 6
p 4 (1) = 1-1-7 + 1 + 6 = 0
p 3 (x) = x 3 - 7x -6
p 3 (-1) = -1 + 7-6 = 0
p 2 (x) = x 2 -x -6 = 0; x 1 = -2; x 2 = 3
Answer: -1; 1; -2; 3 Sum of roots 1 (O)
4 group
Roots: x 1 = -2; x 2 = -2; x 3 = -3; x 4 = -3
Make an equation:
B = -2-2-3 + 3 = -4; b = 4
c = 4 + 6-6 + 6-6-9 = -5; c = -5
D = -12 + 12 + 18 + 18 = 36; d = -36
e = -2 * (- 2) * (- 3) * 3 = -36; e = -36
x 4 +4x 3 - 5x 2 - 36x -36 = 0(this equation is then solved by the 5th group on the board)
Solution. We look for integer roots among the divisors of the number -36
p = ± 1; ± 2; ± 3 ...
p (1) = 1 + 4-5-36-36 = -72
p 4 (-2) = 16 -32 -20 + 72 -36 = 0
p 3 (x) = x 3 + 2x 2 -9x-18 = 0
p 3 (-2) = -8 + 8 + 18-18 = 0
p 2 (x) = x 2 -9 = 0; x = ± 3
Answer: -2; -2; -3; 3 Sum of roots-4 (F)
5 group
Roots: x 1 = -1; x 2 = -2; x 3 = -3; x 4 = -4
Make an equation
x 4+ 10x 3 + 35x 2 + 50x + 24 = 0(this equation is then solved by group 6 on the board)
Solution ... We are looking for integer roots among the divisors of the number 24.
p = ± 1; ± 2; ± 3
p 4 (-1) = 1 -10 + 35 -50 + 24 = 0
p 3 (x) = x- 3 + 9x 2 + 26x + 24 = 0
p 3 (-2) = -8 + 36-52 + 24 = O
p 2 (x) = x 2 + 7x + 12 = 0
Answer: -1; -2; -3; -4 sum-10 (AND)
6 group
Roots: x 1 = 1; x 2 = 1; x 3 = -3; x 4 = 8
Make an equation
B = 1 + 1-3 + 8 = 7; b = -7
c = 1-3 + 8-3 + 8-24 = -13
D = -3-24 + 8-24 = -43; d = 43
x 4 - 7x 3- 13x 2 + 43x - 24 = 0 (this equation is then solved by 1 group on the board)
Solution ... We look for integer roots among the divisors of the number -24.
p 4 (1) = 1-7-13 + 43-24 = 0
p 3 (1) = 1-6-19 + 24 = 0
p 2 (x) = x 2 -5x - 24 = 0
x 3 = -3, x 4 = 8
Answer: 1; 1; -3; 8 sum 7 (L)
3. Solving equations with a parameter
1. Solve the equation x 3 + 3x 2 + mx - 15 = 0; if one of the roots is (-1)
Write the answer in ascending order
R = P 3 (-1) = - 1 + 3-m-15 = 0
x 3 + 3x 2 -13x - 15 = 0; -1 + 3 + 13-15 = 0
By condition x 1 = - 1; D = 1 + 15 = 16
P 2 (x) = x 2 + 2x-15 = 0
x 2 = -1-4 = -5;
x 3 = -1 + 4 = 3;
Answer: - 1; -5; 3
In ascending order: -5; -1; 3. (L N S)
2. Find all the roots of the polynomial x 3 - 3x 2 + ax - 2a + 6, if the remainders of its division by binomials x-1 and x +2 are equal.
Solution: R = P 3 (1) = P 3 (-2)
P 3 (1) = 1-3 + a- 2a + 6 = 4-a
P 3 (-2) = -8-12-2a-2a + 6 = -14-4a
x 3 -3x 2 -6x + 12 + 6 = x 3 -3x 2 -6x + 18
x 2 (x-3) -6 (x-3) = 0
(x-3) (x 2 -6) = 0
The product of two factors is zero if and only if at least one of these factors is zero, while the other makes sense.
2nd group... Roots: -3; -2; 1; 2;Group 3... Roots: -1; 2; 6; ten;
4 group... Roots: -3; 2; 2; 5;
5 group... Roots: -5; -2; 2; 4;
6 group... Roots: -8; -2; 6; 7.
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 solving a 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: | Root types: |
|
1.
Bring the quadratic equation to a general form: General view Аx 2 + Bx + C = 0 Example: 3x - 2x 2 + 1 = -1 Bring to -2x 2 + 3x + 2 = 0 2.
Find the discriminant D. 3.
Find the roots of the equation. |
1.
Valid roots. Moreover. x1 is not equal to x2 The situation arises when D> 0 and A is not equal to 0. 2.
Valid 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 solidify 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
the square root will be denoted 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 coincidence of real roots.
x 2 - 8x + 16 = 0
A = 1, B = -8, C = 16
D = k 2 - AC = 16 - 16 = 0
X = -k / A = 4
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
These are actually all possible cases of solving quadratic equations.
We hope that our online calculator will prove to be of great benefit to you.
If the material was helpful, you can
2x 4 + 5x 3 - 11x 2 - 20x + 12 = 0
First, you need to find one root by the selection method. It is usually a 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 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. The root found by us is put in the first cell of the second line 2. The second line contains the coefficients of the polynomial, which will be the result of division. They are considered as follows:
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In the second cell of the second line, write the number 2, by simply transferring 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 have calculated everything correctly.
2x 4 + 5x 3 - 11x 2 - 20x + 12 = (x - 2) (2x 3 + 9x 2 + 7x - 6)
But it’s not over yet. 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. Divisors of the number -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 empty cells:
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In the second cell of the third line, write the number 2, by simply dragging it out of the corresponding cell in 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 have factorized 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 factorized. To do this, you can solve the quadratic equation through the discriminant, or you can search 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 line, write the number 2, by simply transferring it from the appropriate cell in the third row. | ||||||||||||||||||||||||
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-3 ∙ 2 + 5 = -1 | ||||||||||||||||||||||||
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-3 ∙ (-1) - 3 = 0 |
Thus, we have 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.
