Solve the quadratic equation online. Equations 3 4 solution
Goals:
- To systematize and generalize knowledge and skills on the topic: Solutions of equations of the third and fourth degree.
- To deepen knowledge by completing a series of tasks, some of which are not familiar either in their type or in the method of solving.
- Formation of interest in mathematics through the study of new chapters of mathematics, education of graphic culture through the construction of graphs of equations.
Lesson type: combined.
Equipment: graph projector.
Visibility: table "Vieta's theorem".
During the classes
1. Mental account
a) What is the remainder of the division of the polynomial p n (x) \u003d 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) With what help do we solve the equation of the third and fourth degree?
d) If b is an even number in the 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 tasks are coded) Use the "Vieta Theorem"
1 group
Roots: x 1 = 1; x 2 \u003d -2; x 3 \u003d -3; x 4 = 6
Write 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 - 23 x 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 is the root of the equation. Horner's scheme
p 3 (x) = x 3 -x 2 -24x -36
p 3 (-2) \u003d -8 -4 +48 -36 \u003d 0, x 2 \u003d -2
p 2 (x) \u003d x 2 -3x -18 \u003d 0
x 3 \u003d -3, x 4 \u003d 6
Answer: 1; -2; -3; 6 the sum of the roots 2 (P)
2 group
Roots: x 1 \u003d -1; x 2 = x 3 =2; x 4 \u003d 5
Write 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 \u003d 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) \u003d x 3 -9x 2 + 24x -20
p 3 (2) \u003d 8 -36 + 48 -20 \u003d 0
p 2 (x) \u003d x 2 -7x + 10 \u003d 0 x 1 \u003d 2; x 2 \u003d 5
Answer: -1;2;2;5 sum of roots 8(P)
3 group
Roots: x 1 \u003d -1; x 2 =1; x 3 \u003d -2; x 4 \u003d 3
Write an equation:
B=-1+1-2+3=1;b=-1
s=-1+2-3-2+3-6=-7; s=-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 solved later on the board by group 4)
Solution. We are looking 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) \u003d -1 + 7-6 \u003d 0
p 2 (x) = x 2 -x -6=0; x 1 \u003d -2; x 2 \u003d 3
Answer: -1; 1; -2; 3 The sum of the roots 1 (O)
4 group
Roots: x 1 = -2; x 2 \u003d -2; x 3 \u003d -3; x 4 = -3
Write 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 group 5 on the board)
Solution. We are looking 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) \u003d 16 -32 -20 + 72 -36 \u003d 0
p 3 (x) \u003d x 3 + 2x 2 -9x-18 \u003d 0
p 3 (-2) \u003d -8 + 8 + 18-18 \u003d 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 \u003d -1; x 2 \u003d -2; x 3 \u003d -3; x 4 = -4
Write an equation
x 4+ 10x 3 + 35x 2 + 50x + 24 = 0(this equation is then solved by the 6th group 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) \u003d x- 3 + 9x 2 + 26x + 24 \u003d 0
p 3 (-2) \u003d -8 + 36-52 + 24 \u003d O
p 2 (x) \u003d x 2 + 7x + 12 \u003d 0
Answer: -1; -2; -3; -4 sum-10 (I)
6 group
Roots: x 1 = 1; x 2 = 1; x 3 \u003d -3; x 4 = 8
Write 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 are looking 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) \u003d x 2 -5x - 24 \u003d 0
x 3 \u003d -3, x 4 \u003d 8
Answer: 1; 1; -3; 8 sum 7 (L)
3. Solution of equations with a parameter
1. Solve the equation x 3 + 3x 2 + mx - 15 = 0; if one of the roots is (-1)
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) \u003d x 2 + 2x-15 \u003d 0
x 2 \u003d -1-4 \u003d -5;
x 3 \u003d -1 + 4 \u003d 3;
Answer: - 1; -5; 3
In ascending order: -5;-1;3. (b n s)
2. Find all the roots of the polynomial x 3 - 3x 2 + ax - 2a + 6, if the remainders of its division into binomials x-1 and x + 2 are equal.
Solution: R \u003d R 3 (1) \u003d R 3 (-2)
P 3 (1) \u003d 1-3 + a- 2a + 6 \u003d 4-a
P 3 (-2) \u003d -8-12-2a-2a + 6 \u003d -14-4a
x 3 -3x 2 -6x + 12 + 6 \u003d 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 equal to zero if and only if at least one of these factors is equal to zero, while the other makes sense.
2 group. Roots: -3; -2; 1; 2;3 group. Roots: -1; 2; 6; 10;
4 group. Roots: -3; 2; 2; 5;
5 group. Roots: -5; -2; 2; 4;
6 group. Roots: -8; -2; 6; 7.
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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:
|
In the second cell of the second row, write the number 2, simply by moving it from the corresponding cell of the first row. | ||||||||||||
|
2 ∙ 2 + 5 = 9 | ||||||||||||
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2 ∙ 9 - 11 = 7 | ||||||||||||
|
2 ∙ 7 - 20 = -6 | ||||||||||||
|
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.
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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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