求一元三次方程的根.
求下列一元三次方程的根.
(1) $x^3-x^2-x-2=0$;
(2) $x^3-3x^2+2x^1-2=0$;
(3) $x^3+1=0$;
(4) $x^3-15x-126=0$
(5) $x^3+x^2+x+1=0$
(6) $x^3-6x+2=0$
Answer
问题及解答
1
(1)
>> solve(x^3-x^2-x-2==0)
It is a univariate cubic equation.
x^3-1x^2-1x^1-2 == 0
Answer:
Let y=x-1/3
We get equation of y: y^3+py+q==0
where
p=b-a^2/3, q=c-ab/3+2a^3/27,
and a,b,c,d are coefficients of the equation ax^3+bx^2+cx+d==0
By computation,
p = -4|3
q = -65|27
Now the equation is:
y^3-4|3y-65|27 == 0
Delta = (q/2)^2+(p/3)^3 = 49|36
sqrt(Delta) = sqrt((q/2)^2+(p/3)^3) = 7|6
q/2 = -65|54
-q/2 + sqrt(Delta) = 64|27
-q/2 - sqrt(Delta) = 1|27
u = 4|3
v = 1|3
solution>
x1 = 2
x2 = 4|3w+1|3w^2+1|3 = -1|2+1|2*sqrt(3)*i
x3 = 4|3w^2+1|3w+1|3 = -1|2-1|2*sqrt(3)*i
Where w is
-1/2+i*sqrt(3)/2
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2
(2)
>> solve(x^3-3x^2+2x-2==0)
It is a univariate cubic equation.
x^3-3x^2+2x^1-2 == 0
Answer:
Let y=x-3/3
We get equation of y: y^3+py+q==0
where
p=b-a^2/3, q=c-ab/3+2a^3/27,
and a,b,c,d are coefficients of the equation ax^3+bx^2+cx+d==0
By computation,
p = -1
q = -2
Now the equation is:
y^3-1y-2 == 0
Delta = (q/2)^2+(p/3)^3 = 26|27
sqrt(Delta) = sqrt((q/2)^2+(p/3)^3) = sqrt(26)|3*sqrt(3)
q/2 = -1
-q/2 + sqrt(Delta) = (sqrt(26)-(-3*sqrt(3)))|3*sqrt(3)
-q/2 - sqrt(Delta) = -(sqrt(26)-3*sqrt(3))|3*sqrt(3)
u = sqrtn(1+sqrt(26)|3*sqrt(3), 3)
v = sqrtn(1-sqrt(26)|3*sqrt(3), 3)
solution>
x1 = sqrtn(1+sqrt(26)|3*sqrt(3), 3)+sqrtn(1-sqrt(26)|3*sqrt(3), 3)+1
x2 = sqrtn(1+sqrt(26)|3*sqrt(3), 3)w+sqrtn(1-sqrt(26)|3*sqrt(3), 3)w^2+1
x3 = sqrtn(1+sqrt(26)|3*sqrt(3), 3)w^2+sqrtn(1-sqrt(26)|3*sqrt(3), 3)w+1
Where w is
-1/2+i*sqrt(3)/2
------------------------
3
>> solve(x^3+1==0)
It is a univariate cubic equation.
x^3+1 == 0
Answer:
We get equation of y: y^3+py+q==0
where
p=b-a^2/3, q=c-ab/3+2a^3/27,
and a,b,c,d are coefficients of the equation ax^3+bx^2+cx+d==0
By computation,
p = 0
q = 1
Now the equation is:
y^3+0y+1 == 0
Delta = (q/2)^2+(p/3)^3 = 1|4
sqrt(Delta) = sqrt((q/2)^2+(p/3)^3) = 1|2
q/2 = 1|2
-q/2 + sqrt(Delta) = 0
-q/2 - sqrt(Delta) = -1
u = 0
v = -1
solution>
x1 = -1
x2 = 0w-1w^2-0 = 1|2+1|2*sqrt(3)*i
x3 = 0w^2-1w-0 = 1|2-1|2*sqrt(3)*i
Where w is
-1/2+i*sqrt(3)/2
------------------------
4
(4)
>> solve(x^3-15x-126==0)
It is a univariate cubic equation.
x^3-15x^1-126 == 0
Answer:
p = -15
q = -126
Delta = (q/2)^2+(p/3)^3 = 3844
sqrt(Delta) = sqrt((q/2)^2+(p/3)^3) = 62
q/2 = -63
-q/2 + sqrt(Delta) = 125
-q/2 - sqrt(Delta) = 1
u = 5
v = 1
solution>
x1 = 6
x2 = 5w+1w^2-0 = -3+2*sqrt(3)*i
x3 = 5w^2+1w-0 = -3-2*sqrt(3)*i
Where w is
-1/2+i*sqrt(3)/2
------------------------
5
>> solve(x^3+x^2+x+1==0)
It is a univariate cubic equation.
x^3+x^2+x^1+1 == 0
Answer:
Let y=x+1/3
We get equation of y: y^3+py+q==0
where
p=b-a^2/3, q=c-ab/3+2a^3/27,
and a,b,c,d are coefficients of the equation ax^3+bx^2+cx+d==0
By computation,
p = 2|3
q = 20|27
Now the equation is:
y^3+2|3y+20|27 == 0
Delta = (q/2)^2+(p/3)^3 = 4|27
sqrt(Delta) = sqrt((q/2)^2+(p/3)^3) = 2|3*sqrt(3)
q/2 = 10|27
-q/2 + sqrt(Delta) = (54-10*3*sqrt(3))|27*3*sqrt(3)
-q/2 - sqrt(Delta) = -(54+10*3*sqrt(3))|27*3*sqrt(3)
u = sqrtn(-10|27+2|3*sqrt(3), 3)
v = sqrtn(-10|27-2|3*sqrt(3), 3)
solution>
x1 = sqrtn(-10|27+2|3*sqrt(3), 3)+sqrtn(-10|27-2|3*sqrt(3), 3)-1|3
x2 = sqrtn(-10|27+2|3*sqrt(3), 3)w+sqrtn(-10|27-2|3*sqrt(3), 3)w^2-1|3
x3 = sqrtn(-10|27+2|3*sqrt(3), 3)w^2+sqrtn(-10|27-2|3*sqrt(3), 3)w-1|3
Where w is
-1/2+i*sqrt(3)/2
------------------------
6
(6)
>> solve(x^3-6x+2==0)
It is a univariate cubic equation.
x^3-6x^1+2 == 0
Answer:
p = -6
q = 2
Delta = (q/2)^2+(p/3)^3 = -7
sqrt(Delta) = sqrt((q/2)^2+(p/3)^3) = sqrt(7)*i
q/2 = 1
-q/2 + sqrt(Delta) = (sqrt(7)*i-1)
-q/2 - sqrt(Delta) = -(sqrt(7)*i+1)
u = sqrtn(-1+sqrt(7)*i, 3)
v = sqrtn(-1-sqrt(7)*i, 3)
solution>
x1 = sqrtn(-1+sqrt(7)*i, 3)+sqrtn(-1-sqrt(7)*i, 3)-0
x2 = sqrtn(-1+sqrt(7)*i, 3)w+sqrtn(-1-sqrt(7)*i, 3)w^2-0
x3 = sqrtn(-1+sqrt(7)*i, 3)w^2+sqrtn(-1-sqrt(7)*i, 3)w-0
Where w is
-1/2+i*sqrt(3)/2
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