Using Cardano's method to find an algebraic equation whose root is $sqrt{2} +sqrt[3]{3}$ ...
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Using Cardano's method to find an algebraic equation whose root is $sqrt{2} +sqrt[3]{3}$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What equation intersects only once with $f(x)=sqrt{1-(x-2)^2}$Find all $qinmathbb{Q}$ such that $ qx^2+(q+1)x+q=1 $ has integer solutions.Solve the simultaneous equations $x + frac{3x-y}{x^2+y^2} = 3 $, $ y – frac{x+3y}{x^2+y^2} = 0$Finding a general solution for {sin(x)=sqrt(3)/2 , cos(x)=-0.5} using an algebraic methodFinding transcendental roots to an algebraic equationFind the number of solutions of the equationMinimum degree rational equation with root $a+sqrt{b}+sqrt{c}+sqrt{d}$.About Cardano's formula and the cubic equationFind an equation of the parabola with zeros $0$ and $6$ and a minimum value of $-9$Find the rational solution of the equation
$begingroup$
$sqrt{2} +sqrt[3]{3}$ solution of an algebraic equation $ a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+cdots+a_{0}=0 $
How to find this equation?
I tried Cardano's method, noting that
$$sqrt{2} +sqrt[3]{3} = sqrt[3]{sqrt{8}} +sqrt[3]{3}$$
It means
$$begin{align}
-frac{q}{2}+sqrt{ frac{q^{2}}{4}+frac{p^{3}}{27}}&=3 \[4pt]
-frac{q}{2}-sqrt{ frac{q^{2}}{4}+frac{p^{3}}{27}}&=sqrt{8}
end{align}$$
but this system doesn't have natural solutions.
Maybe it has rational $p$ and $q$?
algebra-precalculus
edited Mar 23 at 13:08
Blue
49.7k870158
asked Mar 23 at 12:19
demspdemsp
103
$endgroup$
add a comment |
$begingroup$
$sqrt{2} +sqrt[3]{3}$ solution of an algebraic equation $ a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+cdots+a_{0}=0 $
How to find this equation?
I tried Cardano's method, noting that
$$sqrt{2} +sqrt[3]{3} = sqrt[3]{sqrt{8}} +sqrt[3]{3}$$
It means
$$begin{align}
-frac{q}{2}+sqrt{ frac{q^{2}}{4}+frac{p^{3}}{27}}&=3 \[4pt]
-frac{q}{2}-sqrt{ frac{q^{2}}{4}+frac{p^{3}}{27}}&=sqrt{8}
end{align}$$
but this system doesn't have natural solutions.
Maybe it has rational $p$ and $q$?
algebra-precalculus
edited Mar 23 at 13:08
Blue
49.7k870158
asked Mar 23 at 12:19
demspdemsp
103
$endgroup$
$begingroup$
Use the method given in Example 4 on p. 3 of this old class handout of mine.
$endgroup$
– Dave L. Renfro
Mar 23 at 12:34
1
$begingroup$
By the way, the context of your question is not clear. Do you simply want to know a method for finding such an algebraic equation (with integer coefficients, something you neglected to say by the way), or do you want to find a method that BEGINS with Cardano's formula? I don't think beginning with Cardano's formula will help much because one can show (with some work) that there is no cubic polynomial with integer coefficients that has $sqrt{2} + sqrt[3]{3}$ as a root.
$endgroup$
– Dave L. Renfro
Mar 23 at 12:44
1
$begingroup$
The minimal polynomial for $sqrt{2}+sqrt[3]{3}$ is has degree 6, so Cardano's method of solving cubic equations won't be any help.
$endgroup$
– Blue
Mar 23 at 13:11
$begingroup$
Thank you very much
$endgroup$
– demsp
Mar 23 at 20:26
add a comment |
$begingroup$
$sqrt{2} +sqrt[3]{3}$ solution of an algebraic equation $ a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+cdots+a_{0}=0 $
How to find this equation?
I tried Cardano's method, noting that
$$sqrt{2} +sqrt[3]{3} = sqrt[3]{sqrt{8}} +sqrt[3]{3}$$
It means
$$begin{align}
-frac{q}{2}+sqrt{ frac{q^{2}}{4}+frac{p^{3}}{27}}&=3 \[4pt]
-frac{q}{2}-sqrt{ frac{q^{2}}{4}+frac{p^{3}}{27}}&=sqrt{8}
end{align}$$
but this system doesn't have natural solutions.
