In the general case, the solution of a fourth-degree equation is carried out using methods for solving equations for higher degrees, for example, the Ferrari method or using the Horner scheme. But some 4th degree equations have a simpler solution.
There are several special types of fourth-degree equations, the methods for solving which you will learn below:
- Biquadratic equation $ax^4+bx^2+c=0$;
- Reciprocal equations of the form $ax^4+bx^3+cx^2 +bx+ a=0$;
- Equations of the form $ax^4+b=0$.
Solving biquadratic equations of the fourth degree
Biquadratic equations $ax^4+bx^2+c=0$ are reduced to quadratic equations by replacing the variable $x^2$ with a new one, for example, $y$. After the replacement, the new resulting equation is solved, and then the value of the found variable is substituted into the equation $x^2=y$. The result of the solution will be the roots of the equation $x^2=y$.
Example 1
Solve the equation $x(x-1)(x-2)(x-3)=24$:
Let's expand the parentheses in the polynomial:
$(x^2-3x)(x^2-3x+2)=24$
In this form, it becomes obvious that we can choose the expression $y=x^2-3x$ as a new variable; let’s substitute it:
$y\cdot (y+2)=24$
Now let's solve two quadratic equations $x^2-3x=-4$ and $x^2-3x=-6$.
The roots of the first equation are $x_1(1,2)=4;-1$, the second has no solutions.
Solving reciprocal equations of degree 4
These equations of the form $ax^4+bx^3+cx^2 +bx+ a=0$ repeat with their coefficients for lower-order terms the coefficients for polynomials with higher degrees. To solve such an equation, first divide it by $x^2$:
$ax^4+bx^3+cx^2 +bx+ a=0|:x^2$
$ax^2+bx+c+\frac(b)(x) + \frac(a)(x^2)=0$
$a(x^2+\frac(1)(x^2))+b(x+\frac(1)(x)) + c=0$
Then replace $(x+\frac(1)(x))$ with a new variable, then $(x^2+\frac(1)(x^2))=y^2-2$, after substitution we get the following square the equation:
$a(y^2-2)+by+c=0$
After this, we look for the roots of the equations $x+\frac(1)(x)=y_1$ and $x+\frac(1)(x)=y_2$.
A similar method is used to solve reciprocal equations of the form $ax^4+bx^3+cx^2 +kbx+ k^2a=0$.
Example 2
Solve the equation:
$3x^4-2x^3-9x^2-4x+12=0$
This equation is a reciprocal equation of the form $ax^4+bx^3+cx^2 +kbx+ k^2a=0$. Therefore, we divide the entire equation by $x^2$:
$3x^2-2x-9 \cdot \frac(2 \cdot 2)(x)+3 \cdot (\frac(2)(x))^2=0$
$3(x^2+\frac(4)(x^2))-2(x+\frac(2)(x)-9=0$
Let's replace the expression $x+\frac(2)(x)$: $3(y^2-4)-2y-9=0$
Let's calculate the roots of this equation, they are equal to $y_1=3$ and $y_2=-\frac(7)(3)$.
Accordingly, now it is necessary to solve two equations $x+\frac(2)(x)=3$ and $x+\frac(2)(x)=-\frac(7)(3)$. The solution to the first equation is $x_1=1, x_2=2$, the second equation has no roots.
Therefore, the roots of the original equation are $x_1=1, x_2=2$.
Equations of the form $ax^4+b=0$
The roots of an equation of this type are found using abbreviated multiplication formulas.
For equations of the fourth degree, all those general schemes for solving equations of higher degrees that we discussed in the previous material are applicable. However, there are a number of nuances in solving binomial, biquadratic and reciprocal equations, which we would like to dwell on in more detail.
Also in the article we will analyze the artificial method of factoring a polynomial, solving in radicals and the Ferrari method, which is used to reduce the solution of a fourth-degree equation to a cubic equation.
Solution of a binomial equation of the fourth degree
This is the simplest type of fourth degree equation. The equation is written as A x 4 + B = 0.
Definition 1
To solve this type of equations, abbreviated multiplication formulas are used:
A x 4 + B = 0 x 4 + B A = 0 x 4 + 2 B A x 2 + B A - 2 B A x 2 = 0 x 2 + B A 2 - 2 B A x 2 = 0 x 2 - 2 B A 4 x + B A x 2 + 2 B A 4 x + B A = 0
All that remains is to find the roots of square trinomials.
