Solution formula of cubic equation of one variable!
What is the root formula of a cubic equation of one variable? ? Just the formula?
Solution to the root formula of a cubic equation of one variable
The root formula of a cubic equation of one variable cannot be obtained through ordinary deductive thinking, but the standard cubic equation can be simplified into a special form x^ by a method similar to the root formula of solving a quadratic equation. 3 px q=0. This method can help us solve the roots of a cubic equation of one variable more conveniently.
The solution to the solution formula of a cubic equation of one variable can only be obtained through inductive thinking. We can summarize based on the forms of root formulas of linear equations of one variable, quadratic equations of one variable and special higher-order equations, and thus obtain the form of the root formulas of cubic equations of one variable. The form obtained by induction is x = A^(1/3) B^(1/3), which is the sum of two open cubes. Then, we need to find the relationship between A and B and p and q. The specific method is as follows:
(1) Cube both sides of x=A^(1/3) B^(1/3) at the same time to get
(2)x^3=(A B) 3(AB)^(1/3)(A^(1/3) B^(1/3))
(3) Since x=A^(1/3) B^(1/3), (2) can be reduced to
x^3=(A B) 3(AB)^(1/3)x, you can get
by moving the terms(4)x^3-3(AB)^(1/3)x-(A B)=0, compared with the cubic equation of one variable and the special type x^3 px q=0, it can be seen that
(5)-3(AB)^(1/3)=p,-(A B)=q, simplify to
(6)A B=-q,AB=-(p/3)^3
(7) In this way, the root formula of the cubic equation of one variable is actually transformed into the root formula of the quadratic equation, because A and B can be regarded as the two roots of the quadratic equation, and (6) is about The Vedic theorem of two roots of a quadratic equation of the form ay^2 by c=0, that is,
(8)y1 y2=-(b/a),y1*y2=c/a
(9) Comparing (6) and (8), we can set A=y1, B=y2, q=b/a,-(p/3)^3=c/a
(10) Since the root formula of a quadratic equation of type ay^2 by c=0 is
y1=-(b (b^2-4ac)^(1/2))/(2a)
y2=-(b-(b^2-4ac)^(1/2))/(2a)
can be changed to
(11)y1=-(b/2a)-((b/2a)^2-(c/a))^(1/2)
y2=-(b/2a) ((b/2a)^2-(c/a))^(1/2)
Substitute A=y1, B=y2, q=b/a,-(p/3)^3=c/a in (9) into (11) to get
(12)A=-(q/2)-((q/2)^2 (p/3)^3)^(1/2)
B=-(q/2) ((q/2)^2 (p/3)^3)^(1/2)
(13) Substitute A and B into x=A^(1/3) B^(1/3) to get
(14)x=(-(q/2)-((q/2)^2 (p/3)^3)^(1/2))^(1/3) (-(q/ 2) ((q/2)^2 (p/3)^3)^(1/2))^(1/3)
Equation (14) is only a real root solution of a three-dimensional equation of one variable. According to Vedic theorem, a cubic equation of one variable should have three roots. However, according to Vedic theorem, as long as one of the roots of a cubic equation of one variable is found, the other two roots will be easy. Out
Root formula of cubic equation of one variable
Solution to the root formula of a cubic equation of one variable
The root formula of a cubic equation of one variable cannot be obtained through ordinary deductive thinking, but the standard cubic equation can be simplified into a special form x^ by a method similar to the root formula of solving a quadratic equation. 3 px q=0. This method can help us solve the roots of a cubic equation of one variable more conveniently.
The solution to the solution formula of a cubic equation of one variable can only be obtained through inductive thinking. We can summarize based on the forms of root formulas of linear equations of one variable, quadratic equations of one variable and special higher-order equations, and thus obtain the form of the root formulas of cubic equations of one variable. The form obtained by induction is x = A^(1/3) B^(1/3), which is the sum of two open cubes. Then, we need to find the relationship between A and B and p and q. The specific method is as follows:
(1) Cube both sides of x=A^(1/3) B^(1/3) at the same time to get
(2)x^3=(A B) 3(AB)^(1/3)(A^(1/3) B^(1/3))
(3) Since x=A^(1/3) B^(1/3), (2) can be reduced to
x^3=(A B) 3(AB)^(1/3)x, you can get
by moving the terms(4)x^3-3(AB)^(1/3)x-(A B)=0, compared with the cubic equation of one variable and the special type x^3 px q=0, it can be seen that
(5)-3(AB)^(1/3)=p,-(A B)=q, simplify to
(6)A B=-q,AB=-(p/3)^3
(7) In this way, the root formula of the cubic equation of one variable is actually transformed into the root formula of the quadratic equation, because A and B can be regarded as the two roots of the quadratic equation, and (6) is about The Vedic theorem of two roots of a quadratic equation of the form ay^2 by c=0, that is,
(8)y1 y2=-(b/a),y1*y2=c/a
(9) Comparing (6) and (8), we can set A=y1, B=y2, q=b/a,-(p/3)^3=c/a
(10) Since the root formula of a quadratic equation of type ay^2 by c=0 is
y1=-(b (b^2-4ac)^(1/2))/(2a)
y2=-(b-(b^2-4ac)^(1/2))/(2a)
can be changed to
(11)y1=-(b/2a)-((b/2a)^2-(c/a))^(1/2)
y2=-(b/2a) ((b/2a)^2-(c/a))^(1/2)
Substitute A=y1, B=y2, q=b/a,-(p/3)^3=c/a in (9) into (11) to get
(12)A=-(q/2)-((q/2)^2 (p/3)^3)^(1/2)
B=-(q/2) ((q/2)^2 (p/3)^3)^(1/2)
(13) Substitute A and B into x=A^(1/3) B^(1/3) to get
(14)x=(-(q/2)-((q/2)^2 (p/3)^3)^(1/2))^(1/3) (-(q/ 2) ((q/2)^2 (p/3)^3)^(1/2))^(1/3)
Equation (14) is only a real root solution of a three-dimensional equation of one variable. According to Vedic theorem, a cubic equation of one variable should have three roots. However, according to Vedic theorem, as long as one of the roots of a cubic equation of one variable is found, the other two roots will be easy. Out
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