Maybe it has rational $p$ and $q$?
algebra-precalculus
edited Mar 23 at 13:08
Blue
49.7k870158
asked Mar 23 at 12:19
demspdemsp
103
$endgroup$
$sqrt{2} +sqrt[3]{3}$ solution of an algebraic equation $ a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+cdots+a_{0}=0 $
How to find this equation?
I tried Cardano's method, noting that
$$sqrt{2} +sqrt[3]{3} = sqrt[3]{sqrt{8}} +sqrt[3]{3}$$
It means
$$begin{align}
-frac{q}{2}+sqrt{ frac{q^{2}}{4}+frac{p^{3}}{27}}&=3 \[4pt]
-frac{q}{2}-sqrt{ frac{q^{2}}{4}+frac{p^{3}}{27}}&=sqrt{8}
end{align}$$
but this system doesn't have natural solutions.
Maybe it has rational $p$ and $q$?
algebra-precalculus
algebra-precalculus
edited Mar 23 at 13:08
Blue
49.7k870158
asked Mar 23 at 12:19
demspdemsp
103
edited Mar 23 at 13:08
Blue
49.7k870158
asked Mar 23 at 12:19
demspdemsp
103
edited Mar 23 at 13:08
Blue
49.7k870158
edited Mar 23 at 13:08
Blue
49.7k870158
edited Mar 23 at 13:08
Blue
49.7k870158
49.7k870158
asked Mar 23 at 12:19
demspdemsp
103
asked Mar 23 at 12:19
demspdemsp
103
asked Mar 23 at 12:19
demspdemsp
103
103
$begingroup$
Use the method given in Example 4 on p. 3 of this old class handout of mine.
$endgroup$
– Dave L. Renfro
Mar 23 at 12:34
1
$begingroup$
By the way, the context of your question is not clear. Do you simply want to know a method for finding such an algebraic equation (with integer coefficients, something you neglected to say by the way), or do you want to find a method that BEGINS with Cardano's formula? I don't think beginning with Cardano's formula will help much because one can show (with some work) that there is no cubic polynomial with integer coefficients that has $sqrt{2} + sqrt[3]{3}$ as a root.
$endgroup$
– Dave L. Renfro
Mar 23 at 12:44
1
$begingroup$
The minimal polynomial for $sqrt{2}+sqrt[3]{3}$ is has degree 6, so Cardano's method of solving cubic equations won't be any help.
$endgroup$
– Blue
Mar 23 at 13:11
$begingroup$
Thank you very much
$endgroup$
– demsp
Mar 23 at 20:26
add a comment |
$begingroup$
Use the method given in Example 4 on p. 3 of this old class handout of mine.
$endgroup$
– Dave L. Renfro
Mar 23 at 12:34
1
$begingroup$
By the way, the context of your question is not clear. Do you simply want to know a method for finding such an algebraic equation (with integer coefficients, something you neglected to say by the way), or do you want to find a method that BEGINS with Cardano's formula? I don't think beginning with Cardano's formula will help much because one can show (with some work) that there is no cubic polynomial with integer coefficients that has $sqrt{2} + sqrt[3]{3}$ as a root.
$endgroup$
– Dave L. Renfro
Mar 23 at 12:44
1
$begingroup$
The minimal polynomial for $sqrt{2}+sqrt[3]{3}$ is has degree 6, so Cardano's method of solving cubic equations won't be any help.
$endgroup$
– Blue
Mar 23 at 13:11
$begingroup$
Thank you very much
$endgroup$
– demsp
Mar 23 at 20:26
$begingroup$
Use the method given in Example 4 on p. 3 of this old class handout of mine.
$endgroup$
– Dave L. Renfro
Mar 23 at 12:34
$begingroup$
Use the method given in Example 4 on p. 3 of this old class handout of mine.
$endgroup$
– Dave L. Renfro
Mar 23 at 12:34
1
1
$begingroup$
By the way, the context of your question is not clear. Do you simply want to know a method for finding such an algebraic equation (with integer coefficients, something you neglected to say by the way), or do you want to find a method that BEGINS with Cardano's formula? I don't think beginning with Cardano's formula will help much because one can show (with some work) that there is no cubic polynomial with integer coefficients that has $sqrt{2} + sqrt[3]{3}$ as a root.
