Goals:
- Systematize and generalize knowledge and skills on the topic: Solutions of equations of the third and fourth degree.
- Deepen your knowledge by completing a number of tasks, some of which are unfamiliar either in type or method of solution.
- Forming an interest in mathematics through the study of new chapters of mathematics, nurturing a graphic culture through the construction of graphs of equations.
Lesson type: combined.
Equipment: graphic projector.
Visibility: table "Viete's Theorem".
During the classes
1. Oral counting
a) What is the remainder when 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 equations of the third and fourth degrees?
d) If b is an even number in a quadratic equation, then what is the value of D and x 1; x 2
2. Independent work (in groups)
Write an equation if the roots are known (answers to tasks are coded) “Vieta’s Theorem” is used
1 group
Roots: x 1 = 1; x 2 = -2; x 3 = -3; x 4 = 6
Make up 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 look for whole roots among the divisors of the number 36.
р = ±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. 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 up 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)
р = ±1;±2;±4;±5;±10;±20.
p 4 (1)=1-8+15+4-20=-8
р 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)
3 group
Roots: x 1 = -1; x 2 =1; x 3 = -2; x 4 =3
Make up an equation:
В=-1+1-2+3=1;В=-1
с=-1+2-3-2+3-6=-7;с=-7
D=2+6-3-6=-1; d=1
e=-1*1*(-2)*3=6
x 4 - x 3- 7x 2 + x + 6 = 0(group 4 solves this equation later on the board)
Solution. We look for whole roots among the divisors of the number 6.
р = ±1;±2;±3;±6
p 4 (1)=1-1-7+1+6=0
p 3 (x) = x 3 - 7x -6
р 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 up an equation:
B=-2-2-3+3=-4; b=4
c=4+6-6+6-6-9=-5; с=-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 look for whole roots among the divisors of the number -36
р = ±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
Write an equation
x 4+ 10x 3 + 35x 2 + 50x + 24 = 0(this equation is then solved by group 6 on the board)
Solution . We look for whole roots among the divisors of the number 24.
р = ±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 (I)
6 group
Roots: x 1 = 1; x 2 = 1; x 3 = -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 group 1 on the board)
Solution . We look for whole 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 equal to (-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. (b N S)
2. Find all roots of the polynomial x 3 - 3x 2 + ax - 2a + 6, if the remainders from its division into 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 -Zx 2 -6x + 12 + 6 = x 3 -Zx 2 -6x + 18
x 2 (x-3)-6(x-3) = 0
(x-3)(x 2 -6) = 0
3) a=0, x 2 -0*x 2 +0 = 0; x 2 =0; x 4 =0
a=0; x=0; x=1
a>0; x=1; x=a ± √a
2. Write an equation
1 group. Roots: -4; -2; 1; 7;
2nd 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.
In this article we will learn to solve biquadratic equations.
So, what type of equations are called biquadratic?
All equations of the form ah 4 +
bx
2
+
c
= 0
, Where a ≠ 0, which are square with respect to x 2, and are called biquadratic equations. As you can see, this entry is very similar to the entry for a quadratic equation, so we will solve biquadratic equations using the formulas that we used to solve the quadratic equation.
Only we will need to introduce a new variable, that is, we denote x 2 another variable, for example at or t (or any other letter of the Latin alphabet).
For example, let's solve the equation x 4 + 4x 2 ‒ 5 = 0.
Let's denote x 2
through at
(x 2 = y
) and we get the equation y 2 + 4y – 5 = 0.
As you can see, you already know how to solve such equations.
We solve the resulting equation:
D = 4 2 – 4 (‒ 5) = 16 + 20 = 36, √D = √36 = 6.
y 1 = (‒ 4 – 6)/2= ‒ 10 /2 = ‒ 5,
y 2 = (‒ 4 + 6)/2= 2 /2 = 1.
Let's return to our variable x.
We found that x 2 = ‒ 5 and x 2 = 1.
We note that the first equation has no solutions, and the second gives two solutions: x 1 = 1 and x 2 = ‒1. Be careful not to lose the negative root (most often they get the answer x = 1, but this is not correct).
Answer:- 1 and 1.
To better understand the topic, let's look at a few examples.
Example 1. Solve the equation 2x 4 ‒ 5 x 2 + 3 = 0.
Let x 2 = y, then 2y 2 ‒ 5y + 3 = 0.
D = (‒ 5) 2 – 4 2 3 = 25 ‒ 24 = 1, √D = √1 = 1.
y 1 = (5 – 1)/(2 2) = 4 /4 =1, y 2 = (5 + 1)/(2 2) = 6 /4 =1.5.
Then x 2 = 1 and x 2 = 1.5.