Example 1
Solve the fourth degree equation 4 x 4 + 1 = 0.
Solution
First, let's factorize the polynomial 4 x 4 + 1:
4 x 4 + 1 = 4 x 4 + 4 x 2 + 1 = (2 x 2 + 1) 2 - 4 x 2 = 2 x 2 - 2 x + 1 (2 x 2 + 2 x + 1)
Now let's find the roots of square trinomials.
2 x 2 - 2 x + 1 = 0 D = (- 2) 2 - 4 2 1 = - 4 x 1 = 2 + D 2 2 = 1 2 + i x 2 = 2 - D 2 2 = 1 2 - i
2 x 2 + 2 x + 1 = 0 D = 2 2 - 4 2 1 = - 4 x 3 = - 2 + D 2 2 = - 1 2 + i x 4 = - 2 - D 2 2 = - 1 2 - i
We got four complex roots.
Answer: x = 1 2 ± i and x = - 1 2 ± i .
Solution of a recurrent equation of the fourth degree
Definition 2The fourth order reciprocal equations are A x 4 + B x 3 + C x 2 + B x + A = 0
x = 0 is not the root of this equation: A · 0 4 + B · 0 3 + C · 0 2 + B · 0 + A = A ≠ 0. Therefore, we can safely divide both sides of this equation by x 2:
A x 4 + B x 3 + C x 2 + B x + A = 0 A x 2 + B x + C + B x + A x 2 = 0 A x 2 + A x 2 + B x + B x + C = 0 A x 2 + 1 x 2 + B x + 1 x + C = 0
Let's change the variables x + 1 x = y ⇒ x + 1 x 2 = y 2 ⇒ x 2 + 1 x 2 = y 2 - 2:
A x 2 + 1 x 2 + B x + 1 x + C = 0 A (y 2 - 2) + B y + C = 0 A y 2 + B y + C - 2 A = 0
So we reduce the reciprocal equation of the fourth degree to a quadratic equation.
Example 2
Find all complex roots of the equation 2 x 4 + 2 3 + 2 x 3 + 4 + 6 x 2 + 2 3 + 2 x + 2 = 0.
Solution
The symmetry of the coefficients tells us that we are dealing with a reciprocal equation of the fourth degree. Let's divide both sides by x 2:
2 x 2 + 2 3 + 2 x + 4 + 6 + 2 3 + 2 x + 2 x 2 = 0
Let's group:
2 x 2 + 2 x 2 + 2 3 + 2 x + 2 3 + 2 x + 4 + 6 + = 0 2 x 2 + 1 x 2 + 2 3 + 2 x + 1 x + 4 + 6 = 0
Let's replace the variable x + 1 x = y ⇒ x + 1 x 2 = y 2 ⇒ x 2 + 1 x 2 = y 2 - 2
2 x 2 + 1 x 2 + 2 3 + 2 x + 1 x + 4 + 6 = 0 2 y 2 - 2 + 2 3 + 2 y + 4 + 6 = 0 2 y 2 + 2 3 + 2 y + 6 = 0
Let's solve the resulting quadratic equation:
D = 2 3 + 2 2 - 4 2 6 = 12 + 4 6 + 2 - 8 6 = = 12 - 4 6 + 2 = 2 3 - 2 2 y 1 = - 2 3 - 2 + D 2 2 = - 2 3 - 2 + 2 3 - 2 4 = - 2 2 y 2 = - 2 3 - 2 - D 2 2 = - 2 3 - 2 - 2 3 + 2 4 = - 3
Let's go back to the replacement: x + 1 x = - 2 2 , x + 1 x = - 3 .
Let's solve the first equation:
x + 1 x = - 2 2 ⇒ 2 x 2 + 2 x + 2 = 0 D = 2 2 - 4 2 2 = - 14 x 1 = - 2 - D 2 2 = - 2 4 + i 14 4 x 2 = - 2 - D 2 2 = - 2 4 - i 14 4
Let's solve the second equation:
x + 1 x = - 3 ⇒ x 2 + 3 x + 1 = 0 D = 3 2 - 4 1 1 = - 1 x 3 = - 3 + D 2 = - 3 2 + i 1 2 x 4 = - 3 - D 2 = - 3 2 - i 1 2
Answer: x = - 2 4 ± i · 14 4 and x = - 3 2 ± i · 1 2 .