$endgroup$
– Dave L. Renfro
Mar 23 at 12:44
$begingroup$
By the way, the context of your question is not clear. Do you simply want to know a method for finding such an algebraic equation (with integer coefficients, something you neglected to say by the way), or do you want to find a method that BEGINS with Cardano's formula? I don't think beginning with Cardano's formula will help much because one can show (with some work) that there is no cubic polynomial with integer coefficients that has $sqrt{2} + sqrt[3]{3}$ as a root.
$endgroup$
– Dave L. Renfro
Mar 23 at 12:44
1
1
$begingroup$
The minimal polynomial for $sqrt{2}+sqrt[3]{3}$ is has degree 6, so Cardano's method of solving cubic equations won't be any help.
$endgroup$
– Blue
Mar 23 at 13:11
$begingroup$
The minimal polynomial for $sqrt{2}+sqrt[3]{3}$ is has degree 6, so Cardano's method of solving cubic equations won't be any help.
$endgroup$
– Blue
Mar 23 at 13:11
$begingroup$
Thank you very much
$endgroup$
– demsp
Mar 23 at 20:26
$begingroup$
Thank you very much
$endgroup$
– demsp
Mar 23 at 20:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $x=sqrt 2 + sqrt[3]{3}$. Then,
$$
x-sqrt 2 = sqrt[3]{3}
$$
Cubing both sides and rearranging we have,
$$
x^3+6x-3=sqrt 2 (3x^2+2)
$$
Squaring both sides and rearranging we have,
$$
x^6-6x^4-6x^3+12x^2-36x+1=0
$$
answered Mar 23 at 13:34
Awe Kumar JhaAwe Kumar Jha
633113
$endgroup$
add a comment |
$begingroup$
In the same spirit as Awe Kumar Jha, let $x=sqrt 2 + sqrt[3]{3}$, that is to say
$$x-sqrt 2 =sqrt[3]{3}implies (x-sqrt 2) ^3=3$$ Expand and group terms to get
$$x^3-3 sqrt{2} x^2+6 x-(2 sqrt{2}+3)=0$$ Use Cardano method; it will show that this is the only real root of the equation.
answered Mar 23 at 15:25
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
$begingroup$
Thank you :))))
$endgroup$
– demsp
Mar 23 at 20:30
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $x=sqrt 2 + sqrt[3]{3}$. Then,
$$
x-sqrt 2 = sqrt[3]{3}
$$
Cubing both sides and rearranging we have,
$$
x^3+6x-3=sqrt 2 (3x^2+2)
$$
Squaring both sides and rearranging we have,
$$
x^6-6x^4-6x^3+12x^2-36x+1=0
$$
answered Mar 23 at 13:34
Awe Kumar JhaAwe Kumar Jha
633113
$endgroup$
add a comment |
$begingroup$
Let $x=sqrt 2 + sqrt[3]{3}$. Then,
$$
x-sqrt 2 = sqrt[3]{3}
$$
Cubing both sides and rearranging we have,
$$
x^3+6x-3=sqrt 2 (3x^2+2)
$$
Squaring both sides and rearranging we have,
$$
x^6-6x^4-6x^3+12x^2-36x+1=0
$$
answered Mar 23 at 13:34
Awe Kumar JhaAwe Kumar Jha
633113
$endgroup$
add a comment |
$begingroup$
Let $x=sqrt 2 + sqrt[3]{3}$. Then,
$$
x-sqrt 2 = sqrt[3]{3}
$$
Cubing both sides and rearranging we have,
$$
x^3+6x-3=sqrt 2 (3x^2+2)
$$
Squaring both sides and rearranging we have,
$$
x^6-6x^4-6x^3+12x^2-36x+1=0
$$
answered Mar 23 at 13:34
Awe Kumar JhaAwe Kumar Jha
633113
$endgroup$
Let $x=sqrt 2 + sqrt[3]{3}$. Then,
$$
x-sqrt 2 = sqrt[3]{3}
$$
Cubing both sides and rearranging we have,
$$
x^3+6x-3=sqrt 2 (3x^2+2)
$$
Squaring both sides and rearranging we have,
$$
x^6-6x^4-6x^3+12x^2-36x+1=0
$$
answered Mar 23 at 13:34
Awe Kumar JhaAwe Kumar Jha
633113
answered Mar 23 at 13:34
Awe Kumar JhaAwe Kumar Jha
633113
answered Mar 23 at 13:34
Awe Kumar JhaAwe Kumar Jha
633113
answered Mar 23 at 13:34
Awe Kumar JhaAwe Kumar Jha
633113
633113
add a comment |
add a comment |
$begingroup$
In the same spirit as Awe Kumar Jha, let $x=sqrt 2 + sqrt[3]{3}$, that is to say
$$x-sqrt 2 =sqrt[3]{3}implies (x-sqrt 2) ^3=3$$ Expand and group terms to get
$$x^3-3 sqrt{2} x^2+6 x-(2 sqrt{2}+3)=0$$ Use Cardano method; it will show that this is the only real root of the equation.