We get x 1 = ‒1, x 2 = 1, x 3 = ‒ √1.5, x 4 = √1.5.
Answer: ‒1; 1; ‒ √1,5; √1,5.
Example 2. Solve the equation 2x 4 + 5 x 2 + 2 = 0.
2y 2 + 5y + 2 =0.
D = 5 2 – 4 2 2 = 25 ‒ 16 = 9, √D = √9 = 3.
y 1 = (‒ 5 – 3)/(2 2) = ‒ 8 /4 = ‒2, y 2 = (‒5 + 3)/(2 2) = ‒ 2 /4 = ‒ 0.5.
Then x 2 = - 2 and x 2 = - 0.5. Please note that none of these equations have a solution.
Answer: there are no solutions.
Incomplete biquadratic equations- it is when b = 0 (ax 4 + c = 0) or c = 0
(ax 4 + bx 2 = 0) are solved like incomplete quadratic equations.
Example 3. Solve the equation x 4 ‒ 25x 2 = 0
Let's factorize, put x 2 out of brackets and then x 2 (x 2 ‒ 25) = 0.
We get x 2 = 0 or x 2 ‒ 25 = 0, x 2 = 25.
Then we have roots 0; 5 and – 5.
Answer: 0; 5; – 5.
Example 4. Solve the equation 5x 4 ‒ 45 = 0.
x 2 = ‒ √9 (has no solutions)
x 2 = √9, x 1 = ‒ 3, x 2 = 3.
As you can see, if you can solve quadratic equations, you can also solve biquadratic equations.
If you still have questions, sign up for my lessons. Tutor Valentina Galinevskaya.
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The concept of equations with two variables is first formed in the 7th grade mathematics course. Specific problems are considered, the process of solving which leads to this type of equations.
However, they are studied rather superficially. The program focuses on systems of equations with two unknowns.
This has become the reason that problems in which certain restrictions are imposed on the coefficients of the equation are practically not considered. Insufficient attention is paid to methods for solving tasks like “Solve an equation in natural or integer numbers.” It is known that Unified State Exam materials and entrance exam tickets often contain such exercises.
Which equations are defined as equations with two variables?
xy = 8, 7x + 3y = 13 or x 2 + y = 7 are examples of equations with two variables.
Consider the equation x – 4y = 16. If x = 4 and y = -3, it will be a correct equality. This means that this pair of values is the solution to this equation.
The solution to any equation with two variables is the set of pairs of numbers (x; y) that satisfy this equation (turn it into a true equality).
Often the equation is transformed so that it can be used to obtain a system for finding unknowns.
Examples
Solve the equation: xy – 4 = 4x – y.
In this example, you can use the factorization method. To do this, you need to group the terms and take the common factor out of brackets:
xy – 4 = 4x – y;
xy – 4 – 4x + y = 0;
(xy + y) – (4x + 4) = 0;
y(x + 1) – 4(x + 1) = 0;
(x + 1)(y - 4) = 0.
Answer: All pairs (x; 4), where x is any rational number and (-1; y), where y is any rational number.
Solve the equation: 4x 2 + y 2 + 2 = 2(2x - y).
The first step is grouping.
4x 2 + y 2 + 2 = 4x – 2y;
4x 2 + y 2 + 1 - 4x + 2y + 1 = 0;
(4x 2 – 4x +1) + (y 2 + 2y + 1) = 0.
Applying the squared difference formula, we get:
(2x - 1) 2 + (y + 1) 2 = 0.
When summing two non-negative expressions, zero will result only if 2x – 1 = 0 and y + 1 = 0. It follows: x = ½ and y = -1.
Answer: (1/2; -1).
Solve the equation (x 2 – 6x + 10)(y 2 + 10y + 29) = 4.
It is rational to apply the estimation method, highlighting the complete squares in brackets.
((x - 3) 2 + 1)((y + 5) 2 + 4) = 4.
In this case (x - 3) 2 + 1 ≥ 1, and (y + 5) 2 + 4 ≥ 4. Then the left side of the equation is always at least 4. Equality is possible in the case
(x - 3) 2 + 1 = 1 and (y + 5) 2 + 4 = 4. Therefore, x = 3, y = -5.
Answer: (3; -5).
Solve the equation in whole numbers: x 2 + 10y 2 = 15x + 3.
This equation can be written as follows:
x 2 = -10y 2 + 15x + 3. If the right side of the equality is divided by 5, then 3 is the remainder. It follows from this that x 2 is not divisible by 5. It is known that the square of a number that is not divisible by 5 must leave a remainder of either 1 or 4. This means that the equation has no roots.
Answer: There are no solutions.
Don't be discouraged by the difficulty of finding the right solution for an equation with two variables. Perseverance and practice will definitely bear fruit.
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
We will denote the square root 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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