Solving a biquadratic equation
Biquadratic equations of the fourth degree have the form A x 4 + B x 2 + C = 0. We can reduce such an equation to a quadratic equation A y 2 + B y + C = 0 by substituting y = x 2 . This is a standard technique.
Example 3
Solve the biquadratic equation 2 x 4 + 5 x 2 - 3 = 0.
Solution
Let's replace the variable y = x 2, which will allow us to reduce the original equation to a quadratic one:
2 y 2 + 5 y - 3 = 0 D = 5 2 - 4 2 (- 3) = 49 y 1 = - 5 + D 2 2 = - 5 + 7 4 = 1 2 y 2 = - 5 - D 2 2 = - 5 - 7 4 = - 3
Therefore, x 2 = 1 2 or x 2 = - 3.
The first equality allows us to obtain the root x = ± 1 2 . The second equality has no real roots, but it has complex conjugate roots x = ± i · 3.
Answer: x = ± 1 2 and x = ± i · 3 .
Example 4
Find all complex roots of the biquadratic equation 16 x 4 + 145 x 2 + 9 = 0.
Solution
We use the replacement method y = x 2 in order to reduce the original biquadratic equation to a quadratic one:
16 y 2 + 145 y + 9 = 0 D = 145 2 - 4 16 9 = 20449 y 1 = - 145 + D 2 16 = - 145 + 143 32 = - 1 16 y 2 = - 145 - D 2 · 16 = - 145 - 143 32 = - 9
Therefore, due to the change of variable, x 2 = - 1 16 or x 2 = - 9.
Answer: x 1, 2 = ± 1 4 · i, x 3, 4 = ± 3 · i.
Solving quartic equations with rational roots
The algorithm for finding rational roots of a fourth-degree equation is given in the material “Solving equations of higher degrees.”
Solving fourth degree equations using the Ferrari method
Quaternary equations of the form x 4 + A x 3 + B x 2 + C x + D = 0 can generally be solved using the Ferrari method. To do this, you need to find y 0. This is any of the roots of the cubic equation y 3 - B y 2 + A C - 4 D y - A 2 D + 4 B D - C 2 = 0. After this, it is necessary to solve two quadratic equations x 2 + A 2 x + y 0 2 + A 2 4 - B + y 0 x 2 + A 2 y 0 - C x + y 0 2 4 - D = 0, whose radical expression is is a perfect square.
The roots obtained during the calculations will be the roots of the original fourth-degree equation.
Example 5
Find the roots of the equation x 4 + 3 x 3 + 3 x 2 - x - 6 = 0.
Solution
We have A = 3, B = 3, C = - 1, D = - 6. Let us apply the Ferrari method to solve this equation.
Let's compose and solve the cubic equation:
y 3 - B y 2 + A C - 4 D y - A 2 D + 4 B D - C 2 = 0 y 3 - 3 y 2 + 21 y - 19 = 0
One of the roots of the cubic equation will be y 0 = 1, since 1 3 - 3 · 1 2 + 21 · 1 - 19 = 0.
Let's write two quadratic equations:
x 2 + A 2 x + y 0 2 ± A 2 4 - B + y 0 x 2 + A 2 y 0 - C x + y 0 2 4 - D = 0 x 2 + 3 2 x + 1 2 ± 1 4 x 2 + 5 2 x + 25 4 = 0 x 2 + 3 2 x + 1 2 ± 1 2 x + 5 2 2 = 0
x 2 + 3 2 x + 1 2 + 1 2 x + 5 2 = 0 or x 2 + 3 2 x + 1 2 - 1 2 x - 5 2 = 0
x 2 + 2 x + 3 = 0 or x 2 + x - 2 = 0
The roots of the first equation will be x = - 1 ± i · 2, the roots of the second will be x = 1 and x = - 2.
Answer: x 1, 2 = - 1 ± i 2, x 3 = 1, x 4 = - 2.
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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 of the division of 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.
The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Man used equations in ancient times, and since then their use has only increased. Solutions of this type of equations can be carried out according to the general scheme for solving equations of higher degrees.
These types of equations have solutions in radicals thanks to the Ferrari method, which allows one to reduce the solutions to a cubic equation. However, in most cases, by factoring a polynomial, you can quickly find a solution to the equation.
Suppose we are given a binomial equation of the fourth degree:
Let's factorize the polynomial:
We determine the roots of the first quadratic trinomial:
We determine the roots of the second trinomial:
As a result, the original equation has four complex roots:
Where can I solve 4th degree equations online?