answered Mar 23 at 15:25
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
$begingroup$
Thank you :))))
$endgroup$
– demsp
Mar 23 at 20:30
add a comment |
$begingroup$
In the same spirit as Awe Kumar Jha, let $x=sqrt 2 + sqrt[3]{3}$, that is to say
$$x-sqrt 2 =sqrt[3]{3}implies (x-sqrt 2) ^3=3$$ Expand and group terms to get
$$x^3-3 sqrt{2} x^2+6 x-(2 sqrt{2}+3)=0$$ Use Cardano method; it will show that this is the only real root of the equation.
answered Mar 23 at 15:25
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
$begingroup$
Thank you :))))
$endgroup$
– demsp
Mar 23 at 20:30
add a comment |
$begingroup$
In the same spirit as Awe Kumar Jha, let $x=sqrt 2 + sqrt[3]{3}$, that is to say
$$x-sqrt 2 =sqrt[3]{3}implies (x-sqrt 2) ^3=3$$ Expand and group terms to get
$$x^3-3 sqrt{2} x^2+6 x-(2 sqrt{2}+3)=0$$ Use Cardano method; it will show that this is the only real root of the equation.
answered Mar 23 at 15:25
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
In the same spirit as Awe Kumar Jha, let $x=sqrt 2 + sqrt[3]{3}$, that is to say
$$x-sqrt 2 =sqrt[3]{3}implies (x-sqrt 2) ^3=3$$ Expand and group terms to get
$$x^3-3 sqrt{2} x^2+6 x-(2 sqrt{2}+3)=0$$ Use Cardano method; it will show that this is the only real root of the equation.
answered Mar 23 at 15:25
Claude LeiboviciClaude Leibovici
126k1158135
answered Mar 23 at 15:25
Claude LeiboviciClaude Leibovici
126k1158135
answered Mar 23 at 15:25
Claude LeiboviciClaude Leibovici
126k1158135
answered Mar 23 at 15:25
Claude LeiboviciClaude Leibovici
126k1158135
126k1158135
$begingroup$
Thank you :))))
$endgroup$
– demsp
Mar 23 at 20:30
add a comment |
$begingroup$
Thank you :))))
$endgroup$
– demsp
Mar 23 at 20:30
$begingroup$
Thank you :))))
$endgroup$
– demsp
Mar 23 at 20:30
$begingroup$
Thank you :))))
$endgroup$
– demsp
Mar 23 at 20:30
add a comment |
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$begingroup$
Use the method given in Example 4 on p. 3 of this old class handout of mine.
$endgroup$
– Dave L. Renfro
Mar 23 at 12:34
1
$begingroup$
By the way, the context of your question is not clear. Do you simply want to know a method for finding such an algebraic equation (with integer coefficients, something you neglected to say by the way), or do you want to find a method that BEGINS with Cardano's formula? I don't think beginning with Cardano's formula will help much because one can show (with some work) that there is no cubic polynomial with integer coefficients that has $sqrt{2} + sqrt[3]{3}$ as a root.
$endgroup$
– Dave L. Renfro
Mar 23 at 12:44
1
$begingroup$
The minimal polynomial for $sqrt{2}+sqrt[3]{3}$ is has degree 6, so Cardano's method of solving cubic equations won't be any help.
$endgroup$
– Blue
Mar 23 at 13:11
$begingroup$
Thank you very much
$endgroup$
– demsp
Mar 23 at 20:26