Why am i getting two real roots of the cubic $x^3+x+5=0$
I was going through the solution of depressed cubic equation $$x^3+x+5=0$$
By Cardano's Method we assume
$$x=u+v$$ Then
we have $$(u+v)^3-3uv(u+v)-(u^3+v^3)=0$$ Comparing with given cubic we get
$$u^3+v^3=-5 tag{1}$$
$$uv=frac{-1}{3}tag{2}$$ Solving $(1)$ and $(2)$ we get
$$u^3-frac{1}{27u^3}=-5$$ Letting $u^3=p$ we get
$$27p^2+135p-1=0$$ which gives two real distinct roots of $p$ which $implies$ two real distinct roots of $u$ and hence two real distinct roots of $v$
Hence two distinct real roots of $x$
what went wrong here?
algebra-precalculus polynomials cubic-equations
asked Nov 24 at 19:45
Umesh shankar
2,53621219
add a comment |
I was going through the solution of depressed cubic equation $$x^3+x+5=0$$
By Cardano's Method we assume
$$x=u+v$$ Then
we have $$(u+v)^3-3uv(u+v)-(u^3+v^3)=0$$ Comparing with given cubic we get
$$u^3+v^3=-5 tag{1}$$
$$uv=frac{-1}{3}tag{2}$$ Solving $(1)$ and $(2)$ we get
$$u^3-frac{1}{27u^3}=-5$$ Letting $u^3=p$ we get
$$27p^2+135p-1=0$$ which gives two real distinct roots of $p$ which $implies$ two real distinct roots of $u$ and hence two real distinct roots of $v$
Hence two distinct real roots of $x$
what went wrong here?
algebra-precalculus polynomials cubic-equations
asked Nov 24 at 19:45
Umesh shankar
2,53621219
1
Did you note that what you wrote occurs when you apply Cardano's method to any reduced cubic?
– José Carlos Santos
Nov 24 at 19:55
add a comment |
I was going through the solution of depressed cubic equation $$x^3+x+5=0$$
By Cardano's Method we assume
$$x=u+v$$ Then
we have $$(u+v)^3-3uv(u+v)-(u^3+v^3)=0$$ Comparing with given cubic we get
$$u^3+v^3=-5 tag{1}$$
$$uv=frac{-1}{3}tag{2}$$ Solving $(1)$ and $(2)$ we get
$$u^3-frac{1}{27u^3}=-5$$ Letting $u^3=p$ we get
$$27p^2+135p-1=0$$ which gives two real distinct roots of $p$ which $implies$ two real distinct roots of $u$ and hence two real distinct roots of $v$
Hence two distinct real roots of $x$
what went wrong here?
algebra-precalculus polynomials cubic-equations
asked Nov 24 at 19:45
Umesh shankar
2,53621219
I was going through the solution of depressed cubic equation $$x^3+x+5=0$$
By Cardano's Method we assume
$$x=u+v$$ Then
we have $$(u+v)^3-3uv(u+v)-(u^3+v^3)=0$$ Comparing with given cubic we get
$$u^3+v^3=-5 tag{1}$$
$$uv=frac{-1}{3}tag{2}$$ Solving $(1)$ and $(2)$ we get
$$u^3-frac{1}{27u^3}=-5$$ Letting $u^3=p$ we get
$$27p^2+135p-1=0$$ which gives two real distinct roots of $p$ which $implies$ two real distinct roots of $u$ and hence two real distinct roots of $v$
Hence two distinct real roots of $x$
what went wrong here?
algebra-precalculus polynomials cubic-equations
algebra-precalculus polynomials cubic-equations
asked Nov 24 at 19:45
Umesh shankar
2,53621219
asked Nov 24 at 19:45
Umesh shankar
2,53621219
asked Nov 24 at 19:45
Umesh shankar
2,53621219
asked Nov 24 at 19:45
Umesh shankar
2,53621219
asked Nov 24 at 19:45
Umesh shankar
2,53621219
2,53621219
1
Did you note that what you wrote occurs when you apply Cardano's method to any reduced cubic?
– José Carlos Santos
Nov 24 at 19:55
add a comment |
1
Did you note that what you wrote occurs when you apply Cardano's method to any reduced cubic?
– José Carlos Santos
Nov 24 at 19:55
1
1
Did you note that what you wrote occurs when you apply Cardano's method to any reduced cubic?
– José Carlos Santos
Nov 24 at 19:55
Did you note that what you wrote occurs when you apply Cardano's method to any reduced cubic?
– José Carlos Santos
Nov 24 at 19:55
add a comment |
1 Answer
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You will get two solutions because both $u^3$ and $v^3$ are solutions of the quadratic, and the second solution simply exchanges the two. The two solutions for the quadratic therefore give the same solutions for the cubic (depending which cube root you take).
answered Nov 24 at 19:50
Mark Bennet
80.1k981179
add a comment |
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You will get two solutions because both $u^3$ and $v^3$ are solutions of the quadratic, and the second solution simply exchanges the two. The two solutions for the quadratic therefore give the same solutions for the cubic (depending which cube root you take).
answered Nov 24 at 19:50
Mark Bennet
80.1k981179
add a comment |
You will get two solutions because both $u^3$ and $v^3$ are solutions of the quadratic, and the second solution simply exchanges the two. The two solutions for the quadratic therefore give the same solutions for the cubic (depending which cube root you take).
answered Nov 24 at 19:50
Mark Bennet
80.1k981179
add a comment |
You will get two solutions because both $u^3$ and $v^3$ are solutions of the quadratic, and the second solution simply exchanges the two. The two solutions for the quadratic therefore give the same solutions for the cubic (depending which cube root you take).
answered Nov 24 at 19:50
Mark Bennet
80.1k981179
You will get two solutions because both $u^3$ and $v^3$ are solutions of the quadratic, and the second solution simply exchanges the two. The two solutions for the quadratic therefore give the same solutions for the cubic (depending which cube root you take).
answered Nov 24 at 19:50
Mark Bennet
80.1k981179
answered Nov 24 at 19:50
Mark Bennet
80.1k981179
answered Nov 24 at 19:50
Mark Bennet
80.1k981179
answered Nov 24 at 19:50
Mark Bennet
80.1k981179
80.1k981179
add a comment |
add a comment |
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1
Did you note that what you wrote occurs when you apply Cardano's method to any reduced cubic?
– José Carlos Santos
Nov 24 at 19:55